The description and use of the trianguler quadrant being a particular and general instrument, useful at land or sea, both for observation and operation : more universally useful, portable and convenient, than any other yet discovered, with its uses in arithmetick, geometry, superficial and solid, astronomy, dyalling, three wayes, gaging, navigation, in a method not before used / by John Brown, philomath.

The Description, and some Uses of the Triangular Quadrant, or the Sector made a Quadrant; being an excel∣lent Instrument for Observations and Operations at Land or Sea, perfor∣ming all the Uses of the Fore-staff, Davis-Quadrant, Gunter's-Bow, Gunter's-Cross-staff, Gunter's-Qua∣drant and Sector, with far more conveniency and as much exactness as any, or all of them will do.

The Description thereof.

1. FIrst, it is a joynted Rule (or Sector) made to what Length or Radius you please, (as to 6, 9, 12, 18, 24, 30, or 36 inches Length, when it is folded or shut together; the shorter of which Lengths is big enough for Land uses, or Paper draughts; the four last for Sea uses, or Observations.) To which is

added, a third Piece of the same length of the Sector, with a Tennon at each end, to fit into two Mortice-holes at the two ends of the inside of the Sector, to make it an Aequilateral Triangle; from which shape, and its use, it is properly called a Triangular Quadrant.

2. Secondly, as to the Lines graduated thereon, they may be more or less, as your use of them, and as the cost you will be∣stow, shall please to command: But to make it compleat for the promised Premises, these that follow are necessary to be inscribed thereon, as in the Figure thereof.

And first you are in order hereunto to consider, The outer-edges of the (Sector or) Instrument, the inner-edges, the Quadran∣tal-side, the Sector-side, and the third or loose-piece, also the fixed or Head-leg, the moving-leg, the head, and the end of each leg, also the head and leg center; of which more in its proper place.

1. And first, on the outer-edge is placed the Lines of Artificial Numbers, Tangents, Sines, and versed Sines, to as large a Radius as the Instrument will bear.

2. Secondly, on the in-side or edge, on short Rules is placed inches, foot measure, the line of 112, or such-like. But on larger Instruments, a Meridian line to one inch,

or half an inch (more or less) for one de∣gree of the Aequinoctial, for the drawing of Charts, according to Mercator, or any o∣ther more useful Line you shall appoint for your particular purpose.

3. Thirdly, in decribing the Lines on the two sides; first I shall speak to the Sector-side, where the middle Lines all meet at the Center at the head where the Joynt is: the order of which (went the head or joynted end lyeth toward your left hand, the Sector being shut, and the Sector-side upermost) is thus:

1. The first pair of Lines, and lying next to you, is the Line of Sines, and Line of Lines, noted at the end with S, and L: for Sines and Lines, the middle Line between them that runs up to the Center, and where∣in the Brass center pricks be, is common both to the Sines and Lines in all Parallel uses, or entrances.

2. The Line next these, and counting from you, is, the Line of Secants beginning at the middle of the Rule, and proceeding to 60 at the end, and noted also with Se for Secants, one of which marginal Lines con∣tinued, would run to the center as the o∣ther did.

3. The next Lines forward, and next the in∣ner-edge on the moving-leg are the Lines of

Tangents; the first of which, and next to you, is the Tangent of 45, being the largest Radius, (as to the length of the Rule:) the other is another Tangent to one fourth part of the length of the other, and proceeds to 76 degrees, a little beyond the other 45: the middle Line of these also is common to both, in which the Center pricks must be. At the end of these Lines is usually set T. T. for Tangents.

4. On the other Leg of the Sector, are the same Lines again, in the same order counting from you; wherein you may note, That as the Lines of Sines and Lines on one Leg, are next the outward-edge; on the other Leg, they are next the inward-edge: so that at every, or any Angle whatsoever the Sector stands at, you have Lines, Sines, and Tangents to the same Radius: and the Secants to just half the Radius, and conse∣quently to the same Radius by turning the Compasses twice; Also any Tangent to the greater Radius above 45, and under 76, by turning the Compasses four times, as after∣wards will more appear: Which contrivance is of excellent convenience to avoid trouble, and save time; and happily made use of, in this contrary manner to the former wayes of ordering them.

5. Fifthly, without or beyond, yet next

to the greater Line of Tangents on the head-leg, is placed the first 45 degrees of the lesser Tangents, which begin from the Center at 45 degrees, because of the strait∣ness of the room next the Center, where they meet in a Point: yet this is almost of as good use, as if it had gone quite to the Center, by taking any parallel Tangent from the middle or common Line on the great Tangents, right against the requisite Number counted on the small Tangent un∣der 45.

6. Sixthly, next to this will not be amiss to adde a Line of Sines, to the same Radius of the small Tangent last mentioned, and figured both wayes for Sine and co-Sine, or sometimes versed Sines.

7. Seventhly, next to this a Line of E∣qual Parts, and Chords, and the Secants in a pricked line beyond the little Tangent of 45, all to one Radius: To which (if you please) may be added, Mr. Fosters Line Soll, and his Line of Latitudes; but these at plea∣sure.

8. Eighthly, on the outermost-part of both Legs next the out-side, in Rules of half an inch thick and under, is set the Line of Artificial versed Sines, laid next to the Line of Artificial Sines, on the outer-edge; but if the Rule be thick enough to bear four

Lines, then in this place may be set the Me∣ridian Line, according to Mr. Gunter, coun∣ting the Line of Lines as a Scale of Equal Parts. Thus much as for the Sector-side of the Instrument.

4. Fourthly, The last side to be descri∣bed is the Quadrantal-side of the Instru∣ment, wherein it chiefly is new. Therefore I shall be as plain as I can herein.

To that purpose I shall in the descripti∣on thereof imagine the loose piece, (or third piece) to be put into the two Mortise-holes, which position makes it in form of an Aequi∣lateral Triangle, according to the Figure annexed, noted with ABCD; where in AB is for brevity and plainness sake called the Moveable-leg, DB the Head or Fixed-leg, DA the loose-piece, B the Head, A and D the ends, C the Leg-center, at the beginning of the general Scale; the center at B the head-center, used only in large In∣struments, and when you please on any o∣other.

For the Lines graduated on this side.

First, On the outer-edge of the moveable-Leg, and loose-piece, is graduated, the 180 degrees of a semi-circle, C being the center thereof.

And these degrees are numbred from 0|60 on the loose-piece toward both ends, with 10, 20, 30, 40, &c. and about on the moveable-leg, with 20, 30, 40, 50, 60, 70, 80, and 90 at the head: Also it is numbred from 60|0 on the moveable-leg, with 10, 20, toward the head; and the o∣ther way, with 10, 20, 30, 40, 50, 60 on the loose-piece; and sometimes also from the Head along the Moveable-leg, with 10, 20, 30, &c. to ••0 on the loose-piece; and the like also from the end of the Head-leg, and sometimes from 60 on the loose-piece both wayes, as your use and occasion shall require.

Secondly, On the Quadrantal-side of the loose-piece, but next the inward-edge is graduated 60 degrees, or the Tangent of twice 30 degrees, whose center, is the cen∣ter-hole or Pin at B, on the Head or Joynt of the Sector.

Which degrees are numbred three wayes, viz. First from D to A for forward Obser∣vations; and from the middle at 30 to A the end of the Moving-leg, with 10, 20, 30; and again, from D the end of the Head-leg to A, with 40, 50, 60, 70, 80, 90, for Observations with Thred and Plummet.

Thirdly, Next to these degrees on the Moving-leg, is the Line of the Suns right Ascention, numbred from 60|0 on the de∣grees, with 1, 2, 3, 4, 5, 6, toward the Head, and then back again with 7, 8, 9, 10, 11, 12, &c. 1, 2, 3, 4, 5, on the o∣ther side of the Line, as the Figure annexed sheweth: The divisions on this Line is (for the most part) whole degrees, or every four minutes of time.

Fourthly, Next above this is the Line of the Suns place in the Zodiack, noted with ♈ ♉ ♊ ♋ toward the Head; then back a∣gain with ♌ ♍ ♎ over 60|0 in the de∣grees, and 12 and 24 in the Line of the Suns right Ascentions▪ then toward the end, with ♏ ♐ ♑; then back again with ♒ and ♓, being the Characters of the 12 Signes of the Zodiack, wherein you have exprest every whole degree, as the num∣ber of them do shew, there being 30 de∣grees in one Sign.

Fiftly, Next above this is a Kalender of Months and Dayes; every single Day being exprest, and three or more Letters, of the name of every Month being set in the Month, and also at the beginning of each Month, and every 10th day noted with a Prick on the top of the Line representing it, as is usual in such work.

Sixtly, Next over the Months, is the Line to find the Hour and Azimuth in a particular Latitude. Put alwayes on smal∣ler Instruments (and very rarely on large Triangular Quadrants for Sea Observations) the lowest Margent whereof, and next the Months, is numbred from the end toward the Head, with 10, 20, 30, 40, 50, 60, 70, 80, 90, 100, 110, 120, 130, near the Head Center. For the Semi-diurnal Ark of the Suns Azimuth, and in the Margent next above this, with 4, 5, 6, 7, 8, 9, 10, 11, 12, near the end, for the Morning hours; then the other way, viz. toward the Head on the other-side the Hour Line, with 1, 2, 3, 4, 5, 6, 7, 8, for the Afternoon hours.

Seventhly, On the same Quadrantal-side, and Moveable-leg on the spare places, beyond the Months toward the end, is set an Almanack; and the Names of 12 or more Stars, to find the hour of the Night; which 12 Stars are noted with 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12. among the degrees in small Figures, as in the Figure.

Eightly, Next of all to the in-side, is the Line of Natural versed Sines drawn to the Center, with his correspondent Line on the other, or Head-leg. Exprest sometimes in a pricked Line, for want of room.

Ninthly, On the Head-leg, and next to

the versed Sines last mentioned, is first the Line of Equal Parts, or Line of Lines: and on the same common Line wherein is the Center, is the Line of Natural Sines, whose length is equal to the measure from the cen∣ter at C to 60|0 on the moveable-leg; so that the Line of degrees is a Tangent, and the measure from C to any Tangent, a Se∣cant, to the same Radius of the Natural Lines of Sines, and Lines: Also beyond the Center C on the same common middle Line is another smaller Line of Natural Sines, whose length is equal to the measure from C to 60 on the loose-piece; then if you count from the Center pin at 60, on the loose-piece, toward the end of the mo∣vable-leg, they shall be Tangents to the same Radius, and the measure from the Center C to those Tangents, shall be Secants to the same Radius, which may be well to be or∣dered, to a third, or fourth part of the for∣mer, from the Center downwards: These two Lines of Sines are best figur'd with their Sines; and Cosines, the other way with a smaller figure, and the Line of Lines from the Center downward from 1 to 10 where 90 is, which Lines of Sines may be called a general Scale for all Latitudes.

Tenthly, Next to this toward the outer-edge is another Line of Natural Sines, fitted

to the particular Line of Hour and Azi∣muths, for one particular Latitude, noted, Pert. Scale of Altitudes; or Sines.

Eleventhly, Next to this is the Line of 29 ½, for so many dayes of the Moon's age, in short Rules of the whole length, but in longer not; being easily known by the single strokes, and Figures annexed to those strokes.

Twelfthly, Next the outer-edge is a Line of 24 hours, 360 degrees, or 12 Signs, or in most Rules inches also, used together with the former Line of 29 ½, and as a Theory of the Sun and Moon, and ready way of finding the Hour by the Moon or fixed Stars.

Thirteenthly, To this Instrument also be∣longs a Thred and Plummet, and Sights, as to other Quadrants; and a pair of Com∣passes as to other Sectors; a Staff and Ball socket also, if you will be curious and ac∣curate.

And for large Instruments for Sea, a Square and an Index, which makes it a per∣fect sinical Quadrant, and two sliding sights also, which makes it a fore and back-staff, and bow, as will appear more at large after∣ward.

Some Uses of the Trianguler Quadrant, for Land and Sea Obser∣vations and Opera∣tions.

CHAP. I. Numeration on the Lines graduated on the Instrument.

IN the first place it will not be amiss to hint a few words, as to the reading the Lines, or (more properly) Numeration on the Lines; wherein take notice, That all Lines of Equal Parts, or Lines applicable to Arithmetick, as the Line of Lines, the Line of Numbers, the Line of Foot-measure, and the like; wherein Fractions of Num∣bers are requisite: they are most commonly accounted in a Decimal way, and as much as may be, the small divisions are numbred, and counted accordingly.

But in the Lines of Sines, Tangents, Se∣cants, and Chords; being Lines belonging properly to a Circle: in regard that the Sexagenary Fraction is still in use, the inter∣mediate Divisions are, as much as may be, fitted to that way of account, viz. by whole degrees, where they come close toge∣ther, (or the Line of no great use.) And if more room is, to half degrees or 30 minuts, and sometimes to quarters of degrees or 15 minuts; but toward the beginning of the Line of Natural Sines, or the end of the Natural Tangents and Secants: where the degrees are largest, they are divided to e∣very 10th minute in all large Rules, as by considering and accounting you may plain∣ly perceive.

Take two or three Examples of each kind.

1. First, On the Line of Lines, to find the Point that represents 15. In the doing of this, or any the like, you must consider your whole Scale, Radius, or length of the Line, may be accounted as 1, as 10, as 100, as 1000, or as 10000; and no further can be applicable to any ordinary Instrument.

Wherein observe, That if the whole Line be one, then the long stroke by every Figure doth represent one tenth of that Integer: and the next shorter without Figures, are

hundredth parts of that one Integer; and a 1000th part is estimated in smaller Instru∣ments, and sometimes exprest in larger: But the hundredth thousand part is alwayes to be estimated by the eye in all Instruments whatsoever.

2. But if the whole Line of Lines shall represent 10, as it usually doth, and as it is figured, then the long stroke at every Fi∣gure is 1, and the next longer are tenths, and the shortest are hundred parts, and the thousand parts as near as can be estimated.

3. But if the whole Line represents a hundred, as here in our present Example, then the long stroke by every Figure repre∣sents 10, and every shorter stroke is one, and the shortest strokes are tenths, and the hundredth parts as much as can be estima∣ted.

4. But if the whole Line shall represent a 1000, then the long stroke by the Figure shall represent a hundred, and every shor∣ter 10, and every of the shortest strokes is one Integer, and a 10th part as near as can be estimated.

5. But lastly, if the whole Line represent 10000, then every long stroke is 100, and every shortest cut is ten, and every single Integer is as near as can be estimated by any ordinary Instrument.

Now our present Example will properly come under the third Rule, by conceiving the whole Line to represent 100; then the first long stroke by 1 is 10, then the next shorter is for 11, the next 12, &c. to 15; which is cut up a little above the Line, for the more ready reckoning without telling the parts: which 15 is the Point required to be found.

Example the second, to find out 1550 on the Line.

This will come under the Notion of the 5th Rule, wherein the whole Line is con∣ceived to represent 10000; then the first 1 is for the 1 thousand, then the fifth longer stroke next is for the 500; and lastly, the middle between the 500 stroke and the 6000 stroke is for the 50, being a little be∣yond the Point for 15 in the first Example.

A third Example of 5025.

This third Example may su••fice for this work, being so plain after a little due con∣sideration: For first, the whole Line is con∣ceived to represent 10000, then the long stroke by 5 is for 5000, then there is no hundreds, therefore the Point required must be short of the next longer stroke, which signifies hundreds, and being it is just 25, which is ¼ of an hundred, the true Point readily sheweth it self; If you re∣quire

a more plainer and larger wording o•• this matter, I refer you to the third Chapte•• of Mr. Windgates Rule of Proportion; 〈◊〉〈◊〉 the first Chapter of the Carpenters Rule, b•• I. Brown.

Lastly, In nameing of any Point found out on the Line, great care and respect mus•• be had as to the true value of the Number according to the rate of the question pro∣pounded: for the same Point that repre∣sents 15, doth represent 150; and also 1500, or 15000, (increasing above the bounds before mentioned) also it signifies one and a half, or 15 of one hundred, which is usually exprest thus in a Decimal Fraction 15/100, or more readily, 0.15.

Also if it should be a Number with a di∣git, two ciphers and another digit, as 2.005, this Number would be found close to the long stroke, by the figure 2: and may represent either two thousand, and 5 of 1000 more; or 20 and 5 of a hundred; or 2 hundred and 5 of another 10 more, or plainly as it is set down, two thousand, no hundred, but five: Thus you see the manner of expressing whole Numbers, or whole Numbers and Decimal Fractions, which on the Lines is one and the same thing; and thus all Decimal Scales are to be accounted, and in the same manner is

the Line of Numbers to be read, as you may see more at large in the two Books be∣fore mentioned.

SECT. II.

But for Numerateon on all Circular Lines, it is much easier: For first, very few Instruments, unless at one part of the Line, can express nearer than minutes of a degree.

Secondly, The whole Radius or Line of Lines is but 90 degrees, or but 45 of the Tangents, or 60 of the Chords or Secants: So that in Instruments of 12 or 18 inches Radius, you may express very well every tenth minute, to 60 on the Line of Sines: and every half degree to 75, and whole de∣grees to 90. And on the Tangents or Chords, every 10th minute quite through: and the Secants as the Sines.

So that any degree or minute being na∣med, t•• find the same on the respective Line, count thus;

First, every 10th degree is noted with a long stroke, and figures set thereunto. Se∣condly, every whole degree is cut between two, or three Lines, and sometimes with a Point or Mark on the end of the stroke; and every 5th degree cut up higher than the rest, and sometimes with three Points, on the end

of the Line, or some other convenient di∣stinction, for readiness sake: and every 10th, 15th, or 30th minute, is cut only be∣tween two Lines and no more; as will ap∣pear very plain with a little practice.

Example, to find the Sine of the Latitude being at London, 51 degrees, 32 minutes.

1. First, look on any Line of Sines, on the Quadrantal, or Sector side, according a•• you have occasion, till you see 50, which i•• 50 degrees; then one degree forward, to∣ward 60 is 51 degrees, then count thre•• 10ths of minutes more for 30 minutes, and then for the odde two minutes, estimate on•• fifth part of the next 10 minutes forwarder and that is the precise Point for the Sine 〈◊〉〈◊〉 51 degrees 32 minutes, the latitu••e of Lon∣don, where sometimes is set a Bra••s Center-Pin.

Example the second.

2. To find the Cosine of the Latitude there are two wayes to count the Comple••ment of any Ark or Angle.

First, by substracting the Ark or Angle out of 90 by the Pen, and count the residu•• from the beginning of the Line of Sines, and

that shall be the Sine Complement of the Latitude required.

Example.

51 32 taken from 90, the remainder is 38 28, now if you count so much from the beginning of the Line of Sines, according to the last Rule that shall be the Point for the Sine of 38 28, the Complement of 51 32, or the Sine Complement of the Latitude.

Or Secondly, If you count 51 32 from 90, calling 80, 10; and 70, 20; and 60, 30; 50, 40; 40, 50, &c. wheresoever the Number whose Complement you would have shall end, that is the Sine Comple∣ment required, which will be at 38 28, from the Center or beginning, for the Co∣sine of 51 32; The like work serves for any other Number, or on any other Line, as on the Degrees, Tangents, or Secants, Natural or Artificial, as by practice will more plainly appear, to the willing Practi∣tioner.

SECT. III.

To find the versed Sine of an Ark or Angle, or the Sine of an Ark or Angle a∣bove 90 degrees, or the Chord above 180 degrees, observe these Rules.

1. First, a right Sine, is the measure on the Line of Sines, from the center or begin∣ning of that Line, to the Point that doth represent the Ark or Angle required.

2. The right Sine of an Ark or Angle above 90 degrees, is equal, to the right Sine of the Complement thereof to 180 degrees, being readily accounted, thus; Count the excess above 90 backwards, from 90 toward the Center; then the measure or distance from the end of the account to the Center, is the Sine of the Ark above 90 required: Example. Let the Sine of 130 be required, first, if you take 130 from 180, the re∣mainder is 50; then I say that the right Sine of 50, is also right Sine of 130; for if you count backwards from 90, calling 80, 100; and 70, 110; and 60, 120; and 50, 130; the measure from thence to 00, or the Center, is the right Sine of 130 de∣grees.

3. The versed Sine of an Ark or Angle, is the measure on the Line of Sines from 90 toward the Center, counted backwards, as the small figures for Complements shew, counting 90 for 00, and the Center for 90, (as the Azimuth Line is figured) opening the Line of Sines to a strait Line, and then counting beyond 90 for the versed Sines a∣bove 90, as on the versed Sines is plainly

seen in the figure of the Rule.

4. For Chords of any Ark or Angle, do thus:

Halve the Ark or Angle required, and take the right Sine thereof, and that shall be the Chord thereof.

Example.

I would have the Chord of 40, the half of 40 is 20; then I say the right Sine of 20 is the Chord of 40, to that Radius that is equal to the right Sine of 30 degrees, at the Radius the Rule stands at.

5. To find a Chord to an Ark or Angle above 180 degrees, you must count as you did the right Sines; for note, the Chord of 180 is equal to the right Sine of 90 dou∣bled, which is the full Diameter of a Circle: and a longer right Line than the Diameter cannot be taken in a Circle; therefore it must needs follow that Chords of above 180, are shorter than the Diameter which is the biggest Chord; therefore the Chord of 260, is equal to the Chord of 100 de∣grees, or right Sine of 50, the Sine of 30 being Radius.

6. In using the Artificial Sines and Tan∣gents, or Secants; if you are to use a Sine above 90, then count 80 for 100, 70 for 110, 60 for 120, &c. But for Secants,

then count after the manner of versed Sines: Thus the Secant of 60 is as far beyond 90, as it is from 30 to 90; so that when you have occasion to use an Artificial Secant, which is not often, Then set the end of the Rule against a Table, and counting back∣wards from 90 to the number of the Secant required, turn that distance beyond 90 on the Board or Table, and that shall be your Secant required, as will be afterward hin∣ted, as they come in use.

CHAP. II. A brief Description of the Lines of a Circle, and the Explanation of some termes used in the following Dis∣course.

FOR the better understanding of the following discourse, it is needful to un∣derstand these Elements or Principles, as the Letters are necessary to be known before reading.

1. A Circle is a figure enclosed in one cir∣cular Line, called the Circumference; in the middle whereof is a Point called the Cen∣ter:

From which Point all right Lines drawn to the Circumference, are equal one to another; as in the Circle ABCD, E is the Center, ABCD the Circumference, the Lines EA, EB, EC, ED are equal.

2. Any right Line crossing the Circumfe∣rence, and passing through the Center of a Circle, is called the Diameter; and it di∣vides the whole Circle into two equal parts, called Semi-circles (or half-Circles.) And the half of that Line is called the Semi-diameter or Radius to that Circle. As the Line AC is the Diameter, and EC the half-Diameter or Radius.

3. Any other Right-line crossing the Cir∣cumference is called a Chord, or Subtence, as the Line FG, which divides the Circle into two unequal parts: And note, that this Subtence belongs both to the lesser, and also to the greater part of the Circumference; that is to say, the Chord of 90 deg. is also the Chord of 270 deg. so that FG is Chord to the Ark FBG 90 deg. and al∣so to the Ark FDG being 270 deg. much more than half the Circle.

4. Half the Chord of any Ark, is the right Sine of half that Ark: thus the right Line HG, the half of FG, is the right Sine of the Ark BG the half of FBG.

5. The Sine Complement or Cosine of

any Ark is the nearest distance from the Cir∣cumference to the Diameter: Perpendicu∣ler to that Diameter from whence you coun∣ted the degrees and minutes of the Ark or Angle. As thus, GI is the Cosine of the Ark BG; and the Right Sine of any Ark is the nearest distance from the Circumfe∣rence to the Diameter you counted the de∣grees from, as GH is the Right Sine of BG.

6. The versed Sine of any Ark or Angle, is the Segment of the Diameter between the right Sine of the same Ark and the Cir∣cumference.

Thus HB is the versed Sine of the Ark BG, and HD the versed Sine of GD. So also is GH the right Sine of the Ark GCD or the Angle GED 45 degrees above 90, viz. 135 degrees.

7. A Tangent is a right Line drawn per∣pendiculer to the Diameter, beginning at one extreme of the given-Ark, and termi∣nated by a right Line drawn from the Cen∣ter to the other extreme, of the given-Ark in the Circumference, till it inter-sect the perpendiculer; Thus CK is the Tangent of the Ark CG, or the Angle CEG, 45 de∣grees.

8. A Secant is a right Line drawn from the Center thorow one extreme of the given-Ark,

till it meet with the Tangent rais'd per∣pendicularly from the Diameter, drawn to the other Extreme of the said Ark; Thus the Line EK is the Secant of the Ark CG, or the Angle GEC.

9. Note, as in a (Natural) Sine, the nearest distance from the Ark to one Diame∣ter, from whence you counted the degrees o•• the Ark or Angle, was the Right Sine; and the nearest distance from the same Point to the other perpendiculer Diameter, is the Cosine of that Ark or Angle.

So likewise the nearest distance from the Point where the Tangent and Secant meets, to one of the Diameters aforesaid, is the Tangent of the Ark or Angle; so the nea∣rest distance from the meeting Point of the same Secant-line is the other Tangent-line to the other Diameter abovesaid, is the Co-Tangent of the Ark or Angle abovesaid.

Thus the Right-line KC is the Tangent of 45, and the Right Line KB the Co-Tangent of 45; Also the Line LC is the Tangent of 53, 30; and the Line MB is the co-Tangent thereof, viz. the Tangent of 36, 30.

Also the nearest distance from L to EB, is the Tangent of 36, 30, to the Radius LC.

10. Every Circle is supposed to be divi∣ded

into 360 degrees; the Semi-circle into 180, the Quadrant or Quarter into 90.

11. Every Degree is supposed to be di∣vided into 60 minutes, and every minute into 60 Seconds, and every Second into 60 Thirds, &c.

12. A Radius, or Semidiameter, is in our Instrumental Practice, supposed to be divided into 10000 parts, and every Chord, Sine, Tangent, or Secant, is to be divided by the Parts of the same Radius, or Radius and Parts more.

13. An Angle is the meeting of two Right Lines, as GE, and EC, meeting at E, do constitute the Angle GEC, called a Right-lined Angle; or when two Circles cross one another, it is called a Spherical Angle, the Anguler Point being noted al∣wayes by the middle Letter of three that shew the Triangle.

14. A Plain Triangle is the meeting of three Right Lines crossing one another; and a Spherical Triangle is constituted by the crossing of three Circles, as in the two Fi∣gures noted II and III, you may plainly see.

15. All Angles, Plain and Spherical, are either Acute, Right, or Obtuce.

16. An Acute Angle hath a sharp Point containing an Angle less than 90 de∣grees,

[illustration]

as the Angle CBA, sheweth in Fi∣gure II.

17. A Right Angle is alwa•••••• just 90 degrees, as you may 〈◊〉〈◊〉 in the Figures II, and III, by the Angles at A in both of them.

18. An Obtuce Angle is alwayes more than 90 degrees, as the Angles at D in both Figures shew.

19. A Parallel Line, is any Line drawn by another Line in such a way, that though it were infinitely produced, yet they would never meet or cross one another, as the Lines AB, CD.

20. A Perpendiculer Line, is when one Line so falleth on another Line, that the Angles on each side are equal, as CA falls of the Line BA, Figure VI.

21. All Triangles are either with three equal sides, as Figure IIII, or two equal sides, as Figure V, or all unequal sides, as Figure VI; the first of which is called Equi∣lateral, the second Isosceles, the third Sca∣lenum.

22. Again, they may be sometimes na∣med from their Angles; thus: Orthigonium, with one Right Angle, and two Acute Angles. Ambligonium, with one Obtuce Angle, and two Acute Angles. Oxigoni∣um, with three Acute Angles only.

23. The three Angles of every Plain Triangle, are equal to two Right Angles.

24. All Four-sided Figures are either Squares, with four sides, and four right Angles all equal; or long Squares (or Ob∣longs) with the two opposite sides equal, or the same crushed together, or not Right-Angled, as the Rombus, and Rombords, or else with four unequal Sides, called Tra∣peziaes.

25. Lastly, many sided-Figures, are some Regular, having every side alike, as 5, 6, 7, 8, 9, 10, &c. Or else unlike, as Fields, and Woods, and Meadows, which being infinite, cannot be comprehended un∣der any Regular Order or Rule.

26. Multiplicator, is a term used in Mul∣tiplication, by which any Number is to be multiplied, as is saying 5 times 6, 5 is the Multiplicator of 6.

27. Multiplicand, is the Number to be multiplied, as 6 by 5, as above named.

28. The Product, is the Issue or Result of two Numbers multiplied one by the o∣ther, as 30 is the Product of 6 multiplied by 5; for 6 times 5 is 30.

29. Divisor, is a term used in Division, and is the Number by which another Num∣ber is to be divided; as to say, How many times 5 in 30? 5 here is the Divisor.

30. Dividend, is the Number to be di∣vided, as 30 abovesaid.

31. Quotient, is the Answer to the How many times (as in the abovesaid) 5 is in 30? 6 times: 6 then is the Quotient.

32. Square, is the Product of two Numbers multiplied together, as the Square of 6 multiplied by 6, is 36.

33. Square-root of any Number, is that number, which being multiplied by it self, shall have a Product or Square equal to the given Number; thus the Square-root of 36 is 6; for 6 multiplied by 6, is 36, equal to the first given Number.

But if it be a Number that cannot be squared, as 72, the content of half a Foot of Board; whose near Square-root is 8: 4852811 of 10000000, then is the Square-root to be exprest as near as you may (or care for) as here the Square-root of 72, which is called a Surd Number, that will not be squared.

34. Cube, is a second Product, or power of two Numbers increasing or multiplied together, as thus; the Square of 6 is 36, the first power: and the Cube of 6 is 216, that is to say 6 times 36, the second power.

35. Cube-root of any Number, is a se∣cond Quotient decreasing between the Num∣ber given to be cubed and 1, as thus; the

[illustration]

Cube-root of 216, is to be found out by the Line of Numbers, the third part of the distance between 216 and 1, is at 2 repeti∣tions found to stay at 6; for if you should have 216 Cubes or Dies, which is a proper Cube, you shall find that 36 laid together one by another, making a Square repeated 6 times, will use or take up 216, the just number, and make one great square Die; and no other Number whatsoever, except 6, will do the like; therefore 6 is the Cube-root of 216.

In Mr. Windgate's Book of Arithmetick, is the way of doing it by Numbers or Fi∣gures, being one of the hardest Lessons in Arithmetick.

CHAP. III. Certain Geometrical Propositions, fit to be known as Preparatory Rudiments for the following Work.

1. To draw a Right Line between two Points.

EXtend a Thred or Hair, from one Point to the other, and that shall be the Line required. But if you use a Rule (being the fittest Instrument) to try your Rule, do thus; apply one end to one Point as to A, and the other end to the other Point at B, and close to the edge draw the Line required: then turn the Rule, and lay the first end to the last Point (yet keeping the same side of the Rule toward the Paper) and draw the Line again, and if the two Lines appear as one, the Rule is streight, or else not. Note the Figure I.

2. To draw a Line Perpendiculer to another, on the middle of a Line.

On the Point E on the Line AB, I would raise the Perpendiculer Line CE, set one point of the Compasses in E, and open them to any distance, as EB and EA, and note the Points A and B, then open the Com∣passes wider, and setting one Point in A, make the part of the Arch by C upwards; and if you have room do the like down∣wards, near D: Then the Compasses not stirring, set one Point in B, and with the other, cross the former Arks, near C & D: a Rule laid, and a Line drawn, by those two crossings, shall cut the Line AB per∣pendiculerly just in the Point E, which was required.

3. To let fall a Perpendiculer from a Point to a Line.

But if the Point C had been given from whence to let fall the Perpendiculer to the Line AB, do thus: First, set one Point of Compasses in C; open the other to any di∣stance, as suppose to A and B; and then (if you have room upon A and B, strike both the Arks by D, which finds the Point D, if

not) the middle between A and B, give; the Point E; by which to draw CED, the Perpendiculer from C desired. Note Figure 2.

Note, That if you can come to find the Point D, by the crossing, it doth readily and exactly divide the Line AB in two e∣qual parts, by the Point E.

4. To raise a Perpendiculer on the end of a Line.

On the end of the Line AB, at B, I would raise a Perpendiculer: First, set one Point of the Compasses in B; open them to any distance, as suppose to C; and set the other Point any where about the middle, between D and E, as suppose at C, then keep that Point fixed there; turn the other till it cut the Line, as at D, and keep both Points fixed there, and lay a streight Rule close to both Points▪ and there keep it; then keep the middle-Point still fixed at C, and turn the other neatly close to the other end and edge of the Rule, to find the Point E; then a Rule laid to the Points E and B, shall draw the Perpendiculer re∣quired.

Or else, when you have set the Com∣passes in the Point C, prick the Point D in

the Line, and make the touch of an Ark near to E; then a Rule laid to DC cuts the Ark last made, at or by E, in the Point E: There are other wayes, but none better than this. Note the Figure 3.

5. From a Point given, to let fall a Per∣pendiculer to the end of a Line, be∣ing the converse of the former.

First, from the Point E, draw the Line ED, of which Line find the middle be∣tween E and D, viz. the Point C: then the extent CE, or CD, keeping one Point in C, shall cross the Ground-line in the Point B, by which, and E, you may draw the perpendiculer Line EB, which is but the converse of the former.

6. To draw a Line Parallel to another, at any distance.

To the Line AB, I would have another Parallel thereto; to the distance of AI, take AI between your Compasses, and setting one Point in one end of the Line, as at A, sweep the Ark EIF; then set the Compasses in the other end, as at B, and sweep the Ark GDH; then just by the Round-side of those Arks, draw a Line,

which shall be the parallel-Line required.

Or thus,

Take BC, the measure from the Point that is to cut the Parallel-line, and one end of the given-Line, viz. B; with this di∣stance, set one foot in A, at the other end of the given-Line, and draw the Arch at K; then take all AB, the given-Line, and setting one Point in C, cross the Ark at K, then C and K shall be Points to draw the Parallel-line by. Note the Figure 4.

7. To make one Angle equal to another.

The Angle BAC, being given, and I would have another Angle equal unto it; set one point of the Compasses in A, and draw the Arch CB; then on the Line DE from the Point D, draw the like Ark EF; then in that Ark make EF equal to CB, then draw the Line DF, it shall make the Angle EDF, equal to the Angle BAC▪ which was required.

8. To divide a Line into any Number of parts.

Let AB represent a Line to be divided

[illustration]

into Eight parts: On one end, viz. A, draw a Line, as AD, to any Angle; and from the other end B, draw another Line Parallel to AD, as BE, then open the Compasses to any convenient distance, and from A and B, divide the Lines AD, and BE, into eight parts; then Lines drawn by a Ruler, laid to every division, in the Lines AD, and BE, shall divide the Line AB in the parts required. Note the Figure marked VI.

This Proposition is much easier wrought by the Line of Lines on the Sector, thus; Take AB between your Compasses, and fit it over parrally in 8, and 8 of the Line of Lines; then the Parallel distance between 1 and 1, shall divide AB into 8 parts re∣quired.

9. Any three Points given, to bring them into a Circle.

Let ABC be three Points to be brought into a Circle; first set one Point on A, and open the other above half-way to C, and sweep the part of a Circle above and be∣low the Point A, as the two Arches at D and E; not moving the Compasses, do the like on C, as the Arks F and G; then set the Compass-point in B, and cross those

Arks in DEF and G; then a Rule laid from D to E, and from F to G, and Lines drawn do inter-sect at H, the true Center, to bring ABC into a Circle.

10. Any two Points given in a Circle, to draw part of a Circle, which shall cut them, and the Circumference first given into two equal parts.

Let A and B be two Points in a Circle, by which two Points, I would draw an Arch, which shall cut the whole Circumfe∣rence into two equal parts. First, draw a Line from A, the Point remotest from the Center, through the Center, and beyond the Circumference, as AD; then draw another Line from A, to a Point in the Circumfe∣rence, perpendiculer to AD, (and cutting the Center C) as the Line AE: Then on the Point E, draw another Line perpendicu∣ler to the Line AE, till it inter-sect AD at D; then these three Points ABD brought into a Circle, or Arch, by the last Rule, shall divide the Circumference into two e∣qual parts. Note the Figure 8, where the first Circle is cut into two equal parts at F and G, by part of a Circle passing through the Points A and B.

11. Any Segment of a Circle given, to find the Diameter and Center of the Circle belonging to it.

Let ABC be the Segment of a Circle, to which I would find a Center; any where about the middest of the Segment, set one point of the Compasses at pleasure, as at B; on the point B (at any meet distance) de∣scribe a Circle, and note where the Circle doth cross the Segment, as at D and E, then (not stirring the Compasses) set one point in D, and cross the Circle twice, as at F and I; and again set one point in E, and cross the Circle twice in G and H: Lastly, by the Points GH, and FI, draw two Lines, which will meet in the point O, the center required.

12. Or else to find the Diameter, thus.

Multiply the Chord (or flat-side) of the half-Segment, viz AK, 12 by it self (which is called Squaring) which makes 144; then divide that Product 144 by 8, the Line KB, called a Sine, the Quotient which comes out will be found to be 18; then if you adde 8 the Sine, and 18 the Quotient together, it shall make 26 for

the Diameter required to be found.

13. Any Segment of a Circle given, to find the Length of the Arch of the Segment.

Lay the Chord of the whole Segment, and twice the Chord of half the Segment, from one Point severally; and to the grea∣test extent, adde one third part of the dif∣ference between the Extents, and that sum of Extents shall be equal to the Arch.

Example.

Let ABC represent the Segment of a Circle; the length of whose Arch I would know, or have a Line equal thereunto Take all the Chord AC, and lay it on any Line, as from D to E; also take the Chord of half the Arch, as AB, or BC, and lay it twice from D to F; then E and F are the two Extents, whose third part FG is to be added to DF the greatest Extent to make 〈◊〉〈◊〉 DG, a Line equal to the Arch ABC which was required to be done. Which operation, may very well be performed o•• a Line of Lines, or inches on your Rule or by Numbers in Figures, thus; Suppos•• AC be 35 inches and 6 tenths; and twic••

[illustration]

AB be 42 inches, 7 tenths: The difference between them you may count, on the Rule, to be 7 inches, and 1 tenth; a third part of which is 2 inches 4 tenths; which added to 42, 7, makes 45, 2. the measure of the Arch ABC, which was required.

14. To draw a Helical Line from any Three Points, to several Radiusses with∣out much gibbiosity; useful for Archi∣tect, Shipwrights, and others.

Let ABCDE be five Points, to be brought into a Helical-Line, smoothly, and even without gibbiosity or bunches, as the under-side of an Arch, or the bending of a Ship, or the like.

First, between the two remote Points of 3, as A and C, draw the Line AC, then let fall a Perpendiculer from B, to cut the Line AC at Right Angles, and produce it to F: draw the like perpendiculer-Line from the point D, to cut the Line CE at Right-Angles produced to F. I say, the Center both for the Arches AB the lesser, and BC the greater, will be found to be in the Line BF; the like on the other-side for DE and CD, the Helical-Circle, or Arch required.

But if you divide the Arch ABCDE into 24 or more parts, the several Centers of the splay-Lines are thus found; Take the measure AG, and lay it from B, or D, or C, on the Line GF; and those Points on GF, shall be the several Points to draw the splay-Lines of the Arch, and Key-stone by.

CHAP. IV. Of the Explanation of certain Terms used in this following Book.

1. RAdius, or Sine of 90,* 1.1 or Tangent of 45, or Secant of 00, are all one and the same thing, yet taken respectively in their proper places, and is the whole Line of Sines, or Tangents, to 45; or more particulary that point at the end of the Natural-Sines, on the Sector-side, and at 90 and 45 on the edge of the Rule for the Artificial Sines and Tangents, or 10 on the Line of Numbers, and 10 and 90 on the Line of Lines, and Sines, on the Quadran∣tal-side of the Instrument.

* 1.22. A Right Sine of any Ark, or Angle, is the measure from the beginning of the Line of Natural-Sines, to that Point on that Line of Sines, which represents the de∣grees and minutes contained in that Ark or Angle required. But on the Artificial-Sines we respect not any measure but the Point only.

* 1.33. The same account is used both for the Right-Tangent, and Secant also; the Na∣tural-Tangent taken from the beginning to the degree and minute required; the Arti∣ficial respecting the Point only.

* 1.44. In the same manner count for the Se∣cants, and Chords, Lines, or versed Sines.

* 1.55. A Cosine, or Sine Complement of any Ark or Angle, is the measure from the Point representing the Ark or Angle, counted from 90, to the beginning of the Line of Sines, being in effect the Right-Sine of the Cosine of the Ark or Angle required: As for Ex∣ample; I would take out the Cosine of the Latitude of London, which is 51, 32; Count 51 32 from 90 toward the begin∣ning, and you shall find your account to end at the Right-Sine of 38 28, which is the Complement of 51 32; for both put together, makes 90, the whole Sine or Radius.

But on the Artificial-Lines count back∣ward

to the Point required, without mind∣ing any distance or measure, till you come to Proportion.

6. A Lateral Sine, Tangent, or Secant,* 1.6 or Scale of Equal Parts, is any Sine Tan∣gent or Secant, taken along the length of any Line, from the beginning onwards, be∣ing a term used only in operation with a Sector, or one Line and a Thred, and op∣posed to a parallel-Sine, Tangent, or Secant, the thing next to be explained.

7. A Parallel Sine, Tangent, or Secant,* 1.7 is any Sine, Tangent, or Secant, taken a∣cross from one Leg to the other of a Sector; or from any degree and minute on one Line to a Thred drawn streight with the other hand, or any other fixed Line whatsoever, at the nearest distance.

8. The Nearest Distance to any Line,* 1.8 is thus taken; When one Point of the Com∣passes stands in any one Point, and the Line being laid, I open or close my Compasses till the other moveable-Foot, being turned about, will but just touch or cleave the Thred. But if you are to lay the Thred to the nearest distance, then one Point of the Compasses being set fast, the other is to be turned about, and the Thred also slipped to and fro, till the Compass-point shall just cleave the Thred in the middest.

* 1.99. To adde one Sine or Tangent, to a Sine or Tangent, is to take the Right-Sine, or Tangent of any Ark or Angle between your Compasses, and setting one Point of the Compasses in the Point of the other Number, and then to see how far the other Point will extend Laterally. Example. To adde the Sine of 20, to the Sine of 30, take the Sine of 20 between your Compasses, and then putting one Point in 30, the other shall reach to the Sine of 51 21; therefore the distance from the beginning to 51 21, is the sum of the Sine of 30 and 20 added together. The like way is to add Tan∣gents.

* 1.1010. To Substract a Sine from a Sine, or a Tangent from a Tangent, is but to take the Lateral least Right-Sine or Tangent be∣tween your Compasses, and setting one Point in the term of the greatest turn, the other toward the beginning, and note the degree and minute that the other Point stayes in, for that is the difference or remainder.

Example.

Suppose I would take the Sine of 10 de∣grees from 25; Take the distance 10 be∣tween your Compasses, and setting one foot in 25, and the other turned toward

the beginning, shall reach to 14 23, the re∣sidue or difference required.

Or, you may sometimes take the distance between the greater and the less, and lay this from the beginning, shall give the re∣mainder in distance on the Sines as be∣fore.

11. The Rectifying-Point,* 1.11 is a Point or Hole on the Head of the Trianguler Qua∣drant in the inter-secting of the hour and Azimuth-line, and the common Line to the Lines and Sines on the Head-leg; in which Point you are, when the Rule is open, to stick a small Pin to look to the object whose Altitude above the Horizon you would have in degrees and minutes.

Of Terms used in DIALLING.

PLain, is that Board, Glass, or flat Su∣perficies you intend to draw the Dial upon, either single of it self,* 1.12 or joyned to some other.

Pole of the Plain,* 1.13 is an imaginary Point in the Horizon (for all upright Dials) di∣rectly

opposite to the Plain, or in all Plains, a Point every way 90 degrees from the Plain.

* 1.14Declination of a Plain, is only the num∣ber of degrees and minutes, that the Pole-point of the Plain is distant from the North and South-points of the Horizon.

* 1.15The Perdendiculer-Line on the Plain, is a Line Square to a Horizontal-line, being part of a Circle passing through the Zenith, and Nadir, and Pole-point of the Plain.

* 1.16The Horizontal-line, is a Line drawn on any Plain, exactly parallel to the true Horizon of the place you dwell in.

* 1.17Reclination, is when a Plain beholdeth the Zenith-point over our heads: But In∣clination, is, when a Plain beholdeth the Nadier; as in a Roof of a House, the Tiled-part reclines, and the Celid-part in∣clines.

* 1.18The Meridian-line, on all Plains is the Hour-line of 12; but the Meridian of the Plain, is the great Circle of Azimuth perpen∣diculer to the Plain, bing the same with the Perpendiculer-line on the Plain, passing through the Points of Declination.

* 1.19The Substile-line on all Dials, is that Line wherein the Stile, Gnomon, or Cock of the Dial doth stand, usually counted from 12, the Meridian-line, or from the Perpen∣diculer-line,

which in all erect Dials is 12.

The Stile of a Dial, is the Angle,* 1.20 between the common Axis of the World and the Plain, upon the Substile-line on the Plain, on all Dials.

The Angle between 12 and 6,* 1.21 is onely the number of degrees and minutes contain∣ed between the Hour-line of 12, and the Hour-line of 6 a clock, on any kind of Plain; especially those having Centers.

The Inclination of Meridians,* 1.22 is the num∣ber of degrees and minutes, counted on the Aequinoctial, between the Meridian or Hour-line of 12: and the Substile being the distance, between the Meridian of the place, viz. 12 a clock, and the Meridian of the Plain, but counted on the Aequinoctial; and doth serve to make the Table of Hour-Arks at the Pole, and to prove your work.

The Lines Parallel to 12,* 1.23 are two Lines peculiar to this way of Dialling by the Se∣ctor, and are only two Lines drawn equi∣distant from, and parallel to the Hour-line of 12.

The Contingent or Touch-line in this way of Dialling with Centers,* 1.24 is a Line drawn parallel to the Hour-line of 6; but in those without Centers, it is drawn alwayes per∣pendiculer to the Substile, and so may it be also, if you please, in those with Centers also.

* 1.25The Vertical Line on the Plain, is the same with the Perpendiculer-line on the Plain, being perpendiculer to the Horizon∣tal-line.

* 1.26By the word Nodus, is meant a Knot or Ball, on the Axis or Stile of the Dial, to make a black-shaddow on the Dial, to trace out the Suns motion in the Heavens; or sometimes an open or hollow-place in the Stile, to leave a light-place to do the same office.

But by Apex is meant the same thing, when the Top-end, or Point of an upright Stile shall shew the Hour and Suns place, as the Spot doth in Celing-Dials, where the Hours and Quarters are all of one length, and distinguished by their tullours or great∣ness only.

* 1.27The Perpendiculer height of the Stile, is nothing else but the nearest distance from the Nodus or Apex to the Plain.

The Foot of the Stile is properly right un∣der the Nodus or Apex at the nearest di∣stance.* 1.28

* 1.29The Vertical-Point, is a Point only used in Recliners and Incliners, being a Point right over, or under the Apex; and yet in the Meridian, being let fall from the Zenith, by or through the Apex or Nodus, to the Plain in the Meridian-line.

The Axis of the Horizon,* 1.30 is only the mea∣sure from the Apex to the Vertical-point last spoken to, being the Secant of the com∣plement of the Reclination to the Radius of the Perpendiculer height of the Stile.

Erect, is when Plains are upright,* 1.31 as all Walls are intended to be.

Direct,* 1.32 is when the Dial-plain beholdeth one of the Four Cardinal Points of the Horizon, as South or North, East or West, that is to say, when the Pole of the Plain, being 90 degrees every way from the Plain, doth lie precisely in one of those Four Car∣dinal Azimuths: Which in an Erect and Direct-Plain will be in the Horizon.

Declining, and Reclining,* 1.33 or Inclining-Plaines, are as the upper or under-side of Roofs at any Oblique Scituation from the Cardinal Points of the Horizon.

Oblique, is only a wry, slanting,* 1.34 crook∣ed; contrary to direct, right, plain, flat, or perpendiculer; and applied variously, as to the Sphear, to Triangles, to Dial-plains, to Discourse and Conversation.

Circles of Position,* 1.35 or rather Semi-circles making 12 Houses, are Circles, whose Pole or Meeting-point is in the Meridian and Ho∣rizon of every Country, dividing the Aequi∣noctial into 12 equal parts, being then cal∣led Houses, when used in Astrologie, and

some times drawn on Sun-Dials.

But when they are used in Astronomy, they require a more near account, as to de∣grees and minutes.

Of certain Terms in Astronomy, and Spherical Definitions of Points and Lines in the Sphear.

NOt to be curious in this matter, a Sphear may be understood to be a united Spherical Superficies, or round Bo∣dy, contained under one Surface; in the middle whereof is a Point or Center, from whence all Lines drawn to the Circum∣ference are equal: Or you may conceive a Sphear to be an Instrument, consisting of several Rings or Circles, whereby, the sen∣sible motion of the Heavenly Bodies are conveniently represented.

For the better Explanation whereof, A∣stronomers have contrived thereon, viz. on the Sphear, ten imaginary Points, and ten Circles, which are usually drawn on Globes and Sphears; besides others not usually drawn, but apprehended in the fancy, for

Demonstrations-sake, in Spherical Conclu∣sions.

The ten Points are, the two Poles of the World, the two Poles of the Zodiack, the two Aequinoctial Points, the two Solstitial Points, the Zenith, and Nadir.

The ten Circles are, The Horizon, the Meridian, the Aequinoctial, the Zodiack; the two Colures, viz. that of the Equi∣noxes, and that of the Solstices; the Tro∣pick of Cancer, and the Tropick of Capri∣corn; and the two Polar-Circles, viz. The Artick or North, the Antartick or South, Polar-Circle.

The first six, are great Circles, cutting the Sphear into two equal parts: And the four last are lesser Circles, dividing the Sphear unequally.

All which Points and Circles shall be represented by the Figure of the Analem∣ma, from whence the Trianguler Quadrant is derived, as a general Instrument, and also by the Horizontal projection of the Sphear fitted for London, being better for the fancy to apprehend the Mystery of Dial∣ling, one thing mainly intended in this Discourse.

Of the 10 Points in the Sphear.

* 1.36THe two Poles of the World, are the two Points P and P in the Analem∣ma, [ 1] being directly opposite one to another; [ 2] about which two Points, the whole frame of the Heavens moveth from East to West; one of which Poles may alwayes be seen by us, called the Artick or North-Pole; represented in the particular Scheam by the Point P. The other being not seen, is not represented in the particular Scheam; but the Line PEP, in the general Scheam, drawn from Pole to Pole, is called the Axis▪ or Axeltree of the World, because the whole Sphear ap∣pears to move round about it.

* 1.37The Poles of the Zodiack are two Points diametrically opposite also, upon which Points the Heavens move slowly from West [ 3] to East, represented by the two Points, I [ 4] and K, 23 degrees and 31 minutes distant from the two former Poles, in the Analem∣ma, and by the Point PZ in the Horizon∣tal projection; but the other Pole of the Zodiack cannot be represented in that par∣ticular Scheam.

* 1.38The Equinoctial Points, are the Points of Aries and Libra; to which two Points, [ 5] [ 6]

when the Sun cometh along the Ecliptick, it maketh the Dayes and Nights equal in all places; at Aries March 10th or 11th; to Libra about the 13th of September, where the Spring, and Autumn begins; being re∣presented in the Analemma by the Point ••, and in the particular Scheam by the Points E, and W.

The two Solsticial Points, are represen∣ted one by the Point ♋,* 1.39 and the other by the Point ♑, in both Scheams; to which [ 7] Points when the Sun cometh along the E∣cliptick, [ 8] it makes the Dayes in Cancer ♋, longest; in Capricorn ♑, shortest; ♋ be∣ing about the 11th of Iune; and ♑ about the 11th of December.

The Zenith is an imaginary Point right over our heads,* 1.40 being every way 90 degrees [ 9] distant form the Horizon; in which Point all Azimuth Lines do meet, represented by the Points Z, in both Scheams.

The Nadir is an imaginary Point under our feet, directly opposite to the Zenith,* 1.41 [ 10] represented by the Point N in the Analem∣ma, but not in the particular Scheam, be∣cause it is not seen at any time.

Of the Circles of the Sphear.

* 1.42THe Horizon is twofold, viz. Rational, and Sensible: The Rational Horizon, is an imaginary great Circle of the Sphear, every where 90 degrees distant from the Zenith, and Nadir; Points cutting, or divi∣ding the whole Sphear into two equal parts, the one called, The upper or visible Hemi∣sphear; the other the lower or invinsible He∣misphear.

This Rational Horizon, is distinguished also into Right, Oblique, and Parallel-Horizon.

1. The Right Horizon is when the two Poles of the World lie in the Horizon, and the Equinoctial at Right Angles to it; which Horizon is peculiar to those that live under the Equinoctial, who have their Dayes and Nights alwayes equal, and all the Stars to Rise and Set, and the Sun to pass twice in a year by their Zenith-point, thereby ma∣king two Winters, and two Summers; Their Winters being in Iune and December, and their Summers, in March and Septem∣ber.

2. The Oblique Horizon is when one Pole-point is visible, and (the other not) having E••evation above, and depression be∣low

the North or South part of the Hori∣zon, according to the Latitude of the place: in which Horizon when the Sun cometh to the Equinoctial, the Dayes and Nights are only then equal; and the nearer the Sun comes to the visible Pole, the Dayes are the longer, and the contrary; also some Stars never set, and some never rise in that Hori∣zon: And all Horizons but two, are in a strict sense Oblique-Horizons, viz. The Right Horizon already spoken to: And

The Parallel Horizon, is that Horizon which hath the Equinoctial for its Horizon, and one of the Pole-points for its Zenith; peculiar only to those Inhabitants under the Pole, (if any be there.) In which Ho∣rizon, ona half of the Sphear doth only al∣wayes appear, and the other half alwayes is hid; and the Sun, for one half year, doth go round about like a Skrew, making it continual Day, and the other half year is continual Night, and cold enough; which Circle in the Analemma is represented by the Line HES, but in the particular Scheam by the Circle NESW.

The visible or sensible Horizon, is that Circle where the Heavens and the Earth seem to touch, where the sight of the Sun and Moon doth seem to begin, or cease to ap∣pear in our sights, being not much differing

[illustration]

in Observation from the true Horizon: and from thence hath been called by Blagrave, and others, The Finitor, or ender of our sight of the Heavenly Bodies.

The Meridian is a great Circle which pas∣seth through the two Pole-points, the Zenith and Nadir, and the North and South-points of the Horizon, and is called Meri∣dian, becuse when the Sun (or Stars) co∣meth to that Circle, it maketh Mid-day, or Mid-night, which is twice in every 24 hours: Also all places, North and South, have the same Meridian; but places that lie Eastward, or Westwards, have several Me∣ridians. Also, when the Sun or Stars come to the South, or North-part of the Meri∣dian, their Altitudes are then highest, and lowest. And the difference of Meridians is the difference of Longitudes of Places, no∣ted by the Circle ZHNS in the Analem∣ma; and NZ ♋ S in Horizontal-projection.

The Equinoctial is a great Circle, every where 90 degrees distant from the two Poles of the World, dividing the Sphear into two halfs, called the North and South Hemisphear; and is called also the Aequa∣tor, because when the Sun passeth by it twice a year, it makes the day and nights equal in all places; noted by WAEE, and AEEAE in both.

The Zodiack, or Signifer,* 1.43 is another great Circle that divides the Sphear and E∣quinoctial into two equal parts,* 1.44 whose Poles are the Poles of the Zodiack, being 90 degrees from it; and it inter-sects the Equinoctial in the two Points of Aries and Libra; and one part of it doth decline Northward, and the other Southward, 23 degrees 31 minutes, as the Poles of the Zodiack decline from the North and South-Poles of the World: The breadth of this Zodiack, or Girdle, is counted 14 or 16 de∣grees, to allow for the wandring of Luna, Mars, and Venus; the middle of which breadth is the Ecliptick-Line, because all Eclipses are in, or very near in this Line. And this Circle is divided into 12 Signs, and each Sign into 30 degrees, according to their Names and Characters, ♈ Aries, ♉ Taurus, ♊ Gemini, ♋ Cancer;, ♌ Leo, ♍ Virgo, ♎ Libra, ♏ Scorpio, ♐ Sagit∣tarius, ♑ Capricornius, ♒ Aquarius, ♓ Pisces. 6 being Northern, and the 6 latter Southern.

The two Colures are only two Meridians, or great Circles,* 1.45 crossing one another at Right Angles;* 1.46 the one Colure passing through the Poles of the World, and the Points of Aries and Libra, there cutting the Equinoctial and Ecliptick: And the other

Colure passeth by the Poles of the World also, and cuts the Ecliptick in ♋, and ♑, making the Four Seasons of the year; that is, the equal Dayes, called the Equinoctial-Colure; and the unequal Dayes, in Iune and December, called the Solsticial-Colures, represented in the Analemma by ZP ♋ NS, and PEP; and in the particular Scheam by WPE, and NPS, the Sol∣sticial-Colure.

* 1.47The lesser Circles are the Tropicks of ♋, and ♑;* 1.48 being the Lines of the Suns moti∣on in the longest and shortest dayes, noted in the Scheams by ♋, ☉, ♋, and 6 ♋ E, and ♑, ♑; and W ♑ 6; to which two Circles when the Sun cometh, it is on the 11th of Iune, and the 11th of December, making the Summer and Winter Solstice.

* 1.49The Polar Circles, are two Circles drawn about the Poles of the World, as far off as the Poles of the Zodiack are,* 1.50 viz. 23 de∣grees, 31 minutes; That about the North-Pole is called the Artick, and that about the South the Antartick, being opposite thereunto, shewed in the Analemma by II, and KK; and by the small Circle about P in the particular Scheam.

Of the other Circle imagined, but not described on Sphears or Globes.

1. HOurs are great Circles,* 1.51 passing through the two Poles, and cut∣ting the Equinoctial in 24 equal parts, as the Lines P1, P2, P3, &c. in the Particular; and P ☉ H in the Analemma; such also are degrees of Longitude, and Meridians; the Meridian being the hour of 12.

2. Azimuths are great Circles,* 1.52 passing through, or meeting in the Zenith and Na∣dir-points, numbred and counted on the Horizon, from the Four Cardinal Points of North and South, East and West, accor∣ding to Four 90ties, or 180 degrees, or according to the 32 Rombs or Points of the Compass, as Z ☉ A, and ZE, the Azi∣muth of East and West, being called the prime Virtical, viz. SE, WZ.

3. Almicanters, or Circles of Altitude,* 1.53 are lesser Circles, all parallel to the Horizon, counted on any Azimuth from the Horizon to the Zenith, to measure the Altitude of the Sun, Moon, or Stars above the Horizon, being the Portion of some Azimuth, between the Center of the Sun, or Star; and the

Horizon, commonly called its Altitude a∣bove the Horizon, showed by A ☉ in the Analemma, and SAE in the particular Scheam.

* 1.544. Parallels of Declination, are parallels to the Equinoctial, as the Almicanters were parallel to the Horizon, as ♋ ☉ ♋, the greatest Declination or Circle of ♋: These parallels have the 2 Poles of the World for their Centers, and in respect of the Sun or Stars are called degrees of Declination; but in respect of the Earth, degrees of La∣titude; being the Arch on the Meridian of any place, between the Pole and Horizon, as 4 ♋ 4 in the Particular, and HP, in the Analemma.

* 1.555. Parallels of Latitude, in respect of the Stars, are Lines drawn parallel to the Ecliptick, as the Almicanters were parallel to the Horizon; so that the Latitude of a Star is counted from the Ecliptick toward the Poles of the Zodiack; but the Sun be∣ing alwayes in the Ecliptick, is said to have no Latitude.

6. Degrees of Longitude, in respect of the Heavens,* 1.56 are measured by the degrees on the Ecliptick, from the first point of Aries forward, according to the succession of the 12 Signs of the Zodiack.

But Longitude on the Earth, is counted on the Equinoctial Eastwards, from some principal Meridian on the Earth, as the Isles of Azores, or the Peak of Tenneriff, or the like.

7. Right Ascention is an Arch of the E∣quinoctial (counted from the first Point of Aries) that cometh to the Meridian with the Sun, Moon, or Stars, at any day,* 1.57 or time of the year, being much used in the following discourse, noted in the Analem∣ma by EH, or the like; but counted as af∣terward is shewed.

8. Oblique Ascention is an Arch of the Equinoctial,* 1.58 between the beginning of A∣ries, and that part of the Equinoctial that riseth with the Center of a Star, or any portion of the Ecliptick in an Oblique-Sphear.

9. Ascentional Difference, is the difference between the Right and Oblique Ascention,* 1.59 to find the Sun or Stars rising before or af∣ter 6.

10. Amplitude is an Arch of the Hori∣zon,* 1.60 between the Center of the Sun and the true East-point, at the very moment of Ri∣sing, represented by ♋ F, in the particular Scheam, and GE, and FE in the Analem∣ma: useful at Sea.

* 1.6111. A Circle of Position is one of the 12 Houses in Astronomy or Astrology.

12. An Angle of Position, is the Angle made in the Center of the Sun, between his Meridian, or Hour, and some Azimuth, as the Prime, Vertical, or the Meridian, or any other Azimuth, being useful in Astronomy, and sometimes in Calculation, represented bp P ☉ Z in the Analemma.

Thus much for Astronomical terms.

CHAP. V. Some Uses of the Trianguler Quadrant.

Use I. And first to rectifie the Rule, or make it a Trianguler Quadrant.

FIrst open the Rule, and put in the loose piece into the two Mortice-holes, (which putting together makes it a Trian∣guler Quadrant) but if you do not use the loose-piece, then open it to an Angle▪ of 60 degrees, which is thus exactly done: Mea∣sure from the Rectifying-point, to any Num∣ber

on the Sines or Lines; then keeping the Point of the Compasses still fixed in the Rectifying-point, turn the other to the Common-Line of the Hour and Azimuth-Line, that cuts the Rectifying-point, and there keep it; then removing the Point of the Compasses from the Rectifying-point, open or close the Rule till the other Point shall touch the distance first measured in the Line of Sines or Lines, then shall you see the Lines on the Head, and Moveable-leg, to meet; and also see quite through the Rectifying-point, to thrust a Pin quite through; and thus is it set to an Angle of 60 degrees, without the help of the loose-piece, or to an Angle of 45, or whatsoever else the Rule is made for.

Use II. To observe the Sun or a Stars Altitude above the Horizon.

Put a Pin in the Center-hole on the Head-Leg, and another in the Rectifying-point, and a third (if you please) in the end of the Hour-line on the Moving-leg. Then on the Pin in the Leg-center, hang a Thred and Plummet; then if the object be low, viz. under 25 degrees high; Look along by the two Pins in the Rectifying-point,

and the Moving-leg, and see that the Plummet playeth evenly and steady, then the degrees cut by the Thred▪ shall be the Altitude required, counting from 60/0 to∣ward the Head, as the smaller Figures shew.

But if the Object be above 25 degrees high, then look by the Pin in the Rectify∣ing-point, and that on which the Plummet hangeth; and observe as before, and the Thred shall shew the Altitude required, as the Figures before the Line sheweth; If you have Sights, use them instead of Pins; and by Practice learn to be accurate in this Work, the ground and foundation in every Obser∣vation; and according to your exactness herein, is the following Work also.

Note also, that this looking up toward the Sun, is only then when the Sun is in a cloud, and may be seen in the Abiss, but will not give a clear shadow: Or else you must use a piece of Red, or a Blue, or Green Glass, to darken the luster that it offend not the eyes.

But if the Sun be clear and bright, then you need not look up toward it, but hold the Trianguler Quadrant so, that the sha∣dow of the Pin in the Center may fall just on the shadow of the Pin in the Rectifying-point, and both those shadows on your

finger beyond them, and the Plummet be∣ing somewhat heavy, and the Thred small and playing evenly by the Rule, then is the Observation so made, likely to be near the very truth.

Note also further, That the shaking of the hand, you shall find will hinder exact∣ness; therefore, when you may, find some place to lean your Body, or Arm, or the Instrument against, that you may be the more steady.

But the surest and best way is with a Ball-socket, and a Three-leg-staff, such as Land Surveighers use to support their In∣struments withal, then you will be at liber∣ty to move and remove it, to and fro, till the Sights or Pins, and Plummet and Thred play to exactness; without which care and exactness, you cannot certainly and know∣ingly attain the Sun's or a Star's Altitude to a minute, either by this nor any other In∣strument whatsoever, though they be never so truly made: Yet I dare affirm to do it, or it may be done as well by this, as by any other graduated Instrument whatsoever: The Line of degrees on this, being only two thirty degrees of a Tangent laid together; of which, that on the in-side of the loose-piece is the largest, and consequently the best, to distinguish the minuts of a degree withal.

Use III. To try if any thing be Level, or Upright.

Set the Moving-leg of the Trianguler Quadrant on the thing you would have to be Level; then if the Thred play just on 60 degrees, or the stroke by 60|0, then is it Level, or else not.

But to try if a thing be upright or not▪ apply the Head-leg to the Wall or Post, and if it be upright, the Thred will play just on the common Line between the Lines and Sines on the Head-leg, and cut the stroke by 90 on the Head of the Instrument, or else not.

Use IV. To find readily what Angle the Sector stands at, at any opening.

First, on the Sector side, about the Head, is 180 degrees, or twice 90 graduated to every two degrees; so that opening the Rule to any Angle, the in-side of the Mo∣ving-leg, passing about the semi-circle of the Head, sheweth the Angle of opening to one degree. But to do it more exactly, do thus:

The two Lines of Sines that issue from the Center in Rules of a Foot, shut, are drawn usually just 5 degrees assunder; or rather the two innermost Lines, on each Leg, are always just one degree from the in∣side, so that if you put a Center-pin in the Line of Tangents, just against the Sine of 30, it makes the two innermost Lines that come from the Center, just 2 degrees as∣sunder, which is easie to remember either in adding or substracting as followeth, two wayes.

1. Take the Latteral Sine of 30, viz. the measure from the Center to 30: the Compasses so set, set one Point in the Cen∣ter-pin in the Tangents just against 30; and turn the other till it cut the common Line, in the Line of Sines on the other Leg, and there it shall shew what Angle the two innermost-Lines make, counting from the end toward the Head, and two degrees less is the Angle the Sector stands at, both on the in-side and out-side, the Legs being parallel; which Number must nearly agree with what the in-side of the Leg cuts on the Head-semicircle, or there is a mistake.

As thus for Example.

Suppose I open the Rule at all adventures▪ and taking the Latteral Sine of 30 from the Sines on the Sector-side, and putting one Point of the Compass in the Center on the Tangents, right against the Sine of 30 on the other Leg (or the beginning of the Se∣cants on the same Leg) and turning the o∣ther Point to the Line of Sines on the other Leg, it cuts the Sine of 60 on the inner∣most Line that comes from the Center; then I say, that the Lines of Sines and Tan∣gents are just 30 degrees assunder, and the in-side or out-side of the Legs but 28, viz. two degrees less, as a glance of your eye to the Head will plainly shew.

2. This way will serve very well for all Angles above 20, and under 80: But for all under 20, and above 80, to 120, this is a better way;

Open the Rule to any Angle at pleasure, and take the distance parallelly (that is, a∣cross from one Leg to the other) between the Center-pin at 30 in the Sines, and that in the Tangents right against it, and measure it latterally from the Center, and it shall shew the Sine of half the Angle the Sines and Tangents stand at; and one degree less

is the Sine of half the Angle the Sector stands at.

Example.

Suppose that opening the Sector at ad∣ventures, or to the Level of any thing, I would know the Angle it stands at: I take the parallel Distance between the two Cen∣ters; and measuring it latterally from the Center, I find it gives me the Sine of 51 de∣grees, viz. the half Angle the Lines stand at; or 50, the Angle the Rule stands at; which doubled, is 102 for the Lines, or 100 for the Legs of the Sector, as a glance of the eye presently resolves by the inner-edge of the Moving-leg, and the divided semi-circle.

3. On the contrary, Would you set the Legs or Lines to any Angle, take the half thereof latterally, or one degree less in the half for the Legs, and make it a Parallel in the two Centers, and the Sector is so set accordingly.

Example.

I would set the Legs to 90 degrees, or a just Square: take out the Latteral Sine of 44, one degree less than 45, the half of 90, and make it a Parallel in the two Centers abovesaid, and you shall find the Legs set

just to a Square, or Right-Angle, as by looking to the Head you may nearly see.

At the same time if you take Latteral 30, and lay it from the Center, according to the first Rule, you shall see a great de∣ficiency therein, as above is hinted.

Use V. The Day of the Month being given, to find the Suns Declination, true place in the Zodiack, Right Ascention, Ascentio∣nal Difference, or Rising and Setting.

1. Lay the Thred to the Day of the Month (in the upper Line of Months, where the length of the Dayes are increasing; or in the lower-Line, when the Dayes are de∣creasing, according to the time of the year) then in the Line of degrees you have his De∣clination; wherein note, that if the Thred lie on the right hand of 60|0, then the Suns Declination is Northwards; the contrary∣way is Southwards: Also on the Line of the Sun's Right Ascention, you have his Right Ascention, in degrees and hours, (counting one Hour for 15 degrees) as the Months proceed from March the 10th, or Equino∣ctial, the Right Ascention being then 00, and so forward to 24 hours, or 360 de∣grees, as the Months and Dayes proceed.

Again, on the Line o•• the Sun's true place, you have the sign and degree of his place in the Ecliptick, Aries, or the Equi∣noctial-point being the place to begin, and then proceeding forward as the Months and Dayes go.

Lastly, on the Hour-line you have the Ascentional-difference, in degrees and mi∣nutes, counting from 6; or the Suns Ri∣sing, counting as the morning hours proceed; or his Setting, counting as the afternoon hours proceed.

Of all which, take two or three Examples.

1. For March the 12^th, lay the Thred to the Day, and extend it streight; then on the Line of degrees, it sheweth near 1 de∣gree, or 54 minutes Northward.

2. The Suns Right Ascention, is in time 8 minutes and better, or in degrees, 2 deg. 5 minutes.

3. The Suns Place, is 2 degrees and 16 minutes in Aries, ♈.

4. The Ascentional Difference, is 1 de∣gree and 10 minutes; or the Sun riseth 4 minutes before, and sets 4 minutes af∣ter 6.

Again, for May the 10th, the Thred laid thereon, cuts in the degrees, 20 deg. 9 min. for Northern Declination; and 57 deg. 24 min. or 3 hours 52 min. Right Ascention; and 29 37 in ♉ Taurus for his true place; and 27 12 for difference of Ascentions, or riseth 11 minutes after 4, and sets 49 minutes after 7.

Again, on the last of October, or the 21 of Ianuary, near the Declination, is 17 22 Southwards, the Right Ascention for October 31, is 225, 53, for Ianuary 21, 314 21: The true place for October 31, is ♏ Scorpio, 18 deg. 22 min; but for Ianuary 21, ♒ Aquarius 11, 52; accor∣ding as the Months go to the end at ♑, and then back again; but the Ascentional diffe∣rence, and Rising and Setting, is very near the same at both times, viz. 23 10, and Riseth 32 minutes, and more, after 7; and Sets 28 minutes less after 4.

Use VI. The Declination of the Sun, or a Star, given, to find his Amplitude.

Take the Declination, being counted on the particular Scale of Altitudes, between your Compasses; and with this distance, set one foot in 90 on the Azimuth-Line, the

other Point applied to the same Line, shall give the Amplitude, counting from 90.

Example.

The Declination being 12 North, the Am∣plitude is 19 deg. 15 min. Northwards. Or the Declination being 20 South, the Am∣plitude is 34 deg. 10 min. Southwards.

Use VII. The Right Ascention and Ascentional-dif∣ference being given, to find his Oblique-Ascention.

When the Declination is North, then the difference between the Right Ascention, and the Ascentional-difference, is the Ob∣lique-Ascention.

But in Southern declinations, the sum of the Right Ascention, and difference of As∣scentions, is the Oblique Ascention.

Example.

On or between the 25 and 26 of Iuly, the Oblique-Ascention is by Substraction 112, 15: On the 30th of October, the Ob∣lique-Ascention is 337, 45 by Addition.

Use VIII. The Day of the Month, or Sun's Decli∣nation and Altitude being given, to find the Hour of the Day.

Take the Suns Altitude, from the parti∣cular Scale of Altitudes, setting one Point of the Compasses in the Center, at the be∣ginning of that Line; and opening the o∣ther to the degree and minute of the Sun's Altitude, counted on that Line; then lay the Thred on the Day of the Month (or De∣clination) and there keep it: Then carry the Compasses (set at the former distance) a∣long the Line of Hours, perpendiculer to the Thred, till the other Point, being tur∣ned about, will but just touch the Thred; the Compasses standing between the Thred and the Hour 12, then the fixed Point in the Hour-Line shall shew the hour and mi∣nute required; but whether it be the Fore or Afternoon, your judgment, or a second observation must determine.

Example.

On the first of August in the morning, at 20 degrees of Altitude, you shall find it to be just 52 minutes past 6; but at the same

Altitude in the afternoon, it is 7 minutes past 5 at night, in the Latitude of 51 32 for London.

Use IX. The Suns Declination and Altitude given, to find the Suns Azimuth from the South-part of the Horizon.

First, by the 4th Use, find the Suns De∣clination, count the same on the particular Scale, and take the distance between your Compasses; then lay the Thred to the Suns Altitude, counted the same way as the Southern-Declination is from 60|0, toward the loose-piece; and when need requires on the loose-piece, then carry the Compasses along the Azimuth-line, on the right-side of the Thred, that is, between the Thred and the Head, when the Declination is Northward; and on the left-side of the Thred, that is, between the Thred and the End, when the Declination is Southward. So as the Compasses set to the Declination, as before, and one Point staying on the Azimuth-line, and the other turned about, shall but just touch the Thred at the nearest distance; then, I say, the fixed-Point shall, in the Azimuth-line, shew the Suns-Azimuth required.

Example 1.

The Sun being in the Equinoctial, and having no Declination, you have nothing to take with your Compasses, but only lay the Thred to the Altitude counted from 60|0 toward the loose-piece, and in the Azimuth-line it cuts the Azimuth required.

Example. At 25 degrees high, you shall find the Suns Azimuth to be 54, 10; at 32 degrees high, you shall find 38, 20, the Azimuth.

Again, At 20 degrees of Declination, take 20 from the particular Scale, and at 10 degrees of Altitude, lay the Thred to 10 counted as before; then if you carry the Compasses on the right-side for North-De∣clination, you shall find 109, 30, from South; but if you carry them on the left-side for South-Declination, you shall find 38, 30, from South.

The rest of the Vses you shall have more amply afterwards.

CHAP. VI. The Use of the Line of Numbers on the Edge, and the Line of Lines on the Quadrantal-side, or on the Sector-side, being all as one.

HAving shewed the way of Numerati∣on on the Lines, as in Chapter the first. Also to add or substract one Line or Number to or from one another, as in Chap∣ter 4th, Explanation the 9th. I come now to work the Rules of Multiplication and Division, and the Rule of Three, direct and reverse, both by the Artificial and Natural-Lines; and first by the Artificial, being the most easie; and then by the Natural-lines both on the Sector and Trianguler Qua∣drant, being alike: and I work them toge∣ther; First, because I would avoid tau∣tology: Secondly, because thereby is bet∣ter seen the harmony between them, and which is best and speediest. Thirdly, be∣cause it is a way not yet, as I know of, gone

by any other. And last of all, because one may explain the other; the Geometrical Fi∣gure being the same with the Instrumental-work by the Natural way.

Sect. I. To multiply one Number by another.

1. By the Line of Numbers on the Edge Artificially, thus:

Extend the Compasses from 1 to the Mul∣tiplicator; the same extent applied the same way from the Multiplicand, will cause the other Point to fall on the Product required.

Example.

Let 8 be given to be multiplied by 6; If you set one Point of the Compasses in 1, (either at the beginning, or at the middle, or at the end, it matters not which; yet the middle 1 on the Head-leg, is for the most part the most convenient) and open the o∣ther to 6, (or 8, it matters not which, for 6 times 8, and 8 times 6, are alike; (but yet you may mind the Precept if you will) the same Extent, laid the same way from 8, shall reach to 48, the Product required;

which, without these Parenthesis, is thus:

• The Extent from 1 to 6, shall reach the same way from 8 to 48. Or,
• The Extent from 1 to 8, shall reach the same way from 6 to 48.

the Product required.

By the Natural-Lines on the Sector-side, or Trianguler Quadrant with a Thred and Compasses, the work is thus;

1. For the most part it is wrought by changing the terms from the Artificial way, as thus;

The former way was, as 1 to 6, so is 8 to 48; or as 1 to 8, so is 6 to 48; but by the Sector it is thus: As the Latteral 6 taken from the Center toward the end, is to the Parallel 10 & 10, set over from 10 to 10, at the end counted as 1; so is the Parallel-distance between 8 & 8, on the Line of Lines taken a-cross from one Leg to the o∣ther, to the Latteral-distance from the Cen∣ter to 48, the Product required.

Or shorter thus.

As the Latteral 8, to the Parallel 10;

So is the Parallel 6, to the Latteral 48.

See Figure I.

2. Another way may you work without altering the terms from the Artificial way, as thus, by a double Radius; Take the Lat∣teral-Extent from the Center to 1, (or from 10 to 9, if the beginning be defective) make this a Parallel in 6 & 6, then the Latteral-Extent from the Center to 8 of the 10 parts between Figure and Figure, shall reach a∣cross from 48 to 48, as before. See Fig. II.

The same work as was done by the Se∣ctor, is done by the Line of Lines, and Thred on the Quadrant-side, that if your Sector be put together as a Trianguler Qua∣drant, you may work any thing by it, as well as by the Sector, in this manner; (or by the Scale and Compass, as in the Fi∣gure I.) and first, as above, Sector-wise.

Take the Extent from the Center to 6 latte∣rally, between your Compasses; set one Point in 10, and with the other lay the Thred in the nearest distance, turning the Com∣pass-Point about, till it will but just touch the Thred, then there keep it; then set one Point of the Compasses in 8, and take the nearest distance to the Thred; this distance laid latterally from the Center, shall reach to 48 the Product.

Or as Latteral 6, to Parallel 10; so is the Parallel 8, to the Latteral 48, the Pro∣duct required.

Or the last way by a double Radius, or a greater and a smaller Scale, as the Latteral Extent from the Center to 1, is to the Paral∣lel 6, laying the Thred to the nearest di∣stance; so is the latteral Extent from the Center to 8 parts, less than the 1 before ta∣ken, carried parallelly along the common-line, till the other Point will but just touch the Thred, it shall on those conditions stay only at 48, the Product required. Observe and note the Figure, by protraction, with Scale and Compass.

3. But if you have an Index and a Square, as is used in the Demonstrative Work of Plain Sayling, as you shall have afterwards, then the representation of this Natural-way will most evidently appear, as thus:

Set the edge of the Square to 1; on the Line of Lines, counted from the Center downwards, where the Figure 1 is, then move the Index till the edge cut 8 of the small parts, counted on the Square, from the Line of Lines toward the end of the Square, and then there keep the Index; then remove the Square downward to 6, on the Line of Lines; then there holding it square, you shall see the Index to cut 48 on the Square, counting after the same rate that the 8 parts was accounted. Note the

[illustration]

Figure III. for illustration sake.

From hence you may observe, That the first and third Numbers must alwayes be ac∣counted alike, and on like Scales; and the second and the fourth in like manner on like Scales and counting; and the Latteral-first Number, must alwayes be less than the Parallel-second, in length or quantity, or you cannot work it; which you must make so, either by changing the terms, or using a less Scale, to begin and end upon.

Here you must except a Decimal grada∣tion, as thus; sometimes the same place which is called 10 in the first, may be coun∣ted 100 in the third, and the contrary; or more or less differing in a Decimal Ac∣count.

But if you would see a Figure of the Sector-way of operation, then it is thus; Let the Line C 6, represent one Leg of the Sector; and the other Line C 6, represent the other Leg of the Sector; then take 1 out of any Scale, as 1 inch, or one tenth part of a foot, or what you please: or the distance from the Center to 1, or 2, on the Line of Lines between your Compasses: put this distance over in 6 & 6, of the Line of Lines. Then is the Sector set to its due Isosocles Angle.

Then take 8 parts, or rather 8 tenth

parts of the former 1, from the Scale from whence you took the first Latteral distance, and carry it parallel between the Line of Lines till it stay in like parts, which you shall find to be at 48, the Product requi∣red.

Or to get the Answer in a Latteral-line, is generally most convenient, by changing the terms; work thus: Take the Latteral-distance from the Center to 8, on the Line of Lines, make it a Parallel in 10 & 10; then the Sector being so set, take the Parallel-distance between 6 & 6, and lay it latteral∣ly from the Center, and it shall reach to 48, the Product required. See Fig. IIII.

Thus you see that the way of the Sector-side, and of the Quadrant-side, is in a man∣ner all one; and the laying of the Thred, or Index to the nearest distance, is the same with setting the Legs of the Sector to their Angle; and the taking the nearest distance from any Point or Number to the Thred, is the same with taking parallelly from Point to Point, or from Number to Number: So that having thus fully explained the Lat∣teral and Parallel-Extent, and laying of the Thred, and setting of the Sector, the fol∣lowing Propositions will be more easie, and ready; and to that purpose, these brief Marks for Latteral, Parallel, and Nearest-distance,

will frequently be used; as thus, for Latteral, thus —; for Parallel, thus =; for Nearest-distance, thus |, or ND, thus: for the Sine of 90, or Radius, or Tangent of 45, thus R; &c.

In all which wayes you may see, that for the want of several Radiusses, which do properly express the unites, tens, hundreds, and thousands, and ten thousands of Num∣bers, there must a due and rational account, or consideration, go along with this Instru∣mental manner of work, else you may give an erroneous answer to the question pro∣pounded; to prevent which, observe, that in Multiplication there must be for the most part, as many figures in the Product, as is in the Multiplicator and Multiplicand, put together; except when the first figures of the Product, be greater than any of the first figures of the Multiplicator, or the Multi∣plicand, and then there is one less;

As for Example.

2 times 2 makes 4, being only one fi∣gure, because 4 is greater than 2; but 2 times 5 is 10, being two figures; where∣in 1, the first figure, is less than 5. Again, in a bigger sum; 52 multiplied by 23, makes 1300, consisting of four figures, as many as

is in the Multiplier and Multiplicand put together; but if you multiply 42 by 22, it makes but 924, which is but three figures; because the first figure 9 is greater than 2 or 4, as in the former, the first figure 1 was less than 5 or 2: And this Rule is general, as to the number of places, or figures, in any Multiplication whatsoever; but note, that no Instrument extant, and in ordinary use, is capable to express above 5 or 6 places: Yet with this help you may come true to 5 places, with a good Line of Numbers.

As thus;

Suppose I was by a Line of Numbers to multiply 168 by 249; the extent from 1 to 168, will reach the same way from 249, to 41832: Now by the Line of Numbers, you can only see but the 418, and estimate at the 3; but the last figure 2, I cannot see by any Line usually put on two foot Rules, therefore the 168, and 249 being before you, say (according to the vulgar Rules of Multiplication) 9 times 8 is 72; therefore 2 must needs be the last figure; and if you can see the former 4, you have the Product infallibly true: if not, multiply a figure more: But by this help you shall be sure

to come right alwayes to 4 figures, or pla∣ces, in any Multiplication whatsoever.

4. Also, by these operations, you may plainly see, That the Line of Numbers, or Gunters-Line, as it is usually called, is the easiest and exactest for Arithmetical opera∣tions, being performed with an Extent of the Compasses only, without any opening or shutting of the Rule, or laying a Thred or Index: But in Questions of Geometry, where a lively draught or representation is required, as to the reason of the work, there the Natural-Lines are more demon∣strative.

In which Natural Work, Note, That the Parallel-distance must alwayes be the greatest, or you cannot work it, unless you make use of a greater, and a lesser Scale; to which purpose, this Instrument is well furnished, with three or four Ra∣diusses, bigger and less, both of Sines, Tangents, and Secants, and Equal-Parts, as in their due places shall be observed; and taken notice of, in the Astronomical-Work.

And note also, That if the Line of Lines were repeated to 4 Radiusses, or 400 instead of 10, you might work right-on to 4 figures; but then the Ra∣diusses

must be very small, or the Instru∣ment very large.

Therefore this of 10, being the most usual, I shall make use of, and work every Question the most convenient way; that by a frequent Practice, the young Beginners, for whom only I write, may see the Reason and Nature of the Work, and the sooner understand it.

For a Conclusion of this Rule of Multi∣plication, take three or four Examples more, both by the Line of Numbers, and Equal-Parts also;

1. First 15 foot, 8 tenths, multiplied 9 foot and 7 tenths, by the Line of Num∣bers; the Extent of the Compasses from 1 to 9 foot, 7 tenths, shall reach the same way from 15—8, to 153, 26.

By the Line of Lines, or Equal Parts.

As the Latteral — 15—8, to the Paral∣lel = 10, at the End counted as 1;

So is the Parallel = 9—7, to the Lat∣teral — 153—26.

Where you may observe, that the first and fourth, are measured on like Scales; and the second and third, also on like Scales.

But note; that as you diminished the ac∣count in the third = work, counting 9—7 less than 10 reckoned as 1:

So likewise in the — fourth, you count 153—26, which is less in Extent than the 15—8, first taken Latterally; yet is to be read as before, viz. 153—26, because 9 times 15, must needs be above 100.

2. As 1, to 9 foot 10 inches; So is 10 foot 9 inches, to 105 foot 6 inches ½.

To work this properly by the Line of ••umbers, you are first by the inches and ••ot-measure on the in-side of the Rule, to ••educe the inches into decimals of a foot; as ••hus: Right against 10 inches, in the Line 〈◊〉〈◊〉 Foot-measure, you shall find 83 ••/••. Also, ••ight against 9 inches, on the Foot-measure, ••ou shall find 75; this being done, which 〈◊〉〈◊〉 with a glance of your eye only, on those two Lines, then the work is thus:

As 1, to 10 75; so is 9 83, to 105 ½, or 105 foot, 6 inches: Now for the odd 6 square inches, you cannot see them on the Rule, but must find them by the help before 〈 math 〉〈 math 〉 mentioned, as thus, Having set down the 9—10, & the 10—9, as in the Margent, say by vulgar Arithmetick thus, Ten times

9 is 90; for which you must set down 7 foot and 6 inches, which is the 6 inches you could not see on the Line of Numbers, and there must needs be 105 foot, and not 10 foot and an half, and better, which is as to the right Number of Figures.

But by the Line of Lines as the — 10— 75, from any Scale, as the Line of Lines doubled, or foot-measure, or the like, is to the = 10, so is the = 9—83, to the Lat∣teral 105—6 ½, as before, though not so quick or plain, as by the Line of Num∣bers.

3. As 1, to 1528, so is 3522, to 5371616, the true answer; which indeed is to more places than possible the Rule can come to without the help last mentioned: But if the Question had been thus, with the same Figures, 15 foot 100/28 parts, by 35 foot 100/22 parts, as so many feet, or yards, and hundred parts: Then the answer would be as before, 537 foot; and cutting off four figures for the four figures of Fractions, both in the Multiplicator and Multiplicand, viz. the 1616, which makes near 2 inches of a foot more, or 16/100 parts of a yard more, which in ordinary measuring is not con∣siderable.

By the Line of Lines.

4. As — 15—28, to = 1, next the Center:

So is = 35—22, to 537 — 1616/100000

To multiply 3 pound, 6 shillings, and 3 pence, by it self; the Product is, 10. l.—19. s.—5. d.—1. f.—7/10: For the Ex∣tent from 1, to 3—3125, the Decimal num∣ber for 3 ••.—6. s.—3. d. shall reach from thence to 10—9726, which reduced again, is as before, 10—19—5—2; as followeth.

Note, That in this way of Multiplication by the Pen, works thus; You must first multiply Pounds by Pounds, one over the other, as 3 by 3: Then the Shillings by the Pounds cross-wise both-wayes, as the black-line sheweth. Then Pounds by Pence, as the long Prick-lines sheweth both-wayes also. Then Shillings by Shil∣lings, as the 6 by 6. Then Shillings by Pence, both-wayes, as the short Prick-lines sheweth. Then lastly, the Pence by the Pence, as 3 by 3; whose true power, or denomination, is somewhat hard to conceive▪ which is thus:

First, 3 times 3 (next the left-hand) is 9 Pounds.

Secondly, 3 times 6, is 18 Shillings.

Thirdly, 3 times 6, is 18 shillings again.

Fourthly, 3 times 3, is 9 Pence, as the long Prick-line sheweth.

Fiftly, 3 times 3, is 9 Pence more.

Sixtly, 6 times 6 is 36, every 20 where∣of is 1 Shilling; and every 5 thereof is 3 Pence; and every 1 is 2 Farthings and 4/10ths of a Farthing: So that 36 make 1 Shilling, 9 Pence, 2 Farthings, 4/10ths of a Farthing.

Seventhly, 6 times 3 is 18; every 5 whereof is a Farthing, and every 1 is two tenths of a Farthing, as the short Prick-line sheweth.

Eightly, 6 times 3 is 18, or 3 Farthings and 6 tenths, as before.

Ninthly and lastly; to the right-hand, 3 times 3 is 9; where note, that there goes 60 to make 1 Farthing; therefore 6 makes one tenth of a Farthing: So that here is 1 tenth and ½: Consider the Scheam and the Decimal-work, to prove it exactly to the hundreds of millions of a Pound, and you will find it to be very near.

Example.

[illustration]

〈 math 〉〈 math 〉

Product is 10. 19. 5. 1. 7. ½.

The same Decimally.

〈 math 〉〈 math 〉

Which Sum, being brought to Shillings, Pence, and Farthings, and tenths of Far∣things, is just as aforesaid, viz. 〈 math 〉〈 math 〉

Or else find the Square of the least De∣nomination in 20 s. and divide the Pro∣duct of the Sums being brought to that least Denomination thereby, and the Quo∣tient shall be the Answer required.

Example.

960, the Farthings in 20 s. squared, is 921600. The sum of 3 l. — 6 s. — 3 d. in Farthings, is 3180; multiplied by it self is 10112400: This Product divided by 921600, the square of the Farthings in 20 s. makes 10 l. 896400/••216•••• in the Quotient, which reduced, is 10 l.—19 s.—5 d.—1 7/10.

To find this Decimal Fraction is very easie thus, by the Line of Numbers; for if 20 shillings be 1000, what shall 6 shillings and 3 pence be? Set one Point in 2, repre∣senting 20; and the other in 1, represen∣ting 1000: then the same Extent laid the same way from 6 and ¼, shall reach to 3.125, the decimal fraction for 6 shillings and 3 pence.

Or by inches and foot-measure; for if you account every 8th of an inch a farthing, then every inch is 2 d. and 6 inches is 12 d. right against which, in the Foot-measure, is the Decimal Fraction required: So that right against 12 farthings, or 1 inch and ½ on the Line of Foot-measure, is 125, the De∣cimal Fraction sought for.

Or if one Pound be the Integer, or whole Number, then every 10th part is 2 shil∣lings; and every 5th is 1 shilling: and the inter-mediate pence and farthings is very near the 5th part; for if you conceive a 5th part, or 50 of an hundred to contain one shilling▪ or 48 farthings; then one of 50 is very near one farthing, for 12 and ⅓ is just 3 d. and 25 is just 6 d. 37 ½ is just 9 d. and 50 just 12 d. So that to set the Compass-point to 3 l. 6 s. and 3 d. is to set the Point on 3.3125, as before, which a little practice will make easie.

By the Line of Lines on the Trianguler-Quadrant, or Sector.

As the Latteral 3.3125 is to the Parallel 10, So is the Parallel 3.3125 to the Lat∣teral 10 l. — 19 s.— 5 d.—2 farthings ferè, or 10.973.

Sect. II. To divide one Number by another.

First, by the Line of Numbers the Rule is,

Extend the Compasses from the Divisor to 1, then the same extent of the Compasses, applied the same way from the Dividend, shall reach the Quotient required.

Or the Extent from the Divisor to the Dividend, shall reach the same way from 1, to the quotient required.

Example the first.

Let 40 be a Number given, to be divided by 5; here 40 is the Dividend, and 5 the Divisor; and the answer to how many, viz. 8 is the Quotient.

Extend the Compasses from 5 to 1, the same Extent shall reach the same way from 40 to 8 the Quotient required; or the Ex∣tent from 5 to 40, shall reach the same way from 1 to 8 the quotient required.

But by the Line of Lines, the work is thus;

As the — Latteral 40, to the = Parallel 5;

So is the = 10 counted as 1 to the — 8.

Or so is the = 1, to the — 8 of the smaller part.

Observe the Figure with the Line AB.

Or, as the — 5, to the = 10;

So is the — 40, to the — 8; As the Line CB in the Figure doth demonstrate, being the manner of working by the Trianguler-Quadrant, the way of the Sector being the same.

A second Example.

Let 1668, be divided by 19.

As 19 to 1, so is 1668 to 87 ••8/100;

Or, 15 of 19: Or,

As 19 to 1668, so is 1 to 87 7••/100;

Or, 15 of 19, as before.

For the Extent from 19 to 1668, shall reach the same way from 1 to 87—78/100; the work by the Lines is as before.

In this work of Division, for most ordi∣nary questions, where there is not above four figures in the quotient, you may come very near with a good Line of Numbers, as that on Serpentine-lines, and the like;

but the difficulty is, to know the Number of Figures, which is thus most certainly done: Write down the Dividend, and set the Di∣visor under it, as in the vulgar way of Di∣vision; and there must alwayes be as many Figures as the Dividend hath more than the Divisor; and one more also, when the first figure of the Dividend is greater than the first figure of the Divisor; as if 152178••365 were to be divided by 365, then there would be 3 figures in the Quotient; for the Divisor would be written 3 times under the Dividend, in the usual way of Division; and those figures be 417 almost: But if 9172318, is divided by 8231, you will have 4 figures, viz. 1115, being one fi∣gure more than 3, the difference of places. In this Rule also you may see the excellency of the Artificial-Lines of Numbers, before and above the Natural-Lines.

Sect. III. To two Lines or Numbers given, to find a third in continual Proportion Geometrical.

By the Line of Numbers work thus:

The Extent from one Number to the o∣ther, shall reach the same way from that second to a third, &c.

Example.

As 5 to 7, so is 7 to 9—82;

So is 9—82 to 13—76, &c. ad infinitum.

By the Trianguler-Quadrant, or Sector.

As the Lateral first Number to the Paral∣lel second, laying the thred to the nearest distance, there keep it:

Then so is the Latteral second, to the Pa∣rallel third.

Sect. IV. Any one side of a Geometrical Figure being given, to find all the rest, or to find a Proportion between two or more Right Lines.

This Proposition is most proper to the Line of Lines, and not to the Line of Numbers; and done thus:

Take the Line given, and make it a Pa∣rallel in its respective Numbers; the Thred so laid to the nearest distance, or the Se∣ctor so set, there keep it: then take out all the rest severally, and carry the Compasses parallelly till they stay in like parts, which shall be the Numbers required. Note the Figure.

Note also, That the Line of Sines, and the Thred will readily lie on all the Angles, and be removed from Radius to Radius more nimbly than any Sector whatsoever, only by drawing the Thred streight, and observing on what degree and part it cuts being so laid.

Let ABCDEFG be the Plot of a Field, whose side ED is only given to be

9 Chains; and I would know all the rest: Take ED, make it a = in 9; lay the Thred to ND, or set the Sector to that gage, and there keep it: then measure every side severally, and you shall find what e∣very one is in the same proportional parts, by carrying the Compasses parallelly, till it stay in like parts by the Sector, or ND, by the Quadrant.

Sect. V. To lay down any Number of parts in a Line, to any Scale less than the the Line of Lines.

Take 10, or any other Number, out of your given Scale, or design any distance to be so much as you please, and set one Point in the same Number, on the Line of Lines; and with the other, lay the Thred to the nearest distance, and there keep it, by no∣ting the degree cut by; then take out any o∣ther Number that you would have, setting one Point in that Number on the Line of Lines; and opening the Compasses, till the other Point will but just touch the middle ••f the Thred, at ND, and that shall be the

other part required; or the length of so much, according to the first Scale given.

Example. Figure I.

Let AB represent a Line which is 100 parts, and I would lay down 65, 30, 42, 83 parts of that 100.

First, take all AB between your Com∣passes, and set one Point in 100 at 10 with the other, lay the Thred to ND, then take out 65, 30, 42, 83, &c. parallelly, and lay them down for the parts required, as here you see.

The like work is by the Sector, making AB a = in 100 & 100; then take out = 30, 42, 65, 83, or any thing else for the parts required.

But note, If the Line be too large for your Scale, or Line of Lines, then take half, or one third, or fourth part of the given Line; then if you take half, you must at last turn the Compasses two times: If you take one third, then turn the Compasses three times; which may prove a very con∣venient help in many cases, in Surveying and Dialling.

Note, That by this Rule you may add to, or take from, any given Line, or Num∣ber, any number of Parts or Lines required;

[illustration]

which is called the increasing, or diminish∣ing a Line, to any Proportion required.

Sect. VI. To divide a Line into any Number of Parts.

Take the whole length of the Line be∣tween your Compasses, and setting one Point in the Number of Parts, you would have the Line divided into; with the other, lay the Thred to ND, and there keep it; then take the ND from 1 to the Thred, and that shall divide the Line into the parts required.

Example.

Let AB be to be divided into 7 parts: Take AB, make it a Parallel in 7, laying the Thred to the ND, there keep it; then the = 1 shall divide the Line into 7 parts. But if the Line were to be divided into ma∣ny parts, as suppose 73: Then first, fit the whole Line in = 73; then take out the = 72, 71, & 70, for the odd 3; then the = 10 s. for every 10^th division, then the = 1 for the smaller parts; or else you shall find it almost an impossible thing, to take at

once any distance, which, being turned a∣bove 50 times over, shall not at last happen to be more or less than the desired Number required.

Note, That if the given Number hap∣pen to be such, that the Part will fall too near the Center; as suppose 11, 12, or any Number under 30; then you may double, treble, or quadruple the Number, and then count 2, 3, or 4, for one of the Numbers required.

As for Example.

Suppose I would divide a Line into 15 parts; multiply 15 by 6, and it makes 90: Now in regard you have multiplied 15 by 6, you must take the = 6, instead of the = 1, to divide the Line into 15 parts, be∣tween your Compasses, because the whole Line is set in = 90, instead of = 15; which is 6 times as much as 15.

Note also, if the Line be too big for your Scale, then take half, or a third, and make it a = in the given Line; then take out the = 1, and turn two or three times, to divide the Line according to your mind, when it is too large for your Scale.

These two last are not to be done by the Line of Numbers, but proper for the Line of Sines only; unless you turn your Lines

to be divided into Numbers, and then work by Proportion, as thus;

As the whole Number of Parts, is to the whole Line, in any other parts;

So is 1, to as many of those Parts as be∣longs to 1.

Sect. VII. To find a mean Proportion, between two Lines, or Numbers given.

A mean Proportion between two Lines or Numbers, is that Number, which being multiplied by it self, shall produce a Num∣ber equal to the Product of the two Num∣bers given, when they are multiplied the one by the other.

Example.

Let 4 and 9 be two Numbers, between which a Geometrical mean is required.

4 and 9, multiplied together, make 36:

So also 6, multiplied by it self, is 36: Therefore 6 is a mean Proportional between 4 and 9. To find this by Arithmetick, is by finding the Square-root of 36.

But by the Line of Numbers, thus;

Divide the distance between 4 and 9 in∣to two equal parts, and the middle-point will be found to be 6, the Geometrical mean proportional required.

But to do it by the Line of Lines, do thus;

First, joyn the Lines, or Numbers, toge∣ther, to get the sum of them, and also the half sum; and substract one from the o∣ther to get the difference, and half the dif∣ference; then count the half difference from the Center down-wards; and note where it ends: then taking the half sum between your Compasses, lay your Thred to 00 on the loose-piece; then, setting one Point in the half-difference, on the Line of Lines: See where, on the loose-piece, the o∣ther Point shall touch the Thred; and mark the place, with a Bead on the Thred, or a speck of Ink, or otherwise: for the mea∣sure from thence to the Center is the mean Proportional required.

Or else use this most excellent way by Geometry

Draw the Line AB, and from any Scale

of Equal Parts, take off 4 and 9, and lay them from C, to A and B; then find out the true middle between A and B, as at E; and draw the half Circle ADB; then on C erect a perpendiculer Line, as CD; then if you take CD between the Compasses, and measure it on the same Scale that you took 4 and 9 from, and you shall find it to be 6, the true mean proportional required: being only the way by the Line of Lines, as by considering the Triangle CDE will ap∣pear.

To do this by the Sector, open the Line of Lines to a Right-Angle (by 3, 4, & 5, or 6, 8, & 10. thus: Take 10 Latterally between your Compasses, make it a Paral∣lel in 6 and 8, then is the Line of Lines opened to a Right-Angle; or if your Rule be large, and your Compasses small, then take Latteral 5, the half of 10, and make it a Parallel in 3 and 4, the half of 6 and 8, and it is rectangle also:) Then set half the difference on one Leg from the Center, then having half the sum between your Compasses, set one Point in the half-differ∣ence last counted, and turn the other Point to the other Leg, and there it shall shew the mean proportional Number required.

1. To make a Square, equal to an Oblong.

Find a mean proportion between the length and the breadth of the Oblong, and that shall be the side of a Square equal to the Oblong.

Example.

Let the breadth of the Oblong be 4, and the length 9, the mean proportion will be found to be 6; Therefore a Square, whose side is 6, is equal to an Oblong, whose breadth is 4, and length 9, of the same parts.

2. To make a Square, equal to a Triangle.

Find a mean proportion between the half Base, and the whole Perpendiculer; and that shall be the side of a Square equal to the Triangle.

Example.

If the half-Base of a Gable-end be 10, and the whole Perpendiculer 11-18; the mean proportion between 10 and 11-18, is 10-575; the side of the Square equal to that Triangle, or Gable-end required.

3. To find a Proportion between the Su∣perfecies, though unlike to one another.

First, to every Superfecies, find the side of his equal Square, whether it be Circle, Oblong, Romboides, or Triangle; then the proportion between the sides of those Squares, shall be the Proportion one to ano∣ther.

Example.

Suppose I have a Triangle, and a Circle, and the side of the Square, equal to the Circle, is 10 inches; and the side of the Square, equal to the Triangle, is 15 inches: The Proportion between these two Squares, as they are Lines, is as 10 to 15; but as Superfecies, as 100 to 45; being thus found out, Take the Extent between 15 and 10, on the Line of Numbers, and repeat it two times the same way from 100, and it shall reach to 45, the Proportion as Superfecies, between that Circle and Triangle, whose Squares equal were 15 and 10.

4. To make one Superfecies, equal to ano∣ther Superfecies, of another shape: but like to the first Superfecies given.

First find a mean proportion between the unequal sides of the given Superfecies, that you are to make one like; and find the mean proportion also between the unequal sides of the Figure that you are to make one e∣qual to.

As thus for Example.

I have a Romboides, whose base is 5, and perpendiculer is 3, (and side is 3-55) the mean proportion between is 3-866: Also, I have a Triangle, whose half-base is 8, and the perpendiculer 4, the mean proportional is 5-6552; and I would make another Romboides as big as the Triangle given, whose Area is 32: Then by the Line of Numbers, say, As 3-866, the one mean proportion, is to 5-6552, the other mean proportion; so is the Sides of the Rom∣boides, whose like I am to make, to the sides and perpendiculer of the Romboides required, to make a Romboides equal to a Triangle given, and like to another Rom∣boides first given.

As thus for Example.

As 3-866 is to 5-6552; so is 5, the base of the Romboides given, to 7-30, the base of the Romboides required. And so is 3, the given perpendiculer of the Romboides, to 4-38, the perpendiculer of the Rom∣boides required: So also is 3-55, the side of the Romboides given, to 5-19, the side of the Romboides required: for, if you mul∣tiply 7-30, the base thus found, by 4-38, the perpendiculer now found, it will make a Romboides, whose Area is equal to 32, the Area of the Triangle, that I was to make the Romboides equal to; and making the side to be 3-55, it will be like the first Rom∣boides propounded.

If it had been a Trapesia, or other form∣ed Figure, it might have been resolved into Triangles, and then brought into Squares, as before: Then all them Squares added into one sum, whose Square-root is the mean proportional or side of a Square, equal to that many-sided Figure, whose like or equal is desired to be made and produced.

5. One Diameter and Content of a Circle given, to find the Content of another Circle, by having the Diameter thereof only given.

The Extent from one Diameter to the o∣ther, being twice repeated the right-way from the given Area, shall reach to the Area required.

If the Area's of two Circles be given, and the Diameter required; then the half-distance on the Numbers, between the two Area's, shall reach from the one Diameter to the other.

Sect. VIII. To find the Square-root of a Number.

To do this by the Line of Numbers, you must first consider, whether the Figures, whereby the Number, whose Root you would have, is expressed, be even or odd fi∣gures, that is, consist of 2, 4, 6, 8, or 10; or 1, 3, 5, 7, or 9 figures.

For if it be of even figures, then you must count the 10 at the end for the unite; and the Root and Square are backwards to∣ward 1.

But if it consist of odd Figures, then the 1, in the middle of the Line, is the unite; and the Root and Square is forwards to∣wards 10: for the Square-root of any Num∣ber, is alwayes the mean proportional, or middle space between 1, and the Number propounded; counting the unite according to the Rule abovesaid: So that the Square-Root of 1728, consisting of four figures, it is at 41 and ••/••, counting 10 for the unite; for the Number 42 & ••/••••, is just in the middest between 1728 and 10.

And to find the Square-root of 144, consisting of three figures; divide the space between the middle 1 and 144, counted forwards, into two equal parts, and the Point shall rest at 12, the Square-root re∣quired.

To do this by the Line of Lines, or Sector.

First, find out a Number, that may part the Number given evenly, or as even as may be; then the Divisor shall be one ex∣tream, and the Quotient another extream; the mean proportional between which two, shall be the Square-root required, working by the last Rule.

Example.

To find the Square-root of 144. If you divide 144 by 9, you shall find 16 in the Quotient: Now a mean proportion be∣tween 9 the Divisor, and 16 the Quotient, is 12 the Root, found by the last Rule, viz. the 7th.

Sect. IX. To find the Cubick-Root of a Number.

The Cubick-root of a Number, is al∣wayes the first of two mean proportionals between 1, and the Number given; coun∣ting the unite with the following cautions: Set the Number down, and put a Point un∣der the 1st, the 4th, the 7th, and the 10th figure; and look how many Points you have, so many figures shall you have in the Root.

Then if the last Point fall on the last Fi∣gure, then the middle 1 must be the unite, and the Root, the Square, and Cube will fall forwards toward 10.

But if the last Point fall on the last but one, then the unite may be placed at ei∣ther end, viz. at 1 at the beginning, or at 10 at the end; and then the Cube will be one Radius beyond the unite, either for∣wards or backwards.

But if it fall on the last but two, then 10 at the end of the Line must be the unite; and the Root, the Square, and Cube will alwayes be in the same Radius, that is be∣tween 10 at the end, and the middle 1.

So that by these Rules, the Cubick-root of 8 is 2; for putti••g a Point under 8, being but one figure it hath but one Point, therefore but one figure in the Root: Se∣condly, the Point being under the last fi∣gure, the middle 1 is the unite; then di∣viding the space between 1 and 8, into three equal parts; the first part ends at 2, the Root required.

So likewise in 1331, there is two Points, therefore two figures in the Root; and the last Point being under the last Figure, the middle 1 is the unite; and the space be∣tween 1 and 1331, being divided into three equal parts, the first part doth end at 11, the Cubick-root of 1331.

Again, for 64 there is one Point, and it falls on the last figure but one; therefore the Root contains but one figure, and 1 at the beginning, or 10 at the end, which you please, may be the unite.

But yet with this Caution, That the Cube must be in the next Radius beyond that which belongs to the unite; so that dividing the space between 10 and 64, beyond the

middle 1, towards the beginning, into three equal parts; the first part falls on 4, the Cubick-root required: Or, if you divide the space between 1 and 64, near the 10, into three equal parts, the first part falls on 4 also.

Again, for 729, there is but one Point; therefore but one figure: Again, it falls on the last but 2; therefore 10 at the end is the unite; and between 10 and the middle 1 backwards, you shall have both Root, Square, and Cube, for the Number required, which will be at 9; For if you divide the space between 10, and 729, into three equal parts, the first part will stay at 9, the Cu∣bick-root required.

Note, if it be a surd Number that can∣not be cubed exactly; yet the Number of figures to be accounted as Integers is as be∣fore; and the residue discoverable by the Line, is a Decimal Fraction.

Example.

For 1750 the Root resulting, is 12 04••/1000, or 12, and near 5 of a 100.

Thus you have a very good and ready way for this hard question in Arithmetick▪ and will come near enough for most uses.

But to perform this by the Natural-lines, at the best it is very troublesome, and cannot come to no such exactness, as by the Line of

Numbers; and therefore I shall omit it a•• inconvenient.

For Application or Use of this last Rule of finding the Cube-Root, observe with me as followeth:

1. Between two Numbers, or Lines given, to find two mean proportional Numbers, or Lines required.

Divide the space on the Line of Numbers, between the two Numbers given, into three equal parts; and the Numbers where the Points of the Compasses stay at each repeti∣tion, (or turning) shall be the two mean proportional Numbers required.

Example.

Let 4 and 32, be two extream Numbers (or the measure of two extream Lines) be∣tween which I would have, two mean pro∣portional Numbers (or Lines) required.

In dividing the space on the Line of Numbers, between 4 and 32, into three e∣qual parts, you shall find the Compasses to stay first at 8, secondly at 16; the two mean proportionals, between 8 and 32, the two extream Numbers first given.

For the Square or Product of 4 and 32, the two extreams, 128, is equal to the Square

or Product of 8 and 16, the two means multiplied together, being 128 also.

2. To apply it then thus for Example: If I have the solid content of a Cube to be 1728 Cubick inches, and the side thereof be 12 inches; I would know what shall the side of the Cube be, whose solid content is 3456, the double of 1728? Divide the space between 1728, and 3456, into three equal parts; then, lay the same distance the Compasses stand at from 12, the side of the Cube given, and it shall reach to 15-12, the side of the Cube required, whose solid content is 3456 inches.

Also, If I have a Shot of Iron, whose weight is 3 pound, and the diameter there∣of 2 inches, and 780 parts of an inch in a 1000; what shall the diameter of a Shot be, whose weight is 71 pound? One third part of the distance, on the Line of Num∣bers, between 3 pound, and 71 pound, shall reach from 2 inches, 780 parts, the gi∣ven diameter, to 8 inches, the true diameter of a cast Iron Bullet, whose weight is 71 pound.

2 Secondly, on the contrary, if the Dia∣meter and Content of one Globe, or Cube be given, and the Diameter of another Globe or Cube, to find the content thereof.

As the diameter of the Globe, whose con∣tent is also given, is to the diameter of the Globe whose content is required; so is the content given, to the content required; by repeating the Extent the same way three times. Example. Suppose the capassity or content of a Globe, whose diameter is 10 inches, be 523 inches solid, and 80 parts; what shall the content of that Globe be whose diameter is 20 inches? the Extent from 10 to 20, being turned three times from 523-8, the content of a Globe of 10 inches diame∣ter, shall reach to 4190 10/1, the Cubick inches contained in a Globe of 20 inches diameter, being 8 times as much as the former.

3. The Proportion between the weights and magnitudes of several Metals, are as fol∣loweth, according to Marinus Ghetaldi.

If 7 pieces of the 7 Metals, are all of one shape, and bigness, either Sphears, or Cubes, or Cillenders, or Parallelepipedons; then their weights are in proportion as fol∣loweth, according to Marinus Ghetaldi.

The Shape and Magnitudes equal, The Weights are in proportion, as,

♃ Tinn	1554
♂ Iron	1680
♀ Copper	1890
☽ Silver	2030
♄ Lead	2415
☿ Quicksilver	2850
☉ Gold	3990

So that if a Cillender of Tinn, whose side is one inch, weigh 1824 grains; What shall a Cillender of Gold weigh, the height and diameter being just one inch, viz. 4682.

For as 1554, is to 3990; So is 1824, to 4682, the grains in one inch of Gold.

The Shapes and Weights of the pieces of the seven several Metals being equal, then the Magnitudes of the sides are as fol∣loweth, according to Mr. Gunter.

☉ Gold	3895
☿ Quicksilver	5433
♄ Lead	6435
☽ Silver	7161
♀ Copper	8222
♂ Iron	9250
♃ Tinn	10000

So that if I have a Sphear of Iron, weigheth 9 pound, whose Diameter is 4 inches; What must the Diameter of a Leaden Sphear, or Bullet, be of the same weight?

Say thus;

One third part of the space between 9250, and 6435, shall reach from 4 the Diameter of the Iron Bullet, to 3 inches 54 parts, the diameter of the Leaden Bul∣let, that weighs 9 pound.

4. So that if I have the weight and mag∣nitude, of a body of one kind of Metal, and would know the magnitude of a body of another Metal, having the same weight: work thus;

The first of two mean proportionals, be∣tween the two Points on the Line of Num∣bers, representing the Numbers in the last Table, for the two Metals, shall reach the right way, from the Magnitude given, to the Magnitude required.

As in the Example before, and illustrated by another thus;

Suppose a Cube of Gold, whose side is 2 inches, weigh 29000 grains; What shall the side of a Cube of Tinn be, having the same weight? Divide the space on the Line of Numbers, between 3895, the Point on the Numbers for Gold; and 10000, the Point for Tinn; and this extent parted into 3 equal parts, and that distance laid from 2, the side of the Cube of Gold, shall reach to 2-74; the side of the Cube of Tinn re∣quired.

5. The Magnitudes of two bodies of several Metals being given, and the weight of the one, to find the weight of the other.

Take the Extent between their Points, on the Line of Numbers, according to the last Table, for each several Metal; and this Extent laid from the given weight, shall reach to the enquired weight, of the other Metal propounded.

Example.

If a Bullet of Iron, of 4 inches Diameter, will weigh 9 pound; a Bullet of Lead, of the same Diameter, will weigh 13 pound.

6. A body of one Metal being given, to make another body like unto it of another Metal, and any other weight, to find the Diameters and Magnitudes thereof.

First, by the 4th last past, find the Mag∣nitude of the side, or Diameter of the Sphear, having equal weight; and note that down, or keep it.

Then find out two mean proportions, be∣tween the weights given; and setting this di∣stance, on the Line of Numbers, the right way, either increasing, or diminishing from

the side given, shall shew the side or dia∣meter required.

Example.

Suppose I have a Bullet of Iron weigh∣ing 60 pound, and the diameter thereof is 7 inches, 57 parts; I would have a Bullet of Lead like unto it, to weigh one third part more; what must his diameter be?

First by finding out of two mean propor∣tionals, between the Points, on the Line of Numbers, for each Metal; and laying it the right way from the side of the body given, it gives the diameter of the other bo∣dy of equal weight, with the first given bo∣dy; as the Iron Bullet being 7 inches, 57 parts diameter, the Leaden Bullet of the same weight is but 6 inches, and 70 parts.

Then find two mean proportionals be∣tween 60, the weight given; and 80, one third part more: and repeat this the right way (viz. increasing) from 6 inches, 70 parts, and it shall shew on the Line of Numbers 7-37, the diameter of a Leaden Bullet, that shall weigh one third part more than an Iron Bullet of 60 pound weight, viz. 80 pound.

The performance of these Propositions by the Sector, without the help of a Line of Superfecies, and solids, is very troublesome to do: And if you have a right Gunters Sector, you may have also his Book of the

Use thereof, but this way by the Numbers is as quick, and more certain, having your proportional Points or Numbers true; but the way by the Pen is best.

Sect. X. To divide a Line, or Number, by ex∣tream and mean proportion.

To part, or divide a Line or Number, by extream and mean proportion, is to part it so, that the greatest part shall bear such proportion to the lesser, as the whole doth to the greatest part; so that the square of the greater part, shall be equal to the Pro∣duct of the whole, and the lesser part mul∣tiplied together: Or, if you add the Square of the whole, and the square of the lesser-part together, it is 3 times as much as the square of the greater-part. See Dig's Theorems.

As if you would part 12, by extream and mean proportion, the greater-part will be 7-42, & the lesser-part 4-58 near; and the whole is 12: as by squaring and dividing you shall find.

For the extent of the Compasses on the Line of Numbers, from 12 to 7-42, being turned once, the same way, from 7-42, shall reach to 4-58, the lesser part.

To find this by Arithmetick, do thus;

First, square the given Number (that is, multiply it by it self) then multiply the Product by 5, and divide this Product by 4; then find the Square-root of the Quoti∣ent, and from it take half the given Num∣ber, the residue is the greater portion, then the greater part taken from the whole, leaves the lesser-part.

By the Sector, work thus;

Open the Sector to a Right Angle, in the Line of Lines (making Latteral 10, a Paral∣lel in 8 and 6) or else make the Latteral 90, a Parallel Sine of 45 (or the Latteral Sine of 45, a Parallel Sine of 30) then upon both Legs count the given Number; then take the Parallel Ex••ent from the whole Number on one Leg, to the half on the other Leg, and lay this from the Center Latterally; and whatsoever the Point reacheth beyond the whole Number, must be added to the half Number, to make up the greater Number; or taken from the half to make the lesser.

Example.

Let the Number given be 12, which may be represented at 6 on the Line of Lines;

then the Sector standing at Right Angles, take the Parallel-distance from 3, the half of 6 (counted as 12) on one Leg, to 6 on the other Leg, and you shall find it reach to 6-71, which doubled is 13-42; from which if you take 6, the half sum, rest 7-42 for the greater part: and if you take 7-42 from 12, there remains 4-58, the lesser-part.

But by the Line of Numbers work thus: following the Arithmetical way.

Extend the Compasses from 1 to 12, and that extent shall reach from 12 to 144; then next the extent from 1 to 144, shall reach from 5 to 720; then the extent from 4 to 1, shall reach from 720 to 180: then to find the Square-root of 180; the half-distance between 180 and 1, you will find to be at 13-42, as before; which used as abovesaid, gives the extream, and the mean proportional parts of 12 required.

Another Example of 26.

Extend the Compasses from 1 to 26, and repeat the same again forward from 26, shall reach to 676.

Again, the Extent from 1 to 676, shall reach from 5 to 3380.

Lastly, the Extent from 4 to 1, shall reach from 3380, to 845; and the Square-root of 845, is 29-07: from which Num∣ber or Root, if you take half the given Number, viz. 13; then there will remain 16-07, one extream: then 16-07 taken from 26-0, rest 9-93, the other extream required.

For the Extent from 26 to 16-07, will reach from 16-07, to 9-93.

Another way by the Line of Sines, Geometrically.

The best and quickest way is by the Line of Sines, thus; Make the given Line a Paral∣lel-Sine of 90; then take out the parallel-Sine of 38 degrees, 10 minutes, and that shall be the greater part.

Also, take out the Parallel-Sine of 22 de∣grees, 27 minutes, and that shall be the lesser extream required: Or, according to Mr. Gunter, use 54 for the whole Line, 30 for the greater part, and 18 for the less.

Also by consequence, having the mean, or greater part, make it a parallel-Sine of 38 degrees, 10 minutes; then Parallel 90 shall be the whole Line, and Parallel 22 degrees, 27 minutes, shall be the lesser part.

And lastly, having the least part, make

it a Parallel in 22 degrees, 27 minutes; then Parallel 90 deg. 10 min. shall be the whole Line; and Parallel 38-10, the greater part.

The Use whereof, you shall have after∣wards in the 11th Chapter, about the cut∣ting off the Platonical Bodies.

Sect. XI. Three Lines or Numbers given, to find a Fourth, in Geometrical Pro∣portion; or, the Rule of Three di∣rect.

1. In all Questions of the Rule of Three, there be three terms propounded, viz. two of Supposition, and one of Demand.

2. Also note, that two of the terms pro∣pounded, are of one denomination, (or at least to be reduced to one denomination) and one of another denomination.

3. Of the three termes propounded, (in direct proportion) that of Demand is al∣wayes the third term, and one of the terms of Supposition, viz. that of the same Deno∣mination, with the term of Demand, is al∣wayes the first; then the other of Suppositi∣on left, must needs be the second term in the Question.

4. In direct proportion Alwayes; As the first term is to the second, so is the third to the fourth term required.

5. Having discovered which be the first, second, and third terms; If the first and third term be of divers Denominations, they must be reduced to one Denomination, if it cannot be done on the Line in the ope∣ration, as many times it may; As thus for instance:

If one pound cost two shillings, what shall 30 ounces cost?

Here you see that the term of Demand, 30 ounces▪ viz. the third term, is not direct∣ly of the same Denomination with one pound, the first term; but is thus to be re∣duced to ounces: Saying;

If 16 ounces cost 2 shillings, what shall 30 ounces cost? 3 s — 9 d.

Thus the first and third terms, are brought to one Denomination: Also you see that the Demand or Question, viz. What shall 30 ounces cost? is joyned to the third term; and also that 16 ounces the first term, is of the same Denomination; there∣fore the 3 s. must needs be the second term, and the Answer to the Question is the fourth.

6. Having thus discovered, which are the first, second, and third terms, and re∣duced

the first and third to the like Deno∣minations; then the work by the Line of Numbers is alwayes thus;

As the first, to the second; so is the third, to the fourth.

Or the Extent of the Compasses upon the Line of Numbers, from the first, to the se∣cond; shall reach the same way, from the third, to the fourth required.

As 16 ounces is to 2 s▪ so is 30 ounces to 2 s ¾, or 9 pence. Or,

As 16 ounces is to 24. d; so is 30 ounces to 45 d; which is 3 s▪ — 9 d. as be∣fore.

7. But by the Line of Lines or Sector, if you will work on one Scale only, you must consider which term of the first or se∣cond is biggest; for you must alwayes or∣der it so, that the Parallel wor•• must be the largest, (or at least so as it may be wrought) and as much as may be, that the fourth term may be a Latteral Extent, as the first alwayes is; for then it is wrought the soonest, and also the exactest.

Yet by this Instrument, you need not much care for these Cautions, having several Scales of Equal-Parts, to begin and end the work on, you are freed from that trouble.

As thus for Example.

When the second term is greater then the first, then the Work is well performed thus, two wayes.

• As the Latteral first 16 or else As Latteral third 30 from a lesser Scale.
• To the Parallel second 02 or else To Parallel first 16
• So is the Latteral third 30 or else So Parallel second 2
• To the Parallel fourth 3 75/••00 or else To Latteral fourth 3 75/100 by the same Scale.

Or, As —second, to = first; So = third, to — fourth; by a less Scale also, if need be.

But when the second term is less than the first, then the work is per∣formed thus:

If 50 Foot of Timber cost 40 s. what shall 20 Foot cost?

• As the Latteral second 40 Or as before; As —3d 20 foot
• To the Parallel first 50 Or as before; To = 1st 50 foot.
• So is the Parallel third 20 Or as before; So = 2d 40 shill.
• To the Latteral fourth 16 Or as before; To —4th 16 shill.

8. Thus you see several wayes of work∣ing: but for Beginners, I would advise thus, briefly.

First, either to observe this Rule of chang∣ing the terms, from the first to the second, viz. To take the second Latterally, and make it a Parallel in the first; then the Parallel third gives you a Latteral Answer. Or else to work directly, as the first to the second, and so be content with a Parallel Answer, which you may alwayes do with the help of a smaller Scale, when need requires it.

Note the Figures of Operation, by the Trianguler Quadrant.

Sect. XII. The Rule of Three inversed.

1. The Rule of Three inversed, or the back-Rule of Three, is, when the term re∣quired, or fourth term, ought to proceed from the second term, according to the same proportion, that the first term proceeds from the third.

As thus for Example.

Car.	Hour.	Car.	Hour.
20.	16.	10.	32.
1.	2.	3.	4.

If 20 Carts carry 60 Square yards of Earth in 60 hours, how many Square yards of Earth shall 10 Carts carry in 16 hours?

Here it is apparent that fewer Carts must have a longer time to carry the like quanti∣ty; therefore to the same time must less work be allotted, as in the work doth fol∣low.

Car.	Hour.	Car.	Hour.
20.	16.	10.	32.
1.	2.	3.	4.

For if you extend the Compasses from 10 to 20, terms of like Denomination, viz. that of Carts; the same extent applyed the contrary way, from 16, the time required, by 20 Carts, shall reach to 32, the time required by 10 Carts, to carry 60 Load.

For Note, as in the former Rule of Three direct:

Look how much the third term is greater

than the first; so much the fourth is greater than the second.

And contrarily,

Look how much the third term is less than the first, by so much is the fourth term less than the second.

As thus in Numbers.

As 2 is to 4, so is 6 to 12, for as 6 the third term, is thrice as much as 2, the first term; so is 12, the fourth term, thrice as much as 4, the second.

And contrarily decreasing.

As 12 is to 6, so is 4 to 2; For as 4 is one third part of 12, so is 2 one third part of 6.

2. But now in this Rule of Three in∣versed, or the back-Rule of three; it is con∣trarily ordered, as thus;

Look how much the third term is greater (or lesser) than the first, by so much is the fourth term lesser (or greater) than the se∣cond.

As thus in Numbers.

As 5 is to 60, so is 30 to 2 1/••; that is, If 5 s. is 60 d. How many shillings is 30

pence? The Answer is, 2 s. ½. For as 30 is greater than 5••; so is 2 ½ less than 60.

Again, as 2 ½ is to 30; so is 60 to 5, in the like manner.

Pion.	Dayes.	Pion.	Dayes.
18.	40.	15.	48.
1.	2.	3.	4.

If 18 Pioneers make a Trench in 40 days, how many Pioneers is needful to perform the same in 15 dayes?

As 40 to 15, so is 18 to 48: Here, as the third is lesser than the first; so is th•• fourth greater than the second.

Hors.	Dayes.	Hors.	Dayes.
12.	30.	24.	15.
1.	2.	3.	4.

Again, if 12 Horses eat 20 bushels of Provender, in 30 dayes; how soon will 24 Horses eat up the like quantity of Pro∣vender? The Answer is in 15 dayes.

3. The manner of working this Rule on the Line of Numbers, is thus.

Extend the Compasses from one term to the other of like Denomination; the same

extent laid the contrary way from the other term, shall reach to the Answer required.

As in the last Example; the extent from 12 to 24, the terms under the denominati∣on of Horses, shall reach the contrary way from 30 to 15, the number of dayes re∣quired.

4. Note, That by due consideration, this back-Rule may be wro••ght by the Precepts, for the direct-Rule,

Thus:

In all Questions of this nature, there be three terms given to find a fourth; of which three terms, two are of one Denomination, and one of a different Denomination; of which, the fourth must alwayes be; which in the first Rule of the tenth Section before going, are called two termes of Supposition, and one of Demand. Now here you are to consider, That

5. When the fourth term required, ought to be greater than that of Demand; which by reason you may certainly know;

Then say,

As the lesser term of Supposition is to the greater;

So is the term of Demand, to his Answer, the fourth.

Example.

Men.	Dayes.	Men.	Dayes.
80.	12.	40.	24.
1.	2.	3.	4.

If 80 Men do a Work in 12 dayes, how soon may 40 Men do the like Work? Here Reason tells me, that fewer Men must have longer time; therefore the fourth term re∣quired must be greater. Therefore,

As 40 to 80, viz. As the lesser term of Supposition 40, to the greater 80;

So is 12, the term of Demand, to 24, the Answer required.

6. But if the required term ought to be lesser, which Reason will discover in like manner; Then thus:

As the greater term of Supposition, is to the lesser; so is the term of Demand to the fourth term required.

As 80 to 40, so is 24 to 12; extending the Compasses the same way from the third to the fourth, as from the first to the se∣cond.

But Note here, That you are not tyed to observe which is the first, second, or third term; but to consider only the nature of

the Question, that you may Answer accor∣dingly; and indeed this way will, general∣ly, take in the direct Rule also. For alwayes in Direct Proportion, you may as well say, As the third term is to the first, so is the second to the fourth; as to say, As the first to the second, so is the third to the fourth.

Also backwards, or inversly; As the third to the first, so is the second to the fourth; extending the Compasses the con∣trary way.

As 80 to 40; So is 12 to 24.

8. To perform this by the Sector, or ge∣neral Scale and Thred, on the Quadrantal∣side, you may generally observe this Rule; Enter the second term taken Latterally, Pa∣rallelly in the first▪ keeping the Sector, or Thred, at that Angle; then the Parallel-third, shall give the Latteral-fourth, LAT∣TERALLY.

9. Or else, As the Latteral-first, to the Parallel-second; so is the Latteral-third, to the Parallel-fourth, PARALLELLY.

And if the second be less than the first, make use of a smaller Scale; or change the terms, as is shewed before▪

Sect. XIII. The Double or Compound Rule of Three, Direct and Reverse.

Having premised the way to bring the back (or inversed) Rule of Three, to be per∣formed by the Rules for the Direct; and considering that the Double and Compound Rules of Three are alike by the Line of Num∣bers; I have therefore joyned them toge∣ther in one Section.

1. The Compound, or Double (Golden) Rule of Three; is, when more than three terms are propounded, or given.

2. The Double Rule of Three, is when five terms are propounded, and a sixt term proportional unto them is demanded.

As thus;

If 6 Men spend 18 l. in three months; How much will serve 12 Men for 6 months?

Or, again.

If two Barrels of Beer serve 12 Men for 14 dayes; How many dayes will 4 Barrels serve 24 Men?

3. The five terms given consi••t of two

parts, viz. a Supposition, and Demand; as in the Rule of Three direct.

The Supposition lies in these three Num∣bers first propounded, viz. If 6 Men spend 18 l. in 3 months; and the Demand lies in the two remaining▪ viz. How much will serve 12 Men 6 mon••••s?

Or in the other Example, viz. If 2 Bar∣rels of Beer serve 12 Men 14 dayes, are the terms of Supposition; and, how many dayes will 4 Barrels serve 24 Men, are the terms of Demand?

4. The next work is to rank the three terms of Supposition, and the two of De∣mand, in their due and proper order, for convenience of Operation; which may be thus:

Of the three terms of Supposition, that which hath the same Denomination with the term required, pla••e in the second place; and the other two, one above another in the first place:

Thus;

6	18	12
3		6

And then place the two terms of Demand one above another in the third place, only observing to keep the Numbers of like De∣nomination

in the same ranks; as 6 Men, and 12 Men in the upper rank; and 3 Months, and 6 Months in the lower rank; as in the Work is exprest.

5. When Questions of this nature are resolved by two single Rules, then the Ana∣logy, or Proportion, is thus;

Operation I.

As the first term, in the upper Rank, is to the second;

So is the third, in the same Rank, to a fourth. Again,

Operation II.

As the first term in the lower Rank, is to the fourth last found;

So is the other term in the lower Rank, to the term required.

As in the first Example; As 6 to 18; so is 12 to 36 a fourth.

Again, as 3 to 36; so is 6 to 72, the term required.

Which by the Line of Numbers, is thus wrought;

Extend the Compasses from 6 to 18; the same extent applyed the same way from 12, shall reach to 36. Then again, extend the Compasses from 3 to 36, the same ex∣tent applied the same way from 6, shall reach to 72, the term required.

By the Trianguler Quadrant, or Sector, thus;

6. As — 18 to = 6; so is = 12 to — 36. Again,

As — 36 to = 3; so is = 6 to — 72, the term required.

Or else work it Parallelly, observing the same order, as by the Line of Numbers, thus;

As — 6 to = 18; so is — 12 to = 36, the fourth term. Again,

As — 3 to = 36; so is — 6, to = 72, the sixt term required.

The Double Rule of Three inversed.

7. In the other Example, is comprehend∣ed the double Rule of Three inverse; which runs thus;

If two Barrels of Beer, serve 12 Men 14 dayes; How many dayes will 4 Barrels serve 24 Men?

If you Rank the terms, according to the former Precept, they will stand thus:

2—14—4 12 24 or thus, 12—14—24 2 4

8. Which if you work according to the back-Rule, the way is thus;

Operation I.

Extend the Compasses from 2 to 4, term•• of like Denomination, viz. of Barrels; th•••• same Extent applied the contrary w••y from 14, shall reach to 7, for a Fourth Propor∣tional.

Operation II.

Again, Extend the Compasses from 12 to 7, the fourth last found; the same Extent shall reach the contrary way▪ from 24 to 14, the number of dayes required.

9. But if you would reduce this, to be wrought by two single direct Rules; you must consider the Precept Rule, the 5th and 6th, of the Eleventh Section; and the terms of Supposition and Demand; and the in∣creasing, or decreasing of the fourth term, which is required.

As thus;

First, I part this into two single Rules, thus:

Operation I.

If 12 Men drink 2 Barrels in 14 dayes, then 24 Men may drink 2 Barrels in 7 dayes.

Operation II.

Again, If 2 Barrels last 24 Men 7 dayes, •• Barrels will last them 14 dayes; the An∣swer to the Question required.

Here by the 6th Rule, where the Number sought is to be less; As 24, the greater term of Men, is to 12 the less of the same Deno∣nomination; So is 14 to 7, the fourth.

Again.

As 2 the lesser term, is to 4 the greater of the same Denomination; so is 7 to 14, the Answer required, by the 5th Rule of the 11th Section.

Or else thus;

As 2 to 7, so is 4 to 14; that is, the Ex∣tent from 2 to 7, shall reach the same way from 4 to 14, the term required.

To work this by the Trianguler Qua∣drant, or Sector, the general Rule in this Section, Rule 6 and 7, giveth sufficient di∣rection.

10. The Rule of Three, compounded of five Numbers, is no other than the double Rule of Three; and is, or for the most part, may be wrought by one Operation, having prepared the Numbers by Multiplication, for that purpose: Which two Multiplicati∣ons by the Line of Numbers, though they

are presently wrought, yet the two Rules of Three are done as soon; so that the Compound Rule, is here of no advantage at all, therefore I might wave it; yet be∣cause the only difficulty lies in the order∣ing the Question, I shall propound it, for the addition sake of another Example, which is this;

If the Carriage of 2 hundred weight, 30 miles, cost 4 s. What will the Carriage of 5 hundred weight cost for 100 miles? The Numbers Ranked, according to the first Precept, will stand thus, as followeth.

〈 math 〉〈 math 〉

11. Then for the Operation, multiply the two first Numbers one by the other; as 2 times 30 is 60, which is the first term; and let the middle Number be the second term; and the Product of the two last (multiplyed together) for the third term: Then the Numbers being so prepared, say,

As 60, the Product of the two first Numbers, is to 4, the middle Num∣ber;

So is 500, the Product of the two last, to 33 ••/••, the Answer required.

By the Line of Numbers, the Extent from 60 to 4, will reach the same way from 500 to 33 ⅓, or, thirty three shillings and four pence, the price of 5 hundred weight, car∣ried 100 miles.

Note, This Rule serves when it is perfor∣med by the Compound Rule of Three di∣rect.

12. But if the inverse, or backer Rule of Three, be used in the work; then Operate thus:

As in this following Example, is manifest.

A Merchant hath received 10 l. 10 s. for the Interest of a certain sum of Money for six Months; and he received after the rate of 6 l. for the use of an hundred pound in a year; the Question is, how much Mone•• was Principal to 10 l. — 10 s. for 6 Months?

First, I range the Numbers, according to the order first propounded, in the 4th Rule of the 12th Section, as followeth.

〈 math 〉〈 math 〉

Then I observe diligently, whether the inverse Proportion be in the first or second Operation•• or Line, as thus in this Questi∣on it is in the lower Line; therefore after the Cross Multiplication, it is to be wrought by the single inversed Rule of Three; but when the inverse Proportion is in the upper Line, it is wrought by the single Rule di∣rect.

〈 math 〉〈 math 〉

Then I multiply the double terms a∣cross; that is, the lowest on the right-hand by the uppermost on the left; and the up∣permost on the right, by the lowest on the left; As thus:

6 by 6, which makes 36, to be set under 6; and 12 by 10-5, or 10l. (which is 126, and 10 s.) and set it under 10: then say by the inversed Rule, thus▪ * 1.62

As 126 to 36, so is 100 to 350, the Answer demanded; So that 350l. as Prin∣cipal will yield 10 l. — 10 s. in 6 Months; Or, the Extent from 126 to 36, shall reach the contrary way from 100 to 350, the Principal Money required.

Which you may more readily prove by reasoning thus:

13. If 3l. be the Interest of 100l. in 6 Months, to how much Money shall 10 l. 10 s. be interest in 6 Months? work thus;

The Extent of the Compasses from 3 to 100, shall reach the same way from 10 l. 10 s. to 350, the Principal Money an∣swering to 10 l. — 10 s. the Answer re∣quired.

By the Line of Lines, work thus;

As — 3 to = 10, counted 100;

So is — 10 ½ at the first 1 next the Cen∣ter, to = 350.

Or,

As — 100, to = 3; so is = 10 1/•• to — 350.

Sect. XIV. The Rule of Fellowship.

1. Rules of Plural Proportion are those, by which those Questions are resolved, which requi••e more Golden Rules than one; and yet cannot be Resolved by the double (Rule of Three, or) Golden 〈…〉〈…〉 was last mentioned.

2. Of these Rules there be ••ivers kinds and varieties, according to the nature of the Question propounded; for here the terms given, are sometimes four, five, or six, or more; and the terms required also more than one, two, or three.

3. The Rule of Fellowship, is to disco∣ver the Gain or Loss of every Partner in the Stock, by their several Stocks, and the whole gain or loss of the whole Stock.

Also observe, That the Rule of Fellow∣ship may be either single or double; of both which in order.

4. The single Rule of Fellowship is, when the Stocks propounded are single Numbers.

As thus for Example.

ABC and D, representing the Names of 4 Men, put into one common Stock 100 l. to trade withal: A puts in 10 l. B puts in 20 l. C 30 l. and D 40 l; and with this Stock, in a certain time, they gained 10 l. or 200 s; Now the Question is, what ought each man to have of the 200 s. that may be proportionable to his particular Stock?

5. The Rule of Operation is, first, by Addition find the total of all the particular Stocks, for the first term; the whole gain (or loss), for the second term; and each

particular Stock for a third term; and re∣peating the Rule of Three as often as there be particular Stocks in the Question, you shall bring forth, or find out, as many fourths for the particular gains (or losses) of each particular Man required.

As thus for Instance.

The sum of the four Stocks are 100 l.

The whole gain is 10 l. or 200 s.

Then,

〈 math 〉〈 math 〉

For the Extent from 100 to 200, shall reach from 10 to 20, and from 20 to 40, and from 30 to 60, and from 40 to 80; the particular gains due to ABCD, which was required.

6. For proof whereof, if you add 20, 40, 60, and 80 together, they make up 200 s, or 10 l; the whole gain of the whole Stock.

7. The double Rule of Fellowship is, when the Stocks propounded are double Numbers.

As thus for Example.

AB and C, holds a Field in common, for which they pay 50 l. a year; and in this Field, A had 25 Oxen went 30 dayes; B had 15 Oxen there 40 dayes; and C had 29 Oxen went there 40 dayes: What ought each man to pay for his part of the Rent, viz. 50 l? Here you see the Stocks propounded are double Numbers, as of Oxen, and their dayes, or time of feeding; as 25 & 30, 15 & 40, 20 & 40, being double Numbers.

8. The Rule of Operation is thus, in the double Rule of Fellowship:

Multiply the double Numbers, severally one by the other, one after another, and take the sum of their several Products, for the first term; and the whole gain or loss, for the second term; and the particular Pro∣ducts of every double Number, for the third term, one after another: This done, repeating the Rule of Three, as often as there be double Numbers, the 4th term produced from those Operations, shall be Answers to the Questions required, viz. the quantity of each mans gain or loss.

Example.

25 & 30, A's Oxen and time of feeding, multiplied, is	750
15 & 40, B's Oxen and time of feeding, multiplied, is	600
20 & 40, C's Oxen and dayes of feeding, multiplied, is	800
The Sum	2150

AS 215, to 50; so is	750 A's	Stock to	17-9 A's	Rent
	600 B's	13-19 B's
	800 C's	18-12 C's

9. To work by the Line of Numbers, the Extent of the Compasses from 1 to 25, shall reach the same way from 30 to 750, the first Product of a A's double Number, or Stock.

And as 1 to 15, so is 40 to 600, the Product of B's double Number, and Stock.

And as 1 to 20, so is 40 to 800, the Product of C's double Number, and Stock.

Which three Products added, make 2150, the first term; and 50 is the second term; and 750, 600, and 800, the three Products severally, the third term. Then,

The Extent from 2150 to 50, shall reach the same way from 750 to 17-45, or 17 l. 9 s. And from 600 to 13-95, or 13 l.— 19 s.

And from 800 to 18-60, or 18 l. = 12 s. the several Answers required; which being added together, make up 50 l. the whole Rent to be paid among them.

There be other Rules of Arithmetick, as the Rule called Allegation, Medial, and Alternate, and the Rule of Position or False∣hood; in the working of which, are so many Cautions in ordering the Numbers, before you come to the proportional work, that it would make the Book more bulky than use∣ful; therefore I shall wave it, and refer you to the particular Books of Arithmetick, as that of Mr. Record, Dee, and Mellis; or that of Mr. Wingate Natural and Artificial, having in it plenty of Examples; and o∣thers also, as Iohnf••us, Iaggers, or Moores Arithmetick, any of which exceed the bounds I intend for this whole discourse; I shall therefore pass on to the Rules of Practice, in several kinds, as measuring Su∣perfecies, and Solids, and Rules of double and treble Proportion and Questions of Interest; which are tedious by the Pen, without the help of particular Tables, and very easie by the Line of Numbers, as will fully appear in the next Chapters.

CHAP. VII. The use of the Line of Numbers in measuring of any kind of Super∣ficial Measure.

THe Measure that is commonly used in this Work, is a Foot-Rule, divided into 100 parts; or else into 12 inches, and those inches into halves, and quarters, or 8 parts; or inches and 10 parts▪ but in regard that the Numbers do most fitly a∣gree to the 100 parts of a Foot, it will be convenient here to shew how to reduce them, or any other Fraction, from 12 s. to 10 s. or any other whatsoever, from one Fraction to the other, which by the Line of Numbers is quickly done; as thus, from 12^• to 10^•.

Reduction.

Extend the Compasses from one Denomi∣nator to the other, the same Extent shall reach the same way from one Numerator to the other.

Example.

As 12 to 10, so is 6 half of 12, to 5 half of 10.

Again.

As 120 to 100, so is 30 a 4th of 120, to 25 a 4th of 100.

Which two Lines of Inches, and Foot-Measure, are usually set together on Rules, for the ready way of Reduction by Occular inspection, only in this manner, as in the Fi∣gure; And the like may be for any thing whatsoever, as Mr. Edmond Windgate hath largely shewed in his Arithmetick. Which Line being next to the Line of Numbers on your Rule, will be very plain and ready in the use of the Line of Numbers for feet and inches, or shillings and pence; and the same Rule of Reduction, serves for all manner of Fractions: For as the Denominator of one Fraction is to the Denominator of the other, (which in the Decimal work is al∣wayes a unite, with one, two, or more Cyphers) so is the Numerator of one, to the Numerator of the other.

And Note, That the operation of Decimal Numbers, and their Fractions, is no other than whole Numbers, except only the cut∣ting off so many Figures as there is Fracti∣ons

in the Multiplicator and Multiplicand, after any Multiplication; as in the follow∣ing Examples will appear.

This being premised, I come next to the Work.

Problem I. The breadth of an Oblong Superficies gi∣ven in Foot▪ Measure, to find how much in length makes one Foot.

The Extent of the Compasses from the breadth to 1, shall reach the same way from 1, to the length required.

Example at 7^10th broad.

As 7 to 1, so is 1 to 1 Foot and 43 parts

The breadth given in inches, to find how much make a Foot.

As the breadth in inches to 12, so is 12 to the length of a Foot in inches, and 10 parts.

Example.

At 8 inches broad, you must have 18 inches to make a Foot; for the Extent from 8 to 12, shall reach the same way from 12 to 18.

To work these two by the Line of Lines.

By Inches. As — 1 to =7, so is =1 to — 1—43, the length in Foot-measure;

By Inches. As — 12 to = 8, so is = 12 to — 18; Or else,

By Inches. As — 8 to = 12, so is — 12 to = 18, the length in inches.

Problem II. Having the breadth of an Oblong Super∣ficies given in Foot-measure, to find how much is in a Foot long.

This is soon wrought; for in every Foot long there is just as much as the breadth is, either in Foot-measure or inches; for a piece of Board half a Foot broad, and a Foot long, is just half a Foot.

Problem III. Having the length and breadth in Foot-Measure, to find the Content in Feet.

The Extent from 1 to the length, shall reach the same way from the length to the Content in Feet.

Example.

As 1 to 1 foot 50, the breadth; so is 11 foot, 10 parts, the length, to 16 foot and 65 parts, the Coment required.

The breadth given in inches, and the length in feet, to find the Content in feet.

As 12 to the breadth in inches, so is the length in feet, to the Content in feet requi∣red.

Example at 9 inches broad, and 11 foot long.

The Extent from 12 to 9, shall reach the same way from 11 to 8 foot, 3 inches, or ¼.

By the Line of Lines.

As — 11, to = 12;

So is — 9, to — 8 ¼.

But Note, That in working this, and ma∣ny such-like, it will be convenient to double your Scale in account, calling 10 at the end 20, and every single figure as much more, as to call 12, and 24, &c.

So that in this Operation, the work runs thus;

As — 11 taken from the Line of Lines, counting 1 for 10 as usually,

To = 6, the half of 12 reckoned double for 12:

So is = 4 ½ counted for 9, to — 8 ¼ be∣tween the Center and 1.

Or else thus;

As — 5 ½ counted for 11, is to = 6 counted for 12;

So is = 9, to — 8 ¼ near the end, and as large as may be.

Thus you may many times vary the man∣ner of work to get the Answer latterally, and as large as may be on the Scale of Lines, by doubling or halfing the Numbers, or ta∣king the whole Number of quarters, or u∣sing a less or a bigger Scale, as hath been hinted, and shall be more in places con∣venient, in the following Discourse, to at∣tain exactness and ease, ••s much as may be, as time and practice will demonstrate to the willing Practitioner in these Operations.

Problem IV. Having the length and breadth given in Inches to find the Content in Superficial-Square Inches.

As 1 inch, to the breadth in inches; so is the length in inches, to the Content in Su∣perficial inches.

Example, 20 inches broad, and 36 inches long.

The Extent of the Compasses from 1 on the Line of Numbers to 20, shall reach the same way from 36 to 72, the true Number of Superficial Square inches in that Oblong.

By the Line of Lines.

As — 36 to = 5, counted as 1;

So is = 10 counted as 20, to — 72, at the largest Extent.

For Note, The reason that the Latteral 72 and 36, are from the same Scale in ac∣count; and the Parallel 1 and 20 counted Decimally, are from the same Scale also, or else according to the Proportion by the Line of Numbers;

As — 1 to = 20, So is — 36 to = 72. Here also is the same advance Decimally from 1 to 20, as before.

Problem V. Having the length and breadth given in Inches, to find the Content in Feet Super∣ficial:

As 144 to the breadth in inches, so is the length in inches, to the Content in Feet Su∣perficial.

Example at 40 broad, and 60 long.

For the Extent on the Line of Numbers from 144, the number of inches in one foot, to 40 the breadth, shall reach from 60 inches the length, to 16 foot ½ and 26 inches. To count so many inches on the Line, ob∣serve with me this way of Reduction, the 16 foot and ½ is very plainly seen. And Note, That there is 10 cuts in this place be∣tween 16 and 17; and 10 times 14 is 140, which is near 144, the inches in a foot; so the Point of the Compasses staying at near 2 10ths beyond the half-foot, I count almost twice 14, which is 26 inches for the Fraction above 16 foot and ½.

By the Line of Lines.

As — 144 (found between 1 and 2 near the Center) is to = 40 (at the figure 4)

So is half = 60, or the measure from the Center to 3, to = 8-35, which is the half of 16 foot ½, and 26 inches: For if you had taken all 60, it would have exceeded the whole Parallel Radius, where the Answer would have been right 16 ½ and better, but taking the half, it gives the half also.

Or else work thus with a Latteral Answer.

As ½ — 60 to = ½ 144, So is all = 40 to — 16 7/10.

Or,

As all — 40, to ½ = 144; So is all = 60 to — 33 and 4/10 the double of 16 and 7/10.

Note, That in both these two last work∣ings, the 144 is at 72, which is the half of 144; to make the work the larger. By these the excellency of the Line of Numbers, over the Line of Lines, is evident in these kind of Proportions. And for discovering the Reason of these Proportions, read the beginning of the 6th Chapter, Section 3d.

Problem VI. The length and breadth of an Oblong Su∣perficies being given, to find the side of a Square equal to it, by the Line of Num∣bers.

Divide the space between the length and breadth into two equal parts, and the middle Point shall be the side of the Square equal to the Oblong given, in quantity.

Example.

If a long Square, or Oblong, be 18 foot one way, and 12 foot the other way; the middle Point between 18 and 12 is 14 and 7/10 ferè; for 18 multiplied by 12, makes 216, and 14-7 multiplied by 14-7, is near 216 also.

To do this by the Line of Lines is shewed at large in the 7th Section of the 6th Chapter.

Problem VII. Having the Diameter, or Circumference of a Circle, to find the Circumference, or Diameter, or Squares equal, or Inscri∣bed, and Content.

For this purpose there are certain Pro∣portional Numbers found out, As thus;

If the Diameter of a Circle be 10, then

the Perifera, or Circumference, is 31 42, the side of the Square equal to the Circle, is 8-862, the side of the Square inscribed is 7-071, and the Superficial Content is 78-54; so that any one of these being gi∣ven, you may find out any of the rest by the Line of Numbers.

Thus having the Diameter, to find the Circumference.

As 10 to 31 42, so is the given Diameter to the Circumference required.

Or,

As 10 to 8-862, so is the given Diameter to the Square equal.

As 10 to 7-071, so is the given Diameter to the Square inscribed.

As 10 to 30, so is 78-54 to the Square of the Area of that Circle, whose Dia∣meter is 30. Or, To the Diameter turning the Compasses twice.

As for Example.

Let the given Diameter of a Circle be 30,

• The Circumference is 94 26 Diameter 10 00
• The qu are equal is 26 58 ½ Circumference ••1 42
• The Square within is 21 21 Square equal 8 862
• The Content or Area is 707 00 Square within 7 071
• The Diameter is 030 00 Area or Content 78 54

Also Note, If the Circumference be first given, then say, As 31 42 is to the Number 10 for the Diameter, or 8-862 for the Square equal; or to 7-071 for the inscribed Square; so is the given Circumference to the rest.

But to find the Area say, As the fixed Diameter 10, is to the given Diameter 30; so is the fixed Area for 10, viz. 78-54, to 707 — by turning the Compasses two times.

Or if the Circumference be given, and I would find the Area.

As 31-42, the fixed Circumference, is to 94-26 the given Circumference; so is 78-54, the fixed Area, answering the fixed Circumference 707, the Area

required, turning the Compasses two times the same way.

Thus by having five Centers at the five fixed Numbers; or four Centers answering to the four fixed Numbers; for a Circle whose Diameter is 10, having any one of those 5 given, you may find any of the o∣ther required.

Thus you have eight Problems couched in one; therefore be the more diligent to understand it. To work these by the Line of Lines, observe the former directions, which for brevity sake I now omit.

Problem VIII. The Content of a Circle being given, to find the Diameter.

Divide the distance on the Line of Num∣bers, between the fixed Content, or the Point 78-54, and the given Content into two equal parts; that distance laid the same way, from the fixed Diameter, shall reach to the required Diameter.

Example. The Content being 707-00.

The half distance between 78-54, and 707, shall reach from 10 to 30, the Dia∣meter required.

Problem IX. The Content of a Circle being given, to find the Circumference.

Divide the distance between the fixed and the given Contents or Area's into two equal parts, the distance laid from the fixed Circumference, shall reach to the required Circumference.

Example.

A Circle, whose Area is 707, shall be 94-26 about. For the half distance between 78-54, and 707-00, shall reach from 31-42 the fixed Circumference, to 94-26, the en∣quired Circumference.

And from 8-862, the fixed Square, e∣qual to 26-58 ½, the inquired Square E∣qual.

And from 7-071, the fixed Square in∣scribed, to 21-21, the inquired Square In∣scribed.

Problem X. Certain Rules to measure several Geome∣trical Figures Superficially.

For the Square, the long Square and Circle, hath been spoken to just before; All other Figures are to be reduced to a Square, or to a long Square; and then measured by Multiplication, as before.

Or thus.

Multiply the Diameter by it self; and then that Product by 11: then lastly, di∣vide this last Product by 14, and the Quo∣tient shall be the Area, or Content, of the Circle required.

* 1.63For a Circle (otherwise) thus; Multiply half the Diameter, by half the Circumfe∣rence, and the Product shall be the Content required.

* 1.64For a Half Circle; Multiply half the Diameter of the whole Circle, and a quar∣ter of the whole Circumference together, and the Product shall be the Content.

* 1.65For a Quadrant, or a quarter of a Circle; Multiply half the Arch, by the half Dia∣meter, or Radius of the Circle and the Pro∣duct shall be the Superficial Content.

* 1.66The like Rule holds for any lesser portion of a Circle, whose Point goeth to the Cen∣ter, viz. to take half the Arch, and the whole Radius, and multiply them together, and the Product shall be the Content.

Any Segment of a Circle given, to find the true Diameter.

Square half the Chord, and divide the Product by the Sine, then add the Quotient and Sine together; the sum is the Diameter.

Chord is 24 ½ 12. Squared is 144; divided by 8, gives 18 in the Quotient, which ad∣ded to 8, makes 26 for the Diameter.

For any other Segment of a Circle,* 1.67 find the true semi-Diameter, and measure it as before; then take out the Triangle, and the remainder is the true Content of the Segment. See Chapter 3.11, 12.

Or else thus, by the Line of Segments joyned to a Line of Numbers, in this manner.

To the Segment given, find the true Dia∣meter, by Chap. III.11, 12. Then having the Diameter, find out the Area, or Content of the Circle, by any of the former Rules, then the Proportion or Analogy is thus;

As the whole Diameer is to 100 on the Segments; So is the Altitude of the Seg∣ment, whose Area is required to a 4^th Num∣ber on the Line of Segments, which you must keep.

Then Secondly,

As 1, to the whole Content of the whole Circle given; So is the 4^th Number, kept, counted on the Numbers, to the Area of the Segment required.

If the Line of Segments is not on your Rule, then this Table annexed, will supply the defect, reasonably well thus:

A Table to divide a Line of Segments, making the whole Circle 10000 parts.

A. Table of Segments.
Seg.	par	Seg.	par	Seg.	par	Seg.	par.
70	1	1127	2	3484		6682
112	2	1151	4	3566		6786
147	3	1177	6	3645		6850
178	4	1201	8	3729		6935
206	5	1224	7	3810	35	7020	75
233	6	1248	2	3892		7106
258	7	1272	4	3971		7193
282	8	1296	6	4050		7281
307	9	1318	8	4131		7370
329	1	1341	8	4211	40	7460	80
350	1	1365	2	4290		7550
371	2	1388	4	4369		7642
392	3	1411	6	4448		7735
412	4	1433	8	4527		7829
431	5	1455	9	4606	45	7924	85
451	6	1478	2	4686		8022
469	7	1500	4	4766		8119
487	8	1522	6	4844		82••2
507	9	1544	8	4922		8327
524	2	1565	10	5000	50	8436	90
558	2	1673	11	5078		8552
592	4	1778	12	5156		8669
626	6	1881	13	5234		8788
657	8	1978	14	5314		8908
688	3	2076	15	5394	55	9029	95
718	2	2171	16	5473		9172
749	4	2265	17	5552		9330
779	6	2358	18	5••31		9505
808	8	2450	19	5710		97••0
083••	4	2540	20	5789	60	10000	100
0864	2	2630	21	5869
0892	4	2719	22	5950
0918	6	2807	23	6029
0948	8	2894	24	6108
0970	••	2980	25	6190	65
1000	2	3065		6271
1027	4	3150		6355
1051	6	3214		6434
1077	8	3318		6516
1102	6	3402	30	6598	70

The Diameter of the Circle, answering to the Segment given, being found out, Say,

As the whole Diameter to 100; so is the Altitude of the Segment to a 4th Num∣ber, which sought in the Table of Segments, or the nearest to it, gives in the parts the Number to be kept. Then again,

As the whole Content of the Circle fixed, viz. 100, is to the whole Content of the new Circle; so is the Number kept, being the Content, of Area, of the fixed Segment, to the Area of the Segment required.

Example.

Let the Segment of a Circle, whose whole Area is 314-2, and whose Diameter is 20, and let the Altitude of the Segment be 5, one 4th part of the whole Diameter.

Then say

As 2-000, the whole Diameter given, is to 10000:

So is the Altitude of the Segment 5, to 2500, the 4th; which sought in the Table of Segments, in the Parts, gives 19-50 for a 5th Number to be kept.

Then again,

As 1, to 314-2, the whole Area; So is 19-50, to 61-30, the Area, or Content of the Segment required.

* 1.68For all manner of Triangles, multiply the longest side (being properly called the base) by half the perpendiculer, and the Product shall be the Content of the Triangle; or as 2 to the base, so is the perpendiculer to the Content.

* 1.69For a Rhombus, being a Figure like a Quarry of Glass, containing 4 equal sides, and two pair of equal Angles: And any Figure having his opposite-sides Parallel one to another; then the length of one side and the nearest distance between the other two opposite-sides multplied together, shall be the true Area required.

* 1.70For all other four-sided-figures, call'd Tra∣peziaes, being irregular Figures; draw a Line from one corner to the other, which makes it two Triangles; then multiply that Line, being the whole base of both the Triangles, by the half sum of both the Perpendiculers, and the Product shall be the Content re∣quired.

Or,

* 1.71For all Regular Polligons, or Figures, with equal sides, the measure from the Center to the middle of one side, and the half sum, of the measure of all the sides multiplied together, shall be the true Area, or Content thereof.

All other Figures whatsoever, of how many sides soever they be, may be reduced to Triangles, or to Trapeziaes, and measured as before; which kind of Figure, Surveyors and Builders oftentimes meet withal, in their Operations.

Problem XI. For the measuring of on Oval, the best way Ovals▪ is to reduce it to a Circle thus;

Divide the distance on the Line of Num∣bers, between the length and the breadth of the Oval into two equal parts; and the middle Point where the Compass stayeth on, shall be the Diameter of a Circle equal in Area to the Oval given.

Example.

Suppose an Oval be 10 foot long (trans∣verse) and 8 foot broad (conjugate); the mean proportion, between 10 and 8, is 8-95: I say, that a Circle whose Diameter is 8-95, is equal to an Oval of 8 broad, and 10 long; And how to measure the Circle, is shewed before.

Of these Figures.

If the Content be 100, then the sides of these Regular Figures are 〈…〉〈…〉, and

also so in proportion, is any other quantity, of content required.

Perpendiculer-Triangle,	13.	123.
Trianguler-side,	15.	2.
Square, its Side,	10.	0.
Pantagon of five Sides,	7.	62.
Hexagon of six,	6.	02.
Heptagon of seven,	5.	26.
Octagon of eight,	4.	55.
Nonagon of nine,	4.	03.
Decagon of ten,	3.	06.
Half Diameter, or Radius,	5.	64.

Example as thus: I would have a Triangle to contain 200, What must the Sides be?

The half distance on the Numbers be∣tween 100 and 200, shall reach from 15-2 to 21-5, the side required.

And from 13-123 the fixed perpendiculer for a Triangle, whose Area is 100, to 18-6, the perpendiculer of an equilatteral Triangle, whose Area is 200.

But if the Sides be given, and you would find the Area, work thus;

The Extent from the fixed-side, to the gi∣ven-side, shall reach at two turnings, from the fixed Area, to the Area required.

The Extent from 15-2, to 21-5, shall reach, at twice repeating, from 100 to 200.

Problem XII. To make an Oval equal to a Circle, having the Diameter of the Circle, and the length or breadth of the Oval given.

S••t one Point of the Compasses in the Diameter of the Circle found out on the Line of Numbers, and the other Point to the Ovals length; then turn that distance the contrary way from the same Diameter-point, and it shall reach to the breadth of the Oval required.

Example.

Let the Diameter of a Circle be 10 foot, I would have an Oval to contain as much as the Circle, and be 12 foot long; the Que∣ry is, how broad must it be?

Set one Point in 10, and the other in 12, that Extent turned the other way from 10, shall reach to 8-34, the breadth of the Oval required.

If you please to alter the breadth or length, you shall soon find the length or breadth accordingly.

To work this by the Line of Lines, you must work by the Directions in the 7th Section of the 6th Chapter, as thus;

First, To find the Content of the Oval, joyn the length and breadth in one sum, to get the sum, the half sum, and difference, and half difference; then open the Sector, (or lay the Thred on 60|0) to a Right Angle; Then count half the difference from the Center downwards, and note the place; then take half the sum between your Com∣passes, and setting one Point in the half-difference, and extending the other to the other Leg, (or perpendiculer Line) and it shall shew a Point, whose distance from the Center is the mean proportional required; which is the Diameter of a Circle, equal in Area, to the Oval, or Elipsis given to be measured; as before is shewed.

To make an Oval equal to a Circle.

Take the guessed half-sum of the length and breadth of the Oval, and setting one Point in the Diameter of the Circle; and on the other Leg, set at a Right Angle, the other Point shall shew half the difference,

between the length and breadth of the O∣val; then if the mean proportional between them be equal to the Diameter, you have wrought right; if not, then resolving upon the length or breadth of the Oval, take more or less, for the breadth or length according∣ly: Herein also is seen the excellency of the Line of Numbers, in many operations.

Problem XIII. The length and breadth of any Oblong Su∣perficies given in Feet, to find the Con∣tent in Yards.

As 9 foot (the number of feet in one yard) to the length in feet and parts;

So is the breadth, in feet and parts, to the Content in yards.

Example at 13 Foot 6 Inches long, and 7 Foot 6 Inches broad.

The Extent of the Compasses from 9 to 13 ½ the length, shall reach the same way from 7 ½ the breadth, to 11 yards and a quarter, the Content.

Note, That if yo•• measure by feet and hundred parts, you shall find this way ex∣ceeding ready; the Answer being given in yards, and hundred parts of a yard. But if you have a yard divided into a 100 parts,

to measure withal; Then the Rule is thus▪

As 1 to the length o•• breadth, so is the breadth or length to the Content in yards.

Example at 3 yards, 72 parts broad, and 5 yards, 82 parts long.

The Extent of the Compasses on the Line of Numbers, from 1 to 3-72, shall reach the same way from 5-82, to 21 yards 65 parts, the Content in square yards, and 100 parts.

By the Line of Lines.

As — 5-82, to = 1 at 10 the end;

So is = 3-72, to — 21-65 yards.

Or in the Example before.

As — 13-6, counting 6 ½ for 13, is to = 9;

So is = 7 ½ to — 11 ¼; as you count∣ed at first.

Problem XIV. The length and breadth of any Wall, being given in feet and 100 parts, to find how many Rods of Walling there shall be at a Brick and an half thick.

First you must Note, That 272 foot and a quarter, makes one Rod, (or so many feet is in a Rod).

Secondly, That let the Walls be half a Brick, one Brick, two Bricks, two and a half, or three Bricks thick; it is to be re∣duced to Brick and a half thick, as a stan∣dard thickness.

Thirdly Note, That this reducing to a Brick and half thick, may be at the mea∣suring, or after the casting-up, as you please, as in the Examples following will plainly appear.

As thus for Instance;

A Front, or side-Wall of a House is to be measured, wherein the Celler-story Wall is 2 Bricks and a half thick; The Shop and first Chamber-story is two bricks thicks; the other Stories 1 Brick and a half thick; and the Gable-ends 1 Brick thick.

The nearest way to measure this Wall, I conceive is thus;

1. The Cellar-story is 10 foot high, but being 2 bricks and a half thick, I make it 16 foot 8 inches high, by adding two thirds of 10 foot, to the 10 foot high, which is 6 foot 8 inches, in all 16 foot 8 inches.

2. The other two Stories, are supposed 22 foot; but in regard they are two bricks thick, I add one third part of 22 foot, which is 7 foot 4 inches, to 22; and it makes 29 foot and 4 inches, the height of the Shop and next Story above.

3. The other two Stories being a brick and half thick, need no alteration, which suppose may be 19 foot.

4. The Gable-end, or Garret-story, if ••ny be, being but one brick thick; you must take away one third part to bring it to a brick and a half. Also if it be a Gable-end, Note, it is a Triangle, and you must mea∣sure but half the height, and the whole breadth, to find the Content; which here may be 5 foot.

The Cellar Story,	16 — 8
Two next Stories,	29 — 4
Two next Stories,	19 — 0
The Garret,	5 — 0
	70 — 0

5. Add all these sums of feet high toge∣ther, and they make 70; then measure the breadth, which is common to every Room, the out-side going upright, which in a double House may be 36 or 40 foot.

6. Then having gotten the Dimentions right by the Line of Numbers, Say,

As 272 ¼ (the feet in one Rod) is to 40 foot, the breadth of the House; so is 70 foot, the whole height of every several Story, (reduced) to 10 Rod and 29 parts; which 29 parts you may call a quarter of a Rod, and 10 foot and a half.

The reason whereof is apparent thus:

As 100 is to 272 ¼; so is 29 to near 79; of which 79-68 is a quarter of a Rod, or 25 of 100 is a quarter likewise, which by the Line of Numbers is apparently seen; then every 10th part is 2 foot, and 72 of a hundred, which is near two and three quar∣ters; so that here 25 being a quarter of a

Rod, there is 4 hundred parts more in 29: Then thus; the double of 4 is 8, or, twice 4 is 8, and four times three quarters is three foot more; of which you must abate some∣what (because 72 ¼ is not 75, which is just three quarters) and all put together, make ten rod, one quarter, ten foot and a half: for if you shall divide the Product of 40, multiplied by 70, which is 2800 by 272 ¼, you shall find the Quotient to be 10 rod, 78 ½, which is, as before, 10 rod, 1 quarter and 10 foot and a half.

But note also by the way, That when you come to take out the deductions for the doors and windows, if any happen in a Wall of two Bricks and a half, or in two Bricks; you must add two thirds, or one third more to the length or bredth one way; and then casting them up severally, when they be of several lengths or breadths, you shall do no wrong to the Work-master nor Work-man: For true Arithmetick and Geometry will lie for no man, or use any kind of partiality.

This I conceive is as near a way, as any such business can be performed. But if you will measure every Story severally, taking account of each Story severally in their thick∣nesses; then, after it is cast up, the best way, by the Rule, to reduce it, is thus;

As 3 half bricks, for a brick and a half, is to any other number of half bricks thick, over or under 3; So is the Content at that rate accordingly, to his Content, at a brick and a half required.

Example.

1269 foot at 5 half bricks thick is 2115, for two thirds of 1269, which is 846, ad∣ded to 1269, makes 2115; For the Extent on the Line of Numbers, from 3 to 5, shall reach the same way from 1269 to 2115, the Number required to be found out.

Otherwise thus;

To bring any kind of thickness, to one brick and a half thick, at one operation, by the Line of Numbers.

For this purpose, you must use several Points, as so many gage Points, as in the short Table following doth appear.

For half a brick, use	3-00000
For 1 brick,	1-50000
For 1 brick & a half, use	1-0000
For 2 bricks,	0-7500
For 2 bricks & a half, use	0-6000
For 3 bricks,	0-5000
For 3 bricks & a half,	0-4285
For 4 bricks,	0-3750
For 4 bricks and a half,	0-3333
For 5 bricks,	0-3000

For 5 bricks and a half,	0-2727
For 6 bricks, &c. ad infinitum.	0-2500

Example at the 6 ordinary thicknesses.

Let a Wall be 30 foot long, and 10 foot high; and let it be supposed of any of these thicknesses following, from half a bricks length, to three bricks length in thickness; then thus in order, increasing, &c.

First, at half a Foot.

For ½ brick. As 3 to 30; so is 10 to 100 foot, at 1 ½.

For 1 brick. As 15 to 30; so is 10 to 200 foot, at 1 brick.

For 1 ½ thick. As 10 to 30; so is 10 to 300 foot, at 1 ½.

For 2 bricks. As 0-75 to 30; so is 10 to 400 foot, at 1 ½.

For 2 ½ thick. As 0-60 to 30; so is 10 to 500 foot, at ½.

For 3 bricks. As 0-50 to 30; so is 10 to 600 foot, at 1 ½.

For 3 ½ thick. As 0-4285 to 30; so is 10 to 700 foot, at 1 ½.

For 4 bricks. As 0-3750 to 30; so is 10 to 800 foot, at 1 ½.

And so for any other thickness, as far as you please; which Points are found thus;

The Exten••▪ from the number of bricks, any Wall is thick to 15 (or 1 and ½) shall reach the same way from 10, or 1, to the Gage-Point required for that Wall, or Walls of that thickness.

Example.

As 2 to 1 ½; so is 10, to 0-750, for 2 bricks thick, &c.

Lastly, having the Number of Feet in the whole work, to find how many Rods there is. Say,

If 272 ¼, be one Rod; what shall any other Number of Feet make in Rods?

The Extent of the Compasses from 172 ¼, to 1, shall reach the same way, from the Number of Feet, to the Number of Rods, and hundred Parts, or Rods, and Quarters, and Feet; as by the 6th, last mentioned.

Example. In 5269 Feet, how many Rods?

The Extent from 272 ¼, to 1, shall reach the same way, from 5269, to 19 Rod, and 36 parts of a 100; or, 19 Rod 1 quarter, and 29 foot, and a quarter of a foot. The 19 Rod, and a quarter, is seen

plainly on the Rule; and 25 being a quar∣ter, 36 is 11 parts more; for w^ch 11 parts more, I say, 2 times 11 is 22 foot, and 11, 3 quarters of a foot is near 8 foot, which put together, makes 29 foot, as before: Or, as the Compasses stand, turn them the con∣trary way, from the Decimal parts, above the even quarter, and it shall reach to the odd feet above the quarter required.

Example.

The Extent from 272 ¼, to 100; or 1, shall reach the contrary way from 10 ½, to 29 foot, the feet above ¼ of a Rod.

8. Observe, That the Tyling, the Roof, the Floors, and Partitions, are measured by the Square; which is 10 foot Square every way, or 100 foot in Area. The Chimneys are usually done by a certain rate for a Chimney; or if to be measured, thus are the height and breadths taken, &c.

If a Chimney stand singly and alone, not leaning against, or in a Wall, the usual way is to girt it about; and if the Jaumes are but a brick thick, and wrought upright over the Mantle-tree to the Floor; then I say, girt it about for a length, and the height of the Story is the breadth, at a brick thick, because of the gathering together, to make room for the next Hearth above in the next Story.

But if the Chimney-back be a Party-Wall, the Wall being first measured, then the brest and the depth of the two Jaumes is one side, and the height of the Story ano∣ther side, to be multiplied together, at a brick and a half thick, or a brick thick, ac∣cording as the Jaumes be, and nothing to be abated for the want between the Hearth and the Mantle-tree, because of the Withs and thickning for the next Hearth.

For measuring the Shafts of the Chimneys.

Girt with a Line, round about the least place of them, for one side; and the height for the other side, at a brick thick, in consi∣deration of the Withs, Pargitting, and Scaffolds.

In measuring of Ceiling a foot broad, and the length of the Vallies is alwayes to be al∣lowed, more than the whole Roof; Also the length of the Rafter feet, above or beyond the Roof.

When Rafters have their usual pitch, which is, when the breadth of the House is 12 foot, the Rafter is 9 foot long, which is 3 quarters of the Floors breadth, be it more or less; then, I say, that the Content of one Floor, and half so much, is the Area of the whole Roof in Squares; to which is to

be added, the Vallies and Rafter-Feet, or Eves, in Tileing.

And also a Deduction for Chimney-room, and Gutters, if any be.

Which work by the Line of Numbers, is done at one Operation, thus;

As 6666, is to the length of the House;

So is the breadth to the Content in the Roof.

Example.

A House 30 foot long, and 20 foot broad, is 900 foot, or 9 square.

For the Extent, from 6666, to 20, shall reach the same way from 30 to 900.

Also in measuring of the Roof, as to Car∣penters work, by the Square, there is to be allowance for those Rafters in the Dormers, and Gable-ends, on which no Tiles are laid, as over-work for a particular use and con∣venience, more than need be in a bare Co∣vering, or Roof.

Also in measuring of Plasterers work in Partitions and Walls; the Timbers and Quarters, are not to be deducted out of the rendring for Work only, except when the Workman finds the Work and Stuff also, then substract a 6th part for the Quarters in the rendring Work: But in Ceilings, the

Summers which are seen, are alwayes abated▪ and Doors and Windows also, unless by a due considerate (or an unconsiderate) bar∣gain of running measure.

Thus you have a brief account of the u∣sual order, used among Workmen, in taking the Dimentions of a House, viz. Brick-work by the Rod; Tileing and Carpenters-work by the Square; Chimneys usually by the Fire; And Plasterers and Painters-work by the Yard; Glasiers, by the Foot.

There are many other things to be taken notice of in the Carpenters Bill, as Lintels, Mantle-trees, and Tassels; Luthern Lights, and other Lights, both Architrave and Plain Lights, Sky-lights, or Cubiloes, Modillean Cornish, and guttering Penthouse Cornish, Timber-Front-Story, Cellar-doors, and Door-cases; the Plank and Curb at the Cellar-stairs, Dogleg-stairs, and Open-Newel-stairs, Canted-stairs, counted either by the step or pair; together with the half Spaces on the Corners of the open Newel-stairs, the Rayles and Ballasters, small and great Cornish, Outside-work and Partiti∣ons, Ceiling Joysts, and the Ashlering, Boarded Partitions, and Chequer-work; back-Doors, and Door-cases; Window-boards, and Wall-timber; Planks in the Foundation, Palcing, Penthouse-floors, and

Penthouse-roof, furring the Platform, Centerings for the Chimney, Trimmers, Girders-ends, Ends of Brest-summers, and Plate; and more the like, which will come in Accompt to be remembred and set down according as the Building is.

Also, with due allowance into the Wall that way the ends of the Joysts are entred or laid in the Wall, as thus;

If it be Framing Work is only mea∣sured, then 9 Inches ought to be allowed into each Wall, that way the Joysts ends are laid; because every Joyst, if well laid, should have 9 inches, at least, hold on the Wall.

But if it be Timber, and Boarding, both to be measured, then 6 inches only is a com∣petent allowance; because the Timber is usually vallued at one third part more than the Boarding is.

Also, As the Workman doth think on this, the Work-master may not forget to deduct for Stairs, and Chimneys also, where Work and Stuff are both measured; though for Work only it may be very well allowed, unless the better Price make an allowance for it.

Note also, That by the Line of Num∣bers, you may readily find the length of the Hips and Rafters, in a Roof of any large∣ness,

at true pitch, by this following Pro∣portion and Table.

The Breadth of the House being 40 Feet, and the Ends Square, the Length and Angles are, as in the Table, at the usual tru•• pitch.

	f••et.	100 par▪
Whole breadth	40	00
Half breadth	20	00
Rafter	30	00
Hip-Rafter	36	00
Diagonal Line	56	57
Half-Diagonal	28	28
Perpendiculer	22	36

		deg.	min.
Hip Angles	at Foot	38	22
at Top	51	38
on the Outside	116	12
Rafter Angles	at Foot	48	10
at Top	41	50

For any other House, by the Numbers thus: as suppose 18 Foot broad.

The Extent of the Compasses from 40, the breadth in the Table, to 18 the breadth gi∣ven, shall reach the same way from 30, the Rafter in the Table, to 13-50, the Rafter

required. And from 36, the Hip in the Table, to 16-22, the Hip required. And from 22-36, the Perpendiculer in the Table, to 10-06, the Perpendiculer required. And from ••6-57, the Diagonal in the Table, to ••5-48, the Diagonal required. The Angles are alwayes the same in all Rooss, small or great, as in the Table, being Square and true pitch.

If you would have Directions for Bevel or Taper Frames, to find the Lengths and Angles of Ra••ters and Hips, you may have it at large, in an Appendix to the Mirrour of Architecture; or, Vincent Stamo••••i, Printed for William Fisher, at the Postern-Gate, 1669.

By which Directions, and the Sector, you may find any thing that is there set down. As also, by the Trianguler Quadrant, Thred and Compasses.

Note also, That having Inches and Foot-measure together, you may presently, by inspection, find the price of one Foot, ha∣ving the price of the Square, and the con∣trary. Also, having the 12 Inches on the o∣ther Foot, divided into 85 parts (near), and figured at every 8 with 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. shall represent pence and half farthings; then at any price the Rod, you have the price of one Foot, & the contrary.

[illustration]

[illustration]

As thus;

Let every Inch, represent one pound; 〈◊〉〈◊〉 ••very 8th part, 2 shillings and 6 pence; or every 10th part, 2 shillings; because 8 half-crowns, or 10 two shillings, is 20 shil∣lings.

Example.

Right against 6 Inches and a half, for 6 l. ••0 s. on this other Line, I find •• pence 3 farthings, the price of one Foot, at 6l. 10s. per Rod: And at 7 farthings per Foot, I find near 40 shillings, or 2 pound per Rod.

Also, at 40 shillings per Square, found at 40, on Foot-measure, is 4 pence 3 far∣things 1/•• per Foot, found just against it on the Inches.

CHAP. VIII. The use of the Line of Num∣bers, in measuring of Land by Perches and Acres.

Problem I. At any length of the Land, to find the breadth of the Acre.

IN the Answering of this Question, it is not amiss, but very needful to premise, how many Square Inches, Feet▪ Yards, Perches; or Chains I mean a Chain of ••6 ••oot long) is contained in a Square Acre of Land; for which purpose, have recourse to the Table annexed, which is drawn with great care and exactness for that purpose.

By which Table you may perceive, That 6272640 Square inches are contained in one Square Acre.

And (100000, or) one hundred thou∣sand Square Links of a 4 Pole Chain, make a Square Acre.

And 43560 Square Feet, make a Square Acre.

And 4840 Square Yards, make a Square Acre.

And 1742, 4 Square Paces, make a Square Acre.

And 160 Square Perch, make a Square Acre.

And 10 Square 4 Pole Chain, make one Acre. As in the Table you may see.

And 3097 1/•• Square Ells, make one Acre of Land, Statute measure.

The Table.

	Inch	Links	Feet	Yards	Pace	Perch	Chain	Acre	Mile
Inch	1	7.92	12	36	60	198	792	7920	63360
Link	62.720	1	1.515	4.56	7.575	25	100	1000	8000
Feet	144	2.295	1	3	5	16.5	66	660	5280
Yard	1296	20.755	9	1	1.66	550	22	220	1760
Pace	3600	57.381	25	2.778	1	3.3	13.2	132	1056
Perch	39204	625	272.25	30 25	10.89	1	4	40	320
Chain	627264	10000	4356	484	174.24	16	1	10	80
Acre	6272640	100000	43560	4840	1742.4	160	10	1	8
Mile	4014489600	64000000	27878400	3097600	1115136	102400	6400	640	1
Square	Inches	Links	Feet	Yards	Pace	Perch	Chain	Acre	Mile

Then as the length of the Land given in Feet, Yards, Paces, Perches, or Chains, is to the number of Square Feet, Yards, Paces, Perches, or Chains in a Square Acre; So is 1 to the breadth of the Land (in that mea∣sure the length was given) to make a Square Acre: See the Examples of all these mea∣sures in their order, viz. of Feet, Yards, Paces, Perches, and Chains.

Suppose a piece of Land be 660 Feet long, or 220 Yards, or 132 Paces, or 40 Perches, or 10 Chains in length▪ which several measures are all of the same quanti∣ty; I would know how much in breadth I must have to make a Square Acre?

Extend the Compasses from the length given, viz. 660 Feet, or 220 Yards, or 132 Paces, or 40 Perches, or 10 Chains; to 43560, for Feet; or to 4840, for Yards; or to 1742, for Paces; or to 160, for Per∣ches; or to 10, for Chains; To the Num∣ber in the Table for that measure in a Square Acre; the same Extent applyed the same way from 1, shall reach to the Feet, Yards, Paces, Perches, or Chains required.

Note the Work.

1. As 660, to 43563, the Feet in a Square Acre;

So is 1, to 66, the breadth in Feet re∣quired.

2. As 220, the length in Yards, to 4840, the Square Yards in a Square Acre;

So is 1, to 22, the breadth in Yards re∣quired.

3. As 132, the length in Paces, to 1742;

So is 1, to 13-2, the breadth in Paces sought.

4. As 40, the length in Perches, to 1••0; So is 1, to 4, the breadth in Perches.

5. As 10, the length in Chains, to 10; So is 1, to 1, the breadth in Chains re∣quired.

6. As 176, the length in Elles, to 3097 ⅓;

So is ••, to 17-6, the breadth in Elles re∣quired.

To work this by the Line of Lines, say;

1. As the — 43560, to = 660; So is = 10, to 66, Latterally.

2: As the Latteral 220, to Parallel 4840;

So is Latteral 1, to Parallel 22, (single, double, or four-fold).

3. As — 132 doubled, is to = 1742 likewise doubled, because it falls near the Center;

So is — 1 quadrupled, viz. 4, to = 13-2 quadrupled, viz. 52-4.

4. As — 160, to = 40;

So is = 10 for 1, to — 4 Perch.

5. As — 10, to = 10;

So is = 1 to — 1, the breadth required.

If you would know how much breadth, at any length, shall make 2, 3, or 4 Acres;

Then say,

As the length given to the quantity of one Acre in that measure, according to the Table;

So is 2, 3, 4, or 5, to the breadth re∣quired.

Example at 30 Perch in length.

The Extent from 30 to 160, shall reach the same way from 4 to 21 Perch, and 34 of 100 (or 5 Foot, 06 Inches) the breadth of 4 Acres, at 30 Perches in length.

Problem II. The length and breadth given in Perches, to find the Content in Perches, of any piece of Land.

The Extent from 1, to the breadth in Perches, shall reach the same way, from the length in Perches, to the true Content in Square Perches.

Example.

As 1 to 50, so is 179 to 8950, the Con∣tent in Square Perches.

Problem III. The length and breadth being given in Perches, to find the Content in Square Acres.

The Extent from 160 to the breadth in Perches, shall ••each the same way, from the length in Perches, to the Content in Square Acres.

Example.

As 160 to 50, so is 179 to 5-58 Acres, or 5 A••res, 2 Rood, and 13 Perches.

Problem IV. The length and breadth of a piece of Land being given in Chains, to find the Con∣tent in Acres.

The Extent from 1, to the breadth in Chains, and 100 parts, which are Links, shall reach the same way from the length in Chains and Links, to the Content in Square Acres.

Example.

As 1 to 5 Chains, 52 Links, the breadth;

So is 8 Chains, 72 Links, to 48 Acres, and 3960 Square Links.

Problem V. Having the Base and Perpendiculer of a Triangle given in Chains or Perches, to find the Content in Acres.

The Extent from 2, if you use Chains; or from 320, if you measure by Perches, to the whole Base, shall reach the same way from the whole Perpendiculer, to the whole Content of the Triangle; or if it be a Tra∣pezia, joyn both the Perpendiculers in one sum.

Example.

As 2 (for Chains) to 3-63, the whole Perpendiculer;

So is 11-80, the whole Base, to 21 A∣cres, 42 Links, the Content of the whole Triangle.

Or in Perches.

As 320 to 14-55, the Perpendiculer in Perches;

So is 47-20, the length, or base Line, in Perches, to 21 Acres, 24 Links, the Content in Acres.

Problem VI. The Area, or Content of a piece of Land given, that was measured by Statute-Perches; to find the Content of the same piece of Land in Wood-land measure, or Customary Acres, or Irish Acres.

For the better understanding of this Pro∣blem, it is necessary to describe the several kinds and quantities of Perch••s, which are spoken of by Authors, and used in several places; together with their proportion to the Statute Perch of 16 Foot and a half square, London measure.

The kinds of Perches, are first Statute-measure of 16 foot ½ to the Perch, according

to the Standard at Guild-Hall, or the King's Majesties Exchequer. Secondly, Woodland-measure, a Perch whereof contains 18 Foot Square of the same London measure. Third∣ly, Irish Acres, of 21 Foot to the Perch or Pole. And lastly, Three sorts of Customary, used in several places of England, of 20, 24, Cheshire measure, and 28 Foot square to the Perch.

As for the Proportions one to another, that is, as 16 ½, to 18, 20, 21, 24, 28, or any the like wha••soever.

But to find their difference in Squares or Scales, the Work is thus; By the Line of Numbers, First appoint what Number in an Inch shall be the Scale for Statute measure, which I shall appoint a Scale of 30 in an Inch.

Then the Extent from 16 ½, to 18, for Woodland measure, shall reach the contrary way from 30, being twice repeated, to 25-2; so, I say, that a Scale made to 25-2 in an Inch, shall be the Scale for a Woodland Perch of 18 Foot Square; and in proporti∣on to that of 16 Foot ½, at 30 parts in an Inch.

Again, For Irish Acres, which are mea∣sured by a Pole of 21 Foot to a Perch, the Extent on the Line of Numbers from 21 to 16 ½, shall reach (being turned twice the

same way) from 30, to 16 ½, the quantity of the Scale for Irish Acres, to be in propor∣tion to a Scale of 30 in an Inch for Statute-measure; and so for the rest, or any other whatsoever, as in the following Table.

16 ½	The Scale that is to it proportio∣nable to 30 for Statute measure is,	16 ½	is	30—00	Statute-Measure.
18	18	25—22	Woodland-Measure.
20	20	20—42	Customary.
21	21	18=50	In an Inch for Irish.
22	22	16—89	Customary.
24	24	14—20	Customary Cheshire-measure.
28	28	10—42	Customary.
30	30	09—08	Or any other.

So that if you have several Scales made upon a Rule (to draw the Plot of your Field withal) to these Proportions (which may be convenient enough for Difference between one another), then for the reducing of

the quantity of Acres found by Statute-measure, to Woodland, Irish, or Customa∣ry, is no more but thus:

Take the Acres, measured by Statute-measure, out of the Scale of 30 in an Inch appointed for Statute measure, and measure it in the Scale of 25-22 in an Inch for Woodland; or by the Scale of 18-55 for Irish Acres; or by the Scale of 16-89 for Customary; and you shall have the quanti∣ty of Woodland, Irish, or Customary Acres required.

Example.

Suppose I have 30 Acres of Statute mea∣sure, how many Acres of Woodland, Irish, or Customary measure will they make? Take 30 from the Scale of 30 in an Inch, and on the Scale of 25-22, it shall give 25-22, for so many Woodland Acres; and on the Scale of 18-55, for Irish Acres, it shall give 18-55 for so many Irish Acres; and on the Scale of 16-89 in an Inch for Customary Acres, it shall give 16-89 for so many Customary Acres, at 22 Foot to the Perch or Pole, &c.

This being thus fully premised, to work these Questions by the Line of Numbers only: the Extent of the Compasses from

1••-5, the Feet in a Statute Perch, to 18 the Feet in a Woodland Perch (or to 21 the Feet in an Irish Perch, or to 22, 24, 28 the Feet in a Customary Perch) shall reach from 30, the Acres in Statute measure, be∣ng twice repeated, to 25-22, the Acres in Woodland measure required, &c. it being a larger-Acre▪ must nee••s be less in quantity.

Which work is performed by the back-Rule of Three in a duplicated proportion.

Problem VII. Having the Plot or Draught of a Field, and its Content in Acres, to find by what Scale it was Plotted; that is, by what parts in an Inch.

Suppose a Triangle, or a Parallellagram, or long Square, do contain 4 Acres and a half, which is set down in figures thus, 4-50; which if I should measure by a Scale of 12 in an Inch, might happen to be 2-25 Chains one way, and 1 Chain, 25 Links ••he other way; which two sums being mul∣tiplied together, make 2-5200, whereas it should be 4-5000;

Therefore by th•• 〈◊〉〈◊〉 of Numbers, to gain 〈…〉〈…〉 do thus;

Divide the dista••ce between 2-5200, and

4-5000, into two equal parts; that distance laid the right way from 12, the Scale I measured by, shall reach to 16, the Scale the Plot was made by.

For Note, That if the Scale I guessed a••, gives more than I should have, then I have too many in an Inch; but if less, I must have more in an Inch, as here, which infal∣libly sheweth which way, which is alwayes the same way as you divided the space, from the guessed Sum or Product, to the true Pro∣duct.

To this Rule may be referred the way to discover the true size of Glasiers Quarries; the method whereof is thus: They are usu∣ally cut to, and called by 8s, 10s, 12s, 15s, 18s, and 20s in a Foot, or any other what you please; that is to say, 8 quarries of Glass of 8s, make a Superficial Foot; and 10 quarries of 10s, make a Foot Superfici∣al; and 12 of the 12s, &c.

Also they are cut in a Diamond form to one sort of Angle for the Squ••re quarries; and another for the Long quarries: The acute Angle of the Squar•• quarries being 77 degrees, and 19 〈…〉〈…〉 and the acute Angle of the Long quar••••es ••7 degrees and 22 minutes: The long ••as being just 6 inches long, and 4 inches broad; and the Square 10s, 6 inches long, and 4

inches, and 80 parts of a 100 broad▪

This being the standing Rule or Method, and those two sizes being known, I would find out any other, as 13s, or 14s, or 17s and the like.

Do thus;

Divide the distance on the Line of Num∣bers, between the Content of some known size, and the Content of the inquired size into two equal parts; and that distance laid the right way from the sides of the known size (increasing for a bigger, and decreasing for a less) shall give the reciprocal sides of the size required.

Example. The Sides, Ranges, Lengths, and Breadt•• of Square 10s, are as in the Table fol∣lowing; and I would have the Ranges, Sides, Length, and Breadth of 14s, an unusual Size.

The Content of a Square quarry of Glass called 10s, is a just 10th part of a Foot, which is 1 inch and 20 parts; or one 10th part of a Superficial Foot, containing 12 long inches.

And the Content of the size called 14s, must be one 14th part of the same measure, or Foot Superficial, which is 0-85714, that is 0-857 parts of one long inch in a 1000 parts.

Then, by the Line of Numbers, divide the space between 1-2000, the Content of the 10s, and 0-857 the Content of the 14s into two equal parts; that Extent, I say, laid the same way from 3-76, the Ranges of Square 10s, shall reach to 3-18, the Ranges for 14s: And from 3-84, the sides of Square 10s, to 3-25 the sides of Square 14s: And from 4-80 the breadth of Square 10s, to 4-05 the breadth of Square 14s: And from 6-00, the length of Square 10s, to 5-07 the length of Square 14s, the requisites of the unknown Size required. And the like for any other whatsoever.

The true size of Glasiers Quarries, both Long and Square, By J. B. 1660.

Square Quarries 77 deg. 19 min.	Long Quarries 67 deg. 22 min.
Sizes	Rang.	Sides.	bredth	length	Content.	Content.	Size.	Rang.	Sides.	bredth	length	Content.	Content.
In. 100	In. 100	In. 100	In 100	Inc. 100	Tens.	In. 100	In. 100	In. 100	In. 100	Inc. 100	Tens.
8	4 20	4 30	5 36	6 70	1 500	1 250	8	4 09	4 41	4 90	7 34	1 500	1 250
10	3 76	3 84	4 80	6 00	1 200	1 000	10	3 65	3 95	4 38	6 57	1 200	1 000
12	3 43	3 51	4 38	5 47	1 000	0 833	12	3 34	3 61	4 00	6 00	1 000	0 833
15	3 07	3 13	3 92	4 90	0 800	0 667	15	2 98	3 23	3 58	5 37	0 800	0 667
18	2 80	2 86	3 57	4 77	0 666	0 555	18	2 72	2 95	3 26	4 90	0 666	0 555
20	2 66	2 72	3 39	4 24	0 600	0 500	20	2 58	2 79	3 10	4 65	0 600	0 500

CHAP. IX. The use of the Line of Numbers in measuring of Solid measure, as Timber, Stone, or the like Solid bodies.

Problem I. A piece of Timber being broader one way than the other, to find the side of a Square that shall be equal thereunto, be∣ing called, Squaring the Piece.

THe side of the Square, that shall be equal to the square of the Oblong, is nothing else but a mean proportion between the length and breadth of the Oblong:

As thu••;

Suppose a piece of Timber is 12 inches in depth, and 16 inches in breadth (and 10 foot in length.)

16 the breadth, and 12 the depth, mul∣tiplied

together, make 192; the Square or Product of 16 and 12 multiplied.

Now the square Root of 192, which is near 13 ••59/1000, is the side of a Square, equal to 12 and 16, the depth and thickness of the piece of Timber propounded. For if you shall multiply 13-859 by 13-859, you shall find 192-071881, the nearest Root, you can express in 5 figures, and an in∣different true mean proportion, between 12 and 16, the depth and breadth; so that in fine, 13-86, is the side of a Square, nearly equal to 12 and 16, whereas the doubling and halfing, the old false way, gives full 14.

To work this by the Line of Numbers, is thus;

Divide the distance on the Line of Num∣bers, between 12 and 16, into two equal parts, and you shall find the Point to stay at 13, and near 86 parts, the Answer re∣quired.

The way of doing it, by the Line of Lines, is shewed in the VI Chapter, and 7th Proposition, either by the Sector, or Tri∣anguler Quadrant, and therefore needs no repetition in this pla••e.

Problem II. At any Breadth and Depth, or Square∣ness, to find how much makes a Foot of Timber.

1. If the Timber be square (or squared) then the way by the Line of Numbers, is thus;

First by Foot measure.

Extend the Compasses from the side of the square to the middle 1, the same Extent ap∣plyed, or turned twice the same way from 1, shall reach to the length that makes a Foot of Timber, at that squareness.

Example.

Suppose a piece of Timber be 50 of a 100 (or 6 inches) or half a Foot Square (which is all as one) Extend the Compasses from 5 to 1 (forwards) the same Extent being turn∣ed two times, the same way from 1, shall reach to 4, being 4 Foot, or 40 such parts, whereof the side of the Square was 5.

2. Secondly, The same again by Inches.

The Extent from 6 to 12, shall reach, be∣ing turned two times the same way from 12, to 48, the number of inches in length that makes a Foot, at that Squareness; being 48

such parts whereof the side of the Square was 6.

So that,

As the side of the Square, in inches, is to 12: so is 12 to a 4th, and so is that 4th to the length of a Foot required, turning the Compasses twice, the same way as you turned from the side of the Square in inches to 12.

3. If the piece of Timber, or Stone, be not Square or Squared,

Then

The Extent from 1 to the depth, shall reach the same way from the breadth to a 4th Number.

Again,

The Extent from that 4th Number to 1, shall reach, being turned once, the same way from 1, to the length of a Foot in Foot-measure required.

Example.

Suppose a piece of Timber be 0-333 deep, and 0-750 broad in Foot-measure; or 4 inches deep, and 9 inches broad, as with a glance of your eye on inches and foot-mea∣sure, you may see how these Numbers agree. The Extent, I say, from 1 to 0-333, shall reach the same way from 0-750 to 2-50.

Again,

I say, The Extent from 250 the 4th, to

1, shall reach the same way, from 1 to 40, or 4 Foot, the length required, to make a Foot at that breadth and depth.

4. By Inch-measure, 〈◊〉〈◊〉 find the length of a Foot in Inches.

As 12 to the breadth in inches, so is the depth in inches to a 4th; then as that 4th to 12, so is 12 to the length in inches re∣quired.

Example.

The Extent from 12 to 9 the breadth, shall reach the same way from 4 the depth, to 3 for a 4th.

Then the Extent from 3 the 4th to 12, shall reach the same way from 12 to 48, the inches in length required, to make a Foot.

5. The breadth and depth given in Inches, to find the length of a Foot of Timber, in Feet and Parts.

Then say, As 1, to the depth; so is the breadth to a 4th.

Again,

As that 4th to 12, so is 12 to the length in feet and parts.

Example.

The Extent from 1 to 4, shall reach the same way from 9 to 36, a 4th;

Then,

The Extent from 36 to 12, shall reach the same way from 12 to 4 foot, the length in feet required.

The reason of this Order and Method, if you consider, you will find thus;

In the 4th way of working, you went thus;

As 12, the inches in a foot, is to the breadth in inches;

So is the depth to 3 Foot.

But in the 5th and last way you say,

As 1 foot to the depth in inches;

So is the breadth to 36 inches, which is 3 foot also.

But altering the Order in the beginning, alters it in the issue, though the same truth, yet in or under divers denominations; for 48 inches, and 4 foot, are the same; yet sometimes one way is more convenient than another.

Problem III. At any Squareness, or Breadth and Depth given in Foot-measure, or Inches, to find how much Timber is in a Foot long, in Foot-measure, or feet, and 100 parts or inches.

1. If the piece of Timber be Square (or Squared) then work thus for Foot-measure.

As 1, to the side of the Square, so is the side of the Square to the quantity of Timber in one Foot long; which multiplied by the length, gives the whole Content required.

Example. At 50, or half a Foot Square, how much is in a Foot long?

Extend the Compasses from 1 to 5, the same Extent turned the same way from 5, reaches to 25, or a quarter of a Foot; then if the Tree be 12 foot long, 12 quarters will make 3 foot, the Content.

2. The Side of the Square given in Inches, to find the Quantity, or Content, in a Foot.

As 12, to the side of the Square, so is the

side of the Square to 3 twelve parts of a Foot Solid, or ¼ of a Foot.

Or,

As 1, to the side of the Square, so is the side of the Square to 36, 144th parts of a Foot Solid.

Example.

The Extent from 12 to 6, the inches Square, shall reach the same way from 6 to 3 inches in a Foot long, which is 3 12th parts of a Foot Solid.

Again,

The Extent from 1 to 6, the inches square, shall reach the same way from 6 to 36, the number of long inches in a foot long; or pieces of 1 inch square, and a foot long, 144 of which makes one foot of Timber.

3. But if the Piece be not square (or squa∣red) then to find how much is in a Foot-long, work thus;

As 1 to the depth, so is the breadth, to the quantity in a Foot.

Example 3 wayes: At 9 and 4 breadth and depth.

1. The Extent from 1 to 0-333, shall reach the same way from 0-75, to 0-25, or a quarter of a Foot; for Foot-measure.

2. The Extent from 1 to 9, shall reach the same way from 4 to 36, the long inches in a foot long; for Inch-measure.

3. The Extent from 12 to 4, shall reach the same way from 9 to 3 inches, or 3 12ths, viz. a quarter of a Foot; for Inch-measure.

Problem IV. The side of the Square, or the breadth and depth given in Inches, or Foot-measure, and the length in F••••t, to find the Quan∣tity, or Conten•• of the whole Piece, in feet and parts.

1. First, for Foot-measure;

As 1, to the side of the Square, in Foot-measure, so is the length in Feet to a 4th, and then that 4 to the Content in feet and parts.

Example.

The Extent from 1 to 0-833, the side of the Square, shall reach from 10 foot 25 parts, the length to 8-54, and from thence to 7-11, the Content in feet and parts re∣quired.

2. For Inch-measure, Say,

As 12 to the side of the Square, in inches, so is the length in feet, to a 4th; and then

that 4th to the Content in feet and parts.

Example at 10 Inches Square, and 10 Foot, 3 Inches in length.

The Extent of the Compasses, on the Line of Numbers, from 12, to 10 inches Square, shall reach the same way from 10 foot ¼, or 3 inches, to 85-4 for a 4th; and from thence to 7-11, or 7 foot 1 inch, and a third part, the Content required. As by look∣ing for 11 on the Line of Foot-measure, right against which, on the inches, is 1 inch and a quarter, and somewhat more.

3. But if the piece of Timber be not square, and you would measure it without squaring, by the first Problem;

Then say first by Foot-measure, thus;

As 1 is to the breadth, so is the depth to a 4th. Then,

As 1 to the 4th, so is the length in feet to the true Content, in feet and parts.

Example.

Let a Timber-tree of one foot 25, or a quarter one way, and one foot 50 the other way, and 12 foot long be measured.

The Extent of the Compasses from 1 to 1-25, shall reach the same way from 150, to 18-74, for a 4th.

Then the Extent from 1 to 18-74, shall reach the same way from 12 foot, the length, to 22-50, for the Content; viz. 22 foot and a half, the whole Content required.

4. When the breadth and depth is given in Inches, and the length in Feet, to find the Content without squaring.

As 12, to the breadth in inches;

So is the depth in inches to a 4th:

Then,

As 12 to that 4th, so is the length in feet and parts, to the Content in feet and parts required.

Example at 15 inches deep, and 18 inches broad, and 13 foot long.

Extend the Compasses on the Line of Numbers from 12 to 15 the depth; the same Extent applied the same way from 18 the breadth, shall reach to 22-50, for a 4th.

Then the Extent, from 12 to 22-50, the 4th, shall reach the same way from 13 foot, the length, to 24 foot 38 parts, or 4 inches and a half, as a glance of your eye to the Inches and Foot-measure will plainly shew.

Thus you have the Solution of any Que∣stion that may concern proper Measuring by Foot-measure, and Inches; using only the Center at 10 for Foot-measure, and at 12 for Inch-measure, without troubling you with 144, or 1728, or 41-57, or the like, as in the little Book of the Carpenters Rule, may be seen.

To work these Questions by the Line of Lines, though it may be done several ways, yet no way so soon, nor so exact, as by the Line of Numbers: Yet I shall shew now in this place, together by themselves, the Three principal Questions, viz. How much makes a Foot in quantity; And, How much is in a Foot long; And, By the length, breadth and depth, the Content in Feet: In the doing whereof, you must conceive the 10 principal parts to be doubled, and then 10 is called 20; and consequently 6 is cal∣led 12, the Point so often used; and 5 is called 10, the Point used for Foot-measure.

1. To find how many Inches makes a Foot at any Squareness.

As the — side of the Square, to = 12;

So is the = side of the Square again, to a — 4th Number.

Again,

As = 12, to that = 4th Number;

So is = 12, to the — Number of Inches that goes to make a Foot of Timber.

Example, at 8 Inches Square.

Take the distance from the Center to 4, accounted as 8; and make it a Parallel in 6, counted as 12; or lay the Thred to the nearest distance, and there keep it. Then, take the nearest distance from 4 to the Thred, and that shall be a Latteral 4th.

Then take the Latteral distance from the Center to 12, according to the usual ac∣count, and make it a Parallel in the 4th last found, laying the Thred to the nearest distance, and there keep it; then take the nearest distance from 6, counted as 12, to the Thred, and that shall reach Latterally from the Center to 27 Inches, the length required, to make a Foot of Timber, at 8 Inches Square.

Which work I more briefly word thus, as formerly is done.

As — 4, counted as 8, to = 6, count∣ed for 12;

So is = 8, to a — 4th.

Then,

As = 12, to = 4th; So is = 12, to — 27, the length in inches re∣quired.

2. If you would use Foot-measure, count the 5 in the midst for 10, or 1 Foot; and work all the rest as before: As thus for Example.

In the same quantity, Square, exprest in Decimals:

As — 0-666, counted double, to = 5, counted double for 10;

So is = 0-666, to — 22 1/••, for a 4th.

Then,

As — 1, to = 22 1/••; So is = 5 count∣ed for 1, to 225, which is 2 Foot ¼, as by the Foot-measure and Inches you may see.

3. If the Piece be not square, then say thus;

As — breadth, to = 12;

So is the = depth, to the — 4th.

Then,

As — 12, to the = 4th;

So is = 12, to — length that goes to make 1 Foot.

Example, at 9 Inches, and 4 Inches, for breadth and depth.

As — 9, to = 12;

So is = 4, to — 150, for a 4th.

As — 12, to = 4th, best taken at 75 for largeness sake;

So is = 12, to — 48 Inches.

Or else thus;

As — 9, to = 1;

So is = 4, to — 1-80, a 4th.

Then,

As — 12, to = 1-80;

So is = 12, to — 4 Foot, the length in Feet, that goes to make 1 Foot of Timber.

4. To find how much is in a Foot-long, at any Squareness.

As the — side of the Square is to = 1, counted double as before;

So is the = side of the Square to the — quantity in a Foot.

Example at 6 Inches, or (5) half •• Foot Square.

As — 5, to = 1; so is = 5, to — 125 for Foot-measure: Or,

As — 6, to = 12; so is = 6, to — 3, for 3 inches, or ¼ of a foot.

5. The side of the Square given in Inches, and the length in Feet, to find the Con∣tent in Feet.

As — side of the Square, to = 12;

So is the = length to a 4th.

Then,

As — 4th; to = 12; so is = side of the Square to — Content required, in feet and parts.

Example, at 9 Inches Square, and 16 Foot long.

As — 9 to = 12, so is = 16 to — 12.

Again,

As — 4th, viz. 12, to = 12;

So is = 9, the square to 9, the true Con∣tent of such a Piece in feet and parts required.

The like Work serves for Foot-measure, using of 1.

6. The Length given in Feet, 〈◊〉〈◊〉 the Breadth and Depth in Inches, to find the Content in feet and parts.

As — breadth, to = 12;

So is = depth, to a 4th.

Then,

As — 4th, to = 12;

So is = length, to — Content in feet.

Example at 5 Inches and a half Deep, and 15 Inches Broad, and 16 Foot Long.

As — 5 ½, to = 12;

So is = 15, to — 69 for a 4th (at 34 ½)

Then,

As — 69 (or 34 ½) that — 4th, to = 12;

So is = 16, taken at 8, to — 9 foot 2 inches; the Content required.

Thus you see the way and manner of working by the Line of Lines, either on the Quadrant, or Sector-side, for the usual Que∣stions; for I have neglected to give the Content of Pieces in Cube Inches, for two Reasons: First, Because it is very seldom

required. Secondly, Because the Line of Numbers at most will shew but 4 figures, which is not sufficient for any Piece above 6 Foot, therefore not fit for Instrumental Work.

And withal you may observe, That al∣wayes the Latter••l Extent first taken, must be less than the distance from the Center to the parallel Point of Entrance, which in these Examples is remedied by calling 6 12: And also, there are so many Cautions in doubling and halfing of Numbers, to make it applicable, that without due considerati∣on, you may soon err; Also, the opening and shutting the Rule, and using of several Scales, makes it far inferior to the Line of Numbers, which may be easily enlarged.

CHAP. X. To measure Round Timber, or Cillenders, by the Line of Numbers.

Problem I. Having the Diameter of a Cillender, given in Inches, or Foot-measure, to find the length of one Foot.

1. AS the Diameter in inches, to 46-90, (at which Diameter one Inch makes a Foot); So is 1 to a 4th, and that 4th to the length in inches.

Example at 10 Inches Diameter.

The Extent from 10 to 46-90, being turned two times the same way from 1, shall reach to 21 inches, 8 10ths, for the length of a Foot, at that Diameter, in Inches.

Or rather work thus;

As the Inches Diameter, to 13-54;

So is 12 twice, to the Inches that make a Foot of Timber.

Or,

The Extent from 10, to 13-54, turned twice the same way from 12, shall reach to 22 Inches.

Or,

The same Extent being turned two times the same way from 1, shall reach to 1-831, which is the Decimal for 22 Inches, as by looking on Inch and Foot measure, you may plainly see.

Again,

2. For the same Diameter in Foot-mea∣sure.

The Extent from 0-833 (the Decimal of 10 Inches) to 1-128, being turned twice the same way from 1, shall reach to 1-83, which is almost 22 Inches, as by comparing Inches and Foot-measure together, is plainly seen.

Problem II. Having the Diameter given in Inches, or Foot-measure, to find how much is in a Foot long.

1. As 13-54 (the Inches Diameter that make a Foot of Timber, at one Foot long), to the Diameter in Inches;

So is 12 to a 4th, and so is that 4th, to the quantity in a Foot long.

Example at 10 Inches Diameter.

The Extent from 13-54 to 10, being re∣peated two times the same way from 12, shall reach to 6 Inches ½, or, 54 of 100, being somewhat more than a half Foot, for the true Content of one Foot long.

2. But if the Timber is great, then it is more convenient to have the quantity of a Foot, in feet and parts.

Then say,

As 13-54, is to the Diameter in Inches;

So is 1, to a 4th, and that 4th to the quantity in a Foot, in feet and parts.

Example, as before, at 10 Inches.

The Extent from 13-54 to 10, the Dia∣meter

in Inches, shall reach, being turned twice the same way from 1, to 0-545, the Content of a Foot long.

Again at 30 inches Diameter.

The Extent from 13-54, to 30, being turned two times the same way from 1, shall reach to 4 foot, 93 parts; which 4-93 multiplied by the length in feet, shall give the whole Content of the Tree.

3. To perform the same, having the Dia∣meter given in Foot-measure,

Do thus;

The Extent of the Compasses from 1-128, (the feet and 10ths Diameter that make a Foot, at one foot in length) to the Diameter in Foot-measure, shall reach, being turned twice the same way from 1, to the quantity in a Foot long.

Example at 1 Foot, 50/100 Diameter.

The Extent from 1-128, to 1-50, shall reach, being turned twice the same way from 1, to 1-77, the true quantity in one Foot long.

Problem III. 1. The Diameter of any Cillender given in Inches, and the length in Feet, to find the Content in Feet.

As 13-••4, to the Diameter in Inches;

So is the length in Feet to a 4th.

Then,

As the length, to the 4th;

So is the 4th, to the Content in Feet re∣quired.

Example at 8 Inches Diameter, and 20 Foot long.

The Extent from 13-54, to 8, being turned twice the same way from 20, the length, shall stay at 6-94, or near 7 foot.

2. The Diameter and length of a Cillender given in Inches, to find the Content in Cube-inches.

The Extent from 1-128, to the Diameter in Inches, being turned twice the same way from the length in Inches, shall reach to the Content in Inches.

Thus the Extent from 1-128 to 10 inches Diameter, shall reach from 24 inches, the

length, to 1888, the Content in inches.

3. The Diameter and Length given in Foot-measure, to find the Content in Feet.

The Extent from 1-128, to the Diameter, shall reach from the length, being twice re∣peated the same way, to the Content in feet required.

Thus the Extent from 1-128, to 1-50, shall reach, being turned twice the same way, from 5-30, to 9-37, the Content in feet required.

Problem IV. Having the Circumference of a Cillender given in Inches, or Foot-measure, to find the length that makes one Foot of Solid-measure.

1. First to find the Inches in length, that makes a Foot.

As the Circumference in Inches, is to 134-50, (because at so many inches a∣bout, one of a Foot in length, is a Foot) so is 12 to a 4th, and so is that 4th to the length of a Foot in inches.

Example at 30 Inches about.

The Extent from 30 to 134-50, being turned twice the same way from 12, shall reach to 24 inches, 13 parts; the inches and parts that make one Foot Solid.

2. To find the length of a Foot in feet and parts.

As the Circumference in Inches, to 134-50;

So is 1 to a 4th, and that 4th to the length in feet and parts, that makes 1 Foot.

For the Extent of the Compasses from 30 to 134-50, being turned twice from 1, the same way, shall reach to two foot, and one tenth, the length that makes one Foot Solid.

3. When the Circumference is given in Foot-measure.

As the Circumference in Feet, or Feet and parts, is to 3-54;

So is that Extent twice repeated the same way from 1, to the length that makes a Foot Solid.

Example.

The Extent from 2-50, to 3-54, being turned two times the same way from 1, doth reach to 2 foot, 001, the length in Foot-measure.

Problem V. The Circumference given in Inches, or Foot-measure, to find how much is in a Foot long.

1. The Circumference of a Tree, when one Foot long makes a Foot of Timber.

As 3 foot, 545 parts, to the feet about;

So is 1 foot to a 4th, and that 4th to the solid Content in one foot long.

Example.

The Extent of the Compasses from 3-545, to 2-50, the feet about, shall reach, being turned twice the same way from 1, to 0-497, the quantity in a foot long, viz. near half a foot.

2. The Circumference given in Inches, to find the Content of one Foot in length, Solid-measure, in Inches.

The Inches a Tree is about, when one 10th of a Foot in length, makes a Foot of Timber in quantity.

As 134-5, to the Inches about;

So is 12 to a 4th, and that 4th to the Con∣tent of one foot long.

Example at 30 inches about.

Th•• Extent from 134-5, to 30, being turned two times from 12, shall reach to near 6 inches for the Content of one foot long, at 30 inches about.

3. The Circumference of a Cillend••r given in Inches, to find the quantity of one Foot long in feet and inches.

As 134-5, to the Circumference;

So is 1 to a 4th, and that 4th to the quantity of one foot long in Feet and Inches.

The Extent from 134-5, to 30, being twice repeated the same way from 1, shall reach to 0-497, or near half a foot, the Content of one foot long, at that Circum∣ference, which being multiplied by the length in feet, gives the true Content of any Cillender whatsoever.

Problem VI. The Circumference, and length of any Cil∣lender given in Inches, or Feet and In∣ches, to find the Content.

1. The Circumference given in Inches, and the length in Feet, to find the Content in feet and parts.

As 42-54 (the Circumference in Inches, that makes 1 foot long, a Foot) is to the Inches in Circumference;

So is the length in Feet to a 4th, and that 4th to the Content in Feet.

Example.

The Extent from 42-54, to 48 the inches about, being twice repeated from 12 foot the length, shall reach to 15-28, the Con∣tent in feet required.

2. The Circumference and length given in Feet, to find the Content in feet and parts.

As 3-545, (because at 3 foot and a half about, and a foot in length, is a Foot) is to the Circumference;

So is the length in Feet to a 4th, and that 4th to the Content in Foot-measure.

Example.

The Extent from 3-545, to 4-0, the Circumference, being turned two times from 12 foot the length, shall reach to 15-28, the Content in feet required.

3. The Circumference and length given in Inches, to find the Content in Inches.

As 3-545, to the Circumference in Inches;

So is the length in Inches to a 4th:

Then,

As the length to that 4th;

So is the 4th, to the Content in Cube-Inches.

Example.

The precise Extent on a true Line of Num∣bers, from 3-545, to 48, being turned two times from 144, the length in Inches, shall reach to 26383, the number of Inches, in a Tree 48 inches about, and 144 inches in length.

This is sufficient for the Mensuration of any solid body, in a square, or Cillender-like form, as Timber or Stone usually is, after the true quantity of a foot, or 1728 Cubical inches; but there is a custome used in buying of Oaken-Timber, and Elm-Tim∣ber, when it is round and unsquared, to take

a Line, and girt about the midst of th•• Piece; and then to double the Line 4 times, and account that 4th part of the Circum∣ference, to be the side of the Square, equal to that Circle; but this is well known to be less than the true measure, by a fifth part of the true Content, be it more or less.

Also in measuring Elm, and Beech, and Ash, whose bark is not peeled off, as Oak usually is; to cast away 1 inch out of the 4th part of the Circumference, which may well be allowed when the Bark is 3 quar∣ters of an inch, or more in thickness, and the Tree about 40 inches about, or the 4th part, 10 inches; but if the Bark is thinner, and the Tree less, then 8 inches-square; then an inch is too much to be allowed.

Also, if the Tree is greater than a foot-square, and the Bark thick, an inch is too little to be allowed, as by this Rule you may plainly see, by the 7th Problem of Superfi∣cial-measure in the 7th Chapter.

Suppose a Tree be 48 inches about, the Diameter will be 15 ¼, the 4th of 48, for the square is 12.

Now if I take away 1 inch ½ from the Diameter, then the Tree will be but 43 inches and 1/•• about, whose 4th part is under 11; so that here I may very well a∣bate 1 inch from the 4th part of the Line;

So consequently, if the Rind be thinner, and the Tree less, a less allowance will serve; and if the Rind be thicker, and the Tree large, there ought to be more, as by cutting the Rind away, and then taking the true diame∣ter, you may plainly see.

This measuring by the 4th part of the Circumference, for the side of the Square, and allowance for the Bark being allowed for, as before, I say will prove to be just one 5th part over-measure.

Especially considering this, That when it is hewed, and large wanes left, then the Tree is marked for more measure, sometimes by 10 foot in 60, than there was before it was hewed; the reason is, because when the Tree is round and unhewn, the girting it, and counting the 4th part for the side of the Square, is but very little more than the in∣scribed Square; and then being hewen, and that scarce to an eight Square, and measur∣ing with a pair of Callipers, to the extremity of that, doth not then allow the Square e∣qual to the Circle for the side of the Square, as by the working by those several Squares, will very plainly appear, which being fore∣told and warned of, let those whom it con∣cerns look to it.

But this being premised, and the Parties agreeing, the difference being as 4 to 5, the

best way to measure round Timber, I con∣ceive, is by the Diameter taken with a pair of Callipers, and the length; which for the just and true measure is largely handled al∣ready.

But if this allowance be agreed on, then the Proportion for it is thus;

As 1-526, to the Diameter;

So is the length to a 4th, and so is that 4th to the Content in feet.

Example.

The Extent from 1-526 to 15-26, shall reach, being twice repeated from 10 foot, the length, to 10 foot the Content required, being all at one Point.

Or, another Example.

The Extent from 1-526, to 20 inches the Diameter, being twice repeated the same way from 10 foot, the length, shall reach to 17 foot ¼ the Content.

Or, if you have the Circumference and length.

Then the Extent from 48, to the inches about, being turned twice the same way from the length in feet, shall reach to the Content required.

The Extent from 48, to 62, the inches a∣bout, being turned twice from 10, the same

way, shall reach to 17 foot ¼, the Content in that measure.

Thus you have full and compleat Di∣rections for the measuring of any round Timber by the Line of Numbers, by ha∣ving the Diameter and length given, after any usual manner, there remains only one general and natural way, by find∣ing the base of the middle, or one end, by the 7th Problem of Superficial measure; and then to multiply that base by the length, will give the true Content in feet or inches.

Thus,

Having found the Base of the Cillender by the 7th or 10th Problem of Superficial-measure; then if you multiply that Base be∣ing found in square inches, by the length in inches you shall have the whole Content in Cube Inches.

Example.

Suppose a Cillender have 10 inches for its Diameter, then by the 7th or 10th above∣said, you shall find the Base to be 78-54; then if you multiply 78-54 by 80, the sup∣posed length in inches, you shall find 2356-20 Cube Inches, which divided by 1728, the inches in a Cube Foot, sheweth how many feet there is, &c. And as to the number of figures, and the fractions cutting off, you have ample Directions in the first

Problem, and the third Section of the six•• Chapter.

Problem VII. How to measure a Pyramis, or taper Tim∣ber, or the Section of a Cone.

1. First, get the Perpendiculer length of the Pyramis or Cone, thus;

Multiply half the Diameter of the Base, AB, by it self; then measure the side AD, and multiply that by it self; then take the lesser Square out of the greater, and the Square root of the residue is the Perpendi∣culer Altitude required, viz. DB.

Example.

Suppose the half Diameter of the Base AC, were 10-25, and the side DA 100, AB 10-25, and 10-25 multiplied together, called Squaring, makes 105, 0625; DC 100, multiplied by 100, called Squaring, makes 10000; then the lesser Square 105, 0625, taken out of 10000, the greater Square, the remainder is 9894, 9375, whose square Root found by the 8th Pro∣blem of the sixt Chapter, is 99-475, the true length of the Line DB, the length or height of the Cone.

Then if you multiply the Area or Con∣tent of the Base AC 20-5, which by the

7th or 10th of Superficial measure is found to be 160-08, by 33-158, a third part of 99-475, the whole height makes 5308, cutting off the Fractions for the true Con∣tent of the Cone, whose length is 99 inches, and near a half, and whose Base is 20 inches and a half Diameter.

2. Then Secondly, for the Segment or Section of a Cone, the shape or form of all round taper Timber, the truest way is thus;

By the length and difference of Diame∣ters, find the whole length of the Cone, which for all manner of Timber as it grows this way is near enough.

As thus;

As the difference of the Diameters at the the two ends, is to the length between the two ends;

So is the Diameter at the Base, to the whole length of the Cone.

Example.

The difference between the Diameters AC, and EF, is 13-70, the length, AE is 66-32. then the Extent on the Line of Numbers from 13-70, the difference of the Diameters; to 66-32, the length between, shall reach the same way from 20-50, the

greater Diameter to 99 and better, the length that makes up the Cone, at that Angle o•• Tapering in the Timber; then if by the last Rule you measure it as a Cone of that length, and also measure the little end or point at his length and diameter; and then lastly, this little Cone taken out of the great Cone, there remains the true Content of the Taper-piece that was to be measured, viz. 5246-71, when 61-30, the Content of the small Cone at the end, is taken out of 5308, the Content of the whole Pyramid.

3. If this way seem too troublesome for the common use, then use this, being more brief:

To the Content that is found out, by the Diameter in the midst of the Timber, and the length, add the Content of a Piece found out, by half the difference of Diame∣ters, and one third part of the length of the whole Piece, and the sum of them two shall be the whole Content required.

4. Or else;

Divide the length of the Tree into 4 or 5 parts, and measure the middle of each part severally, and that cast up by his proper length, shall give the Content of each Piece; then the sum of the Contents of all the Pieces put together, is the true Con∣tent

of the whole Taper Piece, very near.

Note, That this curiosity shall never need to be used, but when you meet with Timber much Taper, and Die-square, or on a Con∣test or Wager; for according to the usual way (and measure) of squaring the Timber, it is well, if the measure of the Square, ta∣ken with Callipers from side to side, in the middle of the length of the Piece, will make amends for half the Timber which is wan∣ting in the wany edges of your squared Timber, and the knots or swellings, & hol∣lows of most round Timber, may well bal∣lance this over-measure found by the Dia∣meter taken in the middle of the length of the Piece. But indeed for Masts of Ships and Yards, being wrought true and smooth, where the price of a Foot is considerable, there exactness is requisite, and necessary to be used; and thus much for Solid-measure in Squares and Cillenders.

Problem VIII. To measure Globes, and roundish Figures.

1. To measure a Sphear or Globe by A∣rithmetick, the ancient way, is to multiply the Diameter by it self, and then that Pro∣duct, to multiply by the Diameter again; which two multiplications is called Cubing

of the Diameter; then multiply this Cube by 11, and then divide this last Product by 21, and the Quotient shall be the Solid Con∣tent of the Sphear, in such measure as the Diameter was.

Example.

Let a Sphear be to be measured, whose Diameter is 10 inches: First, 10 times 10, is 100; and 10 times 100, is 1000; the Cube of 10, that multiplied by 11, makes 11000; which being divided by 21, makes 523-81, for the Solid Content.

Which by the Line of Numbers, you may work thus;

2. The Extent from 1, to the Diameter, shall reach the same way from the Diameter to the Square of the Diameter.

Then again,

The Extent from 1, to the Square of the Diameter, shall reach the same way from the Diameter, to the Cube of the Diameter.

Then,

The Extent from 1, to the Cube of the Diameter, shall reach the same way from 11, to the Product of the Cube of the Dia∣meter, multiplied by 11.

Lastly, This Extent from 21, to this last Product, shall reach the same way from 1, to the Solid Content of the Sphear required.

Or else more briefly thus;

3. The Extent from 1, to the Diameter, being turned three times the same way from 0-5238, shall stay at the Solid Content of the Sphear, or Globe, required.

Example at 12 Diameter.

The Extent from 1 to 12, being turned three times the same way from 0-5238, shall reach to 905-143, the Solid Content re∣quired.

3. The Diameter given, to find the Superficial Content.

Square the Diameter, an•• multiply that by 3-1416, and the Product is the Superfi∣cial Content.

Or, by the Line of Numbers;

The Extent from 1, to the Diameter, be∣ing turned twice the same way from 3-141••, shall reach to the Superficial Con∣tent, of the out-side round about the Gobe, viz. at 12 Diameter, 452-44.

4. Having the Superficial Content, to find the Diameter.

The Extent from 1 to 0-3183, shall reach the same way from the Superficial Content, to the Square of the Diameter, whose

Square-root is the Diameter required. As at 452-44, gives 144.

5. Having the Solid Content, to find the Diameter of a Globe.

The Extent from 1 to 1-90986, shall reach from the Solid Content to the Cube of the Diameter, as ••t 905-143 Solidity gives 1728, the Cube of 12.

6. Having a Segment of a Sphear, to find the Superficial Content.

The Extent from 1, to the Chord of the half Segment, shall reach, being twice re∣peated, from 3-1416, to the Superficial Content of the round part of the Segment, ABC.

Example.

Let the Segment be the half Sphear, ABC; AC being 12, then BC which is the Chord of the Peripheria, BC is 8-485, whose Square is 72.

Then,

The Extent of the Compasses from 1, to 8-485, being turned twice the same way from 3-1416, shall reach to 220-22, the Superficial Content of the round part of the Segment, or half Sphear or Globe, to which if you add the Content of the Circle or Base, you have the whole Superficies round about.

7. To find the Solid Content of a Seg∣ment of a Globe.

First, you must find the Diameter of tha•• Sphear, of which the given Segment to b•• measured is part.

Thus;

Add the Square of the Altitude, and the Square of the Diameter of the Segment to∣gether, and the sum divide by the Altitude of the Segment, the Quotient shall be the whole Sphears Diameter.

Then,

Taking the Altitude of the Segment gi∣ven, from the whole Diameter, there re∣mains the Altitude of the other Segment.

Then;

Extend the Compasses from the whole Diameter of the Sphear, to 1; the same Ex∣tent applied the same way from the Altitude of the given Segment, shall reach to a 4th Number, on a Line of Artificial Solid Seg∣ments joyned to the Line of Numbers, which 4th Number keep.

Then,

The Extent from 1, to the whole Content of the Sphear, shall reach the same way from the 4th Number on the Line of Numbers, to the Solid Content of the Segment required.

[illustration]

Example.

Let the whole Diameter of a Sphear be 14, then the whole Solid Content by the former Rules, you will find to be 1437 ⅓, a Segment of that Sphear whose Altitude or Depth is 4, the Solidity is required.

Extend the Compasses from 14, the whole Sphears Diameter, to 1; that Extent applied the same way from 4, the Altitude of the Segment, shall reach to 2-86 on the Num∣bers, or to 19-88, on the Line of Solid Seg∣ments joyn'd to the Line of Numbers, which 19-88, is the 4th Number to be kept.

Then secondly,

The Extent from 1 to 1437, the whole Content of the whole Sphear, shall reach the same way from 19-88, to 284 2/••, the Content of the Segment required to be found.

If you want the Line of Segments, the Table annexed will supply that defect:

Thus;

Look for the 4th Number, found on the Line of Numbers, among the figures on the Table, and the number answering it in the first Column, is the Solid Segment, or 4th to be kept; as here, on the Numbers, you find 2-86; seek 2-86 in the Table annexed,

and in the first Column, you find near 20 fo•• the 4th in Segments.

8. To perform the same by Arithmetick after the way set forth by Mr. Thomas Diggs, 1574.

To find the Superficial Content of a Globe or Sphear.

Multiply the Diameter by the Circum∣ference, the Product shall be the Superfi∣cial Content round about the Globe.

9. Or,

Multiply the Content of a Circle, having like Diameter, by 4, the Product shall be the Superficial Content.

10. And

If you multiply the Superficial Content, by a 6th part of the Diameter, the Product shall be the Solid Content of the Sphear.

A Table of Segments.

Num. Seg	Num. Segm.	Num. Segm.	Num. Segm.
059	335	506	679
084	342	513	686
104	349	520	694
122	356	527	703
5 137	30 363	55 534	80 712
152	371	540	720
164	378	547	728
176	385	554	737
188	392	560	746
10 197	35 399	60 567	85 753
207	406	574	763
218	413	580	772
228	420	587	782
237	426	594	793
15 245	40 433	65 601	90 803
254	440	608	812
263	447	615	825
272	453	622	836
280	460	629	848
20 288	45 466	70 637	95 865
297	473	644	878
306	480	651	896
314	487	658	916
321	494	665	941
25 328	50 500	75 672	100 1000

11. For the Segment work thus;

Multiply the whole Superficial Content of the whole Globe, by the Altitude of the Segment, and divide the Product by the Sphears whole Diameter, the Quotient shall be the Superficial Content of the Convexity or round part of the Segment.

12. But for the Solid Content, work thus;

First, find the difference between the height of the Segment, and the half Diame∣ter of the Sphear; then multiply this dif∣ference (being found by subtracting the less from the greater) by the Superficial Con∣tent of the Base of the Segment, and the Product subtract from the Product of the Sphears semi-Diameter, and the Convex Su∣perficies of the Segment; then a third part of the remainder shall be the Solid Content of the Segment required.

Example as before.

The Sphears Diameter is 14, the Segments Altitude is 4, the Segments Altitude taken from 7, the half Diameter, the remainder is 3, which multiplied by 126, the Superficial Content of the Base of the Segment, makes 378; then having multiplied 7, the Sphears half Diameter, by the Convex Superficies of

the Segment 176, the Product is 1232, from which number take 378, the Product last found, and the remainder is 854, whose third part 284 ⅔, is the Solidity of the Seg∣ment required.

There are other fragments of Sphears, as Multiformed and Irregular, Cones or Pyra∣mids, and Solid Angles; but the Mensura∣tion of these I shall not trouble my self, nor the Learner with, for whom I only write, intending the Mensuration of things that may come in use only.

13. But yet to conclude this Chapter, take these Observations along with you, con∣cerning the Proportion of a Cube, a Prisma, and a Pyramid, a Cillender, Sphear, and Cone; whose Shapes and Proportions are as in the Figures.

If a Cube be made or conceived, whose side is 12 inches, then the solidity thereof is 1728 Cube inches; and a Prisma, having the same Base and Altitude, contains 864 Cube inches; and a square Pyramis, of the same Base and Altitude, contains 576 Cube inches; and a Trianguler Pyramid, as be∣fore, contains 249-6 Cube inches; A Cil∣lender contains 1357 5/7, being the same Height and Diameter of 12 inches: A Sphear, whose Axis is 12 inches, con∣tains

905 1/7 Cube inches; and a Cone, of the same Diameter and Altitude, contains 452 4/6.

The Superficies of the Cillender about, (excepting the top and bottom) is equal to the Superficial Content of a Globe.

Cube	1728
Cillender	1357 5/7
Sphear	905 2/7
Prisma	864
Piramis	576
Cone	452 4/6
△ Tetrahed.	2496
Octahed. whose Triangle Side is 12	814-6

By the foregoing Proportions, it is evi∣dent that a Cube is double the Prisma, and treble to the Square Pyramis of equal Base and Altitude, or as 3, 2, 1; for 3 times 576 is 1728, and 2 times 864 is 1728.

Also a Cillender is 11/14 of a Cube; and a Globe is 11/•••• of a Cube, or ⅔ of a Cillender, whose Sides and Diameters are equal; and a Cone is ••/•• of a Cillender; so that the Proportion between the Cone, Sphear, and Cillender, is as 1, 2, & 3; for 3 times 452 4/6, makes 1357 5/7; and two thirds

of 452 4/6 makes 905 1/7; the Content of a Sphear. The Trianguler Pyramid is little more than 1/7 of a Cube; so that if any one have frequent occasions for these proporti∣ons, let Centers be put in the Line of Num∣bers, at these proportional Numbers, and then work with those Points from the Cube and Cillender, as is directed before, for the Circumference, and Diameter, and Squares, equal and inscribed in Chap. 8. Prob. 6.

So much for the measuring of regular ordinary Solids; for the extraordinary and irregular, the best Mechanick way is by Weights or Waters to measure their Crassi∣tudes or Solidities, either by Weight or Measure.

A further improvement of the Tri∣anguler Quadrant, as I have made it several times, with a sliding Co∣ver on the in-side, when made hol∣low, to carry Ink, Pens, and Com∣passes; then on the sliding Cover, and Edges, is put the Line of Num∣bers, according to Mr. White's first Contrivance for manner of Opera∣tion; but much augmented, and made easie, by John Brown.

1. THe description thereof for one side▪ being the same with the Line of Numbers on the outter-Edge, except that the first part is sometimes (when required for that particular purpose) divided into 12 parts, for inches, instead of 10 that is to say, The space between the first 1, and the middle 1, on the Rule; (the space I say) between every Figure, on the first half part, is cut into 12 parts, instead of 10, to an∣swer to the 12 inches in a Foot; and the o∣ther half, as the Line of Numbers on the Edge. And in the same manner are White's

sliding Rules made, only for this particular purpose.

2. On the other side, is the Line of Num∣bers drawn double, the one Line to the o∣ther, for the ready measuring of solid Measure at one Operation; the description whereof in brief is thus;

First, The divisions on the sliding-piece in hollow-Rules, or on the right-side in sliding-Rules; when the figures of the Tim∣ber-side stand fit to read, I call the right-side, or single-side, being alwayes toward the right hand, and a single Radius.

The divisions therefore on the fixed-edge of the Rule, must needs be the left-side, and is also divided to a double Radius, or one Radius twice repeated.

So also in sliding-Rules, the double Ra∣dius is on the left-side also. See the Figure thereof, with right and left-side exprest upon it.

For the right reading those Lines, the Method is thus;

The Figures on the right or single-side, do usually begin at 3 or 4, and so proceed with 4, 5, 6, 7, 8, 9, 10, 11, for so many inches of a Foot.

Then 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, &c. for so many whole Feet.

The smaller Cuts between the first Fi∣gures, from 3 inches to 1 foot, being quar∣ters of inches; And the small Divisions be∣tween the Figures, that represent Feet, are only every whole Inch; The halfs, and quarters of Feet, also noted by a longer stroke▪ as in such work is necessary and usual.

3. On the same right-side also, for more ease and readiness in the use, are noted se∣veral Gage-points (as it were); As,

First, At 1 Foot is the word square.

Secondly, At 1 Foot, 1 inch and ½, is a spot; and close to it is set t.d. for true Dia∣meter of a round solid Cillender.

Thirdly, At 1 foot 3 inches ⅜ is another spot, and near to it is set D, for the Diame∣ter of a rough piece of Timber, according to the usual allowance for unhewed Timber, according to the fourth part of a Line girt about and counted for the side of the square.

Fourthly, At 3 foot 6 inches and ½, and near to it is set t: r: for the true Circum∣ference of a round Cillender.

Fiftly, At 4 foot just is set R, for the Circumference, according to the former al∣lowance.

Sixtly, At 1 foot 5 inches 1/7 f••rè, is a spot; and close to it the letter W, as the Gage-point for a Wine-gallon.

Seventhly, At near 19 inches, or 1 foot 7 inches, is another spot; and close to it the letter a, as the Gage-point of an Ale-gallon.

Eightly, At 2 foot 8 inches 8/10 is a spot, and close to it is set B, for the Gage-point of a Beer Barrel; and at 2 foot 7 inches is set A, for an Ale Barrel. The Uses whereof in order follow.

The Figures on the left-side, or fixed-edges, are read and counted as those on the right: For the small, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, are to represent inches, and the cuts between, quarters of inches; Then the 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 Figures next, somewhat bigger, as to represent so many feet, and the cuts between, are whole inches: Then 20, 30, 40, 50, 60, 70, 80, 90, 100, 150, for tens of feet, and the parts between, for single-feet, for the most part; or else whole and half feet, as is usual. The Uses follow.

Use I. A piece of Timber being not Square, (or having its breadth and depth unequal) to make it Square, or find the Square equal;

Set the breadth of the Piece, counted on the right-side, to the same breadth counted

on the left-side; then right against the depth found on the left-side, on the right or single-side, is the inches and quarters square re∣quired.

Example, at 15 inches broad, and 9 inches thick, or deep.

Set 9 inches on the right-side, to 9 on the left; then right against 15 inches, or 1 foot 3 inches, found out on the double or left-side, on the right or single-side, is 11 inches and ⅝ the Square equal required.

Also, if you set 15 to 15, then right a∣gainst 9, found out on the left-side, on the right-side is 11 inches ⅝, the Square equal required.

Use II. The Side of the Square given, to find how much in length will make 1 Foot.

Set the inches (or feet & inches) found out on the right-side, to 1 foot on the left; then right against 1 foot on the right, is the in∣ches, or the feet and inches required, to make a Foot of Timber.

Example at 9 inches square.

Set 9 inches, found out on the right-side, to 1 foot on the left-side; then right against 1, on the right-side, is 1 foot 9 inches 2/•• on the left.

If the Square be so big, that the 1 on the right falls beyond the End at the beginning, then right against 10 foot on the right-side, is on the left, the hundredth part of a foot, that makes a Foot of Timber.

Example, at 4 Foot Square.

Set 4 foot, found out on the right, to 1 foot on the left; then right against 10 foot on the right-side, is 0-063 on the left-side, or against 12 foot you have 9, 12 parts of 1 inch, the length that goes to make 1 foot of Timber required.

Use III. At any (bigness or) Inches, or feet and inches square, to find how much is in 1 Foot long.

Just as the Rule stands even, that is 1 foot on the right, against 1 foot on the left, seek the inches, or feet and inches, the Piece is square on the right or single-side; and just against it on the left or double-side is the Answer required; in inches, or feet and inches.

Example at 19 inches square.

Just against 1 foot 7 inches, or 19 inches, (which is all one) found out on the right-side,

on the left-side is 2 foot 6 inches, the quantity of Timber in 1 foot long, at 19 inches square; which Number of 2 foot 6 inches, multiplied by the length in feet, gives the true Content of the whole Piece of Timber required.

Note, That this is a most excellent way for great Wood, and very exact.

Also Note, That here, by inspection, you may square a small Number, or find the square-root of a small Number.

As thus;

The square of 8 & ½, is near 72;

Or,

The square-root of 72, is near 8 & ½.

Use IV. The side of the Square, and the length of any Piece being given, to find the Con∣tent in feet and parts.

Set the word square, or 1 foot, alwayes to the length, found out on the left-side; then right against the inches, or feet and inches square counted on the right, on the left is the Content required.

Example, at 20 foot long, and 15 inches square.

Set 1 foot on the right, to 20 on the left, then right against 1 foot 3 inches on the

right, is 31 foot 2 inches and ½, the Con∣tent.

Note, That if the Piece be very small, call the feet on the left-side, inches; and the parts between 12^• of inches; then the Answer will be found on the left-side in 144^• of a Foot.

Example, at 2 inches square, and 30 foot long, how much is there?

Set 1 foot on the right, to 30 foot on the left; then right against 2 foot on the right, counted as 2 inches, is 120 parts of a foot divided into 144 parts, being just 10 in∣ches, for 10 times 12 is 120.

But if it be •• great Piece of Timber, then work thus;

Set 1 foot, or the word square, to the length on the left, counting the single feet 10^• of feet; then right against the feet and inches square are the 100^• of feet re∣quired.

Example, at 40 foot long, and 4 foot square.

Set 1 foot to 4 foot, counted as 40, on the left; then right against 4 foot on the right, is 640, the true Content, increasing the 10^• to 100^•. Thus much for square Timber.

Though there be many other wayes and manner of workings, some whereof you may find in a Book set forth under the name of The Carpenters Rule, 1666, by I. Brown, and well known abroad already.

Use V. For Round Timber. The middle Diameter of any Piece given, to find how much is in a Foot long, at true measure.

Set the spot by t.d. to 1 foot on the left, then just against the inches, or feet and in∣ches Diameter, found on the right, is the quantity of Timber in 1 foot long on the left-side required.

Example, at 2 foot 9 inches Diameter.

Suppose a piece of Stone-Pillar, or Gar∣den-Roul, be two foot 9 inches Diameter, set the spot by t.d. just against 1 foot, then right against 2 foot 9 inches, found on the right, on the left is 6 foot, the quantity of solid measure in one foot long; which be∣ing multiplied by the length in feet, gives the true Content of the whole Piece.

Note, That if you would have the usual allowance, set D to 1, instead of t.d.

Use VI. The Diameter of any Piece of Timber gi∣ven, to find how much in length will make one Foot.

Set 1 foot on the left, to the inches Dia∣meter counted on the right; then right a∣gainst t.d. for true measure; or D for the usual allowance, is the Answer required, found on the left-side.

Example, at 9 inches Diameter.

Set 1 on the left to 9 on the right; then just against t.d. is 2 foot 3 ½ on the left; and right against D, is 2 foot 11 inches on the left, the length required to make a foot solid at true measure, or the usual allow∣ance, when the 4th part of the girt about, is counted the side of the Square, equal to the round piece of Timber.

Note, That for great Timber, you must set the left 1 foot, to the feet and inches Dia∣meter as before; but count the t.d. or D, as far beyond 12 foot, as it is placed beyond 12 inches, and you shall have the Answer in 144^• of a foot.

Example, at 5 foot Diameter.

If you set 1 on the left to 5 foot on the right, and count so much beyond 12 foot

on the right, as t.d. is beyond 12 inches, you shall find 7 ¼, that is, 7 144^• and a quarter, to make 1 foot true measure, and 9-12s and ½ for the usual allowance.

But for small Timber, set 1 foot, 2 foot, &c. on the right, counted as 1, 2, 3, 4, and inches, to 1 on the left; then right against t.d. or D, is a Number, that multiplied by 12, is the Number of feet required.

Example, at 2 inches Diameter, how much makes 1 Foot?

Set 2 foot on the right, counted as 1 inch, to 1 on the left; then right against t.d. is 3 foot 10 inches, calling the inches feet, and the feet 10s of feet; which 3 foot 10 inches, multiplied by 12, make 46 foot, for the length of 1 foot of Timber at 2 in∣ches Diameter, the thing required for true measure.

Use VII. The Diameter and Length given, to find the Content.

Set t.d. or D, for true measure, or usual allowance, alwayes to the length counted on the left; then right against the inches Diameter counted on the right, on the left-side is the Content required.

Example, at 6 inches Diameter, an•• 30 Foot long.

Set t.d. to 30; then right against 6 in∣ches, counted on the right; on the left is 5 foot 11 inches, the Content required.

Note, If the Piece be small, then count every foot on the right as inches, and you have the Answer in 144s of a foot, which is easily counted by having 1 set at 12, 2 at 24, 3 at 36, 4 at 48, 5 at 60, &c. to 12 at 144, which little small figures are counted as inches of 12s of a foot.

Example, at 2 inches Diameter, and 20 foot long.

You shall find 64, 144s; that is, 5 in ⅓ true measure.

But for great Pieces, set t.d. or D, to the length, counting 1 foot, or the left for 10 foot, then have you the Answer in 100s of feet.

Example, at 5 foot Diameter, and 30 foot long.

Set t.d. to 3 inches, counting 30 foot for the length; then right against 5 foot on the right, on the left is 592 foot, the Content required.

Use VIII. To measure round Timber, by having the Girt, or Circumference about, and length given.

This being the same in operation with the Diameter, I shall pass it over more briefly; which way of wording, may serve for the Square and Diameter also; only I labour to be plain and brief.

The Circumference given, to find how much in a Foot long.

Set t.r. or R, for true round, or allow∣ance to 1 foot on the left; then against the inches about, on the left is the Answer re∣quired.

Example.

At 2 foot about, you will find 3 inches and 10/12 in a foot long true measure; or just 3 inches at the usual allowance.

The Circumference given, to find how much makes a Foot.

Set the inches, or feet and inches about, to 1 foot on the left; so is t.r. or R, to the length to make a foot.

Example, at 18 inches about.

As single 18, to double 1;

So is t.r. to 5 foot, 7 inches ½ Or,

So is R, to 7 foot 2 inches, for the usual allowance.

The Circumference and length, to find the Content.

As t.r, or R, to the length;

So is the feet and inches about, to the Content.

Example, at 3 foot about, and 30 foot long, true measure.

As t.r. to 30; so is 3 foot, to 21 foot 5 inches, the Content.

For great things, call 1 foot on the left, 10 foot, as before.

For small things, call 1 foot on the right, 1 inch, as before.

Use IX. To Gage small Cask by the mean Dia∣meter and Length.

Set the spot by W, for Wine-gallons, al∣wayes to the length of the Vessel, given in inches, counted on the left-side; then right against the inches, or feet and inches Dia∣meter,

counted on the right; on the left is the Content in Gallons required.

Example, at 24 inches, or 2 foot mean Diameter, and 30 inches long.

As the spot at W, to 30 inches counted at 30 foot;

So is 24 inches, or 2 foot, to 58 Gallons 3 quarters, the Content required.

For greater Vessels, count the feet on the left for 10^s of inches in the length, and you have your desire.

Example, at 60 inches long, and 38 inches mean Diameter.

As W, to 6 foot on the left for 60 inches;

So is 3 foot 2 inches, or 38 inches to 295 Gallons, the Answer required.

If you would have it in Ale-gallons, use the mark at a.

Use X. To gage Great Brewers Vessels, round Tuns.

The Diameter and length being given in feet and inches, to find the Content in Beer-Barrels, at one O∣peration.

Set the spot at BB, to the depth of the

Tun, counted on the left, in feet and inches; then right against the mean Diame∣ter, found out on the right, on the left is the Content in Barrels required.

Example, a Tun 4 foot deep, and 10 foot mean Diameter.

As the spot at BB, to 4 foot;

So is 10 to 53 Barrels, and 2 third parts.

If you would have the Content in Ale-Barrels, use the mark at AB.

Thus much for the Timber-side, the use of the other, or board-side, is the same with that by the Compasses before treated of, and therefore needs here no repetition, unless as to the bare manner of working with it.

The sliding-Rule is only two Rules, or Pieces fitted together, with a short Grove, and Tenon, and two Braces at each end, to keep it from falling assunder; and even so also is the sliding-Cover, and two edges on the inside of the Trianguler Quadrant; and the Numbers graduated thereon, are cut a∣cross the middle Joynt, having the same di∣visions on both sides; that is to say, on each Rule, or on the Cover and Edge of the in∣side of the Rule.

The Reading and description is the same with that in Chap. I. Page 12, 13, 14, 15,

16; and the general Method in use is thus;

That side or part of the Rule, on which you count the first term in the Question, is called alwayes the first-side; then the other must needs be the second.

Then for Multiplication, thus;

As 1, on the first-side, to the Multiplier on the second, or other-side;

So is the Multiplicand, on the first-side, where 1 was, to the Product on the second.

For Division, alwayes thus;

As the Divisor found on any one side, is to 1 on the second, or other-side;

So is the Dividend on the same first-side, to the Quotient on the second.

For the Rule of Three.

As the first term on the first-side, to the second on the other;

So is the third term found on the first-side, to the 4th term on the second-side.

[illustration]

For Superficial Measure, by Inches and Feet.

As 12, to the breadth in inches on the se∣cond;

So is the length in feet, to the Content on the second.

For any thing else, the same Rules and Precepts you find in Chap. VII. will give you ample and plain directions.

The Lines being fitted, as much as may be, to speak out the Answer to the Question, as by well considering the Figure, you may see.

CHAP. XI. To make and measure the Five Regular Platonical Bodies, with their Declinations and Reclinations.

1. For the Cube, being the Foundation of all other.

IT is a Square Solid Body, every way a∣like, and spoken of largely before, as to the Mensuration thereof, and obvious enough to every indifferent Workman, as to the making thereof, and needs no repetition in this place.

2. For the Tetrahedron.

It is a Figure, comprehended of 4 equi∣latteral plain Triangles, or a Trianguler Py∣ramid, last mentioned, the best and nearest way, as I conceive of making, is thus. Ac∣cording to directions of Mr. Iohn Leak:

On any rough Piece, make one side plain and flat, so large, as to contain the Triangle which you intend shall be one side of the Tetrahedron; then set a Bevel to 70 degrees 31 minutes and 42 seconds; and plain ano∣ther side, to fit the former side, and the Be∣vel (secundum Artem); then mark this last plained side, according to the former, and cut away the residue, plaining them away just to the strokes, and fit to the Bevel for∣merly set, and you shall constitute the Te∣trahedron required.

The Superficial Content is the Area of the 4 equilatteral Triangles mentioned before, and the solid Content is found by multiply∣ing the Area of one Triangle by one third part of the Altitude of the Pyramid, or Te∣trahedron, from the midst of one Plain to the Apex, or top of the opposite Solid Angle.

If the measuring the sides, Perpendicu∣ler, and Altitude of the Tetrahedron with Compasses, Callipers, and Scale, serve not to exactness; then proceed thus;

First, for the Perpendiculer, the Triangle being equilatteral.

Multiply one side given by 13, and di∣vide the Product by 15; the Quotient is the Perpendiculer.

Example.

If the side of the Tetrahedron be 12, that multiplied by 13, gives 156; which divided by 15, leaves 10-4, for the length of the Perpendiculer in the equilatteral Triangle, whose side is 12.

Then for the Perpendiculer Altitude, work thus, by the Artificial Numbers and Sines.

As the Sine of 90, to the sine of 70 deg. 31 min. 42 sec;

So is 10-4, the Perpendiculer, to 9-80, the perpendiculer Altitude required.

Or by the Sector, work thus;

Take 12, the side of the Tetrahedron, from (any Scale, or) the Line of Lines, and set the Sector to 60 degrees, by making the Latteral 12, a Parallel 12, then the nearest distance from 12, to the Line of Lines, is the true Perpendiculer; which measured on the same Line of Lines, will be found to be 10-4, as before; then make this 10-4 a Parallel Sine of 90, and 90 the Sector so set, take out the Parallel-sine of 70-31-42, and measure it on the same Scale, and it shall be 9-8, as before.

[illustration]

But if you use the Quadrant-side, then first lay the Thred to 60 degrees, counted from the Head; then take the nearest di∣stance from 12, on the Line of Lines, to the Thred, and it shall be the Perpendiculer of the Triangle, 10-4; then set this Perpen∣diculer in 90, and lay the Thred to the nearest distance, and there keep it; then take the nearest distance from 70-31 in the Sines to the Thred, and that shall be 9-80, for the Perpendiculer Altitude of the Tri∣anguler Pyramid, or Tetrahedron.

Then,

Lastly, This perpendiculer Altitude be∣ing multiplied by the Area of the Base, gives a Number, whose third part is the Solid-Content of the Tetrahedron required.

For 12 the side, and 5-2 the half Per∣pendiculer, makes 62-4, the Superficial-Content of one Triangle, or Base; then 62-4, the Base, multiplied by 9-8, the per∣pendiculer Altitude, gives 611-52, and a third part of 611-52 is 203-86, the solid Content required.

The three Triangles recline from the Per∣pendiculer upright, 19 degr. 28 min. and 18 sec. and decline when the edge is South 60, South East, and South West, and the opposite Plain a just North; but if you

make one South, then the other two are are North-east and North-west 60 deg.

3. For the Octahedron,

Whis is a solid body, comprehended under 8 equilatteral Triangles: The way of making which, is thus;

Make a plain Parallelepipedon, or long-Cube, if the breadth both wayes be 1000, let the length be 1-414; or if the length be 500000, the breadth both wayes must be 3-53553; then to these Measures square it exactly; then divide the length and breadth just in the midst, and draw Lines both wayes on all the 6 sides; then draw the Diagonal-Lines from the midst of the length, to the midst of the breadth; and cut by these Diagonal-Lines, and the Octahedran will appear to be truly made.

For the Mensuration thereof, it is the same as in the Tetrahedron; For, supposing the side of one of the Triangles 12; the Base is 144 in Content, and the Triangles Perpendiculer is 10-4, as before: But the Perpendiculer Altitude is just half the length, viz. 8-49; for if the breadth be 12, then the length must be 16-98, whose halfs are 6 and 8-49; Then if you multi∣ply 144 the Base, by 8-49 the perpendiculer

Altitude, the Product will be 1222-56, whose third part is 407-52, the half of the Tetrahedron, and 815-04 is the whole solid Content of the Tetrahedron required, as near as we can see by Instrumental Ope∣ration; but if you work to a Figure more, you shall find the total Area to be but 814-656 more exact.

To find this Perpendiculer Altitude by the Sector, work thus;

First, The Triangles Perpendiculer being 10-4, as before; Take the Latteral 10-4, from the Line of Lines, make it a Parallel in 90, lay the Thred exactly to the nearest distance, and there keep it; then the Paral∣lel-distance from the Sine of 54 deg. 44 min. 45 sec. the Reclination shall be 8-486, the true Perpendiculer Altitude required.

Then if the Octahedron stand on one Triangle, you have one Horizontal Plain, and one South and North Reclining and In∣clining 19 deg. 28 min. 18 sec. as the Te∣trahedron was; and two South, and two North, declining 60, and reclining and in∣clining 19 deg. 28 min. 18 sec. as afore.

But if it stand on a Point, then you have 4 direct or declining 45, and reclining 54, 44, 45; and 4 incliners, inclining as

much and direct, or declining, as you shall please to set them.

4. For the Dodecahedron,

Which is a regular solid body, contained under (or made up of) 12 Pentagonal Py∣ramids, or Pyramids whose Base hath 5 e∣qual sides, and the perpendiculer Altitudes of those 12 Pyramids equal to half the Do∣decahedrons Altitude, standing on one side, or equal to the semi-diameter of the inscri∣bed Sphear.

To cut this Body, take any round Piece, and if the Diameter be 100000, the length must be 0-81005, or as 4-906 to 3-973, then the Piece being turned round, and the two Ends flat to the former measures of Length and Diameters (which are near ac∣cording to the Sphear inscribed, and to the Circle circumscribing) being measured by Compasses, Callipers, and Line of Lines very carefully and exactly. Then divide the Cir∣cumference of the two Ends of the Cillen∣der into 10 equal parts, and draw Lines Perpendiculer from end to end, and plain all away between the Lines flat and smooth, so that the two Plains on both ends will be∣come a regular ten-sided Figure.

Then making the whole Diameter above∣said, 10000 in the Line of Lines, take out

0-309, and with this measure (as a Radius on the Center) at both ends describe a Circle; and if you draw Lines, from every opposite Line of the 10 first drawn, you shall have Points in the last described circle, to draw a Pentagon by; which is the Base of one of the 12 Pentagonal Pyramids, con∣tained in the body. This Work is to be done at both Ends; but be sure that the Angle of the Pentagon at one end, be op∣posite to a side of the Pentagon at the other end; then these Lines drawn, the two ends are fully marked.

Then to mark the 10 Sides, do thus;

Count the first length 1000, viz. the measure from the top to the bottom, or from Center to Center; and fit this length in 10 and 10, of the Line of Lines; the Sector so set, take out 0-3821, and lay it from the two ends, and either draw, or gage Lines round about from each end; and in the midst between the two Lines will remain 0-2358; then Lines drawn Diagonally on the 10 sides, will guide to the true cutting of the Dodecahedron.

If you set a Bevel to 116 deg. 33 min. 54 sec. and apply it from the two ends, you may try the truth of your Work.

The Declination and Reclination of all the 10 Pantagonal Plains, are as followeth.

First, You have 1 North, reclining 26 deg. 34 min; and 1 South, inclining as much.

Secondly, You have 2 North declining 72, and reclining 26, 34; and 2 South, declining 72, and inclining 26, 34.

Thirdly, You have 2 North, declining 36, and inclining 26, 34; and 2 South, declining 36, and reclining 26, 34; And 1 Horizontal Plain, and his opposite Base to stand on.

As for the measuring of this Body, the Plain and Natural way is thus;

First, find the Superficial Content of the Base of one of the Pentagons, by multiply∣ing the measure from the Center to the middle of one of the Sides, (which is the contained Circles semi-diameter) and half the sum of the measure of all the sides put together; and then to multiply this Product by one third part of half the Altitude of the body, and the Product shall be the Content of one Pentagonal Pyramid, being one twelft part of the Dodecahedron; and this last, multiplied by 12, gives the solid

Content of the Dodecahedron; or 12 times the Superficial Content of one side, is the Superficial Content thereof.

Example.

Suppose the side of a Dodecahedron be 6, then the sum of the sides measured is 30, the contained Circles semi-diameter is 4-12; then 15 the half of 30, and 4-12 multi∣plied together, make 61-80; and 12 times this, makes 741-60, for the Superficial Con∣tent of the Dodecahedron.

Then for the Solid Content, multiply 61-80, the Superficial Content of one side by 2-233, one 6th part of 13-392, the whole Altitude of the body; the Product is 137-99940: Again, this multiplied by 12, the number of Pyramids, makes 16••••▪ 9928, the Solid Content, as near as may be, in such a Decimal way of Computation.

5. For the Icosahedron,

Which is a regular solid body, made up of, or contained under 20 Trianguler Pyra∣mids, whose Base (or one of whose Sides) is an equilatteral Triangle; and the per∣pendiculer Altitude of one of these 20 Py∣ramids, is equal to half the perpendiculer Altitude of the Icosahedron, from any one side, to his opposite side, or equal to the semi-diameter of the inscribed Sphear.

To cut this body, take any round Piece, and if the Diameter thereof be 10000, let the length thereof be turned flat and even to 8075; or if the true Round and Cillen∣drical Form in Diameter be 4910, let the true length, when the ends are plain and flat, be 3964; then divide the Cillendrical part into 6 equal parts, and plain away all to the Lines, so that the two ends may be two 6-sided-figures; then making 5000, the former semi-diameter, 1000 in the Line of Lines, take out 616, and on the Center, at each end, describe a Circle; and by drawing Lines to each opposite Point, make a Triangle, whose circumscribing Circle may be the Circle drawn at each end; but be sure to mark the side of one Triangle op∣posite to the Point of the other Triangle at the other end, as before in the Dodecahe∣dron; thus both the ends shall be fully and truly marked.

Then making the length a Parallel in 1000, of the Line of Lines, take out -379, and -095, and prick those two measures from each end, and by those Points (draw or gage) Lines round about, on the 6 sides. Then Diagonal Lines drawn from Point to Line, and from Line to Point round about, shews how to cut the Body at 12 cuts.

Note, That if you set a Bevel to -138-11-23, and apply it from each end, it will guide you in the true plaining of the sides of the Icosahedron. And a Bevel set to 100 degrees, will fit, being applied from the midst of one side, to the meeting of two sides.

The Reclination of the three Triangles, whose upper sides are adjacent (or next) to the three sides of the upper Horizontal-Triangle is 48 11.23, from the Perpendi∣culer, or 41 48 37, from the Horizon∣tal, and when one corner stands South, the Declination of one of these 3, viz. that op∣posite to the South-corner a direct North; th'other two decline 60 degrees, one South-east, the other South-west; the other 6, a∣bout the corners of the Horizontal-plain, do all recline 19 deg. 28 min. 16 sec. the two that behold the South, decline 22 deg. 14 min. 29 sec. and those two that behold the North, decline 37 deg. 45 min. 51 sec. toward the East and West; the other two remaining, recline as before, and decline one North-east, and the other North-west 82 deg. 14 min. 19 sec.

The other Nine under-Plains, opposite to every one of these, decline and incline, as much as the opposite did recline and decline, as by due consideration will plainly ap∣pear.

For the measuring of this body, do as you did by the Dodecahedron, find the Area of one Triangle, and multiply it by 20, gives the Superficial Content; and the Area of one Triangle, multiplied by one sixt part of the Altitude of the body, gives the solid Content of the Trianguler Pyra∣mid; and that Product multiplied by 20, the number of Pyramids, gives the whole Solid Content of the Icosahedron.

Example.

Suppose the side of an Icosahedron be 12, first square one side (viz. 12, which makes 144); then multiply that Square by 13, and then divide the Product by 30, the Quotient and his remainder is the Superfi∣cial Content of the Equilatteral Triangle, whose side is 12; namely, 62-400; or more exactly, the Square-root of 3888, which is near 62-354; 20 times this, is the Superficial Content, namely, 1247-08.

Then for the Solid Capacity or Content, multiply 3-023, the sixt part of the bodies Altitude, or one third of the Pyramids Altitude, by 62-354, the Area of one Trianguler Base, and the Product will be 188-493229. Lastly, this multiplied by 20, the number of Pyramids in the Body,

the Product is 3769-864380, the true so∣lid Content of the Icosahedron.

Thus you have the way of cutting, and the Declinations and Reclinations and Mea∣sures, Superficial and Solid, of the 5 Re∣gular Bodies, as near as by Decimal Ac∣compt to 100 part of an Integer may be, the exact measuring whereof, requires the help of Algebra, whereof I am ignorant.

The Measures of the Containing, and Contained Sphears, Circles, and Diame∣ters, Sides and Axis's, Diagonal-lines and Altitudes of the five Regular Bodies, ga∣thered in a Table to a Containing Sphear, whose Diameter was 10 inches (or Integers) found out by Geometry, according to this Scheam, taken from Mr. Tho. Diggs.

Let the Line AB be 10 of some Diago∣nal Scale, representing the Diameter of the Containing Sphear. Which Line AB, you must divide into two parts at C, and into three parts at E; AE being one third part, and on the Points C and E, raise two Lines Perpendiculer to AB; and with 5 of your Diagonal Scale, on the Center C, describe the semi-Circle AFDB, and note the Points F and D, in the semi-Circle, with F and D, drawing Lines from either of them to A, and from F to B.

Then, Divide AF by extream and mean Proportion; the greater Segment being AG, (by the 10th Problem of the 6th Chapter) then extend the Line AF to H, making FH equal to FG, and draw the Line HB; and from F, draw another Line Parallel to HB, cutting the Diameter in I, and from I, draw a Line Parallel to CD, as IK; then make IL a third part of IB, and draw ML Parallel to IK; also, draw the Line MB, and divide it into two parts at N, and into 4 parts at 5; then divide the 4th part, MS, by extream and mean Proportion, whose greater Segment (or part) let be SV; then divide FB in 4 parts, making FO the half, and FR the quarter; divide like∣wise FE in two parts, and at the middle set P: The Figure being thus made, then with your Compasses, and Diagonal Scale, you may measure all the Diameters, Sides, and Altitudes, of all the 5 Regular Bodies.

As thus;

AB is in all of them, the Contained Sphears Diameter.

EC, OF, RO, NC, NC, the Contained Sphears Semi-Diameter.

EF, OB, OF, NB, MB, the Containing Circles Semi-Diameter.

EP, OC, CO, VN, MN, the Contained Circles Semi-Diameter.

FB, AF, AD, AG, KB, the Length of the Sides of each Body.

EB, AF, FA, MA, AM, the Altitude of the Bodies.

AD, FB, VB, SB, the Perpendicu∣ler Line of the Bases.

FB, AF, the Diagonal-Line of the Bases, as in the Table.

These Measures and Proportions are for a Sphear of 10 inches Diameter.

If you would have the like for any other, then say by the Line of Numbers, or Line of Lines, or Rule of Three, thus;

As the side (Diameter or Altitude) for 10, as in the Table, is to the given Side, Diameter, or Altitude;

So is any other Number, in the Table, for Diameter, Side, or Altitude, to his Proportional Measure required.

Example. I have a Dodecahedron, whose Side is 6, What shall all his other Sphears, or Circles, Diameters, or Altitude be?

The Extent of the Compasses from 3-570, the Dodecahedrons-side in the Table, to 6 the side given, shall reach from 10, the Containing Sphears Diameter in the Table, to 16, the Containing Sphears Dia∣meter, for a Dodecahedron, whose side is 6: And from 7-970, the Contained Sphears Diameter, to 12-643, the Con∣tained Sphears Diameter. And so for any other whatsoever.

The Table.

The Names of the Bodies	Tetra¦hedron	Cube.	Octa∣hed.	Dode∣cahed.	Icosa∣hed.
Containing Sphear.
The containing Sphears Diame∣ter, that comprehends the body in it, is for ever one of them.	AB	AB	AB	AB	AB
10.000	10.000	10.000	10.000	10.000
Contained Sphear.
The contained SphearS Diame∣ter that is contained in the bo∣dy, called also Axis, is	2.332	8.1648	4 0824	7.970	7.970
EC	OF	RO	NC	NC
The half thereof, is	1.666	4.0824	2.041••	3.983	3.985
Containing Circle.
The containing Circles Diame∣ter, (or the Diameter of that Circle which comprehends one side or base of the body, is	9.420	9.648	9.644	6.070	MB
EF	OB	OF	NB	6.070
The half thereof, is	4.710	4.0824	4.0824	3.035	3.035
Contained Circles.
The contained Circles Diame∣ter, comprehended in the Base of one side, is	4.410	5.7840	3.7840	4.910	MN
EP	OC	OC	VN	3.035
The half thereof, is	2.355	2.8920	2.8920	2.455	1.5175
Sides,
The length of one side of the Triangle Square, or Pentagon, being the base of the figure, is	FB	AF	AD	AG	KB
8.1647	5.774	7.071	3.570	5.260
The half therof,	4.08235	2.887	3.5355	1.785	2.630
Altitude.
The Altitude from side, to the side opposite, or from Side to the Point opposite,	EB	AF	AF	MA	MA
6.666	5.774	5.774	7.960	7.960
The half thereof,	3.3333	2.887	2.887	3.980	3.980
Perpendiculer.
The Length of the Perpendi∣culer Line of any one Side or Base,	AD	FB		VB	SB
7 073	8.1647	6.123	5.485	4.556
The half thereof,	3 5365	4.0823	3.0625	2.7425	2.278
Diagonal-Line.
The Diagonal-Line, from Cor∣ner to Corner of the same Base, is,		FB		AF
none	8.1647	none	5.774	none
The half thereof is,		4.0823		2.887

This Table was gathered from this Geo∣metrical Figure, drawn on a Slate, by a good Diagonal Scale of 6 parts in a Foot, whereby I could very well come to the 100^th part of an Integer; and is true enough for any Mechanick Operation, for whose use I only do it, and I hope it may be as kindly accepted, as it was carefully Calculated, and offered to Publick view.

CHAP. XII. The use of the Line of Num∣bers, in Gaging of Vessels, close or open.

GAging of Vessels, is no other than the Measuring of Solid Bodies; and the former directions for solid Measure, con∣veniently and aptly applied, is fully suffi∣cient; only observing this difference, That the result or issue of the Question is to be rendred in proper terms, according to the

demand of the Question, as thus; in mea∣suring of Timber or Stone, the Question is, How many Feet, or Inches, is there in the Solid Body? But in Gaging, the Question is, How many Gallons, Kilderkins, or Barrels is there in the Vessel to be measured? For which purpose there are fit Numbers, or Gage-Points, requisite to be known, for the more speedy attaining the Answer to the Question, of which in their order, as fol∣loweth;

First, You are to remember, That the solid capacity of a Wine-Gallon, is 231 Cube Inches; a Corn-Gallon 272 ¼ Cube inches; an Ale or Beer-Gallon, is 282 ¼ Cube inches; or as some say, 288 Cube in∣ches; So that when you have found the Content of any Vessel in Cube inches, if you divide that sum in inches, by the re∣spective Number for the Gallons you would have, the Quotient shall be the Content in Gallons required.

Problem I. To measure a Square Vessel.

From hence it follows, That to measure any Square or Oblong Vessel, you must multiply the length and breadth taken in inches, and tenth parts, together; that is to

say, The one by the other; and the Pro∣duct shall be the Content of the Base in in∣ches, superficially: Then multiply this Su∣perficial Content of the Base, by the inches, and tenth parts deep, and the Product shall be the solid Content in Cube inches; then divide this Product by 282, gives the Con∣tent in Ale-Gallons in the Quotient, and the remainder, if any be, are Cube inches.

But if you divide by 10161, the Cube inches in a Beer Barrel; or, by 9032 the Cube inches in an Ale Barrel; the Quotient sheweth the Number of Beer or Ale Bar∣rels, (and the remainder Cube inches.)

Example of a Brewers Cooler.

The length let be 78 inches and 1 tenth, the breadth let be 320 inches and 5 tenths, and the depth 9 inches and 5 tenths, or half an inch; by multiplying and dividing, as above, you will find 843 Gallons, and 68 Cube inches, to be the solid Content of that Cooler; which work is very readily done by the Line of Numbers, in this manner;

Extend the Compasses from 1, to the breadth or length; and the same Extent shall reach from the length or breadth to a 4th, which is the Superficial Content of the Base, or bottom, in Superficial inches.

Then,

The Extent from 282 ¼, to the last Num∣ber found, shall reach the same way from the inches, and tenths deep, to the Content in Gallons.

Example.

The Extent from 1, to 78-1, shall reach the same way from 320-5, to 25031; then the Extent from 282 ¼, to 25031, shall reach the same way from 9-5, to 842-68, the Solid Content in Gallons required.

Indeed, you must Note, You cannot see so many Figures on the Line, as the Product of 4 figures multiplied by 3; yet by the Rules (in Chap. 6. Sect. 3.) you have di∣rections as to the number of Figures, which here is 7; the two last (next the right hand) being Fractions, or parts of an Inch, and is therefore neglected.

Again,

In dividing the Product of 25031, and 9-5, multiplied together, which makes 6 Figures beside the Fraction by 282, there must needs be three Figures in the Quotient, which are the Gallons: This artificial help you have, beside the present view of the Vessel, which will direct you not to call 842 Gallons, only 84 ½, nor 8420, as you must needs do, if you mistake as to the de∣nomination.

Again,

You need not to trouble your self, to know what the 4th Number is; but having found the Point representing it, keep the Compass-point fixed there, and open the o∣ther to 282 ¼, where you may have a Brass Center-pin for more readiness; but let your account go as 282 ¼ to the 4th, for methods sake, and not as the 4th to 282 ¼; for then you must say, so is the depth the contrary way to the Content in Gallons. All this is hinted for plainness and caution sake, in be∣nefit to young Learners.

Also Note, That if you would have had the Answer in Ale or Beer Barrels; then in¦stead of 282 ¼, you must use the Point at 9032, for Ale Barrels; or the Point at 10161 for Beer Barrels, being the number of Cube inches in those Barrels, as 282 ¼ is the number of inches, in a Gallon of Ale or Beer.

Example for the same Cooler.

The Extent from 1 to 78-1, shall reach from 320-5, the same way to 25031; then, the Extent from 10161 to 25031, shall reach the same way from 9-5 to 23 ⅓, the true number of Beer Barrels required. Or,

The Extent from 9032, to 25031, shall reach to 26 Barrels, and near 1 third: which

is as quick and ready a way as can be for Square Vessels.

Problem II. To Gage or Measure any round Tunn or Vessel.

The plain and natural way for measuring of a round Tun, is this; Measure the Dia∣meter in inches and tenths, and set down half thereof; Measure also, the Compass round about the inside, and set down the half of that also, in inches and tenth parts; and multiply those two Numbers together, the Product shall be the Content of the Base, or bottom, in Superficial inches; then this Product multiplied by the depth in in∣ches, gives the solid Content in inches; then lastly, this Product divided by 282, or by 10161, or by 9032, gives the solid Content in Gallons, or Beer, or Ale Barrels, as before.

For, half the Diameter, and half the Circumference, doth reduce the round Ves∣sel to an Oblong Vessel, equal to that round Vessel.

Which Vessel, when it is brought to a Square, by taking of half the Diameter, and half the Circumference; then the Rule

last mentioned, for Square Vessels, performs the work exactly, to Gallons, or Barrels, as you please.

But when the Vessel is Taper, that is to say, the bottom and top of different Diame∣ters, as generally they all are; then the chief care is to come by the true Diameters, which is best done by a sliding Rule applied to the inside, whose regular equal computa∣tion is thus to be ordered;

When the Vessel is taper, and the Sides go streight, like the Segment of a Con••; then you may add the Diameters at top and bottom together, and count the half sum for the mean Diameter of that taper Vessel, and multiply half the Diameter, and half his proportional Circumference, as before; and multiply and divide, to get the solid Content in Gallons, or Barrels.

But when the Staves are bending, as most of your close Cask are, then the readiest way to come to a mean Diameter, is thus;

Say,

As 10 to 7, or as 10 to 6 & 3/10, 4/10, 5/10, 6/10, 7/10, 8/10, 9/10; according as you shall find most true for several Cask:

So is the difference of Diameters to a 4th Number, which is to be added to the least of the Diameters, to make up a mean Diameter.

As for Example.

If the Sides be round or arching, and the less Diameter be 30 inches, and the greater 40 inches; then, As 10 to 7; So is 10, the difference to 7 inches; which makes (being added to 30, the least Dia∣meter) 37, for a mean Diameter.

But Note, It is hinted by Mr. Dary, That Vessels, usually, are between a Sphe∣roid and a Parabolick Spindle; then, if as 10 to 7, be too much to add to the least Diameter; You may say,

As 10, to 6 ½; Or, As 10, to 6 6/10, 7/10, 8/10, 9/10;

So is the difference of Diameters to a 4th Number, which you must add to the least Diameter, to make a mean Dia∣meter.

Having thus gained a mean Diameter, you may work as before; or rather thus more readily and easily, by the Line of Numbers, thus;

As the Gage-point is to the mean Dia∣meter;

So is the Length to a 4th, and that 4th to the Content required.

The Gage-point for Wine, and Oyl-gallons, at 231 Cube inches in a Gal∣lon, is

17—15

The Gage-point for Ale-gallons, at 282 ¼, is,

18—95

The Gage-point for Ale, or Beer-gallons, at 288, is,

19—15

The Gage-point for a Corn-gallon, at 272 ¼, is,

18—62

The Gage-point for a Beer Barrel, at 10161, is,

35—96

The Gage-point for an Ale Barrel, at 9032, is,

33—91

The Extent of the Compasses, on the Line of Numbers, from the Gage-point to the mean Diameter of a Vessel; being turn∣ed two times the same way from the length of a Vessel, shall reach to the Content of the Vessel, in Gallons or Barrels, according to the nature of the Gage-point.

Example.

A mean Diameter being 30, and the Length 40, the Content is in Wine-gallons 123, near.

In the lesser Ale or Beer-gallons, 100-½.

In the greater Ale-gallons, at 288-098 gallons and a half.

In Corn-gallons, at 272-¼, —104 Gal∣lons.

In Beer Barrels, by his Gage-point you will find 2-78, or 2 three quarters: 2—78.

In Ale Barrels, you will find 3 and 11 of a hundred: 3 — 11.

And the like for any other Measure, whose Gage-point is known.

Problem III. To find the Gage-point of any Measure.

The Gage-point of any Solid Measure, is only the Diameter of a Circle, whose Su∣perficial Content is equal to the Solid Con∣tent of the same Measure.

As thus more plainly;

The Solid Content of a Wine Gallon is 231 Cube inches: Now if you have a Circle that contains 231 Superficial inches, the Diameter thereof will be found to be 17 inches, and 15 of a hundred; as by the 7th Problem of the 7th Chapter, is well seen.

These Directions may serve for any round Vessel, either close or open; yet Mr. Ough∣tred, a very able Mathematition, hath a

way accounted somewhat more exact, and consequently more tedious and troublesome to use either by the Pen or Compasses, which is this;

You must measure the Diameters at head and bung, or the top and bottom in inches and 10ths, the length also by the same mea∣sure; then find out the Superficial Con∣tent of the Circles, answerable to those two Diameters, and take two thirds of the grea∣test, and one third of the least, and add them together in one sum; which sum you must multiply by the length in inches and tens, and the Product shall be the Content in Cube inches; which Product divided by 282, gives Ale Gallons; or by 231, gives Wine Gallons, as before.

By the Line of Numbers, this way is more easie and ready thus;

The Extent from 1, to 0-5236, a Num∣ber fit for 2 thirds, of the Circle at the bung;

So is the Square of the Diameter at the bung to a 4th.

Then again;

As 1, to 0-2618, the half of the for∣mer Number, and fit for one third of the Circle at head;

So is the Square of the Diameter at head to a 4th.

These two 4ths add together, then say;

As 231 (for Wine, or 282 ¼ for Ale-Gallons), is to the sum of the two 4ths added together;

So is the length to the Content in Wine-Gallons.

Example, at 18 inches at head, and 32 at bung, that old Example.

The Square of 32, is 1024;

The Square of 18, is 324:

Then,

The Extent of the Compasses from 1, to 0-5236, shall reach from 1024, the Square of 32, to 536-4, two thirds of the bung-Circle.

Again,

The Extent from 1, to 0-2618, shall reach from 324, the Square of 18, the Diameter at head, to 84-9, the sum of 536-4, and 84-9, is 621-3.

Then lastly,

The Extent from 231, the 621-3, shall

reach from 40 the length, to 107-58, or 107 Gallons and a half, and better, the Content in Wine-gallons, as briefly as can be done this way.

But if you take the Diameters at head and bung, with a Line called Oughtred's Gage-line; and set the measure found at the bung by that Line, down twice; and the measure found at the head, found by the same Line, once, and bring them into one sum; then multiply that sum by the length of the Vessel in inches, and 10 parts, and then the Product shall be the Content in Wine-gallons required.

As if I should measure a Cask of 18, and 32, as before: right against 18 inches on Oughtreds-Line, you find 0-367; and right against 32, you shall find 1-161; this last set down twice, and 0-367 once; added, makes 2-689; and then this sum multiplied by 40, makes 107-56, being very near to the former operation, but dif∣fering about 2 Gallons, from the way set be∣fore by the mean Diameter and Gage-point, by reason of the extream swelling of the Cask; But if this way should prove the truest in the Book of the Carpenters Rule, you have a Table to rectifie this difference, which you will very seldom have occasion to use.

Note also, That this Line, called Ough∣tred's Gage-Line, is very excellently impro∣ved to find the Content of Great Vessels, either in the whole, or inch by inch; which you will find at large in the Book before mentioned.

Also, The use of the Lines called Dia∣gonal-Lines, and Lines to find the empti∣ness of Cask, and to measure Corn-mea∣sures by, to which I shall, for the present, refer you.

Problem IV. The Diameter and content of a Vessel being given, to find the length of the Vessel.

Extend the Compasses from the Diameter to the Gage-point, the same Extent twice repeated from the Content, shall give the length required.

Example.

If the Content be 60, and the Diameter 24, then extend the Compasses from 24, to 17-15, the Gage-point for Wine; this Extent turned twice the same way, from 60 the Content, shall reach to 30 inches, and 6 tenths, and a half, the length required.

Problem V. The Length and Content of a Vessel gi∣ven, to find the Diameter.

Divide the space on the Line of Num∣bers, between the Length and the Content, into two equal parts; the Compasses so set, shall reach the same way from the Gage-point to the Diameter of the Vessel.

Example.

The half distance between 31-65 the length, and 60 the Content, shall reach the same way from 17-15 the Gage-point, to 24 the Diameter required.

These two last Problems may be useful for Coopers, to make Cask of any length, diame∣ter, and quantity.

Problem VI. To find what is wanting in any close Cask, at any number of inches and parts, (the Cask lying after the usual manner, with the bung-hole uppermost) from the bung-hole to the superficies of the Liquor given, two wayes.

This Problem I shall resolve two wayes, either of which is experimented to come

near the truth, and will very well serve, till a better comes to light.

The One, by a Line of Segments, joyned to the Line of Numbers, as before in the measuring the Fragments of a Globe; But,

The Other, is by a way found out by Mr. Bennit, a Cooper, that hath long exercised the way of Gaging, which is by comparing a Cask known, and its quantity of empti∣ness, to a Cask unknown, and its inches of emptiness, as followeth.

First, by the Line of Numbers, and Artifi∣cial Line of Segments, to find the quantity of Gallons that any Vessel wants of being full, at any number of Inches, from the inside of the bung-hole, to the superficies of the Liquor, which is usually called Inches dry.

Extend the Compasses, on the Line of Numbers, from the inches and tenths diame∣ter at the bung, to 100 on the Line of Seg∣ments, the same extent applied the same way from the inches and parts dry, shall reach to a 4th Number, on the Line of Artificial Segments; which 4th Number you must keep. (Or, if you will, you may use the inches wet, laying the same extent from the inches wet, and that also will on the Seg∣ments

give a 4th Number, which you must likewise keep.)

Then secondly,

As the Extent from 1, to the whole Con∣tent of the Vessel in Wine or Ale-gallons; So is the 4th Number kept to the Gallons of emptiness, or fullness, that it wants of being full, or the quantity of Gallons in the Ves∣sel.

Example, of a Canary-Pipe, whose Dia∣meter at bung, is 28 inches and 7, and full Content is Gallons 116 ½, at 12 inches dry, or 16 inches, and 7 tenths wet.

The Extent of the Compasses from 28-7, to 100, (at the end of the Line of Seg∣ments) shall reach the same way from 12, the inches dry, to 39 ½ on the Line of Seg∣ments for a 4th; or from 16-7 wet, to 60 2/10 on the Segments, for his 4th also, which two 4ths keep.

Then secondly,

The Extent from 1, to 116 ½, the whole Content in Gallons, shall rea••h from 39, the dry 4th, on the Line of Numbers, to 46 3/10, for the gallons dry or wanting: or the same extent shall reach the same way, on the Line of Numbers, from 60 2/10, the 4th Number for wet, to 70 gallons, and 2 tenths

in the Vessel, at 16 inches and 7 tenths wet; which two Numbers put together, makes up 116 gallons and a half, the full Con∣tent.

The like manner of working serves for any Cask whatsoever, and the nearer the Vessel wants of being half empty, the more near to the truth will your work be, and the most errour in very round and swelling Cask, when the emptiness is not above one or two inches; but in Vessels near to Cil∣lenders, it will give the Answer very true, and as readily as any way whatsoever.

Observe also, That if you use the Seg∣ments in taking the wants, you must abate of the gallons found, till you come to the 2 thirds of the half diameter; that is to say, the Rule sayes, there is more wanting than indeed there is; and that somewhat con∣siderable about the first 6 inches in a vessel of 30 inches diameter: So that I find a Table made as a mean between the Superficial and solid Segments, would do the work the truest and best of any other; Or else, use the mean diameter and mean parts of empti∣ness; found thus.

Take the equaded diameter, from the diameter at the bung; and note the diffe∣rence: then half this difference taken from the inches and parts empty gives the mean

emptiness; then use the mean diameter, and mean emptiness, instead of the other, and the work is more exact.

The other way of Mr. Bennits inven∣tion is thus;

First, you are to fill an ordinary Cask, of a competent magnitude, as 60 or 100 gallons, of a mean form, between a Spheri∣ord (or roundish form) and a Cillenderi∣cal form; or else fill two Casks of each form, and learn the true Content, and Diameter of that mean Vessel, or rather of both those Vessels; and the Vessel being full, draw off with a true gallon-measure, and on the drawing off every gallon, take the exact quantity of inches and 10th parts, that the drawing off of every gallon makes in the emptiness or driness of that mean Vessel, or rather both those Vessels, at least until you have drawn off the half quantity of the Vessel, which number of gallons drawn off, and the inches and tenth parts of emptiness, or fulness, or driness or wetness, you must draw into a Table, or insert them on a Rule, making the inches as equal parts, and the gallons, and his proportional part of a gallon, the unequal parts; then with the Line of Numbers, and this mean Table, or rather two Tables or Scales, which you may

put on a Rule, as Mr. Bennit hath done, you may find out the wants of any Cask what∣soever; either Spherioid, or Cillender-like, as followeth.

This measured Cask on the Scale, or Table, for methods sake, and avoiding tau∣tologie, I shall call the first Cask, and the Vessel or Cask, whose wants you would know, I shall call the second Cask; then the proportion is thus.

As the Diameter at the bung of the se∣cond Cask, is to the bung diameter of the first Cask (which is always fixed);

So is the inches dry of the second Cask to a 4th (on the Line of Numbers) which 4th Number sought on the inches of your Table, or Scale, on the opposite-part of your Scale or Table, gives a 5th Number, which you must keep.

Then,

As the whole Content of the first Cask, is to the whole Content of the second Cask;

So is the first Number kept, to the Num∣ber of Gallons the Vessel wants of be∣ing full, at so many inches dry.

Example.

There is such a Scale made on purpose for Victuallers use, to measure what they want of a Barrel of Ale, being put into a Beer-barrel,

which Scale I shall here use, to try this former Example by.

Suppose, as before, a Canary Pipe want 12 inches of being full, and the Content 116 ½ gallons, and 28 inches and 7 tenths diameter at bung; The Extent on the Line of Numbers from 28-7, to 22-5, shall reach from 12, to 9-4; then just against 9 inches and 4 tenths, on that Barrel Scale, I find 14 gallons of Beer, which is 17 gallons and a half of Wine, being the 5th Number to be kept.

Then the Extent from 44, the Content of a Barrel in Wine-gallons, to 116 ½, the Content of a Canary-Pipe in the same gal∣lons, shall reach the same way from 17 ½ the Number kept, to 46, and near a half, the gallons wanting at 12 inches dry, in the Canary Pipe, and 46 gallons, and 3 quarts, is the Number Mr. Bennit finds in a Canary-Pipe, by measuring at 12 inches dry.

Thus you have an account of the two easie Mechanick wayes, to discover the wants of Cask, very applicable, and ready, and experimented to be Propè verum.

The Gallons wanting in a Barrel, at every inch and quarter.

	Beer Gall.	Wine Gall
	gal. pi. 100	gal. pi. 100	gal. 1000
	0 0 40	0 0 49	0 0612
	0 1 20	0 1 47	0 184	22
	0 2 10	0 2 57	0 321
1	0 3 10	0 3 80	0 475
	0 4 33	0 5 30	0 663
	0 6 00	0 7 35	0 920	21
	0 7 60	1 1 29	1 161
2	1 1 80	1 4 00	1 500
	1 3 90	1 6 56	1 821
	1 6 10	2 1 22	2 153	20
	2 0 66	2 4 34	2 543
3	2 3 50	2 7 98	2 998
	2 6 16	3 3 10	3 388
	3 0 70	3 6 20	3 772	19
	3 3 80	4 2 00	4 250
4	3 6 50	4 5 30	4 663
	4 1 80	5 1 35	5 169
	4 5 25	5 5 60	5 700	18
	5 0 42	6 1 45	6 182
5	5 3 90	6 5 70	6 713

	5 7 20	7 1 70	7 213
	6 2 80	7 6 20	7 777	17
	6 6 50	8 2 65	8 333
6	7 2 20	8 7 20	8 900
	7 5 50	9 3 20	9 400
	8 1 10	9 7 70	9 960	16
	8 4 80	10 4 20	10 525
7	9 0 70	11 1 00	11 125
	9 4 50	11 5 40	11 806
	10 0 40	12 2 20	12 275	15
	10 4 30	12 7 00	12 876
8	11 0 50	13 4 10	13 513
	11 4 30	14 0 80	14 110
	12 0 30	14 5 80	14 725	14
	12 4 29	15 2 80	15 350
9	13 0 30	15 7 70	15 926
	13 4 30	16 4 60	16 577
	14 0 40	17 1 60	17 200	13
	14 4 60	17 6 60	17 827
10	15 0 50	18 3 40	18 425
	15 4 48	19 0 30	19 037
	16 0 80	19 6 50	19 815	12
	16 5 50	20 3 25	20 40••
11	17 2 20	21 1 00	21 225

	17 7 90	22 0 00	21 000
	18 5 49	22 6 98	22 644	11
	19 2 00	23 4 31	23 391
12	19 6 16	24 1 48	24 184
	20 3 00	24 7 30	24 961
	20 7 40	25 4 60	25 575	10
	21 3 10	26 1 36	26 170
13	21 7 40	26 6 38	26 799
	22 3 00	27 3 36	27 130
	22 7 00	28 2 18	28 174	9
	23 3 00	28 5 18	28 648
14	23 7 30	29 2 19	29 275
	24 3 70	29 7 40	29 926
	24 7 40	30 3 90	30 488	8
	25 3 60	31 1 00	31 125
15	25 7 50	31 5 80	31 726
	26 3 30	32 2 60	32 325
	26 7 00	32 7 00	32 875	7
	27 3 00	33 3 80	33 475
16	27 6 40	34 0 30	34 037
	28 2 20	34 4 80	34 600
	28 5 80	35 0 80	35 100	6
	29 1 40	35 5 34	35 668
17	29 4 80	36 1 80	36 225

	30 0 40	36 6 29	36 788
	30 4 10	37 2 29	37 287	5
	30 7 50	37 6 54	37 820
18	31 3 00	38 2 39	38 299
	31 6 10	38 6 64	38 833
	32 1 80	39 2 70	39 338	4
	32 5 00	39 6 00	39 752
19	32 7 80	40 1 80	40 225
	33 2 10	40 4 90	40 614
	33 4 80	41 0 10	41 012	3
	33 7 40	41 3 65	41 457
20	34 2 00	41 6 77	41 848
	34 4 30	42 1 43	42 180
	34 6 20	42 4 00	42 500	2
	35 0 10	42 6 70	42 840
21	35 2 00	43 0 64	43 055
	35 3 60	43 2 70	43 338
	35 4 80	43 4 19	43 524	1
	35 6 00	43 5 42	43 678
22	35 6 80	43 6 51	43 816
	35 7 40	43 7 50	43 938
	36 0 00	44 0 00	44 000

CHAP. XIII. The use of the Line of Numbers, in Questions of Interest and Annuities.

Problem I. At any rate of Interest per annum for a hundred pounds, to find what the Interest of any greater or lesser sum comes to in one year.

EXtend the Compasses from 100 to the increase of 100 l. in one year, the same Extent shall reach from the sum propound∣ed, to its increase for one year, at that rate propounded.

Example. What is the increase or profit of 124 l. 10 s. for one year, at 6 per cent. per annum?

The Extent of the Compasses, from 100 to 6, being laid the same way from 124 l.

10 s. (which is at 124-5) shall reach to 7-47, which is 7 l.— 9 s. — 4 d. the pro∣fit of 124· 10 s in one year.

Problem II. Any sum of Money, and the rate of In∣terest propounded, to find what it will in∣crease to, at any number of years, count∣ing Interest upon Interest.

The Extent of the Compasses from 100, to the increase of 100, being turned as ma∣ny times from the sum propounded the same way, as there be years propounded, shall at last stay at the Principal and Interest re∣quired.

Example. To what sum shall 143 pounds 10 shillings, amount to in 10 years, counting Interest upon Interest, at 6 per cent?

The Extent of the Compasses from 100, to 106, being turned 10 times from 143 ½, shall reach to 257 l. 0 s. the sum of Prin∣cipal and Interest at 10 years end.

Note, That in doing this, you ought to be very precise, in taking the first Extent from 100, to 106; but to cure the uncer∣tainty thereof, you have this very good re∣medy: If you have a Diagonal Scale, equal

to the Radius of the Line of Numbers, then use that; if not, use the Line of Lines on the Sector-side, which should be made fit to (or the double, or the half of) the Radius of the Line of Numbers.

As thus;

Take the Extent from the Line of Num∣bers, between 100, and 106; this Extent measured on the Line of Lines, will be 0253058, could you see so many Figures, but 02531, will serve your turn very well; which Number you must note, is the Lo∣garithm of 106, neglecting the Caracteri∣stick; then this Number multiplied by 10, the Number of years, is 25310; this Ex∣tent taken from the Center, on the Line of Lines, and laid increasing from 143 ½, shall reach to 257 l. 0 s. 0 d. the true Number of the Use and Principal of 143 l. 10 s. put out, or forborn for ten years.

Problem III. A sum of Money being due at any time to come, to find what it is worth in ready Money to be paid presently, at any rate propounded.

This Problem is the contrary to the last, for if you shall turn the Extent between 100 and 106, ten times backward from

257, it will stay at 143 ½, the worth in ready Money.

Or, to make use of the former remedy;

Multiply 0253058, the Logarithm of 106 by 10; then this Extent taken and laid the decreasing way from 257, shall reach to 143 ½.

For Note, That the Line of Lines is the Scale of equal parts, that makes the Line of Numbers, and 10, or 7, or 15, or any other Number multiplied by the Logarithm of 106, taken from that Scale of Lines all at once, is equal to so many repetitions; and consequently more exact, because of the dif∣ficulty of taking the 10, 12, or 15th part of any Number whatsoever; and observe, That so much as you err in the first, it will be 10, 12, or 15, or 20 times so much at last, which may be considerable in this.

Problem IV. A yearly Rent, or Annuity being forborn a certain number of years, to find what the Arrears thereof will amount unto, ac∣cording to any rate propounded.

First, you must find out the Principal-Money, that answers to the Rent, or An∣nuity in question; then find the sum of

that Principal and Use, at the end of the term given, at the rate propounded; then the Principal taken out of this sum, both of Arrears and Principal, the Arrears do re∣main, which is the sum you look for.

Example.

Suppose a Landlord live far from his Tennant, and yet judging his Tennant ho∣nest, and able, is content to take his Rent once in every fourth year, which should be paid every year, or every quarter of the year; and suppose the Rent be 10 l. per an∣num, and the rate of profit, for the for∣bearance, be 8 per cent.

First, to find the Principal for 10 l. per annum, at the rate of 8 l. per cent. Say, If 8 l. have 100 for his Principal, what shall 10 l. have? The Answer will be 125; for the Extent from 8 to 100, shall reach from 10, the same way, to 125; then by the 2d Problem of this Chapter, 125 l. for∣born for four years, will come to 170 l. which is 170 l. 0 s. 0 d. from which sum, if you substract 125 l. there remains 45 l. the Arrears for 10 l. per annum forborn four years, at the rate of 8 per cent.

But if you would have the profit of these Arrearages, supposing 2 l.—10 s. the 4th part of 10 l. per annum to be paid quarterly, and to count Use upon Use at the rate above∣said,

then you will find the Principal and Arrears to be 171 l. 10 s, For if you mul∣tiply 0086, the log. of 102 l. the Interest and Principal of 100 l. for a quarter of a year by 16, the quarters in four years, it will be 1376, which Number taken from the Line of Lines, and laid from 120, on the Line of Numbers, shall reach to 171 ½, or 171 l. 10 s. being 30 s. more than the for∣mer sum, when 150 l. the Principal is ta∣ken away, the residue Arreares is 46 l. 10 s.

Or,

If you turn the distance on the Num∣bers between 100 and 102, 16 times from 125, which you may help thus; turn first 4 times, then take them 4 times in one Ex∣tent, and turn 3 times more, and you will stay at 271 ½, the Answer required.

Problem V. A yearly Rent, or Annuity propounded, to find the worth thereof in ready Money, at any rate whatsoever.

First, by the 4th Problem, find the Ar∣rears that shall be due at the end of the term, and at the rate propounded; then by the 3d Problem, find what those Arrears are worth in ready money, which shall be the worth of the Annuity, or Rent required.

Example. There is a Lease of a House or Land worth 12 l. per annum, and there is 16 years yet to come; which Lease a man would buy, provided he may lay out his money to gain after the rate of 10 l. per cent: the question is, What is it worth?

First, by the last, if 10 l. have 100 for his Principal, What shall 12? the Answer is 120; Then by the second part of the second, 120 l. forborn 16 years, comes to 551 l. the Principal and Interest: from which sum, taking 120 l. the Principal, there remains 431 the Arrears. Then by the third Problem find what 431 due 16 years to come, is worth in ready money; and the Answer will be at 10 in the 100, 93 l. 14 s.

Also herein observe, That if there be any Reversion of a Lease to be expired, before it may be injoyed; then you are to find the worth of 431 l. after so many years more; as suppose it be 5 years before the Annuity begin; then find the worth of 431, for∣born 21 years, which will be 58 l. 4 s.

Problem VI. A sum of Money is propounded, and the rate whereby a man intends to Purchase, to find what Annuity, and how many years to continue, that sum of money will buy.

Take any known Annuity at pleasure, and find by the last, the value of that in ready money, then this proportion holds;

As the value found, is to the Annuity supposed;

So is the sum of money to be improved, to the Annuity required.

Example. What Annuity, to continue 16 years, will 500 l. Purchase, whereby a man may gain after the rate of 10 l. per cent?

By the last Problem I find, That 93 l. 14 s. will purchase 12 l. a year, for 16 years, at 10 per cent.

Therefore,

The Extent of the Compasses from 93 l. 7, to 12 l. per annum, shall reach the same from 500, to 64 l. per annum. For such an Annuity, to continue 16 years, will 500 l. purchase, to gain 10 l. per annum, per cent. for your Money.

Problem VII. Or, first rather; Lands or Houses, sold at any certain num∣ber of years Purchase; to find what the value of the whole will be?

The usual way of valuing Land or Houses, is by the years Purchase, and Land Fee-simple is usually vallued at 20 years Purchase; Coppy-hold-Land, at 15 or 16 years Purchase; and good, strong, and new Houses, at 12, 13, or 14 years Purchase for Fee-simple.

But a Lease of 〈…〉〈…〉 of 21 years about 7 years Purchase; and a Lease of 31 years, about 8 years Purchase, rather less than more; and a Lease of 60, or 100, not worth above 8 ½ years Purchase.

Again,

The usual profit allowed for Land in Fee-simple, is not above 5 l. in the 100 per an∣num, because of the certainty thereof; for Coppy-hold Land, full 6 l. in the 100 per annum; for the best Houses, 7 and 8 l. in the 100 Fee-simple.

But in laying out Money on Leases, either of Land or Houses, Men shall hardly be sa∣vers, if they gain not 8, 9, or 10 in the 100 per annum, for their Money; The reason

and demonstration whereof, you may read at large in Mr. Phillips his Purchasers Pat∣tern.

Thus the number of years Purchase a∣greed on, (which ought to be cleer, from Quit-rent, and Taxes, and the like; the Rent is usually various, according to the place, and time where, and wherein, the Purchase shall happen to be) then to find the quantity of the whole Purchase,

Say,

As 1, to 20, 18, 15, 14, 12, 10, or 8, the number of years Purchase, for Fee-simple, or Coppy-hold Land, or Houses Fee-simple, or Coppy-hold; For Leases of 60, 50, 40, or 30 years, or 21 years;

So is the yearly Rent to the whole value.

Example.

A Parcel of Land worth 10 l. per annum Fee-simple, valued at 20 years Purchase, will amount to 200 l.

For,

The Extent from 1, to 20, will reach the same way from 10 to 200, the whole price of 20 years Purchase, at 10 l. per annum.

CHAP. XIV. The Use of the Line of Numbers IN Military Questions.

Problem I. Any Number of Souldiers being propound∣ed, to order them into a Square Battel of Men; that is, as many in Rank as in File.

FInd the Square-root of the Number of Souldiers, and that shall be the Num∣ber of Men in Rank and File required.

As suppose it were required to order 1770 Men, in the order abovesaid, you shall, by the 8^th Probl. of the 6^th Chapt. find, that the Square-root of 1770 is 42, and 6 over, which here is not consider∣able.

Problem II. Any number of Souldiers propounded, to order them into a double Battel of Men; that is to say, twice as many in Rank as File.

Find the Square-root of half the Num∣ber of Men, and that is the Number of Men in File, and the double the Number in Rank.

As for Example.

If 2603, were so to be placed, the half of 2603, is 1301; whose Square-root by the 8th of 6th, is 36; the number of Men in File: and 72, the double thereof, is the number in Rank. For if you shall multiply 72 by 36, the Product is 2592, almost the number of Men propounded.

Problem III. Any Number of Souldiers being propound∣ed, to order them into a Quadruple Battel of Men; viz. 4 times as many in Rank as File.

Find the Square-root of a 4th part of the Number of Men, and that shall be the Number in File; and 4 times so many the Number in Rank.

So the 4th part of 2603, is 650; whose Square-root is 25 ½, and 4 times 25 is 100, the Number in Rank.

Problem IV. Any Number of Souldiers being given, to∣gether with their Distance one from ano∣ther in Rank and File, to order them into a Square Battel of Ground.

As suppose I would order 3000 Men so, that being 7 foot asunder in File, and 3 foot apart in Rank, the Ground whereon they stood should be Square.

Extend the Compasses from 7 foot, the di∣stance in File; to 3 foot, the distance in Rank; then that Extent applied the same way from 3000, the Number of Souldiers, reaches to 1286, whose greatest Square-root is 35-7; that is, 35, the Number of Men to be placed in File.

Then,

If you divide 3000, the whole Number, by 35-7, the Quotient is 84, the Number in Rank, to use and imploy a Square plat of ground to stand in.

As 7, to 3; so is 3000, to 1286, whose Square-root is 35-7. Then,

As ••5-7, to 1; so is 3000, to 84.

Problem V. Any Number of Souldiers propounded, to order them into Rank and File, accor∣ding to the ratio of any two Numbers given.

This Question is all one with the for∣mer;

For,

As the Number given for the distance in File, is to that for the distance in Rank;

So is the whole Number of Souldiers to a 4th, whose Square-root is the Num∣ber of Men in Rank.

Then.

The whole Number divided by the Num∣ber in Rank, the Quotient is the Number to be placed in File.

Example.

Suppose 3000 Souldiers were to be or∣dered in Rank and File: As 5 is to 10, or as 5 is to 9; that is to say, that the Men in Rank, might be in Proportion to them in File, as 9 is to 5.

Say thus;

As the Extent from 5, to 9;

So is 3000, to 5400, whose Square-root is 73 ½, the Number of Men in Rank.

Then,

As 73 ½, to 1;

So is 3000, to near 41, the Number in File.

Problem VI. There are 8100 Men to be ordered into a Square Body of Men, and to have so many Pikes, as to arm the main Square-Body round about, with 6 Ranks of Pikes; the Question is, How many Ranks must be in the whole Square Bat∣tel? And, How many Pikes and Mus∣quets?

First, the Square-root of 8100, is 90, the Number of Men in, and Number of Ranks and Files; now in regard that there must be 6 Ranks of Pikes round about the Musquetiers, there will be 12 Ranks less of them, both in Front and Flank, than in the whole Body; therefore substracting of 12 from 90, rest 78, whose Square is 6084, the number of Musquetiers; which taken from 8100, there remains 2061, the num∣ber of Pikes.

Problem VII. To three Numbers given, to find a fourth in a doubled Proportion.

For as much as like Squar••••, are in double the Proportion of their answerable sides; therefore you must work by their Squares, and Square-root.

But by the Line of Numbers, in this manner.

If a Fathom of Rope, of 6 inches com∣pass about, weigh 6 pound, 2 ounces, (or 6-••25/1000) what shall a Fathom of Rope of 12 inches compass weigh?

Here Note alwayes, That when the two Numbers of like denomination, which are given, are of Lines, or sides of Squares, or Diameters of Circles; then the Extent of the Compasses upon the Line of Numbers, from one Line to the other, or from one side to the other side; that Extent turned twice the same way from the given Area, or Content, shall reach to the other required:

So here, the Extent of the Compasses from 6 to 12, being turned two times the same way from 6-125, shall reach to 24-50, for 24 pound and a half, the weight required.

But if the two terms given of one deno∣mination, are of Squares, or Superficies, or Areas; then the half distance, on the Line of Numbers, between one Area and the o∣ther, being turned the same way on the Line, from the given Line or Side, it shall reach to the Side, or Line, required.

For the half-distance, between 24-50, and 6-125, shall reach from 12 to 6; or the contrary, from 6 to 12.

An Example whereof, you have in the 4th and 5th Problems of the 12th Chapter: Also, in the 6th and 7th Problems of the 8th Chapter, which treates of Superficial-measure, in measuring of Land.

Note also, That if you have three Lines of Numbers, viz. a Great, a Mean, and a Less; after Mr. Windgates way; then these Questions are wrought without doubling, or halving, and very neatly and speedily.

As thus;

The Extent on the mean Line, from 24-50, to 6-125, the weight of the two Ropes, shall reach on the great Line, from 12 to 6; or from 6 to 12, the inches in com∣pass about of each Rope.

Problem VIII. To three Numbers given, to find a fourth in a tripled Proportion.

For as much as like Solids, are in a tripled Proportion to their answerable side; the Cubes of their sides are proportional one to another; therefore, to work these Que∣stions by the Line of Numbers, do thus;

When the two given terms, of like deno∣mination in the Question, are of Sides, Lines, or Diameters; then the Extent of the Compasses, on the Line of Numbers, from one side to the other, that is, from the side, whose Cube or Solidity is also given, to the other; the same Extent, turned three times from the given Cube, or Solidity, shall reach to the inquired Cube, or Solidity.

As for Example.

If a side of a Cube, being 12 inches, contain in Solidity 1728 cube inches; How many inches is there in a Cube, whose side is 8 inches? The Extent from 12 to 8, be∣ing turned three times from 1728, shall reach to 512, the Solidity required of the Cube, whose side is 8 inches every way.

Again, on the contrary.

When the two terms of the same denomi∣nation, are Cubes or Solidities, then divide the space on the Line of Numbers, between the two Solidities, into three equal parts, and lay that Extent the same way, as the reason of the Question doth require, either increa∣sing or diminishing, from the given Side or Line, and it shall reach to the inquired Side, or Line.

Example.

If 1728, be the Cube of 12, the Root or side; what shall be the Root or side of 864, the half of 1728. being half a foot of Timber? The Extent between 1728, and 864, being divided into three parts, and that third part, laid decreasing from 12, shall reach to 9-525, the side or root requi∣red of half a foot of Timber, though not exactly, yet very near.

Again for another Example.

If an Iron Bullet, of 6 inches Diameter, weigh 30 pound; what shall a Bullet of 7 inches Diameter weigh? The Extent from 6 to 7, shall reach, being turned three times, to 47-7.

Again,

If a Ship, whose Burthen is 300 Tun, be

75 Foot by the Keel; what shall that Ship be, whose Keel is 100 Feet? The Extent between 75 and 100, turned three times from 300, shall reach to 713 Tun Bur∣then.

Again,

If a Ship of 29 Foot and a half at the beam, be 300 Tun Burthen, what shall a Ship of 713 Tun burthen be? The third part of the distance between 300 and 713, shall reach from 29 ½ to 39-35, its mea∣sure at the beam.

Again,

If a Ship of 300 Tun be 13 Foot in hold, what shall a Ship of 713 Tun be in hold? The third part between 300 and 713, shall reach from 13 foot, to 17-35, the Feet in the Hold of a Ship of 713 Tun.

If you have a treble Line, then you may save the dividing, by taking from the little-Line, and measuring on the great-Line, and the contrary, as the nature of the Question doth require.

Lastly,

Know, that by adding of twelve Centers and Points, the Line may be made to speak, as it were, and so made more fit for any mans more particular occasions.

A Brief Touch of the Use of the Lo∣garithms, or Tables, of the Arti∣ficial Numbers, Sines, and Tan∣gents.

See more in Gunter's Works.

IT may happen, that some may meet with this Book, that had rather use the Tables of Logarithms, from whence these Lines are framed, than the Lines on the Rule; or out of curiosity to prove the truth of their work, for whose sakes I have added these following plain Precepts, without Examples.

1. To multiply one Number by another.

Set the Logarithm of the Multiplicator, and Multiplicand, right under one another, and add them together, and the sum is the Logarithm of the Product.

2. To divide one Number by another.

Set down first the Logarithm of the Di∣vidend, and then right under it the Loga∣rithm of the Divisor, and then substract the log▪ of the Divisor, from the log. of the

Dividend, and the remainder is the Log. of the Quotient required.

3. To find the Square-root of a Number.

Half the Logarithm of the Number gi∣ven, is the whole Logarithm of the Square-root of it.

4. To find the Cubick-root of a Number.

One third part of the Logarithm of the given Number, is the full Logarithm of the Cubick-root of the given Number, as a third of 14313637; the logarithm of 27 is 0-4771212, the Log. of 3, the Cube-root of 27, required.

5. To work the Rule of Three direct, or three Numbers given, to find a 4th by the Logarithms.

5. To work the Rule of Three direct, or three Numbers given, to find a 4th by the Logarithms.

Set down the Logarithms of the 1, 2, & 3 Numbers, one right over another; then add the logarithms of the second and third together; and from the sum, substract the logarithms of the first, and the remainders is the logarithms of the 4th required.

6. When in common Arithmetick the se∣cond term is divided by the first, and the Quotient multiplied by the third.

Then by Logarithms,

Take the Logarithm of the first term, from the Logarithm of the second; and add the difference to the log. of the third, and the sum is the log. of the 4th.

7. When in common Arithmetick the se∣cond term is divided by the first, and the third by the Quotient.

Then take the log. of the second, from the log. of the first term; and take the dif∣ference out of the log. of the third, and the remainder is the log. of the 4th term re∣quired.

8. Between two extream Numbers, to find a mean Proportional.

Add the logarithms of the two extream Numbers together 〈…〉〈…〉 sum is the 〈…〉〈…〉.

9. To work the Rule of Three in the Lo∣garithms of Artificial Numbers, Sines, and Tangents.

1. When Radius is the first term.

Add the Logarithms of the second and third terms together, and Radius, or a unite, in the first place, taken from the sum, there shall remain the logarithm of the 4th term required; according to the 5th Precept.

2. When Radius is in the second place, or term.

Then the first term (and second virtually) taken from the third, cutting off a unite in the first place for Radius, is the 4th term.

3. When Radius is in the third place.

Then substract the logarithm of the second term, from the log. of the first term, cutting off a unite for Radius, and the remainder is the 4th term.

4. If Radius be none of the three terms.

Then add the Logarithms of the second and third terms; and from the sum,

substract the logarithm of the first term, and the remainder is the logarithm of the 4th term.

5. Or else.

Set down the Arithmetical complement of the first term, and the logarithms of the second and third term, and add all to∣gether, and the sum cutting off Radius, is the 4th term.

10. When the Number is not to be found in the Canon of Logarithms of Num∣bers, Sines, or Tangents; take the next nearest, or for more exactness use the part proportional.

11. Though Numbers and Sines, or Numbers and Tangents are used together, the work is all one, as with Sines and Tangents, as to the Precept in working.

12. In using the Logarithms, great re∣gard is to be had to the Index, or Char∣racteristick, to rule in the Number of places; the Characteristick being one unite less than the Number of places to express that Number; thus the Characteristick of 3-5932861, the logarithm of 3920 is 3, being one less than the Number of places in 3920, which consists of 4 figures.

CHAP. XV. The use of the Trianguler Qua∣drant, in Geometry, and A∣stronomy.

Use I. The Radius of a Circle, or Line being gi∣ven, to find readily, any required Sine, Tangent, or Secant, or Chord, to that Radius.

And first,* 1.72 to do it by the Quadrantal-side.

FIrst, If your Radius happen to be e∣qual to the greater Scale of (Alti∣tudes or) Sines, issuing from the Center,* 1.73 then the measure of any degree or mi∣nuite, from the Center toward the head, shall be the Sine, the measure from the Center-point at 60|0, on the degrees, to any degree and minuit required, shall be

the Tangent to the same Radius;* 1.74 and the measure from the Tangent to the Center, shall be the Secant,* 1.75 to the same Radius. And if you have an Index, or a Bead upon your Thred, and set the Bead, when the Thred is drawn streight, to the Center at 60|0 on the degrees or Tangents, or to the Sine of 90;* 1.76 then if you lay the Thred to any Number of degrees and minuits, counted from 90, and there keep it; then the extent from the Sine of 90, to the Bead, shall be the Chord of the Angle the Thred is laid to, to the Radius of the greater Scale of Sines, issuing from the Center.

But if this happen to be too large, then the other lesser Line of Sines, issuing up∣wards from the Center, being about one third part of the other, hath first it self for Sines;* 1.77 secondly, the degrees on the loose-piece for Tangents, counting from the Cen∣ter at 60;* 1.78 thirdly, the measure from the Tangent,* 1.79 to the Center, for a Secant; fourthly, the Bead and Thred, for a Chord,* 1.80 as before; all at once to one Ra∣dius, clearly and distinctly, without any interruption, to 75 degrees of the Tangent, or Secant.

But if any other Radius be given, then they will not be had so readily altogether, but thus in order one after another, and

first for the Sine, by the Trianguler Qua∣drant.

Take the Radius between your Compas∣ses, set one foot in 90, with the other lay the Thred to the nearest distance, and there keep it; then take the nearest distance from the Sine of any Ark or Angle you would have, and that shall be the Sine of the Ark or Angle required to the given Radius.

1. But by the Sector-side work thus, be∣ing near alike,* 1.81 fit the given Radius in the Parallel-sine of 90 & 90; then take out the Parallel-sine, of the Ark or Angle re∣quired, and you have your desire.* 1.82

2. Also the Sector being so set, if you take out any Parallel Tangent under 45,* 1.83 you have that also to the same Radius.

3. Also, if you would have any Tangent under 76, as the Sector stands,* 1.84 take out the parallel Tangent thereof, and that shall be the 4th part of the Tangent required to the same Radius, and is to be turned 4 times for that greater Radius.

4. Also,* 1.85 if you want a Secant under 60 degrees; at the same Radius take out the parallel Secant of the Ark or Angle re∣quired; and that shall be the half of the Secant required; for note, the Secant of one degree is more than Radius; and why

I use a half, rather than a 4th part, in time you may well see.

* 1.86By the Artificial Numbers, Sines, and Tangents, this cannot properly be done; only thus you may do by them, counting your given Radius (be it great or little) 10000 parts, you may by them find out readi∣ly how many of them parts will go to make the Sine, Tangent, or Secant, to any Num∣ber of degrees and minuts. As thus;

* 1.87Take the distance from the Sine of 90, on the Artificial Sines, to the Sine of any degree and minuit required; and set the same distance, the same way, from 10 on the Line of Numbers, reading it as a Scale of equal parts, and that shall be the Natural Sine of the degree and minuit required.

Or,

If you lay a Square to the Sine given, on the Numbers, it cuts the Natural Sine re∣quired.

Example.

Right against the Artificial Sine of 30, on the Line of Numbers, you find 5000, which is the Natural Number thereof.

But, if you measure this distance from 10, in the Line of Lines, it will give the Logarithmal Sine thereof, viz. 69897.

* 1.88And the like for the Tangent also, under 45, in the same manner.

But for the Secants, and,* 1.89 for the Tan∣gents above 45, you must count thus;

Measure, as before from 90, to the Co∣sine of the Angle, required, for a Secant; and from 45, to the co-Tangent of 45, for a Tangent; This extent laid the contrary way from 1, in the Numbers, shews how many Radiusses, and also how much above Radius, you must have to make up the Natural Tangent, or Secant required, in Numbers.

Example.

The Secant of 50 degrees, and the Tan∣gent of 57 degrees, 16 minuts,* 1.90 being near alike, is 1 Radius 5556; for the Natural Number thereof: and this distance mea∣sured on the Line of Lines, gives Radius because above 45, and 1919; more for the Artificial Tangent of 57-16, or the Secant of 50 degrees.

This have I hinted, in the first place, that thereby you might see the nature of the Lines, and the making of the Instrument, with its great convenience in the Contri∣vance of the Work on both sides, and the harmony, and proportion, the Natural way hath to the Artificial; also hereby you may readily prove the truth of your Instrument, being an equilatteral Triangle, whether you use the greater or the lesser Sines; For the

measure from the Center,* 1.91 where the Thred is fastened, to the Center-point of Brass on the moveable-leg, and loose-piece at 60 on the degrees, ought to be equal to each Line of Sines; and also to the Tangent of 45 on the Tangent Line: The measure from the Center to the rectifying-point on the Head, at the meeting of the Lines for the Hour and Azimuth, and the Lines for the Sines and Lines, is equal to the Tangent of twice 30 on that piece.

Again,

The measure from the Center, to the rectifying-point on the end of the Head-leg, shall reach from thence to 30 on the loose-piece; and being turned twice, reaches to 0|60 on the loose-piece: Also, the Ra∣dius, or Tangent of 45, turned twice from 0|60 on the loose-piece, shall reach to 75, as by comparing the Natural Numbers to∣gether, will most exactly appear; Though perhaps without this hint, it might not have beeen observed by an ordinary eye.

Having been so large, and plain, in this first Use, I shall be, I hope, as plain, though far more brief in all the rest; for if you look back to Chapt. VI. Probl. I. Sect. 3. you shall there see the full explaining of Latteral and Parallel, and Nearest-distance,

and how to take them; the mark for Latteral being thus —; The mark for Parallel thus =; Nearest-distance thus ND, &c.

Use II. The Sine of any Ark or Angle given, to find the Radius to it.

Take the Sine between your Compasses,* 1.92 and setting one foot of the Compasses in the given Sine; and with the other Point lay the Thred to the nearest-distance, and there keep it; then the nearest-distance from the Sine of 90 to the Thred, shall be the Radius required.

Make the given Sine a Parallel Sine,* 1.93 and then take out the Parallel Radius, and you have your desire.

The Artificial Sines and Tangents, are not proper for this work, further then to give the Natural Number thereof, as before; therefore I shall only add the use of them when it is convenient in the fit place.

Use III. The Radius, or any known Sine being gi∣ven, to find the quantity of any other unknown Sine, to the same Radius.

Take the Radius, or known Sine given,* 1.94

and make it a Parallel in the Sine of 90 for Radius, or in the Sine of the known Angle given, and lay the Thred to ND. Then, take the unknown Sine between your Com∣passes, and carry one Point along the Line of Sines, till the other foot being turned about, will but just touch the Thred; then the place where the Compasses stayes, shall be the Sine of the unknown Angle requir∣ed, to that Radius or known Sine.

* 1.95Make the given Radius a Parallel Radius, or the given Sine a = Sine, in the answer∣able Sine thereof: Then, taking the un∣known Sine, carry it parallelly along the Line of Sines till it stay in like parts, which parts shall be the Numerator to the Sine re∣quired.

Use IV. The Radius being given, by the Sines alone to find any Tangent or Secant to that Radius.

* 1.96Take the Radius between your Compasses, and set one Point in the Sine complement of the Tangent required, and lay the Thred to the ND; then the ND from the Sine of the Tangent required, to the Thred, shall be the Tangent required: And the ND from 90, to the Thred, shall be the Secant required.

Make the given Radius a = in the co-Sine of the Tangent required;* 1.97 then the = Sine (of the inquired Ark or Angle) shall be the Tangent required; and = 90 shall be the Secant required to that Radius.

Use V. Any Tangent or Secant being given, to find the answerable Radius; and then any other proportionable Tangent, or Secant, by Sines only.

First, if it be a Tangent that is given,* 1.98 take it between your Compasses, and set∣ting one foot in the Sine thereof, lay the Thred to ND, then the = Co-sine thereof shall be Radius;

But, if it be a Secant, take it between your Compasses, and set one foot alwayes in 90, lay the Thred to the ND, then the nearest distance from the Co-sine to the Thred (or the = Co-sine) shall be the Radius required.

Take the given Tangent,* 1.99 make it a = in the Sine thereof; then the = Co-sine there∣of shall be Radius.

Or, if it be a Secant given, then

Take the given Secant, make it a = in 90,

then the = Co-sine thereof, shall be the Radius required.

Then having gotten Radius, the 4th Use shewes how to come by any Tangent, or Se∣cant, by the Sines only.

Use VI. To lay down any Chord, to any Radius; less then the Sine of 30 degrees.

Take the given Radius▪ set one Point in the Sine of 30,* 1.100 lay the Thred to the ND (and for your more ready setting it again, note, what degree and minuit the Thred doth stay at, on the degrees) and there keep it. Then the ND from the Sine of half the Angle you would have, shall be the Chord of the Angle required.

* 1.101Take the given Radius, and make it al∣wayes a = in 30, and 30 of Sines; the = Sine of half the Chord, shall be the Chord required.

Use VII. To lay down any Chord to the Radius of the whole Line of Sines.

* 1.102Take the Radius between your Compas∣ses, and setting one Point in 90 of the Sines,

lay the Thred to the ND, observing the place, there keep it.

Then taking the = Sine of the Angle re∣quired, with it set one Point in the Line to which you would draw the Angle, as far from the Center as the Radius is; then draw the Convexity of an Ark, and by that Con∣vexity, and the Center, draw the Line for the Angle required.

Example.

Let AB be a Radius of any length, un∣der or equal to the whole Line of Sines: Take AB between your Compasses, and setting one Point in 90; lay the Thred to ND, then take out the = Sine of 38, or any other Number you please, and setting one Point in B, the end of the Radius from A the Center, and trace the Ark DC, by the Convexity of which Ark, draw the Line AC for the Angle required.

Take the given Radius AB,* 1.103 make it a = in 90, and 90 of Sines; then take out = 38, and setting one foot in B, draw the Ark DC, and draw AC for the Angle required.

Or else work thus;

Take AB, the given Radius, (having drawn the Ark BE) and make it a = in

the Co-sine of half the Angle required; and lay the Thred to ND, (or set the Sector).

Then,

Take the = ND, from the right-sine of the Angle required, and it shall be BE, the Chord required to be found.

Note, That the contrary work finds Radius.

Use VIII. To lay off any Angle by the Line of Tan∣gents, or Secants, to prove it.

* 1.104Having drawn the Ground-Line, AB, at the Point B, raise a Perpendiculer, as the Line BC extended at length, then make AB, the Radius, a = Tangent in 45 and 45; then take out the = Tangent of the Angle required, and lay it from B to C in the Perpendiculer, and draw the Line AC for the Angle required.

Also, If you take out the Secant of the Angle, as the Sector stands, and lay it twice in the Line AE, it will reach just to C, the Point required.

Also Note, That if you want an Angle above 45 degrees, as the Sector stands, take the same from the small Tangent that pro∣ceeds to 75, and turn that Extent 4 times from B, and it shall give the Point required in the Line BC.

Use IX. To lay down, or protract any Angle by the Tangent of 45 only.

First, make a Geometrical Square, a ABCD, and let A be the Anguler Point; then making AB Radius, make AB a = Co-sine of the Angle you would have, and lay the Thred to the nearest distance, then the ND from the right Sine of the Angle to the Thred, shall be the Tangent re∣quired.

Example.

I make AB Radius a = in 50, the co-Sine of 40, then the = Sine of 40 shall be BE.

Again,

If I make AD equal to AB, the = co-Sine of 30, viz. 60; and then take out the = Sine of 30, and lay it from D to F, it shall be an Angle of 60 from AB, or 30 from D to F.

But by the Sector this is more easie;

For making AB, or AD, the side of the Square Radius, lay off the = Tangents of any Angle under 45 from B toward C, and the complements thereof above 45 from D, toward C, calling 40, 50; & 30, 60; and 20, 70; 10, 80, &c.

[illustration]

Use X. To take out readily, any Tangent above 45, by the Tangent to 45 on the Sector-side.

* 1.105Take the given Radius, make it a = in the co-Tangent of the Tangent required; then the = Tangent of 45, shall be the Tangent required.

Example.

I would have a Tangent to 80 degrees; take the given Radius, make it a = in 10, the complement of 80; then the = Tan∣gent of 45, shall be the Tangent of 80 re∣quired.

But if your Radius be so big, that you cannot enter it, then take the half, or a quarter of your Radius, and then = 45 will be the half, or the quarter of the Tan∣gent required.

Use XI. How to work Proportions, in Sines alone, by the Natural Sines.

There are 4 Varieties in this Work, that include all Proportions, viz.

1. When the Sine of 90 is the first term, then the work is thus;

Lay the Thred to the second term, count∣ed on the degrees from the Head,* 1.106 toward the loose-piece; and count the third term on the Line of Sines, from the Center down∣wards; and taking the nearest distance from thence to the Thred, and that distance measured from the Center downwards, on the Line of Sines, gives the 4th term re∣quired.

Example.

As Sine 90, to Sine 23-30;

So is 30, to 11-31.

Take the Latteral second term,* 1.107 make it a = Sine of 90; then take out the = third term, and measuring it from the Center, it gives the 4th term required.

2. When the Sine of 90 is the third term, then work thus;

Take the — Sine, of the second term, from the Center downwards,* 1.108 and make it a = Sine in the first term, laying the Thred to ND; then on the degrees, the Thred shall give the 4th term required.

Example.

As the Sine of 30, to 23-31;

So is the Sine of 90, to Sine of 52-56.

But by the Sector,

* 1.109Take the — Sine of the second, make it a = Sine of the first term; then take out = 90, and measure it from the Center, and it shall give the 4th term required. Example as before.

3. When the Radius, or Sine of 90, is in the second place, work thus;

* 1.110Take — 90 from a lesser Scale, as the uppermost Sine above the Center, or the Line of Right-Ascentions, or the Azimuth-Scale, or the like; and make it a = in the Sine of the first, laying the Thred to ND, then the = third term, taken and measured on the same Scale that 90 was taken from, shall give the 4th term required.

Example.

As — 90, on the Line of Right-Ascenti∣on, is to = 30;

So is = 20, to 43-12, measured on the same Line that 90 was taken from.

Or else secondly, work thus;

As — 30, to = 90;

So is — 20, to = 43-12.

By carrying the Compasses till it so stayes, as that the foot turned about, will but just touch the Thred, at the nearest distance.

Or else thus, thirdly;

By transposing the terms, when the third is not greater than the first: thus;

As the first, to the third;

So is the second term, to the 4th:

Where the Radius being brought to the third place, it is wrought by the second Rule, as before.

By the Sector.

Take a smaller — Sine of 90, make it a = in 30; then the = Sine of 20, taken and measured on the small Sine, gives 43-12, as before.

Again,

As — 90, to = 30;

So is = 20, to — 43-12.

Again,

As — 30, to = 90;

So is = 20, to = 43-12:

Lastly, by transposing.

As — 20, to = 30;

So is = Radius, to — 43-12; as be∣fore.

4. When Radius is none of the given terms.

Then when the first term is greater than the second and third,* 1.111 work thus;

Take the — second term, make it a = in the first, laying the Thred to the ND; then the nearest distance, from the third term, to the Thred measured from the Center downward, give the 4th Sine re∣quired.

Example.

As 20, to 12; so is 18, to 10-50.

By the Quadrant.

As — 12, to = 20;

So is = 18, to — 10-50.

When only the second term is greater than the first, then transpose the terms, and work as before: Or else use a double Ra∣dius, which is on this Instrument very easi∣ly done, having several Radiusses.

Or,

Lastly, use a Parallel entrance, or answer rather, as before, which being carefully wrought, will do very well.

By the Sector.

The same manner of work, is as before by the Quadrant, and the setting the Sector, is all one to the laying the Thred, as will be largely seen in all the following Propositions, wrought both by the Artificial and Natu∣ral Lines, of Numbers, Sines, and Tangents, as followeth.

Use XII. Having the day of the Month, or Suns place given, to find his Declination.

Lay the Thred on the day of the Month in the Kalender,* 1.112 and in the Line of de∣grees, on the Moving-leg, you have his De∣clination, either Northward, or Southward, according to the time of the year, counting from 60|0, toward the Head, for North-declination; or toward the End, for South-declination.

By the Artificial Sines and Tangents on the Edge of the Instrument.

Extend the Compasses from the Sine of 90, to the Sine of 23 degrees 31 minuts, the Suns greatest Declination: The same Ex∣tent applied the same way, from the Sine of the Suns place, or the Suns distance from the next Equinoctial-point, shall cause the Mo∣ving-point to fall, on the sine of the Suns declination; This being the general way of working.

Example.

The Extent from the sine of 90, to the sine of 23-31, shall reach from the sine of 30, to 11 deg. 31 min. the Suns declinati∣on, in ♉ Taurus 30 degrees from ♈ Aries, the next Equinoctial-point, and from 60 degrees, the Suns distance in ♊ Gemini 60 degrees, from ♈ 20 deg. 12 min. the Suns declination then. This being the manner of working by these Lines, by extending the Compasses from the first to the second term: I shall for the rest wave this large repetition of extending the Compasses, and render it only thus by the words of the Cannon-general in all Books;

As Sine 90, to Sine 23-31;

So is the Sine of 30, to Sine 11-31.

Lay the Thred to 23-31,* 1.113 on the degrees on the Moveable-piece, counted from the Head toward the End; then count the Suns place from the next Equinoctial-point, on the Line of Sines from the Center down∣wards, and take the ND from thence to the Thred; then this distance being measured from the Center downwards, shall be the sine of the Suns declination, required for that distance, from the next Equinoctial-point; (by the 1st Rule abovesaid).

By the Sector.

Take — 23-31, from the Sines, make it a = in the sine of 90; then the = sine of the Suns distance from the next Equinoctial-point, shall be the — sine of the Suns de∣clination; Example as before (Rule the 1st).

Use XIII. The Suns Declination being given, to find his true place or distance from ♈ or ♎, the two Equinoctial Points.

Lay the Thred to the Declination count∣ed in the degrees from 60|0,* 1.114 and in the Line of the Suns place, is his true place re∣quired.

Example.

When the Suns declination is 12 de∣grees Northward, the dayes increasing, then the Sun will be 31 deg. and 23 min. from ♈, or 1 deg. 23 min. in ♉, his true place required.

As Sine of 23-31, the Suns greatest de∣clination,* 1.115 to Sine of 90;

So Sine of 12-00, the Suns present de∣clination, to Sine of Suns distance from ♈ or ♎ 31-23.

Which, by considering the time of the year, gives his true place, by looking on the Months and Line of Suns place on the Qua∣drantal-side.

Take the — Sine of the present declina∣tion,* 1.116 make it a = Sine in the greatest de∣clination, laying the Thred to ND; and on the degrees the Thred shall give the Suns distance from ♈, or ♎, required. Example as before.

Make — Sine of the given Suns declina∣tion,* 1.117 a = Sine in the Suns greatest declina∣tion, then = Sine of 90, measured from the Center, is the = Sine of the Suns distance, from ♈ or ♎, required; or count 30 deg. for one sign, and the Center for the next E∣quinoctial-point,

and 90 for the two Tro∣picks of Cancer, and Capricorn. ♋. ♑.

Use XIV. The Suns place, or Day of the Month, and greatest Declination given; to find his Right Ascention from the same E∣quinoctial.

Lay the Thred to the day of the Month,* 1.118 or place given, and in the Line of the Suns Right Ascention, you have his Right Ascen∣tion in degrees, or hours and minutes, coun∣ting 4 minuts for every degree.

Example.

On the 9th of April, near night, the Sun being then entring ♉, the Suns Right Ascen∣tion will be 1 hour 52 min. or 28 degrees of Right Ascention, distant from ♈.

As the Sine of 90,* 1.119 to the Sine comple∣ment of the Suns greatest declination (or C.S.) of 23-31, counting back∣wards from 90, which will be at the Sine of 66-29′.)

So is the Tangent of the Suns distance from the next Equinoctial-point; to the Tangent of the Suns Right Ascen∣tion from the same Equinoctial-point.

* 1.120Take the — co-sine of the greatest de∣clination from the Center downwards, be∣ing the — sine of 66-29′. make it a = sine of 90, laying the Thred to ND; and note what degree and minuit it cuts, for this is fixed to this Proportion: Then take the Tangent of the Suns distance from the next Equinoctial-point, from the Center at 60|0, on the degrees toward the End, and lay it on the sines, from the Center downwards, and note the Point where it stayeth, for the ND from thence to the Thred, shall be the Tangent of the Suns Right Ascention re∣quired.

Note, That if the Suns distance from ♈, or ♎, be above 45 degrees, then the Tan∣gents on the loose-piece, are to be used in∣stead of the Tangents on the moveable-leg.

Or, by Sines only thus;

Or, Take — Sine of the present Suns declination, make it a = in the Sine of the Suns greatest declination, and lay the Thred to ND; then take = Co-sine of the Suns greatest declination, and make it a = in Co-sine of the Suns present declination, and lay the Thred to ND, and in the degrees it cuts the Suns Right Ascention, requi∣red.

Make — Co-sine of 23-31,* 1.121 viz. the right Sine of 66-29, a = sine of 90, then the = Tangent of the Suns distance from ♈, or ♎, is the = Tangent of the Suns Right Ascention from the same Point of ♈, or ♎; as at 30 from ♈, it is 28 degrees, or 1 hour and 52 minuts from ♈, (neer).

Use XV. Having the Suns Right Ascention, and greatest Declination, to find the Angle of the Ecliptick and Meridian.

As Sine 90, to Sine 23-31;* 1.122

So is the Co-sine of the Suns Right As∣cention, to the Co-sine of the Angle of the Ecliptick and Meridian.

Lay the Thred to 23-31,* 1.123 counted on the degrees from the Head; then count the Co-sine of the Right Ascention, from the Center downward, or the Sine from 90 upwards, and take the ND from thence to the Thred, and measure it from the Cen∣ter, and it shall reach to the Co-sine of the Angle required.

Example.

The Right Ascention being 30 degrees, or 2 hours, the Angle shall be 69-50.

* 1.124Make the — right sine of 23-31, a = sine of 90; then the = co-sine of 30, viz. = 60, shall make the — sine of 69-50, the Angle of the Ecliptick and Meridian.

Use XVI. Having the Latitude, and Declination of the Sun or Stars, to find the Suns or Stars Amplitude, at rising or Setting.

* 1.125Take the Suns declination, from the par∣ticular Scale of Sines, and lay it from 6, in the hour or Azimuth-line, and it shall give the Amplitude from South, as it is figured; or from East, or West, counting from 90; observing to turn the Compasses the same way from 90 or 6, as the declination is Northward, or Southwards.

Example.

The Suns declination being 10 degrees Northward, the Suns Amplitude, or Line, is 106-12, from the South, or 16-12 from the East-point.

* 1.126As co-sine of the Latitude, to S. 90;

So is S. of the Suns declination, to S. of the Amplitude.

* 1.127Take the — Sine of the Suns declina∣tion, make it a = in the co-sine of the La∣titude,

and lay the Thred to the nearest distance,* 1.128 and on the degrees the Thred shall shew the true Amplitude required.

Make the — right Sine of the Suns de∣clination, a = in co-sine latitude, then = 90, taken and measured from the Center, gives the Amplitude or Line.

Use XVII. Having the same Amplitude, and De∣clination, to find the Latitude.

As S. of the Suns Amplitude,* 1.129 to S. the Suns Declination;

So is S 90, to Co-sine Latitude.

Take the — sine of the Suns declinati∣on;* 1.130 set one Point in the Sine of the Suns Amplitude, lay the Thred to ND, and on the degrees it sheweth the complement of the Latitude required.

Example.

The Declination being 20 degrees, and the Amplitude 33-15, the complement of the Latitude will be 38-28 —, counting from the Head, toward the End.

Make the right Sine of the Suns Declina∣tion,* 1.131 a = sine in the Suns Amplitude; then the = sine of 90, shall be the — co-sine of the Latitude required.

Use XVIII. Having the Latitude, and Suns Declina∣tion, to find his Altitude at East or West, commonly called the Vertical-Circle; or Azimuth of East or West.

Take the Suns Declination from the par∣ticular Line of Sines,* 1.132 set one Point in 90 on the Azimuth-line, and lay the Thred to the ND, and on the degrees it sheweth the Altitude required; counting from 60|0 toward the End.

As S. latitude S. of 90;

* 1.133So S. of Suns declination, to S. Suns height, at East or West.

* 1.134Take the — sine of the Suns declinati∣on, make it a = in the sine of the latitude, and lay the Thred to ND, and on the de∣grees it shall shew the Suns Altitude, at East and West required.

Example.

Declination 10. Latitude 51-32; the Altitude is 12 degrees, and 50 minuts.

As — S. of the Suns Declination, to = S. of Latitude;* 1.135

So is the = S. of 90, to — S. of Ver∣tical Altitude.

Use XIX. Having the Latitude, and Suns Declinati∣on, to find the time when the Sun will be due East or West.

Having gotten the Altitude by the last Rule, take it from the particular Sine;* 1.136 then lay the Thred to the Suns declination, counted on the degrees; then setting one Point in the Hour-line, so as the other turn∣ed about, shall but just touch the Thred, and the Compass-point shall stay at the hour and minuit of time required.

As Tangent latitude, to Sine 90;* 1.137

So is the Tangent of the Suns declination, to Co-sine of the hour.

Or,

As sine 90, to Tangent Suns declina∣tion;

So is Co-tangent-latitude, to Co-sine of the hour from noon.

Example.

Latitude 51-32, declination 10, the Sun will be due East at 6-32, and West at 5-28.

Take the — Tangent of the Latitude (on the loose-piece,* 1.138 counting from 60 towards the moveable-leg; or else from 60|0, on the

moving-leg, or degrees, according as the Latitude is above or under 45 degrees) and lay it from the Center downwards, and note the Point where it ends. Then take from the same Tangent, the Tangent of the Suns de∣clination, and setting one foot in the Point last noted, lay the Thred to ND, then the = sine of 90, shall be the — sine of the hour from 6.

Or by the Sines only work thus;

Take the — sine of the Suns declination, make it a = in sine of the latitude; lay the Thred to ND, then take ND from the Co∣sine latitude to the Thred; then set one foot in the Co-sine of the Suns declination, lay the Thred to ND, and on the degrees it gives the hour from noon, as it is figured, or the hour from 6, counting from the head, counting 4 minuts for every degree.

Make the small Tangent of the Latitude, if above 45,* 1.139 taken from the Center, a = sine of 90; then the — Tangent of the Suns declination, taken from the same small Tangent, and carried Parallely till it stay in like Sines, shall be the Sine of the hour from 6.

Or, as before, by Sines only.

Make — sine Declination, a = sine La∣titude; then take = Co-sine Latitude, and make it a = Co-sine of the Suns Declinati∣on; then take = 90, and lay it from the Center, it gives the Sine of the hour from 6.

Use XX. Having the Latitude, and Suns Declina∣tion, to find the Ascentional Difference, or the Suns Rising and Setting, and Oblique Ascention.

Lay the Thred to the Day of the Month, (or to the Suns Declination, or true Place,* 1.140 or to his Right Ascention; for the Thred be∣ing laid to any one of them, is then also laid to all the rest) then in the Azimuth-line, it cuts the Ascentional difference, if it you count from 90, or the Suns Rising, as you count the morning hours; or his Setting, counting the afternoon hours.

The Oblique Ascention is found out for the six Northern signs,* 1.141 or Summer half-year, by substracting the Suns difference of Ascentions, out of the Suns Right Ascention. But for the other Winter-half year, or six Southern signs, it is found by adding the

Suns difference of Ascentions to his Right Ascention; this sum in Winter, and the remainder as above-said in Summer, shall be the Suns Oblique Ascention required.

* 1.142As Co-tangent Lat. to sine 90;

To is the Tangent of the Suns declinati∣on, to the sine of the Suns Ascentional difference.

* 1.143Take the — co-tangent latitude, from the loose or moveable-piece, as it is above or under 45 degrees, make it a = in sine 90, lay the Thred to ND, then take the — Tangent of the Suns declination from the same Tangents, and carry it = till it stay in the parts, that the other foot, turned about, will but just touch the Thred, which parts shall be the Sine of the Suns Ascentional dif∣ference required.

Or ••hus, by Sines only;

Make the — sine of Declination, a = Co-sine of the Latitude; lay the Thred to ND, then take the = sine of Latitude, make it a = in Co-sine of the declination, and lay the Thred to ND, and on the de∣grees it shall cut the Suns Ascentional-dif∣ference required; which being turned into time, by counting 4 minuts for every degree,

and added to, or taken from 6, gives the Suns Rising in Summer, or Winter.

Make the — Co-tangent Latitude,* 1.144 a = sine of 90; then take — Tangent of the Suns declination, and carry it = till it stay in like parts, viz. the Sine of the Suns Ascentional difference required.

Example otherwise;

As — sine 90, to = Tangent 38-28;

So is = Tangent of 23-31, the Suns greatest declination, to the — sine of the Suns greatest Ascentional dif∣ference, 33 deg. and 12 min.

Use XXI. The Latitude and Suns Declination given, to find the Suns Meridian Altitude.

When the Latitude and Declination is both alike, viz. both North, or both South; then substract the Declination out of the La∣titude, or the less from the greater, and the remainder shall be the complement of the Suns Meridian Altitude.

But when they be unlike, then add them together, and the sum shall be the comple∣ment of the Meridian Altitude: The con∣trary work serves when the complement of the Latitude and Declination is given, to find the Meridian Altitude.

Lay the Thred to the Declination, coun∣ted on the degrees from 60|0, the right way, toward the Head for North, and to∣ward the End for South declination.

Then,

Take the nearest distance, from the Cen∣ter-prick at 12, in the Hour-line, to the Thred; this distance measured on the Par∣ticular-line of Sines, shall shew the Suns Meridian-Altitude required.

Use XXI. The Latitude, and Hour from the mid∣night Meridian given, to find the Angle of the Suns Position, viz. the Angle be∣tween the Hour and Azimuth-lines in the Center of the Sun.

* 1.145As Sine 90, to Co-sine of the Latitude;

So is the Sine of the Hour from Mid∣night, to the sine of the Angle of Po∣sition.

Example.

As Sine 90, to Co-sine Latitude 38-28;

So is the Co-sine of the Hour from mid∣night, 120, for which you must use 60, to 32-34, the Angle of Position.

* 1.146Take the distance from the Hour to the 90 Azimuth on the Hour-line, and measure it in the particular sines, and it shall shew

the Angle of Position required. This holds in the Equinoctial.

Take — Co-sine Latitude,* 1.147 make it a = in sine 90; then take out the = Co-sine of the Hour from the Meridian, and it shall be the — sine of the Suns Position.

Make — Co-sine Latitude a = sine 90;* 1.148 then = Co-sine of the Hour, shall be — sine of the Suns Position.

Note,

The Angle of the Suns Position may be varied, and it is generally the Angle made in the Center of the Sun, by his Meridian or Hour-circle, being a Circle passing thorow the Pole of the World, and the Center of the Sun; and any other principal Circle, as the Meridian, the Horizon, or any Azimuth, the Anguler-Point being alwayes the Center of the Sun.

Use XXII. The Suns Declination given, to find the beginning and end of Twi-light, or Day-break▪

Lay the Thred to the Declination on the degrees, but counted the contrary way,* 1.149 viz. South-declination toward the 〈…〉〈…〉 North-declination toward the 〈◊〉〈◊〉; then take 18 degrees from the particul•••• 〈◊〉〈◊〉

Sines for Twi-light, or 13 degrees for Day-break, or clear light; Then carry this di∣stance of 18 for Twi-light, or 13 for Day-break, along the Line of Hours on that side of the Thred next the End, till the other Foot, turned about, will but just touch the Thred, then shall the Point shew the time of Twi-light, or Day-break, required.

Example.

The Suns Declination being 12 degrees North, the Twi-light continues, till 9 hours 24 minuts; or it begins in the morning at 38 minuts after 2; but the Day-break is not till 22 minuts after 3 in the morning, or 38 minuts after 8 at night, and last no longer.

To work this for any other place, where the Latitude doth vary, do thus;

Find the Hour that answers to 18 degrees of Altitude, in as much Declination the contrary way, and that shall be the time of Twi-light; or at 13 degrees for Day-break, according to the Rules in the 26th Use, where the way how is largely handled to the 33^d Use, both wayes generally.

Use XXIII. To find for what Latitude your Instrument is particularly made for;

Take the nearest distance from the Cen∣ter on the Head-leg,* 1.150 to the Azimuth-line on the moveable-leg; this distance measu∣red on the particular Scale of Sines, shall shew the Latitude required; or the Extent from 0 to 90, on the Azimuth-line, shall shew the complement of the Latitude, be∣ing measured as before.

Use XXIV. Having the Meridian Altitude given, to find the time of Sun Rising or Setting, true Place, or Declination.

Take the Suns Meridian Altitude from the particular Scale,* 1.151 and setting on Point in ☉ on the Azimuth-line; lay the Thred to the ND, and on the Hour-line it sheweth the time of Rising or Setting; and on the degrees, the Declination; and the rest in their respective Lines.

Example.

The Meridian Altitude being 50, the Sun riseth at 5, and sets at 7.

Use XXV. The Latitude and Declination given, to find the Suns height at 6.

Lay the Thred to the Day of the Month, or Declination,* 1.152 then take the ND from the Hour-point of 06, and 6 to the Thred, and that distance measured on the particu∣lar Scale of Sines, shall be the Suns Alti∣tude at 6 in Summer time, or his depression under the Horizon in the Winter time.

* 1.153As sine of 90, to sine of the Suns Decli∣nation;

So is sine Latitude, to sine of the Suns Altitude at 6.

* 1.154Count the Suns declination on the de∣grees from 90, toward the End, and there lay the Thred; then the least distance from the sine of the Latitude to the Thred, mea∣sured from the Center downwards, shall be the sine of the Suns Altitude at 6.

Make the — sine of the Declination a = sine of 90;* 1.155 then the = sine of the La∣titude, shall be the — sine of the Suns height at 6.

Example.

Latitude 51-32, Declination 23-31, the height at 6, is 18 deg. 13 min.

Use XXVI. Having the Latitude, the Suns Declina∣tion and Altitude, to find the Hour of the Day.

Take the Suns Altitude,* 1.156 from the parti∣cular Scale of Sines, between the Compasses; then lay the Thred to the Day of the Month, or Declination; then carry the Compasses along the Line of Hours, between the Thred and the End, till the other Point (being turned about) will but just touch the Thred, and then the fixed Point shall shew the true hour and min. required, in the Fore, or After-noon; if you be in doubt which it is, then another Observation pre∣sently after, will determine it.

Example.

May 10th, at 30 degrees of Altitude, the hour will be 32 minuts after 7 in the Morning, or 28 minuts after 4 in the After∣noon.

This Work being somewhat more diffi∣cult than the former, I shall part it thus;

1. First, to find the Hour the Sun being in the Equinoctial.

Take the — sine of the Suns Altitude,* 1.157

make it a = Co-sine of the Latitude; lay the Thred to ND, and on the degrees it shall give the Hour from 12, as it is figured, counting 15 degrees for an hour, or from 6, counting from the Head at 90.

Example.

Latitude 51-30, Altitude 20, the hour is 8 & 12′ in the forenoon, or 3-48′ in the afternoon.

The same by Artificial Sines & Tangents.

As Co-sine Latitude, to sine 90;

So is the sine of the Suns Altitude, to sine of the hour from 6.

* 1.158Make — S. ☉ Altitude, a = S. in ☉ Latitude; then take out = S. 90, and it shall be the — sine of the hour from 6.

2. The Latitude, Declination, and Alti∣tude given, to find the Hour at any time.

* 1.159First by the 25th Use, find the Suns Al∣titude or depression at 6; then in Summer-time, lay this distance from the Center downwards; and in Winter-time, lay it upwards from the Center toward the End of the Head-leg; and note that Point for that day, or degree of Declination; for by taking the distance from thence to the Suns Altitude, on the General Scale, you have

added, or substracted the Altitude at 6, to, or from the present Altitude.

(For by taking the distance from that noted Point, over, or under the Center, to the Suns present Altitude, you have in Summer the difference between the Suns pre∣sent Altitude, and his Altitude at 6. And in Winter you have the sum of the present Altitude, and the Altitude at 6.)

This Operation is plainly hinted at, in the 4th Chapter, and 9th and 10th Section, which being understood, take the whole Operation in shorter terms, thus;

Count the Suns Declination from 90, to∣ward the end, and thereunto lay the Thred; the nearest distance from the sine of the Latitude to the Thred, is the Suns height, or depression at 6: In Winter use the sum of, in Summer the difference be∣tween, the Suns Altitude at 6, and his pre∣sent Altitude; with this distance between your Compasses, set one Point in the co-sine of the Latitude; lay the Thred to ND, then take the ND from 90, to the Thred; then set one foot in the Co-sine of the Suns declination, and lay the Thred to ND, and on the degrees it gives the hour required; from 6 counting from 90, or from 12, as it is figured.

Example.

On April 20, at 30 deg. 20 min. of Al∣titude, Latitude 51-32, the hour will be found to be just 2 hours from 6, or just 8.

Again,

On the 10th of November, at 8 deg. 25 min. high, it is just 3 hours from 6, or 9 a clock in the forenoon, or 3 afternoon.

Or somewhat differing thus;

Take the — sine of the sum, or difference, of the Suns present Altitude, and Altitude at 6, and make it a = in the co-sine of the Latitude, and lay the Thred to the nearest distance; then take out the = Secant of the declination beyond 90^d, and make it a = sine of 90; and laying the Thred to the nearest distance, on the degrees it shall shew the hour from 6 required.

* 1.160First, by Use 25, find the Suns height at 6, or depression in Winter; then by the former 2d, find the sum or difference be∣tween the Altitude at 6, and the Suns pre∣sent Altitude; but if you have Tables of Natural Sines and Tangents; then in Win∣ter, add the Natural Sines of the two Alti∣tudes together; and in Summer, substract the lesser out of the greater, and find the Ark of difference more exactly.

Then,

As the Co-sine of the Latitude, to the Secant of the Declination (counted be∣yond 90, as much forward as from 90 to the Co-sine of the Suns Declina∣tion);

So is the Sine of the sum, or difference, to the hour from 6▪ required.

Or else ••hus;

As the Co-sine of the Latitude, to the Sine of the sum, or difference;

So is sine of 90, to a 4th.

Then,

As the Co-sine of the Suns declination, to that 4th;

So is sine 90, to the hour from 6.

By the Sector.

Take the — secant of the Suns declina∣tion, make it a = in the co-sine of the La∣titude; then take out the = sine of the sum or difference, and turn it twice from the Center lattera••ly, and it shall be the sine of the hour from 6, required.

Example.

April 20, the Suns Declination is 15 de∣grees; and the Suns Height at 6, then is, 11 deg. 42 min. now the Natural sine of 11-42, 20278, taken from the Natural sine of 30 deg. 20 min. 50502, the Suns present Altitude, the residue is 30224, the sine of 17 deg. 35 min. and a half.

Then,

The — Secant of 15 made a = sine of 38-28, and the Sector so set, the = sine of 17-35 ½, turned latterally twice from the Center, shall reach to 30, the sine of 2 hours from 6, the hours required.

Use XXVII. Having the Latitude, the Suns Declina∣tion, and Altitude, to find the Suns Azimuth.

* 1.161Take the Declination from the particular Scale of Sines, for the particular Latitude the Instrument is made for; Then, count the given Altitude on the degrees from 60|0 toward the loose-piece, and sometimes on the loose-piece also; and thereunto lay the Thred, then carry the Compasses, so set, a∣long the Azimuth-line on the right-side of the Thred in Northern-declinations, and

on the left-side in Southern-declinations, till the other foot, turned about, will but just touch the Thred; then the fixed-point shall stay at the Suns true Azimuth required.

Take two or three Examples.

1. First, When the Sun is in the Equi∣noctial and hath no Declination, then there is nothing to take between your Compasses, but just to lay the Thred to the Suns Alti∣tude, counted from 60|0 on the loose-piece toward the End: then 〈◊〉〈◊〉 the Azimuth-line, it cuts the Azimuth from the South re∣quired.

Example.

At 00 degrees high, the Azimuth is 90 from South; and at 10 degrees high, it is 77-5; at 20 high, the Azimuth is 62-45; at 30 degrees high, it is 43-30; at 34 de∣grees high, it is 32 degrees of Azimuth from South; and at 38-28 degrees high, it is just South.

2. Secondly, at 10 degrees of Declina∣tion Northward, and 20 degrees of Alti∣tude, take 10 degrees from the particular Scale, and lay the Thred to the Suns pre∣sent Altitude, as before, and carry the Com∣passes on the right-side of the Thred on the Azimuth-line, till the other foot, being turned about, will but just touch it; then

shall the Point rest at 80 degrees, 42 min. of Azimuth from the South.

3. But if the Declimation be the same to the Southwards, and the Altitude also the same; then carry the Compasses on the left-side of the Thred, on the Azimuth-line, till the other foot, turned about, will but just touch it, and you shall find the Point to stay at 41 deg. 10 min. the true Azimuth from the South required.

Note, That any thing, as thick as the Rule, laid by the Rule, and the Thred drawn over, it will keep the Thred steady, till you get the nearest distance more truly.

First, by the 18th Use, find the Suns Al∣titude in the Vertical Circle, or Circle of East and West, thus;

* 1.162Take the sine of the Suns Declination, and set one foot in the sine of Latitude, lay the Thred to ND, and in the degrees you shall have the Altitude at East and West re∣quired.

Which Vertical Altitude in Summer or Northern Declinations, you must substract out of the Suns present Altitude; or take the lesser from the greater, to find a dif∣ference; but in Winter, you must add this depression in the Vertical Circle, to the Suns

present Altitude to get a sum, which must be done on a Line of Natural Sines, or by the TABLE of Natural Sines, as be∣fore, in the Hour, by laying it over or un∣der the Center, and taking from that noted Point to the Suns present Altitude all that day. Then take the distance from the Cen∣ter to the Tangent of the Suns present Al∣titude on the loose-piece, which is the Se∣cant of the Suns present Altitude, and lay it from the Center on the Line of Sines, and note the place; then take the distance from 60, on the loose-piece, to the co-tangent of the Latitude (by counting 10, 20, 30, &c. from 60, toward the moveable-leg) be∣tween your Compasses; then setting one Point on the Secant of the Suns Altitude last found, and noted on the Line of Sines; and with the other, lay the Thred to the nearest distance, and there keep it, (by noting what degree, day of the month, or hour & minut, or Azimuth it cuts).

Then take the — distance on the Sines, from the sine of the Suns Vertical Altitude, to his present Altitude, for a difference in Summer; Or,

The distance from a Point made beyond the Center, (equal to the sine of the Suns Vertical depression) to the Suns present Al∣titude, for a sum in Winter.

Then having this — distance of sum or difference, for Winter or Summer, between your Compasses; carry one Point parallelly on the Line of Sines, till the other, being turned about, shall just touch the Thred at the ND, the place where the Point stayeth, shall be the Azimuth from East or West, as it is figured from the Center; or from North or South, counting from 90.

Which work in brief, may be sufficiently worded thus;

As — co-tangent of the Latitude, to the = secant of the Suns present Altitude, laying the Thred to ND;

So is the — sine of the sum, or difference, of the Suns present Altitude, & Verti∣cal depression in Winter, or the dif∣ference between his Vertical and pre∣sent Altitude in Summer; to the = sine of the Suns Azimuth, at that Alti∣titude and Declination.

Yet again, more short.

As — C.T. Lat. to = Sec. ☉ Alt.

So — S. of sum or difference, to = S. ☉ Azimuth.

But note, That in Latitudes under 45, when the complements of the Latitude are too large, then work thus;

As the — co-sine of the Suns Altitude, to = Tangent of the Latitude, taken from the degrees on the moveable-leg, laying the Thred to ND, then the — sine of the sum or difference carried parallelly, shall stay at the Suns Azimuth required.

If the Tangents are too small, on the Sector-side is a larger; and if the Sines are too great, on the Head-leg there is a less.

Find the Vertical Altitude by Use 18, and the sum or difference of the present and Vertical Altitude by the Table,* 1.163 or Line of Natural Sines, as before shewed; then the Canon or Proportion runs thus;

As the Co-sine of the Suns Altitude, to the Tangent of the Latitude;

So is the sine of the sum or difference, to the sine of the Azimuth, from East or West.

Or,

As Co-tangent Latitude, to Secant of the Suns Altitude;

So is the sine of the sum or difference, to the sine of the Azimuth.

* 1.164Make the — Secant of the Suns Alti∣tude, a = Co-tangent of Latitude; then the — sine of the sum or difference, shall be half the — sine of the Azimuth; or being turned twice from the Center, the whole sine.

Or else thus;

Make the — Tangent of the Latitude, a = Co-sine of the Suns Altitude; then the = sine of the sum or difference, shall be the — sine of the Azimuth, measured on the Sine, equal to the Radius of the Tangents first taken.

Example.

In Latitude 51-32, Declination North and South 13-15, the Vertical Altitude or Depression being 17-01, and the present Altitude 20; the Azimuth for South-de∣clination will be found to be 31-45, from South, the Depression at East and West be∣ing 17-01; and the sum of the present Al∣titude and Depression 39-25.

Again,

For North-declination, or Summer-time, the difference between the Vertical and pre∣sent Altitude, is 2-54; and the Azimuth from South, will be found to be 86 degrees and 15 minuts.

Use XXVIII. To make a Scale, whereby to perform all th••se Propositions, by the former Rules, agreeable to the Trianguler Quadrant, being added chiefly as a Demonstration of the Instrument, and former Operations.

First, Draw an Equilatteral Triangle, as ABC, at any largeness you please, by drawing first the Line AB; then take the Extent AB between your Compasses, set one Point in A, and with the other draw a touch of an Ark about C, then removing the Point to B, cross the former Arch in the Point C; then the drawing the Lines AC, and BC, will constitute the Equilatteral-Triangle. Then consider whether one Ra∣dius of your Scale shall be double, tripple, or quadruple one to the other, and accor∣dingly divide the Line AB into 3, 4, or 5 parts, as here it is into 3 parts, to make one double to the other, (and for Sea-Instru∣ments into 5 parts is best to make the Scales quadruple one to the other) whereof AD is one; Then make BH equal to BD, and AG and GI equal to DA; Also, make D 90, equal to DF, the nearest distance from D to AC; and D 9••▪ equal to DE, the nearest distance from D to CB. Also,

[illustration]

make BE the half of BH, and AF the half of AG. Again, make F 45, equal to FD, and E 45 equal to ED, at nearest di∣stance. Further, if you lay the Radius DF once from I; which is 60, it shall reach to 69-54 near C; and being repeated again, it shall reach to 75; for 1-0000 the Radius or Tangent of 45, once added to 1-732, the Tangent of 60, makes 2-732, the Tangent of 69-54; but if you add it twice, it makes 3-732, the Tangent of 75 just.

Then making DE Radius, describe the Circle 90 EI, and divide it into 180 equal degrees; Also▪ draw the lesser Circle 90 F to the Radius DF, then a Rule laid to the Center D, and every one of the 180 de∣grees, shall divide the Tangent Lines AC, and BC, into 180 degrees; and if you work right, you will meet with all the for∣mer Points, F, G, 45, I, 69-54, 60-45, H, and E, in their true places, as first drawn.

Also, Perpendiculers let fall from every degree in the Circle 90 EI to the Line DB, shall divide the Line of Sines, D 90, to the the greater Radius; and the like Perpendi∣culers from the degrees in the lesser Circle, to the Line DA, shall divide the lesser Line of Sines; Also, the Extent from the Center D, to the Tangent of any Ark or Angle in

the Line AC, counting from F, shall be the Secant to that Ark or Angle, to the les∣ser Radius; and the measure from the Cen∣ter D, to the Tangent of any Ark or Angle in the Line CB (but counted from E) shall be the Secant to that Ark or Angle, to the greater Radius.

This little Instrument thus made, and a Thred fastened at D, will perform any Pro∣position by the Rules here inserted, and is the very making of the Trianguler Qua∣drant; or you may put these Lines on a Rule as a plain Scale, and use them thus:

As for Example, for the Azimuth last treated on.

First draw a streight Line, as AB, re∣presenting the Line AB in the Trianguler-Quadrant; then appoint in that Line any Point for a Center, as C; then for this Pro∣position of finding the Azimuth, the Sines and Tangents being on a streight Scale, work thus;

First, to find the Suns Altitude, or De∣pression in the Vertical-Circle.

Take the Sine of the Latitude, and lay it from C to 51-30; then take out the Sine

of 13-15, between your Compasses, and setting one Point in the Point 51-30, last made in the Line AB, and strike the touch of the Arch at D, and draw the Line CD; also, on the Line CB, lay down from C the sine of 90 out of the Scale, then the nearest distance from the Point for 90 in CB, to the Line CD, shall be the Sine of the Suns Altitude in the Vertical, in Summer or Northern declination, or his depression in Winter, viz. 17-01.

Then, as before, on the Line of Sines, find a sum for Winter, or a difference in Summer, between the Vertical and present Altitude; Now supposing the Altitude 15, the sum is 33-30, or the difference is 1-58, which you must remember.

Then take the Secant of 15, the Suns present Altitude from the Scale, lay it from C to E; then take out the Co-tangent of the Latitude between your Compasses, set one Point in E, and strike the touch of an Ark, as at F, and draw the Line CF; then take the sine of 33-30, the sum, if it be Winter, or 1-58, if it be Summer, between your Compasses, carry one Point in the Line CB, higher or lower, till the other foot, turned about, will but just touch the Line CF; then the measure from thence to the Point C, shall be the Sine of the Azi∣muth

required, viz. in Winter 43-50; and in Summer 92-30, from the South, because the present Altitude is less than the Vertical, or East and West.

But when the Co-tangent of the Lati∣tude is too large for a Parallel entrance, then prick off first the Tangent of the Latitude, and take the Co-sine of the Suns Altitude to work in a Parallel way which will reme∣dy the inconveniences; Thus you see that by drawing three Lines only this work is done; yet not so soon by far, as by the In∣strument with the Thred, which represents those Lines more certainly and exactly, after the same way of Operation.

To find the Suns Azimuth in Southern Declinations.

As the Co-sine of the Latitude, to the Sine of the Suns present Altitude;* 1.165

So is the Sine of the Latitude to a 4th sine; which 4th sine is to be added to the Suns Amplitude, for that time, on a Line of Natural sines, and the sum observed, as a 5th.

Then,

As the Co-sine of the present Altitude, is to the sine of the sum last found;

So is the sine of 90, to the sine of the Suns Azimuth, from East or West, re∣quired.

For the Amplitude, work thus;

As Co-sine Lat. to S. Suns declination;

So is S. 90, to the sine of Amplitude.

Use XXIX. Having the Latitude, Suns Altitude, and Vertical Altitude, to find the Azimuth.

And first for Northern-Declinations.

* 1.166First, find the Vertical Altitude by the former Rule, and find the difference be∣tween it and the present Altitude, by the Line of Sines: then take this difference from the general Sines between your Compasses, and setting one foot in the Co-sine of the Latitude, lay the Thred to the ND, then take the ND from the sine of the Latitude to the Thred; having this distance, set one one foot in the Co-sine of the Suns Alti∣tude, and lay the Thred to ND, and on the degrees it shall shew the Suns true Azi∣muth at that Altitude and Declination re∣quired.

Example.

The Suns Declination being 7, the Vir∣tical Altitude is 8-57; the Suns present Altitude being 30, the difference or residue in Sines will be 20-13, and the Suns Azi∣muth found thereby will be 60-12′.

The same by Artificial Sines and Tan∣gents, in Summer.

As Co-S. Lat. to S. of residue;

So is S. 90, to a 4th sine.

Then,

As Co-S. ☉ Alt. to the 4th sine;

So is S. 90, to S. of ☉ Azimuth, from East or West.

Secondly, in Southern-Declinations, work thus;

First, find the Suns Amplitude for that Declination, thus; Take the — sine of the Declination, make it a = in the C••-sine of the Latitude; lay the Thred to ND, and on the degrees it gives the Suns Am••litude for that Declination, which you must re∣member.

Then,

Take the — sine of the Suns present Altitude, make it a = in the Co-sine of

Latitude, lay the Thred to the ND, then take the ND from the sine of the Latitude to the Thred, and as the Compasses so stand, set one foot in the sine of the Suns Amplitude first found, and turn the other foot onward toward 90; then take from thence to the Center. Thus have you added the Amplitude, and last found distance to∣gether on Sines, then this added Latteral-distance, must be made a Parallel in the Suns Co-altitude, and the Thred laid to the nearest distance in the degrees, gives the A∣zimuth required.

Example.

At 15 degrees of Declination, and 10 degrees of Altitude, the Azimuth will be found to be 49 degrees 46 minuts from the South, and the Amplitude 24-30 in 51-32 of Latitude.

The same, work by Artificial Sines and Tangents, in Winter.

As co-S. of Lat. to S. of ☉ present Alt.

So is S. of Lat. to a 4th; which you must add to the Suns Amplitude on Natural Sines, and keep it as a sum;

Then,

As co-S. of ☉ Alt. to S. of the sum;

So is S. 90, to S. Azim. from East or West.

Use XXX. Having the Latitude, the Suns Declina∣tion, his Meridian and present Alti∣tude given, to find the Hour.

Make the — Secant of the Latitude,* 1.167 a = in the Co-sine of the Suns declination, laying the Thred to ND (and note the place); then take the — distance on the sines, between the Suns Meridian and pre∣sent Altitude, and lay it from the Center toward 90; then the ND from that Point to the Thred (as before laid) shall be the versed Sine of the Hour, measured on a Line of versed Sines, equal in Radius to the Line of Secants first taken, as the Sines above 90 are.

Make the — Secant Latitude, a = Sine of Co-declination;* 1.168 then the — distance between the Suns Meridian and present Alti∣tude, laid on both Legs from the Center latterally, and the = distance between, measured on versed Sines, equal to the Se∣cants, shall give the hour required; as the great Line of Sines on the Sector are, by turning the Compasses twice, because the Line of Secants is half the Radius of those Sines, as at first was hinted.

Example.

Latitude 51-32, Meridian Altitude 50, present Altitude 40, Declination North 11-30, the Hour will be found to be 9-32 in the forenoon, or 2^h 28′ in the after∣noon.

Note, That if the Suns Meridian Alti∣tude be above 90, count the excess from 90 toward the Center, and take from thence to the present Altitude.

Use XXXI. To find the Suns Azimuth, by having the Latitude, Declination, and Suns present Altitude.

* 1.169First, make the — Secant of the Lati∣tude, a = Sine in the Suns Co-altitude, and lay the Thred to ND, and note the exact place where it is laid; then find by Addition the sum of the complements of the Latitude, and Suns Altitude; and observe whether the sum of them, be above or un∣der 90.

For when under 90, count it from 90 toward the Center, as is usual, and take the distance from thence to the Center, and substract this on the Line of Sines, out of the declination, or take the Sine of the decli∣nation,

out of this, (and lay the residue from the Center); and note the place.

Or shorter thus;

If the sum of the complements of the Co-latitude and Co-altitude be under 90, then count the same on the Line of Sines from 90, toward the Center, and take the di∣stance from thence to the Sine of the Suns Declination, and lay this from the Center; for the = distance from thence to the Thred (as first laid) shall be the versed Sine of the Azimuth from the South required.

Example.

Latitude	51	32
Declination	15	00
Altitude	50	00
Co-altitude	40	0
Co-latitude	38	28
Sum	78	28
Azimuth is,	33	0

But if the sum of the complements of the Latitude and Suns Altitude be above 90, then count from 90 to the Center as 90,

and reckon the excess above 90, from the Center toward 90, and take from thence to the Center, and add this distance on the Sines to the Suns declination toward 90, and take from thence the nearest distance to the Thred, and that shall be the versed Sine of the Suns Azimuth from noon.

But when the complements are under 90, then the ND from the noted place to the Thred, shall be the versed Sine of the Azimuth required.

But in Winter, when the sum of the complements are above 90, and are counted backwards, from the Center, towards 90; take the — distance from thence to the Sine of the Suns declination, the lesser from the greater, and set this distance, or residue, from the Center downwards; then the nearest distance from thence to the Thred, shall be the versed Sine of the Azimuth.

But when the Latitude is less than the Suns Declination, and the same way; then take the — distance (on the Sines) from the sum of the Suns Altitude, and Co-latitude, found by Addition, when under 90, and counted from the Center to the declination, and lay that from the Center, as before is shewed.

But if the sum of the Suns Altitude, and the complement of the Latitude, be above

90; then, having counted forwards from the Center to 90, count the excess from 90 toward the Center, and take the — distance from thence, to the Sine of the Suns declina∣tion, and lay it from the Center, as before; then the ND from thence to the Thred, shall give the versed Sine of the Suns Azi∣muth on the small Sines beyond the Center.

The very same manner of Operation that serves for the General-Quadrant,* 1.170 serves al∣so for the Sector, and this way being more troublesome than the rest, I shall say no more to it, but proceed to others.

Use XXXII. The Suns Altitude, the Latitude, and Declination given, to find the Hour.

Add the Co-latitude, Co-altitude, and Suns distance from the Elevated Pole toge∣ther, for a sum; and find the half sum,* 1.171 and the difference between the half sum and the Co-altitude.

Then say;

As Sine 90, to Co-sine Latitude;

So is the Sine of Suns distance from the Pole, to a 4th Sine.

Again,

As the 4th Sine, to the Sine of the half-sum;

So is the Sine of the difference, to the versed Sine of the Hour, if you have them on the Rule; if not, to a 7th Sine, whose half-distance on the Sines towards 90, gives a Sine, whose com∣plement doubled, and turned into time, is the Hour from South re∣quired.

Example, at 36 deg. 42 min. Altitude, and 23 deg. 31 min. Declination, La∣titude 51-32 North.

53-18, the Co-altitude; 38-28, the Co-latitude; and 66-29, added together, makes 158-15 for a sum; then the half-sum is 79-07, and the difference between 79-07 and 53-18, is 25-49 for a difference.

Then,

The Extent from sine 90, to the Sine of 38-28, will reach the same way from the sine of 66-29, to the sine of 34-47, for a 4th Sine.

Again,

The Extent from Sine 34-47, to Sine 79-7, shall reach the same way from the

Sine of 25-49, the difference, to the Sine of 48-34, a 7th Sine, right against which, on the versed Sines, is 60, viz. 4 hours from noon.

Or else,

The half-distance, between Sine 48-34, and the Sine of 90, is the Sign of 60 de∣grees, whose complement, viz. 30 doubled is 60 degrees, or 4 hours in time, from noon.

Use XXXIII. To find the Suns Azimuth, having the same things given, viz. Co-latitude, Co-altitude, and Suns distance from the Pole.

Add, as before,* 1.172 the three Nunbers toge∣ther, and thereby find the sum, and half-sum, and the difference between the half-sum, and the Suns distance from the Eleva∣ted Pole.

Then say,

As the sine of 90, to the Co-sine of the Latitude;

So is the Co-sine of the Altitude, to a 4th sine.

Again,

As the Sine of the 4th, to the Sine of the half-sum;

So is the Sine of the difference, to the versed Sine of the Suns Azimuth, from South, (or to a 7th sine, whose half-distance, toward 90, gives a sine, whose complement doubled, is the Azimuth from South).

Example, Latitude 51-32, Altitude 41-53, Declination North 13.

The 3 Numbers, viz. 38-28, 49-7, and 77-0, added together, makes 163-35; whose half is 81-47 ½, and the difference between the half-sum, and the Suns distance from the Pole, is 4-47 ½.

Then,

As sine 90, to sine 38-28;

So is sine 48-7, to sine 27-36.

Then,

As sine 27-36, to sine 81-47;

So sine 4-47 ½, to (V.S. of 130, the A∣zimuth from the North:) the sine of 10-15, a 7th sine, whose half-di∣stance toward 90, is 25, whose com∣plement 65 doubled, is 130, the Azi∣muth from the North, whose comple∣ment to 180, viz. 50, is the Azimuth from South.

Having the same complements, to find the Hour, and Azimuth, by the General-Quadrant and Sector; and first for the Azimuth.

First,* 1.173 of the complements of the Lati∣tude, and Suns present Altitude, by sub∣straction find the difference.

Secondly, Count this difference on the Line of Natural Sines from 90, toward the Center, as the smaller figures are counted.

Thirdly, Take the distance on the Sines, from thence to the — sine of the Suns de∣clination. But note, That when the Lati∣tude and Declination differ, viz. one North, and the other South, as it is with us in Winter; you must count the Suns Decli∣nation beyond the Center, and call it the Suns distance from the Elevated Pole, and take from thence.

Fourthly, Make this — distance, a = in the Co-sine of the Latitude, laying the Thred to ND, or keeping the Sector at that opening. Then,

Fiftly, Take out the = sine of 90, And

Sixtly, Make it a = sine in the Suns Co-altitude, setting the Sector, or laying the Thred to the (nearest distance) ND.

Seventhly, Take out the = sine of 90.

And,

Eightly, Measure it from the sine of 90, towards (and if need be beyond) the Center, and it shall reach to the versed sine of the Suns Azimuth from North or South, when you count from 90; or from East or West, if you count from the Center, on a Line of Sines, or middle of the Line of versed Sines.

Note, That if the general Sines are too big, you have a less adjoyning, whereon to begin and end the Work; as sometime the Hour-Scale, and sometimes the Line of Right Ascentions.

Example.

In the Latitude of 51-32, the Suns De∣clination 18-30, the Suns Altitude 48-12, you shall find the Suns Azimuth to be 130 from the North, or 50 from South.

Secondly, for the Hour, by the same data, or things given.

1. First of the complement of the Lati∣tude, and the Suns distance from the Eleva∣ted Pole, find the difference by Substracti∣on.

2. Count it on the Line of Sines from 90

toward the Center, (or beginning of the Sines).

3. Take the — distance from thence, to to the sine of the Suns present Altitude.

4. Make this — distance, a = in the Co-sine of the Latitude, setting the Sector, or laying the Thred to the ND, and there keep it.

5. Then take out the = sine of 90;

And,

6. Make that a = in the Co-sine of the Suns Declination, laying the Thred to ND.

7. Then take out the = sine of 90 again.

And,

8. Measure it from 90, toward the Cen∣ter, and it shall shew the versed sine of the Hour from Mid-night, or the contrary from noon; or from 6, if you count from the Center of the Sines, or the middle on versed Sines.

Example.

Latitude 51-32, Declination North 20-14, Altitude 50-55, you shall find the Hour to be 150 from North, viz. 10 in the fore-noon, or 30 degrees short of South.

Use XXXIV. Having the Latitude, Suns Altitude, and distance from the Elevated Pole, to find the Hour, by the Line of versed Sines, on the Sector.

First, By Addition, find the sum of, and by Substraction, the difference between the complement of the Latitude, and the Suns distance from the Elevated Pole.

Secondly, Count this sum and difference from the Center, or the versed Sines on the Sector, (or the beginning of the Azimuth-Line, if you use that, or any other, which is not drawn from a Center) and with Com∣passes take the — distance between them.

Thirdly, Make this — distance, a = versed Sine of 180.

Fourthly, Take the — distance between the versed sine of the sum, and the comple∣ment of the Suns Altitude, and carry parallelly till it stay in like versed Sines, which shall be the versed Sine of the Hour from the North Meridian, or mid-night.

Or,

If you take the — distance from the difference to the Co-altitude, and carry that = till it stay in like sines, it shall be the hour from noon; counting the Center 12

at noon, the middle at 90, the two sixes and 180 at the end, for 12 at night.

Use XXXV. Having the Latitude, the Suns Altitude, and Distance from the Elevated Pole, to find his true Azimuth from South or North, by Natural versed Sines.

First, Of the Co-altitude, and Co-lati∣tude, find the sum and difference, by Ad∣dition and Substraction.

Secondly, Count the sum and difference from the Center, and take the — distance between them with Compasses on the versed Sines.

Thirdly, Make it a = versed sine of 180, and so keep the Sector.

Fourthly, Take the — distance, be∣tween the sum, and the Suns distance from the Pole, (counting the Center the Eleva∣ted Pole, and 90 the Equinoctial) and carry it = till it stay in like parts, which shall be the Azimuth from South.

Or,

If you take the — distance from the difference, to the Suns distance from the Pole, and carry it as before, it shall stay at the versed sine of the Azimuth, from the North part of the Horizon.

These five general wayes of finding the Hour and Azimuth, are not all needful to be learned by every one, but to delight the ingenious, and to hold forth the usefulness of the Instrument, and to supply defects that at some times may happen by Excursi∣ons, and as a four-fold Testimony, to shew the harmony in several wayes of Operation; the first particular way, and this last by versed Sines, being most easie and compre∣hensive of any other.

Use XXXVI. To work the last without the Line of versed Sines.

Note, That if for want of room, the versed Sines be set but on one Leg, then it is to be laid at the nearest distance instead of like parts, after the manner of using the Thred on the General Quadrant.

Also, If you have it not at all, then the Azimuth-line for the particular Latitude; and if that be too large, the little Line of Sines beyond the Center, will supply this defect very well thus;

First, Turn the Radius, or whole length of that Line of Sines, two times from the Center downwards, (which in Sea-Instru∣ments, will most conveniently stay at 30 on

the large Line of Sines, or general Scale, as was hinted in the 28th Use, being just 4 times as much one as the other). For a Point representing 180 of versed Sines, to set the Compasses in, when you lay the Thred to ND, and to take any versed sine above 90 degrees; this being premised, the Operati∣on is thus:

Example.

Lat. 51-32, ☉ Dist. from Pole 80, ☉ Alt. 25, to find the Hour;* 1.174

The sum of Co-lat. 38-28, and 80, is 118-28;

And the difference is 41-32.

Now in regard the sum is above 90, count the Center 90, 10 on the smaller Sines 100, and 20 on the same Sines, 110, and 28 deg. 28 min; 118 deg. 28 min. turn this distance the other way from the Center downwards, and note that place, for the Point, representing the sum on the ver∣sed sines.

Then,

The — Extent between this sum, and the difference 41-32, as the smaller figures reckon it, being taken between your Com∣passes, set one Point in 180, the Point first found, and lay the Thred to ND, and there keep it (or observe where it cuts), then ta∣king

the — distance between the versed sine of the difference, counted as the small figures are reckoned, and the sine of the Suns Altitude 25, as the greater figures are reckoned from the Center toward the End; and carrying this Extent parallelly along the greater Line of sines, till the other Point will but just touch the Thred at ND; Then, I say, the measure from that Point to the Center, measured on the small sines, as versed sines, shall be the versed sine of the Hour required, viz. 62 from South, or 7 hours 52 minuts from mid-night.

This Rule, or Use, is longer far in word∣ing, than the Operation need be in work∣ing; for if you shall approve of this way, the adding of two brass Center-pins will shew you the two Points most used very readily, and the Thred is sooner laid, than the Legs can be opened or shut, and the In∣strument keeps its Trianguler form as it is in, during the time of Observation.

Use XXXVII. Having the Latitude, Suns Declination, and Hour, to find his Altitude.

This Problem being not of such use as the contrary, viz. having the Altitude, to

〈◊〉〈◊〉 the hour, it shall suffice to hint only two ••ayes, the most convenient. And,

First by the Particular Quadrant.

Lay the Thred to the Day, or Declination, then the ND from the Hour to the Thred, measured in the particular Scale of Alti∣tudes, shall shew the Suns Altitude re∣quired.

Secondly, by the versed Sines.

1. First, of the Co-latitude, and Suns distance from the Pole, find the sum and difference.

2. Take the — distance between them, and make it a = versed sine of 180, by set∣ting the Sector, or laying the Thred to ND.

3. Then take the = versed sine of the Hour, and lay it latterally from the sum, and it shall give the complement of the Al∣titude required.

This work is the same, both by the Sector, or General Quadrant, as is shewed in Use ••he 36th, and is nothing else but a back∣ward working; but the Altitude at any Azimuth, is not so to be done.

To do the same by the Natural-Sines.

First, having the Latitude, and the Suns Declination, find the Suns Altitude, or Depression at 6; and note the Point, either below, or above, or in the Center, as is large∣ly shewed in Use the 26th, where the Al∣titude is given, to find the hour in any La∣titude.

Then,

Lay the Thred to the Hour, counted in the degrees either from 12, or 6;

Then,

Take the ND from the Co-sine of the Suns Declination, and make it a = in the sine of 90, laying the Thred to the ND; then the ND from the sine complement of the Latitude to the Thred, shall reach from the noted Point, for the Suns Altitude or Depression at 6, to the Suns Altitude re∣quired.

Example.

Latitude 51-32, Declination 23-71, a 8 or 4, viz. 2 hours from 6 Southwards the Altitude will be found to be 36-42′.

1. For the Altitude at 6, at any time of the year, say;

As the sine of 90;* 1.175 to sine of the Lati∣tude;

So is the sine of the Suns Declination, to the sine of the Suns Altitude at 6.

2. For the Suns Altitude, at any hour or quarter, in Aries or Libra, (the Equi∣noctial).

As sine 90, to Co-sine Latitude;

So is the sine of the Suns distance from 6 in degrees, to sine of the Suns Alti∣tude.

3. For the Suns Altitude at all other hours, or times of the year.

As sine 90, to Co-tangent Latitude;

So is sine of the Suns distance from 6, to the Tangent of a 4th Ark, in the Tangents.

Which 4th Ark being taken from the Suns distance from the Elevated Pole, then the residue is the 5th Ark; but for hours before and after 6, add the 4th Ark, and the Suns distance from the Pole together, to make a 5th Ark.

Then say,

As the Co-sine of the 4th Ark, to the sine of the Latitude;

So is the Co-sine of the residue (or sum) being the 5th Ark, to the sine of the Suns Altitude at that hour.

Use XXXVIII. The Latitude, Suns Azimuth and De∣clination given, to find the Altitude, or height thereof.

First, to find the Suns Altitude at all Azi∣muths in the Equinoctial.

* 1.176As sine 90, to Co-tangent Latitude;

So is the Co-sine Azimuth from South, to the Tangent of the Suns Altitude in Aries.

Or,

As sine 90, to the Co-sine of the Azi∣muth from South;

So is Co-tangent Lat. to the Tangent of the Suns Altitude, at that Azimuth in the Equinoctial, which you must ga∣ther into a Table for every single de∣gree.

Then,

As the sine Lat. to the sine of the Suns Declination;

So is the Co-sine of the Suns Altitude in Equinoctial, to the sine of a 4th Ark.

Then,

When the Latitude and Declination are alike, as both North, or South; then add the 4th Ark and the Altitude (in the Equa∣tor) together, and the sum is the Altitude required.

But in Winter-time, when the Latitude and Declination is unlike, take the 4th Ark out of the reciprocal Altitude in the E∣quator, and the residue is the Suns Alti∣tude required.

Also, in all Azimuths from East and West Northwards, in Summer-time also, you must use Substraction also, and not Additi∣on; as the Rule before-going suggests.

By the Particular Quadrant, work thus;

Take the Sun or Stars Declination from the particular Scale, and setting one Point in the Suns Azimuth, on the Azimuth Line, and with the other lay the Thred to the ND, the right way, and on the degrees the Thred cuts the Altitude required.

By the General Quadrant.

As the — Co-tangent Latitude, taken from the Moving-leg, or Loose-piece, to = sine of 90, laying the Thred to ND;

So is the = Co-sine of the Suns Azi∣muth from South, to the — Tangent of the Suns Altitude in the Equator, at that reciprocal Azimuth.

Which being remembred, or gathered into a Table together, then say;

As the — Co-sine of the Suns Altitude in the Equator, to the = sine of the Latitude, laying the Thred to the ND;

So is the = sine of the Suns Declination, to the — sine of the 4th Ark.

Which 4th Ark is to be added, or sub∣stracted, as immediately before is directed, and the sum or residue, shall be the true Al∣titude required.

Example.

At 60 degrees of Azimuth from South, the Equinoctial Altitude will be found to be 21-40, for London latitude of 51-32; and the 4th Ark in ♋ or ♑ is 28-16.

Then,

21-40, the Suns Altitude at 60 in ♈, and 28-16, the reciprocal 4th Ark in ♋ added, makes 49-56, the Suns Altitude at 60 de∣grees from the South in ♋.

The same way of working serves for the Sector, as is used for the General Quadrant,* 1.177 only observing to set the Sector, instead of laying the Thred to the nearest distance, as the Ingenious will soon perceive.

Use XXXIX. Having the Latitude, Declination, Azi∣muth, and Altitude, to find the Hour.

As the — Co-sine of the Suns Altitude,* 1.178 to = Co-sine of the Suns Declina∣tion;

So is the = sine of the Suns Azimuth, to — sine of the Hour.

Or else thus;

First find the Altitude, at that Azimuth; and then at that Altitude, and Declination,* 1.179 the Hour.

As — Co-sine of Declination, to = sine of the Azimuth;

So is the — Co-sine Altitude, to = sine of the Hour.

* 1.180As Co-sine Declination, to the Sine of the Azimuth;

So is Co-sine Altitude, to Sine of the Hour.

Use XL. Having the Latitude, Declination, Hour, and Altitude, to find the Azimuth.

* 1.181As — Co-sine of Declination, to = Co-sine of the Suns Altitude;

So is = sine of the Hour, to — sine of the Azimuth.

* 1.182First, find the Altitude at that Hour, and then the Azimuth at that Altitude, as before.

* 1.183As Co-sine of Altitude, to sine of the Hour;

So is Co-sine Declination, to sine of the Azimuth from South, or North, as the Hour is counted; that is to say, from South, if the Hour is between 6 at morning, and 6 at night; and from the North if the contrary; that is to say, between 6 at night, and 6 next morning, or next to midnight.

Use XLI. Having the Latitude, and the Suns De∣clination, to find the Suns Azimuth at 6.

As the sine of 90,* 1.184 to the Co-sine of the Latitude;

So is the Tangent of the Suns Declinati∣on, to Co-tangent of the Suns Azimuth from the North, at the hour of 6.

First find the Suns height at 6, and then the Suns Azimuth at that Altitude.* 1.185

Make the — Tangent of the Declina∣tion, a = sine of 90,* 1.186 laying the Thred to ND, then the = Co-sine Latitude shall be the — Co-tangent of the Suns Azimuth from the North at 6.

Use XLII. To find the Amplitude, Azimuth, Ri∣sing, Setting, and Southing of the fixed Stars, having the Latitude, Altitude, and Declination, or time of the year given.

First for the Amplitude,* 1.187 Take the Stars Declination, out of the particular Scale of Altitudes, and measure it from 90 in the

Azimuth-line; and count the same way, and the other Point shall shew the Stars Amplitude required.

Example.

The Declination of the Bulls Eye, being 15-48; if you take 15-48 from the parti∣cular Scale, and lay it from 90 in the Azi∣muth-line, it will reach to 26 degrees, counting from 90 towards either end, the same as for the Sun in Use 16. But in other Latitudes, work as you do for the Sun by the Rules in the 16th Use abovesaid.

For a Stars Azimuth.

The work here is the same as for the Sun, thus; Take the Stars Declination from the particular Scale of Altitudes, or Sines, be∣tween your Compasses, and lay the Thred to the Stars Altitude, counted from 60|0 toward the Loose-piece; then carry the Compasses, or the right-side of the Thred, for Northern-stars; and on the left-side for Southern-stars, along the Azimuth-line, till the other foot, being turned about, will but just touch the Thred; then the fixed Point on the Azimuth-line shall shew the Stars Azimuth, from the South, required.

Example.

The Bulls Eye being 30 degrees high, shall have 77 degrees and 10 minuts of A∣zimuth from the South▪

If you be in other Latitudes, use the ge∣neral wayes, as for the Sun in all respects, having the same Declination that the Star hath North or South.

To find the Stars Rising, or Setting.

Count the Stars Declination on the de∣grees, as you count the Suns, North, or South, and there lay the Thred; and in the Line of Hours is the Stars Rising, or Set∣ting, when the Stars Right Ascention and Declination are equal.

But at other times, you must reckon thus;

First, find the Suns Right Ascention, by Use 14, and set down the complement thereof to 12 Hours, and the Stars Right-Ascention, and the hour of Rising the Thred cuts, and add them into one sum, and the sum, if under 12, is the time of his Rising in common hours; or if you add the hour of Setting that the Thred sheweth, it shall give his setting.

Example.

If you lay the Thred to 15-48, the De∣clination of the Bulls Eye, in the Hour-line it cuts 4 hours 36 min. for Rising; or 7-24, for his Setting; then if you work, for April the 23d, the Suns Right Ascenti∣on, then is 2-44, and the complement there∣of to 12, is 9-16; and the Stars Right-Ascention is 4 hours and 16 minutes; and the Hour cut, is 4-36 for Rising; and the three Numbers, viz. 9-16, the complement of the Suns Right Ascention, and 4-16, the Stars Right Ascention, and 4-36, the Hour of Rising the Thred cuts, being added, makes 18-8; from which, taking 12, rest 6-8, the time that the Bulls Eye Riseth on April 23; and if you add 7-24, the time of Setting that the thred cuts, there comes forth 8-56, viz. one hour and 32 min. after the Sun.

To find the time of a Stars coming to South.

Substract the Right Ascention of the Sun, from the Right Ascention of the Star, in∣creased by 24, when you cannot do with∣out, and the remainder, if less than 12, is the time between 12 at noon, and 12 at

night; but if the remainder be more than 12, it is the time between mid-night, and mid-day, following.

Example.

The Lyons-Heart, whose Right Ascenti∣on is 9-50, will come to the South on March 10, at 9-48, the Suns Right Ascen∣tion, being then only 2 minuts.

By the Line of 24 hours (say, or) work thus;

Extend the Compasses from the Suns Right Ascention, to the Stars Right Ascen∣tion; that distance laid the same way from 12 at the middle, or at the beginning, shall reach to the time of the Stars coming to South.

To find the time of the Stars continuance above the Horizon.

First, find what the Suns semi-diurnal-Ark is, having the same declination, and that doubled, is the whole time of continu∣ance; Or, if you shall add and substract it to, or from the time of the Stars coming to South, you shall find the time of Setting or Rising.

Or else,

By laying the Thred to the Stars Decli∣nation, it sheweth the Ascentional difference

in this Latitude, which added in those Stars that have North declination, or sub∣stracted in Southern to 6 hours, gives the semi-diurnal Ark of the Star above the Ho∣rizon.

Example.

The Bulls Eye's Ascentional-difference, is one hour and 24 minuts; which added to 6 hours, because of Northern declination, makes 7-24, for the semi-diurnal-Ark, or 14^• 48′, for the whole time of being above the Horizon.

Note, That to work this for other Lati∣tudes, the Suns Ascentional-difference is to be found for that Latitude you are in, and the Operation is general for all places.

To find a Meridian Line by the Sun.

On any flat Horizontal-Plain, set up a streight Wyre in the Center of a Circle; or hold up a Thred or Plummet, till the shadow of the Thred cut the Center, and any where in the Circumference, which two Points you must note; then immediate∣ly take the Suns Altitude, and find the Suns Azimuth, and count so many degrees in the Circle the right way, as the Suns Azi∣muth comes to, from the Points of the sha∣dow marked in the Circumference, and draw that Line for a true Meridian-line. This Work is best done before 10 in the

morning, and after two afternoon; or in the night, by two Plumb-lines, set in a right-Line with the North-Star, at a right scitua∣tion.

Use XLIII. To find the Hour of the Night by the Fixed Stars.

First, find the Stars Altitude, by looking along the Fixed or Moveable-leg, to the middle of the Star, letting the Thred, with a weighty Plummet, play evenly by the de∣grees, between your Thumb and Fore∣finger, to the end you may command the Thred, and know whether it playeth well or no by feeling.

Then,

Take the Altitude found, from the par∣ticular Scale of Sines, and laying the Thred over the Stars declination, which for readi∣ness sake is marked with 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, according to the Figures set to the 12 Names of the 12 Stars on the Rule; and then carrying the Compasses as you do in finding the hour by the Sun, you shall find how much the Star wants, or is past the Meridian, which is called the Stars-Hour; And note, That if the Star be past the South, it is an aft••rnoon hour; if not

come to the South a morning hour, which you must remember.

Also, knowing the Suns Right Ascention, set one Point of the Compasses in the Suns Right Ascention, (counted in the Line of twice 12, or 24 hours, on the outward-leg of the fixed-piece, next to the particular Scale of Sines) and open the other to the Stars Right Ascention, noting which way you turn the Compasses; for the same Ex∣tent, applied the same way, from the Stars hour last found, shall shew the true hour of the night required.

Example.

Suppose on the 10th of Ianuary, I should observe the Altitude of the Bulls Eye to be 20 degrees; if you take 20 degrees, the Al∣titude, from the particular Scale, and lay the Thred on 15-48, the Stars declination Northward, and measure from the Hour-scale the nearest distance to the Thred, you shall find the Compass-point to stay at 6-49 on the East-side of the Meridian; (suppose)

Also, The Suns Right Ascention, the same day, is 8 hours and 12 minuts.

Then,

The Extent from 8 hours 12 minuts (on the Line of twice 12 hours) the Suns Right Ascention, to 4-16, the Stars Right Ascen∣tion,

shall reach the same way from 6-49, the Stars hour, to 2-53, the true hour.

Use XLIV. To find the Hour of the Night by the Moon.

First, by an Almanack, or Ephemerides, find the Moons Age, and true Place for the present time; then, by laying the Thred on the Moons place, you may have her Right Ascention, and also the Suns Right Ascen∣tion; and by the Moons Altitude, taken from the particular Scale, and the Thred laid over the Moons place, you find what the Moon wants, or is past coming to South, which is called the Moons hour.

Then, by the Line of 24 Hours, say;

As the Suns Right Ascention, is to the Moons Right Ascention;

So is the Moons hour last found, to the true hour.

Example.

Suppose that on the 8th of Ianuary, a∣bout 40 min. after 3, there is a New Moon; then note, That the Suns true place, is the Moons true place; and consequently, their Right Ascentions; and the Moons Hour and Altitude is the same with the Suns.

Therefore,

As 8 hours 04 min. the Suns Right As∣cention, is to 8-04, the Moons Right Ascention;

So is the Moons hour at any Altitude, to the Suns true hour.

Again,

Suppose that on the 1st Quarter-day, the Moon being gone 90 degrees from the Sun, to find her place;

Then do thus;

Set one Point in the Moons place the Change-day, and open the other to the be∣ginning or the end of the Line of 24 hours,

Then,

The same Extent applied the contrary way from 6 hours, or 7 dayes and a half, the Moons Age, shall give 28 deg. 58 min. ♈; to which you must add 7 degrees and 30 minuts (the Suns place) between, and the sum shall be the Moons true place re∣quired, viz. 6-28 degrees in ♉.

Example.

If the Moon Change on the 8th day, the First Quarter being 7 dayes and a half af∣ter, will be on the 15th day later at night; then the difference between the Sun and Moons Right Ascention, will be found to be near 6 hours; for the Suns Right Ascention,

Ianuary 15, is 8-32; and the Moons Right Ascention, the same day, being about 8 de∣grees and a half in ♉, is 2 hours and 28 minuts; if you take the distance between them, on the 24 hours, it is near 6 hours; which is the difference of time between the Moon and the Suns hour.

Again,

For the Full Moon; on the 22 day, near 4 hours after noon, the Moons Age being 14 dayes ¾; if you add 12 hours, or 6 signs, to the Moons place a•• the Change, you shall find ♋ 29-0; to which if you add 14-45, the dayes between the New and Full, you shall find ♌ 13 deg. 45 min. for the Moons place; the Suns Right Ascention the 22 day is 9 hours, and the Moons the same day at 1 afternoon, is 9 hours also (or rather 12 difference) so that the Suns hour and the Moons is equal; only one is North, and the other South.

Again,

For the Last Quarter 22 ¼ dayes, or 18 hours added, and 22 degrees also together, makes ♏ 22 deg. 11 min. for the Moons place, by help of which, to find the Moons hour by her Altitude above the Horizon found by observation.

Or,

Without regarding the Sun or Moons Right Ascention, having her true Age, and Hour,

Say thus;

As 12 on the Line of 24 hours, is to the Moons Age in the Line of her Age;

So is the Moons hour, to the true hour.

For,

The Extent from 12 in the middle, to the Moons Age under or over the middle, shall reach the same way, on the same Line, from the Moons hour, to the true hour.

The like work serves to find the hour of the night by any Planets, as Saturn, Mars, or Iupiter, which are seen to shine very brave and bright in Winter evenings; and having learned their Place by their distance from the fixed Stars, or by the Ephemerides, then their Altitude and Place will find their hour from the Meridian, and the compa∣ring their Right Ascentions with the Suns, gives the true hour, as before, in the Fixed-Stars.

Use XLV. To find the Moons Place and Declination, without the Ephemerides, some∣what near.

First, observe when the Moon is in the Meridian, and then find her Altitude, and take the same from the particular Scale be∣tween your Compasses; then set one Point in the hour 12, and lay the Thred to ND, and on the degrees it shall shew the Moons declination; and in the Line of the Suns Place, the Moons present Place, counting her Progress orderly from the last Change-day, or New Moon, when she was with the Sun.

Otherwise thus;

Observe what Hour the Moon sheweth on any Sun-dial, at the same instance by the Fixed Stars, or other wayes, find the true Hour;

Then,

The Extent from the Moons Hour, to the the true Hour, shall reach the same way from 12, to the Moons Age, right against which is her coming to South, at which time you may find her true Altitude, and so come by her Declination.

Yet again, for her Age and Place, accor∣ding to Mr. Street, and Mr. Blundevil.

Add the Epact, the Month, and Day of Month in one sum, counting the Months from March, by calling March the first Month, April the second, &c. then that sum, if under 30, is the Moons Age; but if the sum be above 30, then substract 30, and the remainder is the Moons Age, when the Month hath 31 dayes; but if the Month hath but 30, or less than 30 dayes, then sub∣stract but 29, and the remainder is the Moons Age.

Or thus;

Add to the Epact for the present year, and in Ianuary 0, in February 2, in March 1, in April 2, in May 3, in Iune 4, in Iuly 5, in August 6, in September 8, in October 8, in November 10, in December 10; and the sum, if under 30, or the excess above 30, added to the day of the Month, abating 30, if need be, gives the Moons Age that day; but substracted from 30, leaves the day of her Change in that Month, or from the 〈…〉〈…〉 ••onth.

Example. July 10. 1668.

The Epact that year is 26, and the Num∣ber for Iuly is 5, the Excess above 30, is 1; which added to any day of the Month as to 10, gives 11, for the Moons Age, Iuly 10. 1668.

Then for the Moons Place.

Multiply the Moons Age by 4, and the Product divided by 10, the Quotient giveth the signs; and the remainder multiplied by 3, gives the degrees, which you must add to the Suns place that day, to find out the Moons place for that day of her Age.

Example.

On Iuly 10. 1668, the Moons Age is 11, which multiplied by 4, makes 44; and 44 divided by 10, gives 4 signs in the Quoti∣ent; and 4, the remainder, multiplied by 3, makes 12 degrees more; which added to Cancer, 29 degrees, the Suns place on the 10th day of Iuly, makes 11 degrees in Sagittarius, the Moons place the same day, propè verum.

Or rather by the Rule thus, on the Line of 24 hours by particular Scale, having the Moons place, to find her Age by the Line of 24 hours.

The Extent from the Suns true place, to the Moons true place, shall reach the same way, from 0 day, to the day of her Age.

Or contrarily, having the Moons true Age, to find her true Place.

The Extent from 0 day old, to the Moons true Age, shall reach the same way from the Suns true Place to the Moons.

Or, having the Moons true Place at the New Moon, to find her Place any day of her Age after.

The Extent from ♈, to the Moons true Place at the Change, shall reach the same way, from the day of her true Age, to her true Place, adding as many degrees to the Number found, as the Moon is dayes old.

Then,

Having her Place, and Age, it is easie to find the Moons Hour, and then her true Hour; but I fear I spend herein too much time on an uncertain subject.

Use XLVI. The Right Ascention and Declination of any Star, with the Suns Right Ascen∣tion, and the Hour of the Night given, to find the Altitude and Azimuth of that Star, and thereby to know the Star, if you knew it not before.

Set one Point of the Compasses in the Stars Right Ascention, found in the Line of twice 12 hours; and open the other to the Suns Right Ascention, found in the same Line; then this Extent shall reach, in the same Line, from the true hour of the Night, to the Stars hour from the Meridian; then laying the Thred to the Stars Declination, the ND from the Stars hour, in the Line of hours, to the Thred, measured on the par∣ticular Scale of Altitudes, gives the Stars Altitude; then by his Declination and Al∣titude, you may soon find his Azimuth, by Use 27.

And if the Instrument be neatly fixed to a Foot, to set North and South, and turn to any Azimuth and Altitude, you may find any Star, at any time convenient and visible.

Use XLVII. The Altitude and Azimuth of any Star being given, to find his Declination.

Lay the Thred to the Altitude on the de∣grees, counted from 60|0 toward the end, then setting one Point on the Stars Azimuth, counted in the Azimuth Line, and take the ND from thence to the Thred; which di∣stance measured from the beginning of the particular Scale of Altitudes, shall give the Declination.

If the Compasses stand on the right-side of the Thred, then the Declination is North; if on the left, it is South; according as you work for the Suns Azimuth in a particular Latitude.

Use XLVIII. The Altitude and Declination of any Star, with the Right Ascention of the Sun, and the true Hour of the Night given, to find the Right Ascention of that Star.

First, by the 43d Use, find the Stars Hour, viz. How many hours and minuts it wants of coming to, or is past the Meridi∣an; then the Extent of the Compasses (on

the Line of 24 hours on the Head-leg) from the Stars hour to the true hour, shall reach the same way from the Suns Right Ascenti∣on, to the Stars Right Ascention, on the Line of twice 12, or 24 hours.

Use XLIX. To find when any Fixed-Star cometh to South, by the Line of twice 12, or 24 hours.

In Use 42, Section 4, you have the way by Substraction, with its Cautions: But by the Line of twice 12, or 24 hours, work thus;

Count the Suns Right Ascention on that Line, and take the distance from thence to the next 12 backward, viz. that at ♈, at the beginning of the Line, when the Suns Right Ascention is under 12 hours; or, to the next 12 in the middle of the Rule at ♎, when the Suns Right Ascention is above 12 hours, (which is nothing but a rejecting 12 for more conveniency).

Then,

The same Extent laid the same way from the Stars Right Ascention, shall reach to the Stars coming to South.

Or,

The Extent from the Sun, to the Stars Right Ascention, shall reach the same way from 12, to the Stars coming to South.

Example, for the Lyons-Heart, August 20.

The Suns Right Ascention the 20th of August, is 10 hours 36 minuts; the Right Ascention of the Lions-Heart, is 9 hours and 50 min.

Therefore,

The Extent from 10 hours 35 min. to the beginning, shall reach the same way from 9 hours 50 min. (by borrowing 12 hours) because the Suns Right Ascention is more than the Stars) to 11 hours 13 min. of the next day, viz. at a quarter past 11; or, at 11 hours and 13 min. the same day; where you may observe, that the remainder being above 12, if you add 24 hours, the time of Southing is between mid-night, and mid-day next following.

Use L. To find what two dayes in the year are of equal length, and the Suns Rising and Setting.

Lay the Thred on any one day in the upper Line of Months and Dayes, and at the same time the Thred cuts in the lower-Line of Months the day that is answerable to it in length, rising, setting, and declina∣tion, and other requisites.

Example.

The 1st of April, and the 21 of August, are dayes of equal length; and the Suns Rising and Setting is the same on both those dayes; only in the upper-Line, the dayes are increasing in length, and in the lower-Line they are decreasing.

Use LI. To find how many degrees the Sun is un∣der the Horizon at any Hour, the De∣clination and Hour being given.

Count the Suns Declination on the de∣grees, the contrary way, viz. for North De∣clination, count from 60|0 toward the end; and count for Southern Declination toward the Head, and thereunto lay the Thred;

then take the nearest distance from the hour given to the Thred; this distance mea∣sured in the particular Scale of Altitudes, shall shew the Suns Depression under the Horizon at that hour.

Example. January the 10th at 8 at Night, how many degrees is the Sun under the Horizon.

On that Day and Hour, the Suns Decli∣nation is about 20 degrees South; then if I lay the Thred to 20 degrees of Declination North, and take the nearest distance from 8 to the Thred, that distance, I say, mea∣sured in the particular Scale, gives 34 de∣grees and 9 min. for the Suns Depression un∣der the Horizon of 8 afternoon.

To do this in other Latitudes, you are to find the Suns Altitude at 8 in Northern Declination, by Use 37.

CHAP. XVII. The use of the Trianguler Qua∣drant, in finding of Heights and Distances, accessable or inaccessable.

Use I. To find an Altitude at one Station.

FIrst, The Trianguler Quadrant being rectified, and fixed to a Ball and Socket and three-legged-staff, being necessary in these Operations to perform them exactly, especially for Distances; look up to the ob∣ject as you would to a Star; and observe what degree and minut the Thred cuts, and set it down: Also, observe the place where you stand at the time of Observation, and the distance from your Eye to the ground, and the place on the object that is level with your eye also; as the playing of the Thred and Plummet will plainly shew.

Also, you must have the measure from the place where you stood observing, to the Point exactly right under the object, whose height you would have in Feet, Yards, Perch, or what you please, to Integers, and Fractions in Decimals, if it may be.

Also Note, That in all Right-Angle-Triangles, one Acute Angle is alwayes the complement of the other; so that observing or finding one by Observation, by conse∣quence you have the other, by taking that from 90.

These things being premised, the Opera∣tion followes, by the Artificial Numbers, Sines and Tangents, and also by the Na∣tural.

Note also by the way, That in regard the complement of the Angle observed is fre∣quently used, if you count the degrees the contrary way, that is to say from the Head, you shall have the complement required; as hath been oftentimes hinted before.

Then,

As the sine of the Angle, opposite to the measured side, is to the measured side, counted on the Numbers;

So is the sine of the Angle found, to the Altitude or Height required on Num∣bers.

Example at one station.

Standing at C, I look up to B the object, whose Height is required, and I find the Thred to fall on 41 degrees and 45 minuts; but if you count from the Head, it is 48-15, the complement thereof, as in the Figure you see.

Also, the measure from C to A, is found to be 218 foot.

Then,

As the sine of 48-15, the Angle at B, being the complement of the Angle at C, is to 218 on the Line of Numbers;

So is the sine of the Angle at C, 41-45, to 195 the Altitude of AB the height required, found on the Line of Num∣bers.

To which you must add the height of your eye from the ground, in the time of Observation; or on rising grounds from a mark on the Building, or any other object that is level with your eye in time of Ob∣servation.

[illustration]

A second Example standing at D.

* 1.188But if I were standing at D, 129 foot and a half from A, and would find the height AB, the complement of the Angle at D, that is to say, the Angle at B is 33-30.

This being prepared, then say;

As the sine of 33-30, the Angle at B, to the measured-side DA, 129 ½ counted on the Numbers;

So is 56-30, the sine of the Angle at D, to 195, the Altitude required, AB, and 5 foot more, the usual height of the eye from the Level to the ground, makes 200, the whole height re∣quired.

To work this by the Trianguler Quadrant, say thus;

As — 129 ½, taken from any Scale, is to the = sine of 33 deg. 30 min. lay∣ing the Thred to the nearest di∣stance;

So is the = sine of 56-30, the Angle at D, to the — measure of 195 on the Scale you took 129 ½ from.

The like manner of work is by the Sector, as thus, in the foregoing Example.

As 218, taken from the Line of Lines, to the = sine of 48 deg. 15 min.

So is the = sine of 41-45, to 195 on the Line of Lines latterally.

And yet further,

So is the = sine of 90, to 291, the Line CB.

Use II. To find an Altitude at two stations.

But if you cannot come to measure to the foot of the object,* 1.189 then you must observe at two places.

As thus for Example.

First, as before, find the Angle at D, or rather the complement thereof, viz. 33-30; then go further backward in a right Line with the object and first station, any com∣petent Number of feet, as suppose 88 ½ to C; there also observe the Altitude or Com∣plement, viz. the Angle ABC, 48-15.

Then,

Find the difference between 48-15, and 33-30, and it is 14-45.

Then,

As the sine of the difference last found, viz. the Angle CBD, 14-45, to 88 ½, on the Line of Numbers;

So is the sine of the Angle at C, 41-45, to the measure of the side DB, 233, on the Line of Numbers.

Again, for the second Operation.

As the sine of 90, the Angle at A, to the Hypothenusa DB, 233;

So is the sine of 56-30, the Angle at D, to 195, the Altitude required.

The same by the Trianguler Quadrant, or Sector.

As — 88 ½, the measured distance CD, to the = sine of 14-45, CBD;

So is = sine of 41-45, to the — measure of 233, the opposite-side DB.

Again,

As — 233, taken from the Line of Lines, to = sine of 90;

So is the = sine of 56-30, the Angle at D, to — 195, on the Line of Lines, the height required.

Use III. Another way to save one Operation from IC.

First,* 1.190 observe the complement of the Angle at D, and also the complement of the Angle at C; then count these two comple∣ments on the Line of Natural Tangents, on the loose-piece, or moving-leg, and take the distance between them, and measure it on the same Tangent-line from the beginning thereof, and note what Tangent the Com∣pass-point stayeth at, and count that for the first term, in degrees and minuts.

Then,

As the Tangent of this first term, to the measured distance CD, 88 ½, on the Line of Numbers;

So is the Tangent of 45, to the Altitude required.

Thus in our Example;

The distance measured is 88 ½, the two complements 33-30, and 48-15; the di∣stance between them makes the Tangent of 24-34, to be used as a first term.

Then,

As the Tangent of ••4-34, the first term last found, to 88 ½ on the Numbers;

So is the Tangent of 45, to 195, ferè, on the Numbers, the height required.

But if the distance from D or C, to A, the foot of the Object, were required, then the manner of Calculation runs thus;

As the Tangent of the difference of the Co-tangents first found, 24-34, is to the distance between D and C 88 ½;

So is the Co-tangent of the greater Ark 48-15, to the greater distance CA 218.

Or,

So is the Co-tangent of the lesser Ark 33-30, to the lesser distance DA, 129 ½.

But if the Hypothenusaes be required, then reason thus;

As the Tangent of the difference first found is 24-34, to the distance be∣tween the stations D and C, 88 ½;

So is the Secant of the Angle at B the greater, viz. 48-15, counted beyond 90, to CB 291.

Or,

So is the Secant of 33-30, the lesser Angle at B, to 233 the lesser distance DB, the Hypothenusa required.

To work these two last by the Trianguler Quadrant.

First, prick off the Tangents and Secants to be used parallelly, from the loose-piece, on the greater general Scale; and note those Points for your present use.

As thus;

The Tangent of 24-34, taken from the loose-piece from 60, counted as 00 will reach to the sine of 10-40, on the general Scale.

Secondly,

The Secant of 33-30, being the measure from the Tangent of 33-30, on the loose-piece (counting from 60) to the Center, will reach on the general Scale from the Center, to 28-50.

Thirdly,

The measure from the Tangent of 48-15, on the loose-piece, to the Center, being the Secant of 48-15, will reach from the Cen∣ter to 32-5, on the general Scale.

This being prepared, the work is thus;

As — distance between the two stati∣ons, to = Tangent, of the first term, at 10-40;

So is = Tangent of 45, to the Altitude required.

Again, for the Distance.

As — distance between the two stati∣ons, to the = Tangent of the first term;

So is the = Tangent of the greater Angles complement, at 26-36, to the greatest distance CA 218. Or,

So is the = Tangent of the lesser Angles complement, at 15-25, to the lesser distance DA, 129 ½; Or,

So is the = Secant of the greater Angles complement, at 32-5•• to the greater Hypothenusa CB, 291. Or,

So is the = Secant of the lesser Angles complement, at 28-50, to the lesser Hypothenusa DB, 233.

Use IV. Another way for Altitudes, by the Line of Shadows, either accessable or unaccessable, by one or two stations.

If this way be desired, it may be put on this, as well as any other Quadrants.

Then the use is thus; Figure II.

Suppose that AB be the height of a Tree, or other Object to be found; go so far back

from it, as suppose to C, till looking up by the two Pins put for sights, the Thred falls on 45 degrees on the Quadrant, or on 1 on the Line of Shadows; then, I say, that the height AB, is equal to the distance CA, more by the height of your eye from the ground.

But if you go further back still to D, till the Thred falls on 2 on the Line of Sha∣dows; that is to say, at 26 deg. 34 min. the Altitude will be but half the distance from A; but if you remove to E, the Thred fal∣ling on 3 on the Shadows, the Altitude will be but one third part of the distance EA.

From hence you may observe, that obser∣ving at C, and at D, where the Thred falls on 1, and on 2, the distance between C and D, is equal to the Altitude; so likewise at D and at E, and so by consequence at 1 ½ and 2 ½ and 3 ½, or any other equal parts. This is an excellent easie way.

The like will be if you observe at D and C, looking up to F, where the Altitude AF is twice the distance AC.

Use V. Another way, by the Line of Shadows, at one station.

Measure any distance, as feet, yards,* 1.191 or

the like from any object; as suppose from A to D were 200 foot, and looking up to B, the Thred cuts the stroke by 2 on the Line of Shadows.

Then by the Line of Numbers, say;

As 2, the parts cut, is to 1;

So is 200, the distance measured, to 100 the height.

Or,

Suppose I measured any other uneven Number, and the Thred fall between 00 on the Loose-piece, and 1 on the Shadows, commonly called contrary Shadow.

The Rule is alwayes thus;

As the parts cut by the Thred, are to 1;

So is the measured distance, to the height required, being less than the measured distance.

But when the Thred falls between 1 and 90 at the Head, called right Shadow; then the Rule goes thus;

As 1, to the parts cut by the Thred;

So is the measured distance, to the height, being alwayes more than the measured distance from the foot of the object, to the station.

Use VI. Another way by the Line of Shadows, and the Sun shining.

When the Sun shineth, find his Altitude, and also as the Thred lies, see what division on the Line of Shadows is cut by the Thred, and then straightway measure the shadows length on the ground; and if the Sun be under 45 degrees high, the shadow is lon∣ger than the length of that object which causeth the shadow; but if the Sun be a∣bove 45 degrees high, then the object is longer than the shadow; and the Operati∣on is thus by the Line of Numbers, only with a pair of Compasses.

The Height of the Sun being un∣der 45, say;

As the parts cut by the Thred on the Sha∣dows, is to 1;

So is the Shadow measured, to the height required.

The Height of the Sun being a∣bove 45, say;

As 1, to the parts cut by the Thred on the Line of Shadows;

So is the measure of the shadow, to the height in the same parts.

Use VII. To find an inaccessable Altitude, by the Quadrat and Shadows, otherwise.

Observe the Altitude at both stations, and count the observed Altitudes at both stati∣ons, on the Quadrat or Shadows, according as it happens to be either above or under 45 degrees; and take the lesser out of the greater, noting the remainder for the first term; and the Divisor to divide the di∣stance between the stations, increased with Cyphers, if need be; and the Quotient is the Answer required.

But by the Line of Numbers, work thus;

The Extent from the difference to 1, shall reach the same way from the measured di∣stance, to the height required.

Example. Figure II.

Let ABCDE represent the Object and three Stations; let the Line AC represent the Altitude; the Point B one station, 50 foot from A; D another station, 100 foot from A, or 50 from B; and E another sta∣tion, 73 foot from D, or 173 foot from A; all which measures you need not know be∣fore, but only BD, and DE; Also, the

Angle at B, 63-27, and his complement, counting the other way, being the Angle at C 26 degrees 33 minuts; the Angle at D 45, and his complement so also; the Angle at E 30, and his complement 60. Now mind the Operation by either of these, First lay the Thred on 26-33, and in the Qua∣drat it cuts 50; lay the Thred on 45, and in the Shadows, or Quadrat, it cuts 100, or 1; or, if you lay the Thred to 60, then in the shadows it cuts 173.

The difference between 173, and 100, is 73.

Then,

As 73, the difference in Tangents be∣tween the two observations, is to the distance in feet 73;

So is Radius 100, or the side of the Qua∣drat, to 100, the hight required.

Again, for the two nearest Observations, whose difference of Tangents, is 50.

As 50, the difference in Tangents, to 50 foot the measured distance▪

So is 100, the side of the Quadrat, to 100 the height.

Again, lastly by the observations at B & E, the difference of Tangents being 123.

As 123, the difference in Tangents, to 123, the measured distance;

So is 100, the Radius or side of the Quadrat, to 100, the height required.

Or,

In the first Figure, the Angles at the top being 33-30, and 48-15; and the mea∣sured distance 88 foot and a half, the dif∣ference in Tangents will be 45-8.

Then,

As 45-8, to 100, the side of the Qua∣drat;

So is 88 ½, the measured distance to 194, the Altitude required.

This way is general for any Station, though both of right shadow, or both of contrary, or mixt of right and contrary, and done by the Line of Numbers, or by Multiplication and Division.

Also Note, That you may find this dif∣ference in Tangents or Secants, by the Na∣tural Tangents, or Natural Secants on the Sector, and the Scale of equal parts belong∣ing to them.

Thus;

Take the distance between the comple∣plement of the two observations, on the greater or lesser Line of Tangents, (as is most convenient) and measure this distance in the Line of Lines, or equal parts equal to that Radius; and that shall be the diffe∣rence in Tangents required. The like for the Secants.

Also, By the Artificial Numbers, Sines, and Tangents, you may come by this diffe∣rences in Tangents, or Secants, very well thus;

Just right against the Tangent of the Co-altitude, counted on the Line of Tangents, in the Line of Numbers, is one Number; and against the Tangent of the complement of the other Angle, is the other Number; only with this Caution, That if the Tangent be above 45, then take the distance from 45 to the Tangent, as it is counted backward, with Compasses, and set the same the in∣creasing way from 1, on the Numbers, to the other Number required; then the lesser taken from the greater, leaves the difference in Tangents that was required. In the same manner, the Sines counted from 90, and laid the contrary way from 1 increasing, will give the difference in Secants, to measure the Ba••••, and Hypothenusa by Numbers only.

Use VIII. Another pretty way by Scale and Compass, without Arithmetick, from T. S.

[illustration]

On any plain Boards end, or Trencher, draw a right Angle, and in the meeting-Point, and on one Lines-end, knock two Pins, or small Nails, as near as you can up∣right; then on the Pin that stands in the right Angle, hang a Thred and Plummet; then lift up the Board, with the right Angle toward the Object, whose height you would have, till the two Pins and the Object are brought to a streight Line, the Plummet playing even and truly.

Then draw the Line, that the Thred maketh, on the Board;

Then measure from your standing, to the foot of the Object, and take the number of feet, or yards from any Scale, and lay it from the right Angle on the other Line, and raise a Perpendiculer from thence to the Plumb-line made by the Thred, and that shall be the Altitude required, being mea∣sured on the same Scale.

Example.

Let ABGD represent the Boards end, or Trencher, and on that, let AB be one streight Line, and AG another Perpendi∣culer

to it; in the Point A, knock in one Pin; and in B, or any where toward the end, another; On the Pin at A, hang a Thred and Plummet; and standing at I, any convenient station, look up by the two Pins at B and A, till they bourn in a right Line with the Point H, the object whose height is to be measured; then the Plummet playing well and even, make a Point just therein, and draw the Line AD, as the Thred shewed.

Then, having measured the distance from G the foot of the Object, to I the station, take it from any first Scale, and lay it from A to G; then on the Point G, raise a Perpendi∣culer to AG, till it intersect the Plumb-line AD; then, I say, the distance CD, mea∣sured on the same Scale you took AC from, shall be equal to the Altitude GH, which was required.

Use IX. The same work at two stations.

But if you cannot come to measure from I, the first station to G;* 1.192 then measure from I to K; and having observed at I, and drawn the Plumb-line AD, take the mea∣sure between I and K, the two stations, from any fit Scale of equal parts, and lay it on the

Line AC, from A to C, viz. 79 parts and in the Point C, knock another Pin, and hang the Thred and Plummet thereon, and observe carefully where this last Plumb-line doth cross the other, as suppose at E; then from E, let fall a Perpendiculer to the Line AC, which Line AC shall be the height GH required; (or thus, the nearest di∣stance from E to AC is the height re∣quired) viz. 120 of the same parts that IK is 79; Note the Figure, and behold that ACFE, the small Figure on the Board is like and proportional to AA, GH, the greater Figure.

* 1.193Other wayes there be, as by a Bowl of Water, or a Glass, or a Plash of Water, or a Square; but these set down, are as convenient and ready as any whatsoever; As in the next Figure you may see the way by the Glass, and Square.

As thus;

Let C represent a Glass, a Bowl, or Plash of Water, wherein the Eye, at A, sees the picture or reflection of the Object E.

Then, by the Line of Numbers;

As CB, the measure from your foot to the Glass, is to AB, the height from your eye, to the ground at your foot;

So is the measure from C to D, to the height DE. See Figure VI.

Again, to find a distance by the Square, that is not over-long.

Let C represent the upper-corner of a Square, hung on a staff at F; then the one part of the Square directed to E, and the other to A.

The Proportion will hold, by the Line of Numbers.

As FA 11-37, to FC 50;

So is FC 50, to FE 220.

That is,

So many times as you find AF in FC;

So many times is FC in FE, and the like.

Note, That you must conceive AFE to be the Ground, or Base-line in this Opera∣tion by the Square; C being the top of an upright Staff, 5 foot long, called 50 for Fraction sake.

Use X. To find a Distance not approachable by the Trianguler Quadrant.

Let A represent the place of standing, and AC be the distance required.

[illustration]

First, I plant my Trianguler Quadrant, set upon a three legged Staff and Ball socket, right over the place A; and then bring the Index with two sights in it, laid or fastened to the Center of the Trianguler Quadrant, right over the Lines of Sines, and Lines cutting 90 at the Head; the Index and sights so placed, hold it there, and bring it and the Instrument together, till you see the mark at C, through the two sights, by help of the Ball-socket, and then there keep it; then remove the Index only to 0-60 on the loose-piece, which makes a right Angle; and set up a mark in that Line, at any conveni∣ent distance; as suppose at B, 102 foot from A; then remove the Instrument to B, and laying the Index on the Center, and 0-60 on the loose-piece, direct the sights to A, the first station, by help of a mark left there on purpose; Then remove the sights till you see the mark at C, and note exactly on what degree the Index falleth, as here on 60, counting from 0|60 on the loose-piece; or on 30, counting from the Head, which is the Angles at B, and at C.

Then by the Artificial Numbers, Sines and Tangents on the edge, say;

As the sine of 30, the Angle at C, to 102, the measured distance counted on the Numbers;

So is the sine of 60, the Angle at B, to 117, on the Numbers, the distance re∣quired.

So also is 90, the Angle at A, to 206, the distance from B to C.

Or, by the Lines and Sines on the Qua∣drant-side, as it lies, thus;

As the — measure of 102, taken from any Scale, as the Line of Lines doub∣ling, to the = sine of 30, laying the Index, or a Thred, to the nearest di∣stance;

So is the = sine of 60, to 117, measured latterally on the same Line of Lines.

And,

So is the = sine of 90, to 206, the di∣stance from B to C.

So also, If you observe at B, and at D only, you must be sure to set your Instru∣ment at one station, at the same scituation▪

as at the other, as a looking back from sta∣tion to station will do it, and the same way of work will serve.

For,

As the Sine of 20, to 110;

So is the Sine of 40, to 206.

And,

So is the Sine of 120, to the Line DC 278, &c.

Use XI. To find a Breadth and a Distance at any two Stations.

* 1.194Let AB be two marks, as two corners of a House or Wall, and let the breadth be∣tween them be demanded, and their distance from C and D, the two stations; First, set up two marks at the two stations, then set∣ting up the Instrument at C, set the fidu∣cial Line on the Rule to D, the other mark; then direct the sights exactly to B, and to A; observe the Angles DCB 45, and DCA 113-0, as in the Figure.

Secondly, Remove the Instrument to D, the other station, and set the fiducial-Line of the Quadrant (viz. the Line of Lines and Sines) directly to C; then fix it there, and remove the Index and sights to A, and to B, to get the Angles CDA 42-30, and

CDB 109-0; Then observe, that the 3 Angles, of every Triangle, being equal to 180 degrees; having got the Angles at C 113, and the Angle at D 42-30, by consequence, as you take 155, the sum of the Angles at C and D, out of 180, then there remains 24-30, the Angle at A.

So also, Taking 109 and 45 from 180, rests 26, the Angle at B; then also, taking 45, the Angle BCD, out of 113, the Angle DCA, rests 68 degrees; the Angle BCA, in like manner, taking 42-30 from 109, the Angles at D, rests 66-30, the Angle ADB; and let the distance mea∣sured, between the two stations, be 100, viz. CD. These things thus prepared by the Artificial Numbers, Sines and Tangents on the edge,

Say,

As the Sine of 24-30, the Angle at A, to 100, on the Numbers, the mea∣sured side CD;

So is the Sine of the Angle at D 42-30, to 164, on the Numbers, the side CA.

So is the Sine of 113, the Angle ACD, to 222, on the Numbers, the distance from C to B.

Also, for the other Triangle, at the o∣ther Station D.

As the Sine of 26, the Angle CBD, to 100, on the Numbers, the mea∣sured distance CD;

So is the Sine of 45, to 161, on the Line of Numbers, the distance from D to B;

So is the Sine of 109, the Angle CDB, to 216, on the Numbers, the distance from D to A.

Then, having the Sides DB 161, and AD 222, and ADB the Angle in∣cluded 66-30, to find the Angles DAB, or ABD, use this Propor∣tion.

As the sum of the two sides given, is to the difference between the two sides;

So is the Tangent of half the sum of the two Angles sought, to the Tangent of half their difference.

Example.

222, and 161, make 383 for a sum; and 161, taken from 222, rest 61 for a diffe∣rence.

Again,

66-30, taken from 180, rest 113-30, for a sum of the Angles sought, whose half 56-45, is the third Number in the propor∣tion.

As 383, the sum of the two known sides, is to 061, the difference on the Num∣bers;

So is the Tangent of 56-45, the half-sum of the two Angles sought, to the Tangent of half the difference 13-40; which half-difference, 13-40, added to 56-45, makes 70-25, the greater Angle required at B, viz. ABD.

Then also,

If you take 13-40, from 56-45, the half-sum of the Angles inquired, rest 43-05, the Angle BAB; the like may you do with the other Triangle ABC, being needless in our Proposition.

Thus having found the Angles, and one side, the Sines of the Angles, as proportional to their opposite sides.

As the Sine of 44-33, the Angle ABC, is to the side AC 146, on the Num∣bers;

So is the Sine of 68, the Angle a•• C, to 217, the distance between the marks required.

Or,

As Sine 43-05, the Angle at A, to 161;

So is Sine of 66-30, the Angle at D, to 217, the distance between the marks required.

Also note,

That if this manner of Calculation be tedious or difficult, then on a Slate, or sheet of Paper, you may do it by protraction, by the Line of Lines and Chords, or half Sines, very near the matter with care;

Thus: Draw CD the Station-Line, or measured-distance; and make AD 100, from any fit Scale. Then, on C and D draw a Circle, and on that Circle lay off from C and D the Angles, found by obser∣vation, and draw those Lines, and where they cross one another, as at A and B, draw the Line AB: those Lines and Angles mea∣sured on the same Scales and Chords, shall be the Sides, breadth, and distances required; as you see in the Figure.

Use XII. Another way for a long Distance.

* 1.195Let C be your standing place to set your

Instrument, and let E be the mark afar off, whose distance from you C would know: first, move in a right Line between C and E to A, any number of Yards or Perches, as suppose 50 Perch, and set a mark at A; Then move in a Perpendiculer-Line to CE, from A to B any distance, and there set up a mark at B, as suppose 66 Perches from A.

Then come back again to C, and remove in a Perpendiculer-Line to CE, till you see the mark set up at B, and the enquired point at the distance E in a Right-Line; and note that place at D, getting the exact distance thereof from C, suppose 76.

Then substract the measured distance AB from the measured distance CD, and note the difference 10. Then, by the Line of Numbers, or by the Rule of Three, say,

As the Difference between AB and BC, 10, is to the distance between A & C 50:

So is the measured distance CD 76, to the distance CE 380.

Or,

So is the measured distance AB 66, to the distance AE 330, the distance required.

Note, That you must be careful and ex∣act in measuring the Distances AC, AB, & CD, and the Answer will be the more ex∣act accordingly.

Use XIII. To find an Altitude of a House or Tower, by knowing part of it.

If you divide the inside-edge of the Loose-peice into inches,* 1.196 or any equal parts, such as the nearest distance from the Rectifying-Point to that inside-edge may be 1000, and for this use two small sliding sights may be convenient: Then the use is thus for any Angle under 30 degrees;

Fix the Instrument to the Ball-socket and Staff, and turn it toward the Object, causing the Plummet to play on 30 degrees; for then the Loose-piece is perpendiculer. Then one pin or sight set in the Rectifying-Point, slip on a sight along the inner-edge of the Loose-piece, till you see the Object at the upper part of the Altitude, and another sight at the lower part of the Altitude known; and observe the precise distance in parts between the two sights, on the Loose-piece; Or, the several parts cut by the In∣dex at each Observation:

Then,

As the distance between the two sights, is to the distance between the remotest sight from the middle of the Loose-piece;

So is the height of the known part, to the

whole height required above the level of the eye.

Example.

Let CI represent the Altitude of a Pyra∣mid on the Tower of a Steeple 30 foot high, and, standing at B, I would know the height of IA from the level of the eye upward.

Fix the Trianguler Quadrant on the Staff and Ball-socket, with the Head-Center at B, with the Plummet playing on 30 degrees, and the Loose-piece perpendiculer: Then slip two sights on the Loose-piece, one in a Right-Line to C, the other to I; and note the parts between, and the parts the furthest sight cuts, from the middle stroak on the Loose-piece, from whence the parts are num∣bred, which in our Example let be 500, the sight of H, and the sight at G to cut 359; then the distance between the sights will be 143, and the remotest from the middle of the Loose-piece to be 500; and the known Altitude, being part of the whole, to be 30 foot.

Then, by the Line of Numbers, say,

As 143, the distance betwixt the sights at G & H, to 500 the remotest sight from the level or middle, viz. FH:

So is 30 foot, part of the Altitude known, CI, to 105, the whole Altitude un∣known, AC.

Or,

[illustration]

As 357, the parts cut at G, to 500 the parts cut at H, the remotest sight:

So is 75, the height of the lower part, to 105 the whole height AC.

Or,

As 143, the distance between the sights, to IC the part of the height known 30:

So is 357, the parts cut between F and G, to 75 the height AI unknown, &c.

Use XIV. Having the Height, to find a Distance.

Let CA be the Altitude given, and AB the distance required. Then I standing at C, observe the Angle CAB, by setting the end of the Head-leg to my eye, and the Head-end downwards, and set down, as the Thread cuts, numbring both wayes, for the Angle at C and at B his complement.

Then say,

As the Angle at B, 30 deg. 40 minutes, counted on the Sines, to 105 the height of the Tower:

So is 59 deg. 20 min. the Angle at C on the Sines, to 176 the distance required on the Numbers.

Also note by the way,

That if you take an Altitude at two sta∣tions,

as suppose at E and at B; if the Angle observed at B, be found to be the half of the Angle at E; as here in Figure VIII, the Angle at E, being 61-20, and the Angle at B 30-40, the just half thereof; then, I say, that the distance between the two stations, is equal to the Hypothenusa EC, at the first station, viz. EB is equal to EC; which being observed, say;

As the sine of 90, to 120, on the Numbers.

So is 61-20 on the Sines, to 105, the height required on the Numbers.

A further proof hereof, take in this following Figure IX.

Let AB be a breadth of a Wall, or Fort, not to be approached unto; then by the degrees on the in-side of the loose-piece, to find that breadth one way, is thus; Put two Pins into the two holes in the Head and Moving-leg, (or set the sights there in large Instruments); then move nearer or further from the objects, till your eye, fixed at the rectifying Point, can but just see the marks A and B by the two Pins in each Leg, which will only be at the mark C, at an Angle of 60 degrees; for so the Rule is made to that Angle: then the Instrument being still fixed at C, look backward in a right Line from the middle of the loose-piece, and rectify∣ing

Point toward D, putting up a mark ei∣ther in, or over, or beyond the Point D; and also be sure to leave a mark at C, the first place of observation▪ Then remove the sights to 15 degrees, the half of 30, counting from the middle, and go back in a right Line from C, toward D, till you can just see the marks by the two sights set at 15 degrees each way; for then, I say, that the measure between the two stations, C and D, shall be exactly equal both to AB, the breadth required, and also to CB, or CA, the Hypothenusaes; then, having the sides CB, and CD, and the Angles BCE, and CBE, and BDC, it is easie to find all the other Sides and Angles, by the Rules before rehearsed, by the Lines of Artificial Nmmbers and Sines.

For,

As the Sine of 15 degrees, the Angle at D, viz. BDC, to 108 on the Num∣bers;

So also is the Angle at B, viz. DBC 15, to 108 on the Sines and Numbers.

So also is the Sine of 150, the Angle at C, viz. DCB, to BD 208 ½ on the Numbers.

Note also, That if the Angles of 60 and 30 be inconvenient, then you may make use of 52 and 26, or 48 and 24, or 40 and 20, or any other, and the half thereof; and

then the measured distance, and the Hypo∣thenusa BC, at the nearest station, will al∣wayes be equal; but not equal to the breadth at any other Angle, except 30 and 60, as in the Figure. But having the Angles, and those Sides, you may soon find all the others by the Artificial Numbers, Sines and Tan∣gents, by the former directions.

The End of the First Part.

Notes

• * 1.1

  Radius.
• * 1.2

  Right-Sine.
• * 1.3

  Tangent.
• * 1.4

  Secant, Chord.
• * 1.5

  Cosine.
• * 1.6

  Lateral-Sine.
• * 1.7

  Parallel.
• * 1.8

  Nearest-Distance.
• * 1.9

  Addition on Lines.
• * 1.10

  Substra∣ction on Lines.
• * 1.11

  Rectifying Point.
• * 1.12

  Plain.
• * 1.13

  Pole of the Plain.
• * 1.14

  Declina∣tion.
• * 1.15

  Perpendi∣culer-line on the Plain.
• * 1.16

  Horizon∣tal-line.
• * 1.17

  Reclina∣tion and Inclina∣tion.
• * 1.18

  Meridian-line.
• * 1.19

  Substile.
• * 1.20

  Stile.
• * 1.21

  Angle be∣tween 12 and 6.
• * 1.22

  Inclina∣tion of Meridians
• * 1.23

  Parallels▪
• * 1.24

  Contin∣gent.
• * 1.25

  Vertical-line.
• * 1.26

  Nodus or Apex.
• * 1.27

  Perpendi∣culer height of the Stile.
• * 1.28

  Foot of the Stile.
• * 1.29

  Virtical-point.
• * 1.30

  Axis of the Hori∣zon.
• * 1.31

  Erect.
• * 1.32

  Direct.
• * 1.33

  Declining Reclining or Inclin∣ing-plains
• * 1.34

  Oblique.
• * 1.35

  Circles of Position.
• * 1.36

  Poles.
• * 1.37

  Poles of the Zo∣diack.
• * 1.38

  Equinocti∣al-Points.
• * 1.39

  Solsticial-Points.
• * 1.40

  Zenith.
• * 1.41

  Nadir.
• * 1.42

  Horizon. 1.
• * 1.43

  Zodiack
• * 1.44

  4.
• * 1.45

  Colures.
• * 1.46

  5. & 6.
• * 1.47

  Tropicks.
• * 1.48

  7. & 8.
• * 1.49

  Polar-Circles.
• * 1.50

  9. & 10.
• * 1.51

  Hours.
• * 1.52

  Azimuths
• * 1.53

  Almican∣ters.
• * 1.54

  Declina∣tion.
• * 1.55

  Latitude.
• * 1.56

  Longitude.
• * 1.57

  Right-As∣cention.
• * 1.58

  Oblique-Ascention.
• * 1.59

  Ascentio∣nal Diffe∣rence.
• * 1.60

  Ampli∣tude.
• * 1.61

  Circles and Angles of Position.
• * 1.62

  In both these the inversed Proporti∣on is in th lower line
• * 1.63

  Circle.
• * 1.64

  Half-Circle.
• * 1.65

  Quadrant, or the quarter.
• * 1.66

  Lesser-parts.
• * 1.67

  Segments.
• * 1.68

  Triangles.
• * 1.69

  Rhombus.
• * 1.70

  Trape∣ziaes.
• * 1.71

  Regular-Polligons.
• * 1.72

  By th' Tri∣anguler-Quadrant.
• * 1.73

  Sines.
• * 1.74

  Tangent.
• * 1.75

  Secant.
• * 1.76

  Chord.
• * 1.77

  Sines.
• * 1.78

  Tangents.
• * 1.79

  Secants.
• * 1.80

  Chords.
• * 1.81

  By the Sector-side.
• * 1.82

  Sine.
• * 1.83

  Tangent.
• * 1.84

  Tan. to 76.
• * 1.85

  Secant.
• * 1.86

  By the Lines on the Edge.
• * 1.87

  Sines.
• * 1.88

  Tangent.
• * 1.89

  Secants & Tangents beyond 45 degrees.
• * 1.90

  Artificial-logarithms
• * 1.91

  The proof of the truth of the Instru∣ment.
• * 1.92

  Quadrant.
• * 1.93

  Sector.
• * 1.94

  Quadr.
• * 1.95

  Sector.
• * 1.96

  Quadrant.
• * 1.97

  Sector▪
• * 1.98

  Quadr.
• * 1.99

  Sector.
• * 1.100

  Quadr.
• * 1.101

  Sector.
• * 1.102

  Quadr.
• * 1.103

  Sector.
• * 1.104

  Sector.
• * 1.105

  Sector.
• * 1.106

  Quadr▪
• * 1.107

  Sector.
• * 1.108

  Quadr.
• * 1.109

  Sector.
• * 1.110

  Quadr.
• * 1.111

  Quadr.
• * 1.112

  By the Quadr. particu∣larly.
• * 1.113

  Quad. Ge∣nerally.
• * 1.114

  Quadr· Particu∣larly.
• * 1.115

  Artificial-S. & T.
• * 1.116

  Quad. Ge∣nerally.
• * 1.117

  Sector.
• * 1.118

  Particular Quadr.
• * 1.119

  Artificial-S. & T.
• * 1.120

  Quad. Ge∣nerally.
• * 1.121

  Sector▪
• * 1.122

  Art. Sine.
• * 1.123

  Quadr.
• * 1.124

  Sector·
• * 1.125

  Partic. Quadr.
• * 1.126

  Art. Sines.
• * 1.127

  Quadr. generally.
• * 1.128

  Sector.
• * 1.129

  Art. Sine.
• * 1.130

  Quadr.
• * 1.131

  Sector
• * 1.132

  Partic. Q.
• * 1.133

  Artific. S.
• * 1.134

  Gen. Quad.
• * 1.135

  Sector.
• * 1.136

  Part. Q.
• * 1.137

  Artificial S. & T.
• * 1.138

  Gen. Quad.
• * 1.139

  Sector.
• * 1.140

  Partic. Q.
• * 1.141

  Oblique-Ascention.
• * 1.142

  Artificial S. & Tan.
• * 1.143

  G. Quad.
• * 1.144

  Sector.
• * 1.145

  Artificial-S. & T.
• * 1.146

  Partic. Q.
• * 1.147

  Gen. Quad
• * 1.148

  Sector.
• * 1.149

  Partic. Q.
• * 1.150

  Particular Quadr.
• * 1.151

  Particular Quadrant.
• * 1.152

  Particular Quadrant.
• * 1.153

  Artificial-S. & T.
• * 1.154

  Gen. Quad.
• * 1.155

  Sector.
• * 1.156

  Particular Quadrant▪
• * 1.157

  Gen. Quad.
• * 1.158

  Sector.
• * 1.159

  Gen. Quad
• * 1.160

  By Artifi∣cial Sines & Tang.
• * 1.161

  Particular Quadrant.
• * 1.162

  General-Quadr.
• * 1.163

  By the Artificial-Sines and Tangents.
• * 1.164

  By the Sector.
• * 1.165

  By Artifl. S. & T.
• * 1.166

  General-Quadr▪
• * 1.167

  Gen. Quad,
• * 1.168

  By the Sector.
• * 1.169

  Gen. Q••ad
• * 1.170

  Sector.
• * 1.171

  Artificial S. & Tat.
• * 1.172

  Artificial S. & T.
• * 1.173

  By the General-Quadrant & Sector.
• * 1.174

  Gen. Quad.
• * 1.175

  By Artifi∣cial Sines & Tang.
• * 1.176

  By Artifi. S. & T.
• * 1.177

  By the Sector.
• * 1.178

  General-Quadr. Or, Sector.
• * 1.179

  Particular Quadrant.
• * 1.180

  By the Artificial-Sines and Tangents.
• * 1.181

  General-Quadr.
• * 1.182

  Particular Quadrant.
• * 1.183

  Artificial-S. & T.
• * 1.184

  Artificial-S. & T.
• * 1.185

  Partic. Q.
• * 1.186

  Gen. Quad
• * 1.187

  Particular Quadrant.
• * 1.188

  Fig. I.
• * 1.189

  Fig. I.
• * 1.190

  Fig. I.
• * 1.191

  Fig. II.
• * 1.192

  Fig. V▪
• * 1.193

  Fig. VI.
• * 1.194

  Fig. IV.
• * 1.195

  Fig. VII.
• * 1.196

  Fig. VIII.