Nine geometricall exercises, for young sea-men and others that are studious in mathematicall practices: containing IX particular treatises, whose contents follow in the next pages. All which exercises are geometrically performed, by a line of chords and equal parts, by waies not usually known or practised. Unto which the analogies or proportions are added, whereby they may be applied to the chiliads of logarithms, and canons of artificiall sines and tangents. By William Leybourn, philomath.

GEOMETRICALL PROPOSITIONS and THEOREMS, Necessary to be known and practised for the more easie understanding of the subsequent EXERCISES. The First EXERCISE.

THE ARGUMENT.

THE following Propositions, Theorems and Problems, are such as will come in continuall use in the Practice of the subsequent EXERCISES, and there∣fore I have here promiscuously inser∣ted such as do relate to the following Treatises, and ought throughly to be understood and often practised; by which means nothing in the ensuing Cases, Propositions and Problems, will be difficult to be understood, but that you may gradually proceed, in your Practice, from Exercise to Exercise, without be∣ing referred to any other Book or Books.

GEOMETRICALL PROPOSITIONS. PROP. I. A right Line being given, to divide the same into two equal parts at right Angles.

[illustration] geometrical diagram

LET the right Line, to be divided, be A B.—First, open your Compasses to any distance greater then the length of half the given Line; then setting one foot in A, with the other foot describe the Arches C C, both above and below the Line A B.—Se∣condly, (the Compasses still resting at the same distance,) set one foot in B, and with the other describe the Arches D D, cutting the former Arches C C in the Point E a∣bove, and F below the given Line A B. —Lastly, draw the Line E F, which will divide the given Line A B in two equal parts, in the Point G, and at right Angles.

PROP. II. Ʋpon a right Line given, to erect a Perpendicular upon any part thereof.

LET the given Line be H K, and from the Point L let it be required to erect a Perpendicular.—First, open your Compasses to any convenient distance, and setting one foot in the given Point L, with the other make a Mark or

Point at pleasure, as M. Then keeping the Compass-point in M, with the other de∣scribe the Arch N N above the Point L, and also another Arch at O, cutting the given Line in O.—Lastly, lay a Ruler from O to M, which will cut the Arch N N before drawn in the Point P. So a Line drawn from P to L shall be a Perpendicular to the given Line H K, and from the Point L.

[illustration] geometrical diagram

PROP. III. From a Point above, to let fall a Perpendicular upon a right Line given.

LET the Point given be Q, and the Line upon which the Perpendicular is to fall be R S.—First, opening the Compasses to any distance greater then Q X, set one foot in Q, and with the other cross the given Line R S in the Points T and V.—Secondly, (the Compasses unaltered,) set one foot in T, and with the other describe the Arch w w, below the given Line R S.— Thirdly, remove the Compasses to V, (being still at the same distance) and cross the Arch w w in the Point Z—. Lastly, a Ruler laid from Q to Z will cut the Line R S in the Point X. So a Line drawn from Q to X shall be perpendicular to the given Line R S.

[illustration] geometrical diagram

PROP. IV. A right Line being given, to draw another right Line which shall be parallel thereto at any distance required.

LET A B be a Line given, and let it be required to draw another right Line which shall be parallel thereunto, and at the distance of the length of the Line C.—First, take the length of the Line C in your Compasses, and setting one foot to∣wards one end of the given Line, as at D, describe the Arch E.— Secondly, set one foot of the Com∣passes towards the other end of the given Line, as at F, and describe the Arch G.—Lastly, lay a Ruler to the Arches E and G, so that the Ruler onely touch the Ar∣ches, and not cut or cross them in any part. So a Line drawn thereby shall be parallel to the given Line A B, and at the distance of the Line C.

[illustration] geometrical diagram

PROP. V. A right Line being given, to draw another right Line parallel thereunto, which shall pass through a given Point.

[illustration] geometrical diagram

LET the given Line be H K, to which let it be required to draw a parallel Line, which shall pass through the Point L.— First, take in your Compasses the distance K L, and setting one foot of the Compasses in

H, with the other describe the Arch M M.—Secondly, take the Line H K in your Compasses, and setting one foot in the given Point L, with the other cross the Arch M M in the Point O. So a Line drawn from L to O shall be paral∣lel to the given Line H K, and shall pass through the Point L.

PROP. VI. Three right Lines being given, to make a Triangle, whose three Sides shall be equal to the three given Lines.

LET the three Lines given be N, P, Q.—First, take the Line N in your Compasses, and lay that down from R to S.—Secondly, take the Line P in your Com∣passes, and setting one foot in S, with the other describe the Arch V V.—Thirdly, take the Line Q in your Compasses, and setting one foot in R, with the o∣ther cross the Arch V V, in the Point T.—Lastly, draw the Lines T R, and T S. So shall you have constituted the Triangle T R S, whose three Sides are equal to the three given Lines, N, P, Q.

[illustration] geometrical diagram

PROP. VII. Three Points (which lie not in a straight Line) be∣ing given, to finde the Centre of a Circle, which being described shall pass through the three given Points.

LET the three given Points be A, B, C.—First, open your Compasses to any distance greater then half the distance between A and B: and setting one foot in B,

with the other describe the Arch G D. Then remove the Compasses, set one foot in A, and with the other cross the former Arch in the Points D and F, and draw the Line D F.—Secondly, (the Compasses still continuing at the same distance,) set one foot in the Point C, and with the other cross the Arch (before drawn) in the Points E and G, and draw the Line E G, crossing the other Line D F in the Point H. So shall H be the Centre of the Circle, which be∣ing described shall pass directly through the three given Points, A, B, C.

[illustration] geometrical diagram

PROP. VIII. Two Points within any Circle being given, how to de∣scribe the Arch of another great Circle which shall pass through those two given Points, and also divide the Circumference of the given Circle into two equal parts.

LET the two Points given be E and F, within the Circle A B C D.—First, through either of them (as through E) draw the right Line E D, passing through the Centre of the Circle at K.—Secondly, draw the Line A C at right Angles to B D; so shall the Circle be divided into four equal parts or Quadrants, by the Lines A C and B D. —Thirdly, draw the Line E A, and upon the Point A (by the II. Prop.) erect the Perpendicular A G, cutting the Line B D (it being extended) in the Point G; so have you three Points, E, F, and G, through which (by the last Prop.) ou may draw a Circle to pass, whose Centre will be at H:

upon which Point if you describe an Arch of a Circle, at the di∣stance H E or H F, it will pass through the two given Points E and F, and divide the Circle A B C D into two equal parts, in the Points L and M, which was requi∣red. And that this Arch, thus drawn, doth divide the Cir∣cle into two equal parts, is evident, for a Line drawn from L to M will pass direct∣ly through the Centre K.

[illustration] geometrical diagram

These are such Geometricall Propositions as are absolutely necessary for the working of the severall Conclusions in the fol∣lowing Exercises. More might have been added, but these well understood and practised will be sufficient to carry you through this Work.

And now I will describe unto you the making of the slight: Instrument by which all contained in this Book is performed, namely, The Line of Chords, and shew you the general Ʋse thereof, in the protracting or laying down of Angles of any quantity: Or if any Angle be already laid down, to finde (thereby) the quantity thereof.—Then will I give you some few usefull and necessary Theorems, chiefly appertaining to Tri∣gonometry, or the Solution of Triangles, and so conclude this first EXERCISE.

How to make a Line of Chords.

ACcording to the largeness of your Line of Chords you intend to make, draw a right Line, as A B, and upon the Point A (by the II. Prop.) erect the Perpendicu∣lar A C, and upon A (as a Centre) describe the Quadrant B D E C, which you must divide into 90 equal parts or De∣grees. Which that you may readily doe, your Compasses being opened to the distance A B, set one foot in B, and the other will reach to E; also set one foot in C, and the o∣ther will reach to D: so is your Quadrant divided in∣to three equal parts, each part containing 30 degr. This done, divide each of these three parts into three more; so shall you have divided your Quadrant in∣to 9 equal parts, each con∣taining 10 degr. and each of these 9 parts, being di∣vided into halves, will con∣tain 5 degr. and (if you make your Line large enough) you must divide those into 5 equal parts, which you may very well doe, if the Line A B be but two inches long, as all the Schemes and Figures in these Exercises are drawn by a Line of Chords of that length.

[illustration] geometrical diagram

Your Quadrant being thus divided into 90 degr. draw the Line B C, and parallel thereto two other Lines, one pretty close to B C, to contain the small Divisions, and the other at a larger distance, to set the Figures in. Now it is the Line B C which is called the Line of Chords, (possibly for this Reason,

the Arch or Ark B D E C representing Arcus, a Bow, and B C the String or Chord thereof) the divisions whereof are to be transferred from the degrees of the Quadrant B D E C, in this manner.—First, setting one foot of your Compasses in B, extend the other to 80 degr. in the Quadrant, and from the division of 80 degr. in the Quadrant draw the Arch 80, 80, which will cut the Chord-Line in 80; doe so with 70, 60, 50, &c. and the like with every fifth degree, as you see in the Fi∣gure. And if your Line be very large, you may doe so to every single degree, and part of a degree. And by this means have you reduced the degrees of the Quadrant B D E C to the straight Line B C, more commodious to be set upon a Ruler, then the crooked Arch B D E C.

The Uses of the Line of Chords.

THE Ʋses of this Line are principally two. The one is, To protract or lay down upon Paper an Angle of any quantity (that is, of any number of degrees) required.— The other Ʋse is, If an Angle be already protracted or laid down, to finde how many degrees and parts of a degree it containeth. —In both which I would have the Reader very perfect, because very much contained in this Book hath dependence thereupon.

And here it will be necessary that I give you the Definiti∣on of an Angle. Know therefore that an Angle is the Inclination or bowing of two right Lines the one to the o∣ther.—As the two right Lines C A and B A incline the one to the other, and touch or meet each other in the Point A, in which Point, by rea∣son of the inclination of the said Lines, is made the Angle C A B.

[illustration] geometrical diagram

And here note that an Angle is commonly signed by three Letters, the middlemost whereof signifies the angular Point. As in this Figure, when we say the Angle C A B, you are to understand the very Point at A.

I. How to protract (or lay down upon Paper) an Angle containing any number of Degrees and Minutes by the Line of Chords.

DRaw a right Line at pleasure, as A B, and from the Point A let it be required to protract or lay down an Angle containing 40 degrees.—First, open your Compasses alwaies to 60 degr. of your Line of Chords, (which is equal to the Line A C of the Quadrant,) and with this distance, set∣ting one foot of the Compasses upon the Point A, with the other foot de∣scribe the Arch B C.—Secondly, take in your Compasses 40 degr. (which is the quantity of the Angle to be laid down) out of the Line of Chords, from the beginning thereof, and setting one foot in B, the other will reach to C upon the Arch: where∣fore through the Point C draw the Line C A. So shall the Angle at A contain 40 degr. as was required.

[illustration] geometrical diagram

II. An Angle that is already protracted, how to finde the quantity of Degrees it containeth.

SUppose C A B were an Angle already protracted, and it were required to finde the quantity thereof.—First, open your Compasses to 60 degr. of your Line of Chords, and setting one foot in A, (the angular Point) with the other describe the Arch B C.—Secondly, take in your Com∣passes

the distance between B and C, which distance apply to your Line of Chords, (by setting one foot in the beginning thereof) and you shall finde the other to fall upon 40 degr. which is the quantity of the Angle at A.

Thus have you the Ʋses of your Line of Chords in protra∣cting and finding the quantities of Angles. And now it will not be impertinent, if in this place I shew you how Angles may be protracted and laid down, and also their quantities found, by an Instrument which I shall make use of towards the end of this Book, which I call a Protracting Quadrant.

Its Description.

IT is no other then a Quadrant made upon a piece of very thin Brass, and divided into 90 degr. the Brass being cut away close to the divisions of the degrees on the out-side, and also the hollow within, so that there remains nothing but the Limb and the two Sides, as you may discern by the Figure. In the protracting or laying down of Angles, and in finding of the quantity of Angles already laid down, this is

Its Ʋse.

[illustration] geometrical diagram

SUppose you were to finde the quantity of the Angle C A B. Hold a Pin or Needle upon the angular Point at A, to which bring the Centre of your Quadrant, (noted also with A) and there turn it about, till the Me∣ridian Line thereof, A B, lie upon the Line A B of the Angle: then see un∣der what degrees of the Quadrant the Line A C

lieth, which you shall find to lie just under 40 degr. And such is the quantity of the Angle C A B.—And if upon the Line B A you were to protract such an Angle of 40 degr. Lay the Meridian Line of the Quadrant upon the Line A B, (the Centre of the Quadrant upon the Point A) and with your Needle make a prick or point just against 40 degr. of the Quadrant's limb. So a Line drawn from A through this Point shall make an Angle of 40 degr.

Trigonometricall Theorems.

1. A Triangle is a Figure consisting of three Sides, and as many Angles; as is the Figure C A B.

2. Any two Sides of a Triangle are called the Sides of the Angle contained by them; as the Sides C B, and A B, are the Sides containing the Angle C B A.

3. The measure of an Angle is the quantity of the Arch of a Circle, described upon the angular Point, and cutting both the Sides containing the Angle. As in the Triangle A C B, the Arch f g is the measure of the Angle at C, the Arch d e is the measure of the Angle at B, and the Angle at A is a right Angle containing 90 degr.

[illustration] geometrical diagram

4. A Degree is the 360. part of any Circle. Therefore,

5. A Semicircle con∣tains 180 degr. And

6. A Quadrant (or right Angle) contains 90 degr.

7. The Complement of an Angle less then 90 degr. is so much as that Angle wanteth of 90 deg.

As the Angle A C B of the Triangle containeth 53 degr. 7. min. the Complement whereof is 36 degr. 53 min. which is so much as 53 degr. 7 min. wanteth of 90 degr. For if you subtract 53 degr. 37 min. from 90 degr. (or from 89 degr. 60 min. for ease in subtracting) the remainder will be 36 degr. 53 min.

8. The Complement of an Angle to a Semicircle is so much as that Angle wanteth of 180 degr. So the Angle C being 53 degr. 7 min. take 53 degr. 7 min. from 180 degr. (or from 179 degr. 60 min.) and the remainder will be 126 degr. 53 min. which is the Complement of the Angle C to 180 degr.

9. An Angle is either right, acute, or obtuse.

10. A Right Angle is that whose measure is 90 degr. or a Quadrant.

11. An Acute Angle is less then a right Angle, and alwaies contains less then 90 degr.

12. An Obtuse Angle is greater then a right Angle, and al∣waies contains more then 90 degr.

13. A Triangle is either right-angled, or oblique-angled.

14. A right-angled Triangle is such a Triangle as hath one right Angle. As the Triangle C A B hath one right Angle, namely, that at A, which containeth just 90 degr.

15. In every right-angled Triangle, that Side which sub∣tendeth (or lieth opposite to) the right Angle is called the Hypotenuse; and of the other two Sides the one is called the Perpendicular, and the other the Base, at pleasure: But most commonly the shorter side is called the Perpendicular, and the longer the Base. Thus in the Triangle C B A, B C is the Hypotenuse, C A the Perpendicular, and A B the Base.

16. In every right-angled Triangle, if you have one of the acute Angles given, the other is also given, it being the Complement thereof to 90 degr. As in the Triangle C A B, if you have the Angle at C 53 degr. 7 min. given, you have also the Angle at B given, it being the Complement of that at C to 90 degr. wherefore take 53 degr. 7 min. from

90 degr. and there will remain 36 degr. 53 min. which is the quantity of the Angle at B.

17. In all right-lined Triangles whatsoever, (either right-angled or oblique-angled) the three Angles together are

[illustration] geometrical diagram

equal to two right Angles, or contain 180 degr. Therefore, if you have any two Angles of a Triangle given, you have also the third given, it being the Complement of the other two to 180 degr.—Thus in the Triangle C D B, if there were given the Angle C D B 43 degr. 20 min. and the Angle C B D 14 degr. 40 min. I say, by consequence you have the third Angle D C B also given, it being the Complement of the other two to 180 degr. For the two given Angles B D C 43 degr. 20 min. and C B D 14 degr. 40 min. being added together, make 58 degr. which being taken from 180 degr. there will remain 122 degr. the quantity of the obtuse Angle D C B.

18. In all Triangles whatsoever, the Sides are in proportion one to the other as the Sines of the Angles opposite to those Sides. So in the Triangle C D B, the Sine of the Angle at D is the proportion to the Side C B, which is opposite to it; as the Sine of the Angle at B is to the Side C D, or the Angle at C to the Side D B.

THE SOLƲTION Of Right-lined TRIANGLES By the LINES of EQƲAL PARTS, & CHORDS. The Second EXERCISE.

WHAT a Line of equal parts, and a Line of Chords, is, is not unknown to any conver∣sant in the least with Mathematicall Practi∣ces. For a Line of equal parts is onely a Line of any length equally divided into any number of equal parts, as 10, 20, 30, 60, 100, 1000, &c.—And a Line of Chords is no other then the Degrees of a Quadrant, or quarter of a Circle, (which contains 90 degr.) transferred from those degrees to as straight Line; as in the foregoing Exercise I have shewed. These

Lines of equal parts and Chords, being put upon a plain Ruler, (which any that make Mathematicall Instruments know how to doe) will, by the Precepts following, measure all manner of Triangles, whether plain or spherical. But before I come to the Solution of the severall Cases appertaining to Triangles, I would have you take notice, That every Triangle, whether plain or sphericall, consisteth of six parts, namely, of three Sides, and as many Angles; any three of which being given, a fourth may be found. Now in the resolving of right-lined Triangles, if they be right-angled, I call the Side opposite to the right Angle the Hypotenuse; and of the other Sides, comprehending the right Angle, I call the longer of them the Base, and the shorter, the Perpendicular.—But in the Solution of oblique-angled plain Triangles, I call the longest Side of the Triangle the Base; and the other two, the two Sides, without any other distinction of Denomination or Name. These things being premised, I come now to the Solution of plain Triangles, both right and oblique-an∣gled. And

I. Of Right-angled plain Triangles.

THE Triangle which I shall make use of in the severall Cases belonging to a right-angled plain Triangle shall be this following, C A B, in which

		parts.
A B, the Base,	contains	180
C A, the Perpendicular,	135
C B, the Hypotenuse,	225
And		deg.	m.
A, the right Angle,	contains	90	00
C, the Angle at the Per.	53	07
B, the Angle at the Base,	36	53

[illustration] geometrical diagram

CASE I. The Base B A 180, and the Perpendicular C A 135, being given, to finde the Angles B and C.

DRaw a Line A B, and from your Scale of equal parts take 180, and set them from A to B; then on the Point A raise the Perpendicular A C, and because it contains 135, take 135 parts from your Scale of equal parts, and set them from A to C; then draw the Line C B: which three Lines will constitute the Triangle C A B.

Now to finde the Angles C and B, take in your Compasses 60 degr. of your Line of Chords, and setting one foot in C, describe the Arch f g; also setting one foot in B, describe the Arch d e: then take in your Compasses the distance from f to g, which measured upon your Line of Chords will reach from the beginning thereof to 53 degr. 7 min. and such is the quantity of the Angle at C.—In like manner take the distance between d and e in your Compasses, that distance applied to your Line of Chords will reach to 36 degr. 53 min. —Or, when you had found the quantity of the Angle C to be

53 degr. 7 min. if you had subtracted that from 180 degr. the remainder would have been 36 degr. 53 min. the quantity of the Angle at B, without drawing of the Arch d e, and measuring it upon your Chord.

For such as have a Canon of artificial Sines, Tangents and Logarithms, and would resolve this Case by them, this is

The Analogie or Proportion.

As the Logarithm of A B is to the Logarithm of A C,

So is the Radius to the Tangent of B.

CASE II. The Hypotenuse C B 225, and the Base A B 180, being given, to finde the Angles B and C.

DRaw a right Line A B containing 180 parts of your Scale of equal parts, also out of the same Scale of equal parts take 225, your Hypotenuse, and setting one foot of your Compasses in B, with the other describe the obscure Arch h k; then on the Point A raise the Perpendicular A C, which will cut the obscure Arch h k in C; then draw the Line C B, so have you the Triangle C A B: then may you measure the quantity of the Angles at C and B as in the last Case. And so will C be 53 degr. 7 min. and B 36 degr. 53 min.

The Analogie or Proportion is,

As the Log. of C B is to the Radius,

So is the Log. of the Side A B to the Sine of C.

CASE III. The Base A B 180, the Angle C 53 degr. 7 min. and the An∣gle B 36 deg. 53 min. being given, to finde the Perpendicular C A.

DRaw a right Line A B containing 180 parts of your Scale, for the Base of your Triangle; then taking 60 degr. from your Line of Chords, on the Point B describe the Arch

d e, and (because the Angle at B contains 36 degr. 53 min.) take 36 degr. 53 min. from your Chord, and set it from d to e, and from B, through the Point e, draw the Line B C. Also upon the Point A erect the Perpendicular A C, crossing the Line B C in C. So have you formed the Triangle C A B. Lastly, take the length of the Line A C in your Compasses, and measuring it upon your Line of equal parts, you shall find it to contain 135. And that is the length of the Perpendicular C A.

The Analogie or Proportion is,

As the Sine of the Angle at C is to the Log. of A B,

So is the Sine of the Angle B to the Logar. of C A.

Or,

As the Radius is to the Logar. of A B,

So is the Tangent of B to the Logar. of C A.

CASE IV. The Hypotenuse C B 225, the Angle C 53 degr. 7 min. and the Angle at B 36 degr. 53 min. given, to finde the Base B A, and the Perpendicular C A.

[illustration] geometrical diagram

DRaw a right Line C B containing 225 of your Line of equal parts; then taking 60 deg. out of your Line of Chords, set one foot of the Compasses in B, and with the other describe the Arch e d; also (the Compasses continuing at the same distance) place one foot in C, and with the other describe the

Arch g f. Then from the Point B, and through the Point d, draw a right Line; also from the Point C, and through the Point f, draw another right Line: these two Lines will intersect or cross each other in the Point A, forming the Triangle C A B. Lastly, take the Line A B in your Com∣passes, and applying it to your Scale of equal parts, you shall finde it to contain 180; and that is the length of the Base A B. Likewise A C being taken in the Compasses, and measured upon the Line of equal parts, will be found to contain 135, which is the length of the Perpendicular C A.

The Analogie or Proportion is,

As the Radius is to the Logarithm of C B,

So is the Sine of C to the Logarithm of A B,

And the Sine of B to the Logarithm of C A.

CASE V. The Hypotenuse C B 225, and the Base A B 180, being given, to finde the Perpendicular C A.

DRaw a right Line A B containing 180 of your Scale of equal parts, and upon the end A erect a Perpendicular A C. Then take out of your Scale of equal parts 225, (the length of your Hypotenuse given,) and setting one foot of the Compasses in B, with the other describe the Arch h k, cut∣ting the Perpendicular A C in C, then draw the Line C B: so have you constituted the Triangle C A B. Lastly, take in your Compasses the length of the Line A C, and apply it to your Line of equal parts, where you shall finde that it will contain 135: and that is the length of the Perpendicular C A.

The Analogie or Proportion is,

1. Operation.

As the Logarithm of C B is to the Radius,

So is the Logarithm of A B to the Sine of C.

2. Operation.

As the Radius is to the Logarithm of C B,

So is the Sine of B (the Complement of C) to the Log. of C A.

CASE VI. The Base A B 180, the Angle C 53 degr. 7 min. and the Angle B 36 degr. 53 min. being given, to finde the Hypotenuse C B.

DRaw a right Line A B containing 180 parts of your Scale, for the Base of your Triangle, and on the end A erect a Perpendicular A C. Then take 60 degr. out of your Line of Chords, and upon the Point B, with that distance, describe the Arch d e; and (because the Angle at B is 36 degr. 53 min.) take 36 degr. 53 min. from your Line of Chords, and set it upon the Arch from d to e. Then from B, through the Point e, draw a right Line, till it meet with the Perpendicular before drawn, which it will do in the Point C. And thus have you protracted your Triangle C A B. Lastly, take in your Compasses the length of the Hypotenuse C B, and measure it upon your Scale of equal parts, and you shall finde it to contain 225.

The Analogie or Proportion is,

As the Sine of C is to the Logarithm of A B,

So is the Radius to the Logarithm of C B.

CASE VII. The Base A B 180, and the Perpendicular C A 135, being gi∣ven, to finde the Hypotenuse C A.

DRaw a right Line A B containing 180 equal parts, and upon the end A erect the Perpendicular A C, and out of your Scale of equal parts take the length thereof 135, which set from A to C, and draw the Line C B, which con∣stitutes

the Triangle C A B. Lastly, take the length of the Hypotenuse C B in your Compasses, and measuring it upon your Line of equal parts, you shall finde it to contain 225.

The Analogie or Proportion is,

1. Operation.

As the Logarithm of A B is to the Logarithm of C A,

So is the Radius to the Tangent of B.

2. Operation.

As the Sine of B is to the Logarithm of C A,

So is the Radius to the Logarithm of C B.

These are the severall Varieties or Cases that can at any time fall out in the Solution of Right-angled plain Triangles, wherefore we will now proceed to the Solution of Oblique plain Triangles.

II. Of Oblique-angled plain Triangles.

THE Triangle which I shall make use of in the Solution of the severall Cases appertaining to an Oblique-angled plain Triangle shall be this following, C D B, in which

		parts
D B, the Base,	contains	335
C B, the longer Side,	271
D C, the shorter Side,	100
And		deg.	m.
C, the obtuse Angle,	contain	122	00
D, the 2 acute Angles,	43	20
B, the 2 acute Angles,	14	40

[illustration] geometrical diagram

CASE I. Two Sides, as the Base D B 335, and the Side C B 271, and the Angle D 43 degr. 20 min. opposite to C B, to finde the Angle at C, opposite to the Base D B.

DRaw a right Line D B representing the Base of your Triangle, which, by help of your Scale of equal parts, make to contain 335. Then upon the Point D, with the distance of 60 degr. of your Line of Chords, describe the Arch k l, and from your Chords take 43 degr. 20 min. the quantity of the Angle at D, and set it upon the Arch∣line from l to k, drawing the Line C D. And because your other given Side B C contains 271 parts, take 271 out of your Line of equal parts, and setting one foot in B, with the other describe the Arch m n, crossing the for∣mer Arch k l in the Point C: then draw the Line C B. So shall you have constituted the Triangle C D B. Lastly, be∣cause it is the Angle at C that is required, take 60 degr. of your Chords, and upon C describe the Arch g h, and taking the distance between g and h, apply it to your Line of Chords,

and you shall finde it to reach from the beginning thereof be∣yond the end of the Line; wherefore take 90 degr. the whole Line, and set that distance from g to o; then take the remainder of the Arch o h, and measure that upon your Chord, and you shall finde it to contain 32 degr. which added to 90 degr. make 122 degr. and that is the quantity of the An∣gle at C, which was required.

The Analogie or Proportion is,

As the Logarithm of B C is to the Sine of D,

So is the Logarithm of D B to the Sine of C.

CASE II. The Base D B 335, and the Side D C 100, with the Angle D, 43 degr. 20 min. contained between them, to finde either of the other Angles at B and C.

DRaw a right Line, as D B, containing 335 of your Scale of equal parts, which shall be the Base of your Triangle. Then with 60 degr. of your Line of Chords, upon the Point D describe the Arch k l; and because the given Angle at D contains 43 degr. 20 min. take 43 degr. 20 min. from your Line of Chords, and set it from l to k, drawing the Line D k. Again, because the given Side D C contains 100, set 100 of your Line of equal parts from D to C; then drawing a right Line from C to B, you shall by that means find the oblique-angled Triangle C D B. Lastly, being the other two Angles at B and C are to be found, with 60 degr. of your Chord on the Point B describe the Arch e f; also upon the Point C describe the Arch g o h. Then if you take the distance be∣tween e and f in your Compasses, and measure it upon your Line of Chords, you shall finde it to contain 14 degr. 40 min. And that is the quantity of the Angle at B. Then being the Angle at C, which is also required, is obtuse, and contains a∣bove 90 degr. take 90 degr. out of your Line of Chords,

[illustration] geometrical diagram

and set that distance upon the Arch g h, from g to o: and ta∣king the other part of the Arch o h in your Compasses, mea∣sure that upon your Chord, and you shall find it to contain 32 degr. which added to 90 degr. makes in all 122 degr. And such is the quantity of the other enquired Angle at C.

The Analogie or Proportion is,

As the Log. of the Sum of the two Sides given, C D and C B, is to the difference of those Sides,

So is the Tang. of half the Sum of the two unknown Angles, C and B, to the Tangent of half their difference.

CASE III. The three Sides, D B 335, C B 271, and D C 100, being given, to finde any of the Angles, as B.

DRaw a right Line C D, containing 100 of your Line of equal parts. Then the other side of the Triangle being 271, take 271 out of your Scale of equal parts, and setting one foot of the Compasses in C, with the other describe the Arch r s. Also take the length of your Base 335 out of your Line of equal parts, and setting one foot of the Compasses in

D, with the other describe the Arch p q, crossing the former Arch r s in the Point B. Then draw the Lines C B and D B, and they shall form your Triangle C D B. Lastly, because the Angle at B is sought, take 60 degr. of your Line of Chords, and setting one foot of the Compasses in B, with the other describe the Arch e f. Then take the distance between e and f, and measure it upon the Line of Chords, and you shall find it to contain 14 degr. 40 min. which is the quantity of the Angle at B. And in the same manner might any of the other An∣gles at C or D have been found.

The Analogie or Proportion is,

As the Log. of the greater side D B is to the Sum of the other two Sides, D C and C B,

So is the difference of the two Sides to a fourth Sum.

Which fourth Sum being taken from the Base, will leave another number, the half whereof will be the place in the Base where a Perpendicular let fall from the obtuse Angle would fall upon the Base: and so the oblique Triangle is re∣duced into two right-angled, and may be resolved by the Precepts of right-angled Triangles.

CASE IV. The three Angles, C 122 degr. D 43 degr. 20 min. and B 14 deg. 40 min. being given, to finde any of the Sides, as B C.

IN this Case, where the three Angles are given, and none of the Sides, you are to take notice, that the Sides cannot be absolutely found themselves, but the Proportionality of them. Wherefore

Draw a Line, as D B, of any length, and taking 60 degr. of your Line of Chords, set one foot of the Compasses upon D, and with the other describe the Arch l k; also set one foot in B, and with the other describe the Arch e f. Then, be∣cause the Angle at D contains 43 degr. 20 min. take 43 degr.

[illustration] geometrical diagram

20 min. from your Line of Chords, and set them from l to k. Also the Angle at B being 14 degr. 40 min. take them likewise from your Line of Chords, and set them from e to f. This done, draw the Lines B f and D k, extending them till they meet one with another, which they will doe in the Point C. So have you constituted the Triangle C D B, the Sides whereof will be in proportion the one to the other as the Sides of this Triangle are.

CASE V. The two Sides D C 100, and C B 271, with the Angle at C 122 degr. being given, to finde the Base D B.

DRaw a right Line, as C B, containing 271 of your Line of equal parts, and on the Point C, with 60 degr. of your Line of Chords, describe the Arch g o h. Then, because the given Angle C contains 122 degr. it being 32 degr. above 90 degr. first take 90 degr. from off your Line of Chords, and set it upon the Arch from g to o; and then take 32 degr. from your Chord also, and set them upon the same Arch from o to h, and draw the Line C h: then take 100, the length of the other given Side, and set it from C to D, and draw the Line D B,

which will contain 335 of your Scale of equal parts. And that is the length of the Base D B, which was required.

The Analogie or Proportion.

You must first finde the two Angles at D and B. Then make choice of one of the Sides, as C D, to work your Pro∣portion by. Then

As the Sine of B is to the Sine's Complement of C,

So is the Log. of C D to the Log. of D B.

The End of the Doctrine of the Dimension of right-lined Triangles.

THE DIMENSION Of Sphericall TRIANGLES BY A LINE OF CHORDS. The Third EXERCISE.

AS the Sides of Right-lined Triangles were measured by Equal parts and their Angles by a Line of Chords; so the Sides of Sphe∣ricall Triangles being Arches of great Cir∣cles of the Sphere, both they and the An∣gles also may be measured by Chords onely.

Of Sphericall Triangles there be two kinds, as there were of Right-lined; namely, right-angled and oblique-angled: of both which there are 28 Cases, viz. 16 of right-angled, and 12 of oblique-angled. And to

observe the same Method in these as I did in plain Triangles, I will begin with the Solution

I. Of Right-angled Sphericall Triangles.

THE Right-angled sphericall Triangle which I shall make use of for my Examples shall be this following, A B C, whose Sides and Angles contain as followeth.

			deg.	m.
The Side	A B	containeth	30	00
B C	11	30
A C	27	54
The Angle at	A	containeth	23	30
B	69	22
C	90	00

[illustration] geometrical diagram

CASE I. The Base A C 27 degr. 54 min. and the Perpendicular C B 11 degr. 30 min. being given, to finde the Hypotenuse A B.

The Analogie or Proportion is,

As the Radius is to the Co-sine of B C 78 degr. 30 min.

So is the Co-sine of A C 62 degr. 6 min. to the Co-sine of A B.

First, take the Sum and the Difference of the second and third Terms in the Analogie or Proportion; namely, the Sum and Difference of the Complements of the Perpendicular B C 11 degr. 30 min. and the Base A C 27 degr. 34 min.

	degr.	m.
The Base A C is	27	54.	its Comp.	62	06
The Perpendicular B C is	11	30.	its Comp.	78	30
			Sum	140	36
			Differ.	16	24

HAving found this Sum and Difference, draw a right Line B A C; then take 60 degr. out of your Line of Chords, and setting one foot of the Compasses in A, with the other describe the Semicircle B D C, and upon the Centre A erect the Perpendicular A D. This done, take 16 degr. 24 min. (the Difference) out of your Line of Chords, and set them from B to e; also take the Sum 140 degr. 36 min. (or rather 39 degr. 24 min. the Complement thereof to 180 degr.) and set them from C to g, and from the Points e and g let fall the two Perpendiculars e f and g h: then divide the space be∣tween f and h into two equal parts in M, and set the distance

[illustration] geometrical diagram

M h or M f from A to k, and draw the Line k l parallel to A B, cutting the Semicircle in l: so shall D l be the quantity of the Hypotenuse A B, which if you measure upon your Line of Chords, you will finde to contain 30 degr.

CASE II. The Hypotenuse A B 30 degr. and the Angle at the Base A 23 degr. 30 min. being given, to find the Perpendicular B C.

The Analogie or Proportion is,

As the Radius is to the Sine of the Hypotenuse A B 30 degr.

So is the Sine of the Angle at A 23 degr. 30 min. to the Sine of the Perpendicular B C, 11. degr. 30 min.

Take the Sum and Difference of the Angle at A 23 degr. 30 min. and the Hypotenuse A B 30 degr.

	degr.	m.
The Sum is	53	30
The Difference is	6	30

HAving drawn your Semicircle B D C, as in the last Case, take 6 degr. 30 min. the Difference, out of your Line of Chords, and set them upon the Semicircle from B to n; also take 53 degr. 30 min. from your Line of Chords, and set them upon your Semicircle from B to o, and from the Points n and o let fall two Perpendiculars n p and o q. Then divide the space between p and q into two equal parts in s; then take the distance q s, and set it from A to t, and through the Point t draw a Line t r parallel to A B: so shall B r, be∣ing measured upon your Line of Chords, contain 11 degr. 30 min. which is the quantity of the Perpendicular B C.

CASE III. The Base A C 27 degr. 54 min. and the Angle at the Base A 23 d. 30 min. being given, to find the Angle at the Perpendicular B.

The Analogie or Proportion is,

As the Radius is to the Co-sine of A C 62 degr. 6 min.

So is the Sine of the Angle at A 23 degr. 30 min. to the Co∣sine of the Angle at B.

Take the Sum and Difference of the Co-sine of A C 62 d. 6 min. and of the Sine of the Angle at A 23 degr. 30 min.

	degr.	m.
The Sum is	85	36
The Difference is	38	36

BEing thus far prepared, and having drawn your Semicir∣cle B D C, out of your Line of Chords take 85 degr. 36 m. and set them upon your Semicircle from B to 2; also take the Difference 38 degr. 36 min. from your Chord, and set them from B to 3, and let fall the two Perpendiculars, 3 4, and 2 M. Then divide the distance between M and 4 into two

equal parts in the Point 5, and taking the distance M 5 in your Compasses, set it upon the Line A D, from A to 6. Lastly, draw the Line 6 7. parallel to B A; so shall D 7 being mea∣sured upon your Line of Chords contain 69 degr. 22 min. the quantity of the Angle at B required.

These three Cases are all (in Sines alone) that have the Ra∣dius in the first place of the Analogie or Proportion, and so con∣sequently all that can be wrought by this Artifice: wherefore those which follow must be resolved by other means.

CASE IV. The Perpendicular B C 11 degr. 30 min. and the Angle at the Base A 23 degr. 30 min. to finde the Hypotenuse A. B.

The Analogie or Proportion is,

As the Sine of the Angle at A 23 degr. 30 min. is to the Sine of the Side B C 11 degr. 30 min.

So is the Radius to the Sine of A B.

FIrst, draw a right Line B A, and upon the Point A, with 60 degr. of your Line of Chords, describe the Quadrant A B C.

Then take 23 degr. 30 min. the quantity of the given Angle at A, out of your Line of Chords, and set them upon the Quadrant from B to E. Also take 11 degr. 30 min. the quan∣tity of the given Side B C, out of your Line of Chords, and set them from B to D; and draw the Lines E F and D G both parallel to B A.

Then with your Compasses take the distance between A and F, and setting one foot in C, with the other describe the Arch M, and from A draw the Line A L so that it may onely touch the Arch M. Then taking the distance A G in your Com∣passes, set one foot of them upon the Line A C, and draw it

[illustration] geometrical diagram

gently and softly along the Line A C, till the other foot, being turned about, will onely just touch the Line A L; and when it so toucheth, mark where the other Point resteth upon the Line A C, which you shall finde it to doe at the Point 6. Lastly, through the Point 6 draw the Line 62, parallel to A B. So shall the distance 2 B 6, being measured upon the Line of Chords, contain 30 degr. the quantity of the Hypotenuse A B, which was the Side required.

CASE V. The Perpendicular B C 11 degr. 30 min. and the Angle at the Base A 23 degr. 30 min. being given, to finde the Angle at the Perpendicular B.

The Analogie or Proportion is,

As the Co-sine of B C 78 degr. 30 min. is to the Co-sine of A 66 degr. 30 min.

So is the Radius to the Sine of the Angle at the Perpendicu∣lar B.

[illustration] geometrical diagram

HAving drawn your Quadrant A B C, take from your Line of Chords 78 degr. 30 min. the Complement of the given Side B C, and set it from B to a. Also take from your Chords 66 degr. 30 min. the Complement of the given An∣gle at A, and set them from B to b, and through the Points a and b draw the Lines a c and b d parallel to B A. Then ta∣king in your Compasses the distance A c, set one foot in C, and with the other foot describe the Arch p, and draw the Line A O so that it onely touch the Arch p. Likewise, take in your Compasses the distance A d, and with that distance, setting one foot upon the Line A C, move it gently along that Line till the other foot, being turned about, do onely touch the Line A O. So shall you finde the point of the Com∣passes to rest in the Point e, through which Point draw the Line e o parallel to B A. Then measure the distance B o up∣on your Line of Chords, and you shall finde it to contain 69 degr. 22 min. the quantity of the enquired Angle at B.

CASE VI. The Angle at the Base A 23 degr. 30 min. and the Angle at the Perpendicular B 69 degr. 22 min. being given, to finde the Base A C.

The Analogie or Proportion is,

As the Sine of the Angle at A 23 degr. 30 min. is to the Co-sine of the Angle at B 20 degr. 38 min.

So is the Radius to the Co-sine of the Base A C.

YOur Quadrant being drawn, take 23 degr. 30 min. the quantity of the Angle at A, and set it from B to E. Also take 20 degr. 38 min. the Complement of the Angle at B, out of your Line of Chords, and set them upon your Qua∣drant from B to R, and through the Points E and R draw the Lines E F and R P parallel to A B.

Then take in your Compasses the distance A F, and setting one foot in C, with the other describe the Arch M, and draw the Line A L so that it onely touch the Arch M. Then in your Compasses take the distance A P, and setting one foot upon the Line A C, move it along till the other, being turned about, do onely touch the Line A L; and when it so toucheth, note upon what part of the Line A C the Compass-point resteth, which you will finde it to doe at the Point T; through that Point draw the Line T S parallel to B A. So shall the distance B S, being measured upon your Line of Chords, contain 62 d. 6 min. the Complement of the enquired Side: or, S C will give 27 degr. 54 min. the Side it self.

CASE VII. The Base A C 27 degr. 54 min. and the Hypotenuse A B 30 degr. being given, to finde the Angle at the Perpendicular B.

The Analogie or Proportion is,

As the Sine of the Hypotenuse A B 30 degr. is to the Radius, So is the Sine of the Base A C 27 degr. 54 min. to the Sine of the Angle at the Perpendicular B.

HAving drawn your Quadrant, take from your Line of Chords 30 degr. the quantity of the Hypotenuse, and set them upon your Quadrant from B to 2; also take 27 degr. 54 min. the quantity of the Base, from your Chords, and set them from B to K, and draw the Lines 2 6 and K H, both parallel to the Line B A. Then taking in your Compasses the distance A 6, set one foot of them in C, and with the other de∣scribe the Arch Z, and draw the Line A X, so that it onely touch the Arch Z. Then taking in your Compasses the di∣stance A H, set one foot upon the Line A C, moving it along till the other foot, being turned about, will onely touch the Line A Z; and where the point of the Compasses resteth up∣on the Line A C, which it will doe at e, through that Point draw the Line o e parallel to B A. So shall B o, being mea∣sured upon your Line of Chords, give you 69 degr. 22 min. the quantity of the enquired Angle at B.

CASE VIII. The Base A C 27 degr. 54 min. and the Hypotenuse A B 30 degr. being given, to finde the Perpendicular B C.

The Analogie or Proportion is,

As the Co-sine of the Base A C 62 degr. 6 min. is to the Radius, So is the Co-sine of the Hypotenuse 60 degr. to the Co-sine of the Perpendicular B C.

HAving drawn your Quadrant A B C, take out of your Line of Chords 62 degr. 6 min. the Co-sine of the Base A C, and set them from B to S. Also take from the Chords 60 degr. the Co-sine of the Hypotenuse, and set them from B to m: and draw the Lines S T and m n both parallel to B A. Then taking the distance A T in your Compasses, set one foot in C, and with the other describe the Arch V, and draw the Line A W so that it may onely touch the Arch V. Then taking A n in your Compasses, move one foot thereof gently along the Line C A, till the other, being tur∣ned about, doth onely touch the Line A W; and where the Point resteth upon the Line C A, which you will finde to be at c, there make a mark, and draw the Line c a paral∣lel to B A. Lastly, take the distance from B to a; and mea∣sure it upon your Line of Chords, where you shall finde it to contain 78 degr. 30 min. the Complement of the Perpendi∣cular; or, C A measured upon the Chord will give you 11 d. 30 min. the Perpendicular it self.

These Five last are all the Cases in a Right-angled Spheri∣call Triangle that are resolvable by Sines alone. Those which follow are to be resolved by Sines and Tangents joyntly, and so will require another manner of Operation then the former.

CASE IX. The Hypotenuse A B 30 degr. and the Angle at the Base A 23 d. 30 min. being given, to finde the Angle at the Perpendicular B.

The Analogie or Proportion is,

As the Radius is to the Co-sine of the Hypotenuse A B 60 deg. So is the Tangent of the Angle at the Base A 23 degr. 30 min. to the Co-tangent of the Angle at the Perpendicular B.

FIrst, draw a right Line, as C H, and upon one end there∣of (as at C) erect the Perpendicular C A, and with the distance of 60 degr. of your Line of Chords, upon the Cen∣tre C describe the Quadrant C A D. Also upon the Point A with 60 degr. of your Chord describe the Quadrant A B C.

[illustration] geometrical diagram

Being thus prepared, First, take 60 deg. the Co-sine of the Hypotenuse A B, and set them from A to O; also take 23 d. 30 m. the quan∣tity of the An∣gle at A, and set them from C to r; and draw the Line A M F, and the Line O G, parallel to A C.

Then take in your Compasses the distance be∣tween C and G, and setting one foot in D, with the other de∣scribe the Arch K, and draw the Line A L so that it onely touch the Arch K. Then placing one foot of the Compasses in F, take the least distance you can to the Line C L, which set from C to E. Lastly, draw the Line A E, cutting the Quadrant C B in N. So the distance

C N measured upon your Chords shall give you 20 degr. 38 min. the Complement of the Angle at B, which was re∣quired; or, the distance B N will give you 69 degr. 22 min. the Angle it self.

CASE X. The Hypotenuse A B 30 degr. and the Angle at the Base A 23 d. 30 min. being given, to finde the Base A C.

The Analogie or Proportion is,

As the Radius is to the Co-sine of the Angle at A 66 d. 30 m. So is the Tangent of A B the Hypotenuse 30 degr. to the Tan∣gent of the Base A C.

HAving prepared your Quadrants, out of your Line of Chords take 66 degr. 30 min. the Complement of the Angle at A, and set them from A to a, and draw the Line a b parallel to A C: also take 30 degr. the Hypotenuse from your Chord, and set them from C to c, and draw the Line A c, prolonging it to d. This done, take in your Compasses the distance from C to d, and setting one foot in D, with the other describe the Arch P, and draw the Line C Q so that it may onely touch the Arch P. Then setting one foot of the Compasses in b, take the least distance between b and the Line C Q, which distance will reach from C to e. Lastly, draw the Line A e, cutting the Quadrant A C B in the Point M. So shall the distance C M, being taken in the Compasses and measured upon the Line of Chords, contain 27 degr. 54 m. which is the quantity of the Base required.

CASE XI. The Base A C 27 degr. 54 min. and the Angle at the Base A 23 degr. 30 min. being given, to finde the Perpendicular B C.

The Analogie or Proportion is,

As the Radius is to the Sine of the Base A C 27 degr. 54 min.

So is the Tangent of the Angle at A 23 degr. 30 min. to the Tangent of the Perpendicular B C.

[illustration] geometrical diagram

YOur Qua∣drants be∣ing prepared, out of your Line of Chords take 27 d. 54 m. the quantity of the given Base A C, and set them from A to S, drawing the Line S R paral∣lel to C D. Al∣so take out of your Chords 23 degr. 30 mi∣nutes, the quan∣tity of the gi∣ven Angle at A, and set them from C to r, and draw the Line A r, prolonging it to F.

This done,

take in your Compasses the distance A R, and setting one foot in D, with the other describe the Arch T, and by the side thereof draw the Line C V onely to touch it. Then set one foot of your Compasses in F, and with the other take the nearest distance to the Line C V, which distance will reach from C to X. Lastly, draw the Line A X, which will cut the Quadrant A B C in the Point Z. So shall C Z, being mea∣sured upon the Line of Chords, contain 11 degr. 30 min. which is the quantity of the Perpendicular B C, which was required.

CASE XII. The Base A C 27 degr. 54 min. and the Perpendicular B C 11 d. 30 min. being given, to finde the Angle at the Base A.

The Analogie or Proportion is,

As the Sine of the Base A C 27 degr. 54 min. is to the Ra∣dius,

So is the Tangent of the Perpendicular B C 11 degr. 30 min. to the Tangent of the Angle at the Base A.

DRaw the two Quadrants A B C and C A D, as before. Then take out of your Line of Chords 27 degr. 54 m. the quantity of the given Base A C, and set them upon the Quadrant from B to E, and draw the Line E F parallel to C D. Also take 11 degr. 30 min. the quantity of the Per∣pendicular B C, and set them upon the Quadrant from C to N, and draw the Line A N, cutting the Line C H in G.

This done, take in your Compasses the distance A F, and setting one foot in D, with the other describe the Arch M, and close by the out-side of it draw the Line C L. Then

take in your Com∣passes the distance C G, and setting one foot of them upon the Line C D, move it along till the other foot, be∣ing turned about, will onely touch the Line C L; and when it so toucheth, mark where the other foot resteth upon the Line C D, which it will do at K. Last∣ly, draw the Line A K, cutting the Quadrant C B in the Point O. So shall C O, being measu∣red on the Line of Chords, contain 23 degr. 30 min. the quantity of the en∣quired Angle at A.

[illustration] geometrical diagram

CASE XIII. The Base A C 27 degr. 54 min. and the Angle at the Base A 23 d. 30 min. being given, to finde the Hypotenuse A. B.

The Analogie or Proportion is,

As the Co-sine of the Angle at A 66 d. 30 m. is to the Radius, So is the Tangent of the Base A C 27 degr. 54 min. to the Tangent of the Hypotenuse A B.

THE Quadrants being drawn, take the quantity of the given Base A C 27 degr. 54 min. out of the Line of Chords, and set them upon the Quadrant from C to V, and draw the Line A V, continuing it till it cut the Line C D in S. Also take 66 degr. 30 min. the Complement of the given Angle at A, out of the Chords, and set them from B to O, and draw the Line O P parallel to C D or A B.

This done, take the distance A P in your Compasses, and setting one foot in D, with the other describe the Arch R, drawing the Line C Q onely to touch it. Then take in your Compasses the distance C S, and placing one foot of them upon the Line C D, move it along the same, till the other, being turned about, will onely touch the Line C Q; and where the Compass-point resteth upon the Line C D, (which you will finde it to doe at T,) there make a mark, and from it draw the Line A T, cutting the Quadrant A B C in the Point X. So shall C X, being measured on the Line of Chords, give you 30 degr. the quantity of the Hypotenuse required.

CASE XIV. The Perpendicular B C 11 degr. 30 min. and the Angle at the Base A 23 degr. 30 min. being given, to finde the Base A C.

The Analogie or Proportion is,

As the Tangent of the Angle at A 23 degr. 30 min. is to the Tangent of the Perpendicular B C 11 degr. 30 min.

So is the Radius to the Sine of the Base A C.

HAving drawn your Quadrants, take out of your Line of Chords 23 degr. 30 min. the quantity of the Angle at A, and set them from C to O, and draw the Line A O, continu∣ing it till it cut the Line C D in K. Also take out of your Line

of Chords 11 degr. 30 min. the quantity of the Perpendicu∣lar given, and set them from C to N, drawing the Line A N, and prolonging it to G.

[illustration] geometrical diagram

This done, take in your Compasses the distance be∣tween C and K, and setting one foot in D, with the other describe the Arch M, and draw the Line C L. Then take in your Com∣passes the distance C G, and setting one foot upon the Line C D, move it gen∣tly along the same, till the other, being turned about, do on∣ly touch the Line C L; and when it so toucheth, keeping the Compasses at the same distance, set it from A to F. Lastly, draw the Line F E parallel to A B. So shall the distance B E, being measured on your Line of Chords, contain 27 degr. 54 min. the quantity of the Base A C, which was required.

CASE XV. The Base A C 27 degr. 54 min. and the Hypotenuse A B 30 degr. being given, to finde the Angle at the Base A.

The Analogie or Proportion is,

As the Tangent of the Hypotenuse A B 30 degr. is to the Tan∣gent of the Base A C 27 degr. 54 min.

So is the Radius to the Co-sine of the Angle at A.

THE Quadrants being drawn, out of your Line of Chords take 30 degr. the quantity of the Hypotenuse given, and set them upon the Quadrant from C to X. Also take 27 degr. 54 min. the quantity of the given Base, from the Line of Chords, and set them upon the same Quadrant from C to V, and draw the Lines A X and A V, prolonging them to T and S.

This done, take in your Compasses the distance between C and T, and setting one foot in D, with the other describe the Arch Z, and draw the Line C Y onely to touch it. Then take in your Compasses the distance C S, and placing one foot upon the Line C D, move it gently along, till the other, being turned about, do onely touch the Line C Y; and where the Point rest∣eth upon the Line C D make a mark, which will be at*. Then take the distance C*, and set it from A to P, and draw the Line P O parallel to C D. So shall B O, being measured up∣on the Line of Chords, give you 66 degr. 30 min. the Com∣plement of the Angle at A, which was required; or, the distance O C upon the Chord shall give you 23 degr. 30 min. the Angle it self.

CASE XVI. The Angle B at the Perpendicular 69 degr. 22 min. and the An∣gle at the Base A 23 degr. 30 min. being given, to finde the Hy∣potenuse A B.

The Analogie or Proportion is,

As the Co-tangent of the Angle at B 20 degr. 38 min. is to the Tangent of the Angle at A 23 degr. 30 min.

So is the Radius to the Co-sine of the Hypotenuse A B.

HAving drawn the two Quadrants, as before, take out of your Line of Chords 23 degr. 30 min. the quantity of the Angle at A, and set them from C to O. Also take 20 degr. 38 min. from your Chord, (the Complement of the Angle at B) and set them from C to a, and draw the Lines A O and A a, continuing them to c and K upon the Line C D.

This done, take in your Compasses the distance C K, and setting one foot in D, with the other describe the Arch M, and draw the Line C L that it onely touch the Arch M. Al∣so take in your Compasses the distance C c, and setting one foot upon the Line C D, move it gently along the same till the other, being turned about, do onely touch the Line C L; and where the Compass-point resteth upon the Line C D make a mark, as at g. Lastly, take the distance C g, and set it from C to m, and draw the Line m n parallel to A B. So shall the distance D n, measured on the Line of Chords, contain 60 d. the Complement of the Hypotenuse required; or, A n shall contain 30 degr. the Hypotenuse it self.

These 16 Cases are all that can be proposed in a Right-an∣gled Sphericall Triangle. There are 12 other Cases which belong to Oblique-angled Sphericall Triangles, which we now come to resolve.

II. Of Oblique-angled Sphericall Triangles.

THE Triangle that I shall make use of in the Solution of the 12 Cases of an Oblique-angled Sphericall Triangle shall be this, A B E: whose Sides and Angles are as followeth.

			degr.	m.
The Side	A B, the Base,	contains	30	00
B E	18	47
A E	42	51
The Angle	A	contains	23	30
B	122	36
E	38	15

[illustration] geometrical diagram

CASE I. The Angle at E 38 degr. 15 min. the Angle at A 23 degr. 30 m. and the Side A B 30 degr. being given, to finde the Base B E.

The Analogie or Proportion is,

As the Sine of the Angle at E 38 degr. 15 min. is to the Side A B 30 degr.

So is the Sine of the Angle at A 23 degr. 30 min. to the Sine of the Side B E.

FIrst, draw the Quadrant A B C: then out of your Line of Chords take 38 degr. 15 min. the quantity of the Angle at E, and set it upon the Quadrant from B to F. Also take 30 degr. the quantity of the given Side A B, out of your Line of Chords, and set them upon the Quadrant from B to E, and draw the Lines F G and E H both of them parallel to A B.

[illustration] geometrical diagram

This done, take in your Compasses the distance A H, and

setting one foot of that extent in G, with the other describe the Arch L, and draw the Line A M so that it may onely touch the Arch L. Then setting one foot of the Compasses in I, with the other take the nearest distance to the Line A M: this di∣stance set from A to K, and draw the Line N K parallel to A B. So shall the distance B N, measured upon the Line of Chords, contain 18 degr. 47 min. the quantity of the enqui∣red Side B E.

CASE II. The Side A B 30 degr. the Angle at E 38 degr. 15 min. and the Side B E 18 degr. 47 m. being given, to finde the Angle at A.

The Analogie or Proportion is,

As the Sine of the Base A B 30 degr. is to the Sine of the An∣gle at E 38 degr. 15 min.

So is the Sine of the Side B E 18 degr. 47 min. to the Sine of the Angle at A.

YOur Quadrant being drawn, from your Line of Chords take 30 degr. the quantity of the Side A B, and set them from B to E. Also take 38 degr. 15 min. the quantity of the Angle at E, and set them from B to F. Likewise take 18 degr. 47 min. the quantity of the given Side B E, from your Line of Chords, and set them from B to N; and draw the Lines E H, and F G, and N K, all parallel to A B.

This done, take in your Compasses the distance A H, and setting one foot in G, with the other describe the Arch L, draw∣ing the Line A M onely to touch the Arch. Then take the di∣stance A K in your Compasses, and setting one foot of them upon the Line A C, move it along that Line gently, till the other Point, being turned about, will onely touch the Line A M; and where the Compass-point resteth upon the Line A C, which it will doe at the Point I, through I therefore draw

the Line I D parallel to B A. So shall B D, being measured on your Line of Chords, contain 23 degr. 30 min. the quan∣tity of the enquired Angle at A.

CASE III. The Side A B 30 degr. and the Side B E 18 degr. 57 min. and the Angle at A 23 degr. 30 min. being given, to finde the Base A E.

The Analogie or Proportion is,

As the Co-sine of A B 60 degr. is to the Co-sine of B E 71 d. 13 min.

So is the Co-sine of A C 62 degr. 6 min. the place where a Perpendicular would fall from B, to the Co-sine of C E 75 degr. 3 min. Which C E, being added to A C, will give the quantity of the Side A E.

HAving drawn your Quadrant, take 60 degr. the Com∣plement of A B, out of your Line of Chords, and set

[illustration] geometrical diagram

them from B to P. Also take 71 degr. 13 m. the Complement

of B E, and set them from B to S. In like manner take 62 d. 6 m. and set them from B to Q; and draw the Lines P O, S T, and Q R, all of them parallel to B A.

This done, take in your Compasses the distance A O, and fetting one foot in T, with the other describe the Arch Y; and draw the Line A Z onely to touch the Arch Y. Then take the distance A R, and setting one foot upon the Line A C, move it gently along the same, till the other foot, being tur∣ned about, do onely touch the Line A Z; and where the Com∣pass-point resteth, which it will doe at V, through that Point draw a Line V X parallel to B A. So shall the distance C X, measured upon your Line of Chords, contain 14 degr. 57 min. which added to the former Segment of the Side A C 27. degr. 54 min. the Sum will be 42 degr. 51 min. the quantity of the whole Side A E, which was required.

CASE IV. The Angle at 23 degr. 30 min. the Angle at B 122 degr. 36 m. and the Side A B 30 degr. given, to finde the Angle at E.

By the foregoing Cases in Rectangled Sphericall Triangles (if you let fall a Perpendicular from the Angle B upon the Side A E, as B C,) you may finde the Angle A B C 69 degr. 22 min. and the Angle C B E 53 degr. 14 min. which being obtained,

The Analogie or Proportion is,

As the Sine of the Angle A B C 69 degr. 22 min. is to the Sine of the Angle C B E 53 degr. 14 min.

So is the Co-sine of A 66 degr. 30 min. to the Co-sine of the Angle at E.

THE Angles being found and the Quadrant drawn, take 53 degr. 14 min. out of your Chord, and set them from B to E. Likewise take 69 degr. 22 min. and set them from B

to G. Also take 66 degr. 30 min. and set them from B to F; and draw the Lines G H, F I, and E K, all parallel to B A.

This done, take in your Compasses the distance between A and K, and setting one foot in H, with the other describe the Arch M, and draw the Line A N onely to touch the Arch. Again, setting one foot of the Compasses in I, with the other take the least distance to the Line A N, which distance set from A to L. Lastly, draw the Line L D parallel to B A, cutting the Quadrant in D. So shall the Arch B D, measu∣red upon your Chords, contain 51 degr. 45 min. the Com∣plement of the enquired Angle at E; or, D C shall give you 38 degr. 15 min. the Angle it self.

CASE V. The Angle at A 23 degr. 30 min. the Angle at E 38 degr. 15 m. and the Side A B 30 degr. to finde the Angle A B E.

By the preceding Cases of Right-angled Sphericall Tri∣angles, the Perpendicular B C being let fall, finde the Angle A B C, which will be found to be 69 degr. 22 min. This Angle being obtained,

The Analogie or Proportion is,

As the Co-sine of the Angle at A 66 degr. 30 min. is to the Co∣sine of the Angle at E 51 degr. 45 min.

So is the Sine of A B C 69 degr. 22 min. to the Sine of the Angle C B E.

THE Angle A B C being found, and the Quadrant drawn, take 66 degr. 30 min. out of your Line of Chords, and set them from B to F. Also take 51 degr. 45 min. from the Chord, and set them from B to D. Likewise take 69 degr. 22 min. (the quantity of the Angle found A B C) from your

Chord, and set them from B to G, and draw the Lines F I, D L, and G H, all parallel to A B.

This done, take in your Compasses the distance A L, and setting one foot of that extent in I, with the other describe the Arch M, and draw the Line A N onely to touch the Arch. Again, set one foot of the Compasses in H, and with the other take the least distance to the Line A N, which had, it will reach from A to K; through which Point K draw the Line K E parallel to B A, cutting the Quadrant in E. So shall the distance B E, being measured on your Line of Chords, contain 53 degr. 14 min. the quantity of the remaining part of the Angle A B E; which being added to the Angle A B C (before found) 69 degr. 22 min. the Sum of them will be 122 degr. 36 min. the quantity of the Angle A B E, which was required.

These Five last Cases are all that in Oblique-angled Sphericall Triangles are resolvable by Sines onely, except the Three last, which we resolve by another kind of Artifice. The 6. 7. 8. and 9. fol∣lowing have Tangents ingredient in the Proportion, and so do require a different way of resolving, which is as followeth.

CASE VI. The Side A B 30 degr. the Base A E 42 degr. 51 min. and the Angle at A 23 degr. 30 min. being given, to find the Angle at E.

The Perpendicular B C being let fall from the Angle at B, you must find the Segment of the Base C E 14 degr. 57 min. by the preceding Cases of Right-angled Sphericall Triangles: Which had,

The Analogie or Proportion is,

As the Sine of the Side C E found 14 degr. 57 min. is to the Sine of A C 27 degr. 54 min.

So is the Tangent of the Angle at A 23 degr. 30 min. to the Tangent of the Angle at E.

YOur Quadrants A B C and C A D being drawn, take 14 degr. 57 min. the quantity of C E found, out of your Line of Chords, and set them from B to E. Also take 27 d. 54 min. the Segment of the Base A E, and set them from B to H; and draw the Lines E F and G H both parallel to A B. Also, take out of your Line of Chords 23 d. 30 min. the quantity of the Angle at A, and set them from C to N; and draw the Line A N, continuing it to I.

[illustration] geometrical diagram

This done, take the distance A G, and set it from C to O. Also take the distance A F, and setting one foot in O, with the other de∣scribe the Arch L; and draw the Line C M only that it may touch the Arch L. Then take in your Compasses the distance C I, set one foot upon the Line C D, moving it along the same till the other, being turned about, do onely touch the Line C M; and where the other Point of the Compasses rest∣eth upon the Line C D, which you will find it to doe at K, draw the Line A K, cutting the Quadrant in P. So shall C P,

being measured upon your Line of Chords, contain 38 degr. 15 min. the quantity of the Angle E, which was required to be found.

CASE VII. The Angle at A 23 degr. 30 min. the Angle at B 122 degr. 36 m. and the Side A B 30 degr. being given, to find the Side B E.

Having let fall the Perpendicular B C, and found the Angle B C E by the former Cases, then

The Analogie or Proportion is,

As the Co-sine of C B E 38 degr. 46 min. is to the Co-sine of A B C 20 degr. 38 min.

So is the Tangent of A B 30 degr. to the Tangent of B E.

THE Quadrants drawn, out of your Line of Chords take 38 degr. 46 min. the Complement of the Angle C B E, and set them from B to S. Also take from your Chords 20 degr. 38. min. the Complement of the Angle A. B C, and set them from B to Q, and draw the Lines P Q and R S both parallel to C D. Again, take in your Compasses the di∣stance A R, and set it from C to T. Likewise take the distance from A to P, and setting one foot of that distance in T, with the other describe the Arch Y, and by it draw the Line C M onely to touch the Arch Y. Then take 30 degr. from your Chord, and set them from C to X, drawing the Line A X, and prolonging it to Z. Lastly, from the Point T take the nearest distance to the Line C M, which distance set from C to V, and draw the Line A V, cutting the Quadrant in AE. So shall C AE contain 18 degr. 47 min. of your Line of Chords, which is the quantity of the enquired Side B E.

CASE. VIII. The Side A B 30 degr. the Side B E 18 degr. 47 min. and the Angle at A 23 degr. 30 min. being given, to finde the Angle at B.

Having let fall the Perpendicular B C, and found the Angle A B C 69 degr. 22 min. by the foregoing Cases of Right-an∣gled Sphericall Triangles,

The Analogie or Proportion is,

As the Tangent of B E 18 degr. 47 min. is to the Tangent of A B 30 degr.

So is the Co-sine of A B C 20 degr. 38 min. to the Co-sine of C B E.

[illustration] geometrical diagram

HAVING drawn your Quadrants, out of your Line of Chords take 18 degr. 47 min. the Side B E, and set them from C to M. Also take 30 d. the quantity of the Side A B, from your Chord, and set them from C to N; and draw the Lines A M and A N, continuing them to G and H. Likewise take 20 degr. 38 min. out of your Line of Chords, and set them from B to F, and draw F E parallel to A B.

This done, take in your Compasses the distance C C, and setting one foot in the Point H, with the other describe the Arch K, and draw the Line C L onely to touch it. Then take in your Compasses the distance A E, and setting one foot of the Compasses upon the Line C D, move it along the same gently, till the other, being turned about, do onely touch the Line C L; and where the Compass-point resteth, which you will finde it to doe at O, make a mark, and take the distance O C in your Compasses, and set it from A to P, and draw the Line P Q parallel to A B. So shall B Q, measured upon the Line of Chords, contain 36 degr. 46 min. the Comple∣ment of the Angle C B E; or, the distance C Q, measured upon the Chords, will give 53 degr. 14 min. the Angle C B E it self; which being added to 69 degr. 22 min. the Angle A B C, the Sum will be 122 degr. 36 min. the quantity of the whole Angle A B E required.

CASE IX. The Angle at A 23 degr. 30 min. the Angle at E 38 degr. 15 m. and the Side B A 30 degr. being given, to find the Side E A.

The Perpendicular being let fall, and A C 27 degr. 54 min. a part of the Side A E; then

The Analogie or Proportion is,

As the Tangent of the Angle at E 38 degr. 15 min. is to the Tangent of the Angle at A 23 degr. 30 min.

So is the Sine of A C 27 degr. 54 min. to the Sine of E C.

HAving drawn your Quadrants, take out of your Life of Chords 38 degr. 15 min. the quantity of the Angle at E, and set them from C to a. Also take 23 degr. 30 min. the quan∣tity of the Angle at A, and set them from C to b, drawing the Lines A b and A a, and continuing of them to T and V.

This done, take in your Compasses the distance C T, and setting one foot of them in V, with the other describe the Arch X, and draw the Line C Y onely to touch the Arch X. Then take in your Compasses the distance A R, and set it from C to AE, and from the Point AE take the least distance to the Line C Y, which distance set from A to m, and draw the Line m n parallel to A B. So B n, being measured upon your Line of Chords, shall contain 14 degr. 57 min. a por∣tion of the Side A E; which being added to the other portion A C 27 degr. 54 min. the Sum of them is 42 degr. 51 min. the whole Side A E, which was required.

[illustration] geometrical diagram

CASE X. The Side A B 30 degr. the Side B E 18 d. 47 m. and the Angle at B 122 d. 36 m. contained by them being given, to find the Base.

Take the Sum and the Difference of the two Sides A B 30 d. and B E 18 degr. 47 min.

	degr.	m.
Their Sum is	48	47
Their Difference is	11	13

FIrst, draw a right Line A O C, and upon the Centre O de∣scribe the Semicircle A B C, drawing the Semidiameter or Perpendicular B O.

[illustration] geometrical diagram

Being thus prepared, take out of your Line of Chords 48 d. 47 min. the Sum of the two Sides given, and set them upon the Semicircle from A to E. Also take 11 degr. 30 min. the Difference of the two given Sides, and set them from A to D. Again, take 122 degr. 36 min. the quantity of the given An∣gle, and set them from A to K; [but A B being 90 degr. set the residue, namely, 32 degr. 26 min. from B to K,] and draw the Lines D I, E H, and K L, all of them perpendicu∣lar to the Line A C.

This done, take in your Compasses the distance between I and H, and setting one foot of that extent in C, with the other describe the Arch M, and draw the Line A N so that

t may onely touch the Arch M. Then setting one foot of the Compasses in L, with the other take the nearest distance you can from L to the Line A N, and set that distance from I to G. Lastly, upon the Point G erect the Perpendicular G F, cutting the Semicircle in F. So shall the distance A F, being measured upon your Line of Chords, contain 42 degr. 51 m. the quantity of the Base of the Triangle A E, which was re∣quired.

To work this by the Canon.

The Analogie or Proportion is,

1. As the Radius is to the Co-sine of the Angle given,

So is the Tangent of one of the given Sides to the Tangent of a fourth Arch.

2. As the Co-sine of this fourth Arch is to the Co-sine of the former given Side,

So is the Co-sine of the other given Side, the fourth Arch be∣ing subtracted therefrom, to the Co-sine of the Side sought.

CASE XI. The three Sides A B 30 degr. B E 18 degr. 47 min. and A E 42 d. 51 min. being given, to finde the Angle at E.

First, find the Sum and the Difference of the Sides B E and A E.

	degr.	m.
Their Sum is	61	38
Their Difference is	24	04

THE Sum and Difference of the two Sides being taken, out of your Line of Chords take 24 degr. 4 min. the Difference of them, and set them from A to P. Also from your Line of Chords take 61 degr. 38 min. the Sum of the two Sides, and set them from A to X. Again, take 30 degr.

the quantity of the third Side A B, and set them from A to Q; and draw the Lines P S, Q T, and X Y, all three per∣pendicular to A C.

[illustration] geometrical diagram

This done, take the distance between Y and S, and setting one foot of the Compasses in C, with the other describe the Arch Z, and draw the Line A AE so that it may onely touch the Arch Z. Then take in your Compasses the distance be∣tween S and T, and setting one foot thereof upon the Line A C, move it gently along the same, till the other foot, being turned about, do onely touch the Line A AE; and where the Compass-point resteth, which you will find it to doe at the Point Y, upon this Point Y erect the Perpendicular Y R. So shall the distance A R, being measured upon your Line of Chords, give 38 degr. 15 min. the quantity of the Angle at E, which was required to be found.

To work this by the Canon.

The Analogie or Proportion is,

As the Rectangle contained under the Sines of the Sides is to the Square of the Radius,

So is the Rectangle contained under the Sines of the half Sum of the three Sides, and the Difference between this half Sum and the Base, to the Square of the Co-sine of half the An∣gle required.

CASE XII. The three Angles A 23 degr. 30 min. B 122 degr. 36 min. and the Angle E 38 degr. 15 min. being given, to finde any of the Sides.

IF for either of the Angles next the Side required we take its Complement to 180 degr. those Angles will be turned into Sides, and the Sides into Angles: And then will the man∣ner of resolving the Triangle be the same as in the last Case.

Postscript.

BY the Precepts before in this Treatise delivered you have the whole Doctrine both of plain Right-lined Triangles, and also of Sphericall both Right and Oblique-angled, wrought by a way not usually practised, which if it be carefully performed, you may by a Line of Chords of 3 or 4 inches Radius come within a few minutes of Calculation; which will be of suffici∣ent use for any Practice either in Astronomicall, Geographicall or Nauticall Questions, wrought thereby. And if in the wor∣king of any of the Cases there be any difficulty, it is in the moving of the Compasses along one Line, while the other foot, being turned about, doth only touch another Line. But this is not a real, but a seeming, difficulty; for in 3 or 4 times trial you will finde it very expeditious and easie: and although it be not so Geometricall as the way by drawing of Parallels, yet it is alto∣gether as exact; more ready and speedy in performance, and avoids the drawing of multiplicity of Lines, which would much cumber the Diagrams to no purpose. If the Schemes that are here drawn seem to be cumbred with multiplicity of Lines, the reason is, because many of them serve for 3, 4, or 5 Cases; but if a particular Diagram had been drawn for every Case, then none of them would have consisted of above 5 Lines besides the Quadrant, as you will see by trying of any Case. And here note, that every one of these Cases, both in Right-lined and Sphericall Triangles, is applicable to some one thing or other in the Practice of the Mathematicks. As, by plain Right-lined Tri∣angles all Propositions are resolved that concern the taking of Heights, Depths and Distances, Measuring of Land, Sailing by the plain Sea-Chart, and also by Mercator's, and divers other particulars. The Solution of Sphericall Triangles resolves Que∣stions in Astronomy, Geography, Dialling, Sailing by the Arch of a great Circle, with many other particulars too tedious here

to be recited; of some of which (especially those that concern the taking of Distances in Astronomy, Geography, Navigation, &c.) I shall in the following Treatises exemplifie their Ʋse.

And because the Doctrine of Sphericall Triangles is more difficult then that of Plain Triangles, I will in this next Treatise following shew you how (by your Line of Chords onely) you may draw or project the Globe or Sphere upon a Plain; by which means you may see before you how the Sphericall Triangles lie in the Sphere it self; which will be a great help to the un∣derstanding of the nature of a Sphericall Triangle. The Sphere being thus projected upon a Plain, I will shew how (upon that Plain) to measure both the Sides and Angles of a Spheri∣call Triangle, and, in so doing, resolve some Questions in Astro∣nomy; by which you may discern, in that particular, how sub∣servient Trigonometry is to Astronomy, and, as the Proverb is, judge of Hercules by his Foot: for to instance in the Variety of things that the Doctrine of Triangles is assistent to, were an end∣less Work.

To project the SPHERE Upon the Plain of the MERIDIAN BY A LINE OF CHORDS. Whereby the Sides and Angles of Sphericall Triangles are naturally laid down in Plano, as they are in the Sphere it self; By which the nature of them is dis∣covered, and their Sides and Angles measured with speed and exactness. The Fourth EXERCISE.

BEing to treat of the Projection of the Sphere in Plano, I suppose the Reader to be acquainted with the Doctrine of the Sphere or Globe, and with the Circles thereof; the nature of them, how they are there situate one from another in respect of distance, and to what use each of them is appro∣priate. But if there be any that hath a desire to make use of this

Treatise, and is ignorant of the Sphericks, let them reade the Books of such as have written of the Sphere or Globe. In Latin there are divers, as Theodocius, Orontius, Clarius, &c. In English there is Record's Castle of Knowledge, Hill's School of Skill, Blundevil's Exercises, Hews of the Globes, Newton, Moxon, &c. But that this Treatise may not be accounted de∣ficient in that which is so absolutely necessary for the understan∣ding and practice of what is herein contained; I will in this place give you a brief and plain Description of the Names, Pro∣perties, Distances, &c. of such Circles of the Sphere as in this Book we shall have occasion to project for the delineating of Sphericall Triangles in Plano, and that in a correspondent Po∣sition to their situation on the Globe or Sphere it self.

The Circles therefore chiefly made use of in this Projection are these:

• 1. The Meridian
• 2. The Horizon
• 3. The Aequinoctial
• 4. The Ecliptick
• 5. The Prime Verticall, or Circle of East and West
• 6. The Hour-Circles
• 7. Azimuths, or Verticall Circles
• 8. The two Tropicks
• 9. Parallels or Circles of Declination
• 10. Circles or Parallels of Altitude.

Of these Circles all but the three last are great Circles of the Sphere, which divide it into two equal parts; and the two Tro∣picks, the Circles of Altitude and Declination, are smaller Circles, and divide the Sphere into two parts unequally.

Besides these great and small Circles, there are severall Points of note upon the Globe: as (1.) The Zenith, which is the Point in the Heavens directly over our heads, in what part of the Earth soever we be; (2.) The Nadir, which is the Point directly un∣der our feet; (3.) The Poles of the World, about which the Heavens are moved; (4.) The Poles of the Ecliptick; (5.) The Poles of all other Circles.

I. Of the MERIDIAN.

THE Meridian is a great Circle of the Sphere, which pas∣seth through both the Poles of the World, and also through the Zenith and Nadir Points, and crosseth the Horizon in the North and South Points thereof. Unto this Circle (any Day in the year) when the Sun cometh it is Noon or Mid-day; and when the Moon, Stars or Planets, in the Night come to touch this Circle, they are then said to be upon the Meridian, or at the highest they will be that Night. This Circle in the Scheme of this Projection is noted by the Letters Z H N O.

II. Of the HORIZON.

THE Horizon also is a great Circle of the Sphere, and it is that Circle which divideth the visible part of the Heavens which we see from the not visible, that is, it divideth the Sphere into two Hemispheres, the lower and the higher. To this Circle when either the Sun, Moon, Stars or Planets, come on the East part, they are then said to rise; and when they have passed from the Easterly Point, by the Meridian, and descended to the Western part of this Circle, they are then said to set. This Circle is represented in the Projection by the right Line H A O.

III. Of the AEQƲINOCTIAL.

THE Aequinoctial is a great Circle, and in the Sphere it is elevated above the Horizon (upon the Meridian Cir∣cle) so much as is the Complement of the Latitude of the Place. As at London, where the Latitude is 51 degr. 30 min.

there the Aequinoctial is elevated 38 degr. 30 min. (which is so much as 51 degr. 30 min. wants of 90 degr.) and it cutteth the Horizon in the Points of East and West. Unto this Circle when the Sun cometh (which is twice every year, namely, about the 10. of March and the 12. of September) it causeth the Daies and Nights to be of equal length all the World over. This Circle is noted in the Scheme with AE A ae, and cuts the Horizon in the Point A, which represents both the East and West Points thereof.

IV. Of the ECLIPTICK.

THIS also is a great Circle of the Sphere, and (in the Northern Hemisphere, where the North Pole is visible above the Horizon, and the South Pole not visible) is elevated above the Aequinoctial Circle so much as is the Sun's greatest Declination, which is 23 degr. and about 30 min. and is as much depressed below the Aequinoctial in the Southern Hemi∣sphere. This Circle is called by some The Way of the Sun, for that the Sun in his motion never swerveth or goeth out there∣of, and so his Longitude or Place is counted in this Line. It cutteth the Horizon in the East and West Point A, as the Aequi∣noctial did. It is represented in the Scheme by the Line ♋ A ♑, and hath charactered upon it the 12 Signs of the Zodiack; the six Northern Signs, ♈ ♉ ♊ ♋ ♌ and ♍ being on that half which is above the Horizon, and the six Southern Signs ♎ ♏ ♐ ♑ ♒ and ♓, on the other half, which is below the Horizon.

V. Of the PRIME VERTICALL.

THE Prime Verticall, or Circle of East and West, (general∣ly called the Aequinoctial Colure, and then (as the Sphere is here projected) the Meridian representeth the Solstitial

Colure) is a great Circle passing through the Zenith and Na∣dir Points, and also through the East and West Points of the Horizon. Unto this Circle when the Sun, Moon, Stars or Planets, do (in their Motions) arrive, they are then due East or West. It is in the Projection signified by the right Line Z A N, passing through Z, the Zenith, N, the Nadir, and A, the East and West Point of the Horizon: and also cutteth the Aequinoctial in the Points ♈ and ♎.

VI. Of the HOƲR-CIRCLES.

THE Hour-Circles are great Circles of the Sphere, mee∣ting together in the Poles of the World, and crossing the Aequinoctial at right Angles, dividing it at every 15 de∣grees; and then every of those Divisions is one Hour of time: but if they pass through other parts of the Aequi∣noctial, dividing it unequally, then do those Hour-Circles re∣present unequal Spaces of time, according to the distance they are from the Meridian, or one from another. Of these Cir∣cles in the Scheme of the Projection there are four, thus no∣ted P B S, P A S, P C S, and P D S.

VII. Of the AZIMƲTH CIRCLES.

THE Azimuth or Verticall Circles are great Circles of the Sphere, meeting together in the Zenith and Nadir Points, as the Hour-Circles do in the Poles of the World, and divide the Horizon at right Angles, either equally, or unequally, as the Hour-Circles did the Aequinoctial. In the Scheme of the Pro∣jection there are four of these Verticall Circles, thus noted, Z O N, Z F N, Z A N, and Z G N.

VIII. Of the TROPICKS.

THE Tropicks are lesser Circles of the Sphere, dividing it unequally, and are drawn parallel to the Aequino∣ctial, at 23 degr. 30 min. distance therefrom, equal to the Sun's greatest Declination on either side. That Tropick which is on the North-side is called The Tropick of Cancer, to which when the Sun cometh (which is but once in a year, about the 10. of June) it maketh the longest Daies to all the Northern In∣habitants of the World, and the shortest Nights. The other Tropick, which is on the South-side of the Aequinoctial, is cal∣led The Tropick of Capricorn, to which when the Sun cometh, which is about the 11. of December, it maketh the shortest Daies and the longest Nights to all Northern Inhabitants, and the con∣trary to all the Southern Inhabitants of the World. In the Projection the Tropick of Cancer is signfied by ♋ I ♋, and the Tropick of Capricorn by ♑ K ♑.

IX. Of the CIRCLES or PARALLELS of DECLINATION.

THESE also are smaller Circles of the Sphere, and are drawn parallel to the Aequinoctial, towards both the Tropicks, and up to them. Those that are on the North-side of the Aequinoctial are called Parallels of North Declination, and those that are on the South-side of the Aequinoctial are called Parallels of South Declination. Of these Parallels there are in the Scheme of the Projection two, one towards the Tropick of Cancer, the other towards the Tropick of Capricorn, and either of them 20 degrees distant from the Aequinoctial. The Nor∣thern Parallel of Declination is noted with ♊ ☉ ♌, and the Southern with ♒ E ♐.

X. Of the CIRCLES or PARALLELS of ALTITƲDE.

THE Circles of Altitude are likewise small Circles of the Sphere, and are drawn parallel to the Horizon, as the Circles of Declination were to the Aequinoctial. These Paral∣lels are drawn from the Horizon towards the Zenith Point, and upon occasion, in many Cases, quite up unto it. By these Parallels are measured the Altitude or Height of the Sun, Moon and Stars. In the Scheme there is onely one of them, and that is expressed by the Letters M E L.

Thus have I given you a brief and plain Description of the Circles, both great and small, which we shall have occasion to use in this following Treatise. And here note, that every Circle of the Sphere (both great and small) hath his proper Poles, which Poles (of all the great Circles) are 90 Degrees, or a Quadrant of a Circle, distant from the Circle it self. The Poles of the Circles in this Projection are as followeth.

• Z and N Are the Poles of H A O, the Horizon.
• P and S Are the Poles of AE A ae, the Aequinoctial.
• O and H Are the Poles of Z A N, the Prime Verticall.
• Q and R Are the Poles of the Ecliptick.
• AE and ae Are the Poles of P A S, the Axis of the World.

The Poles of these five Circles are all in the Meridian, and so there needeth no farther Precept for the finding of them; and the Pole of the Meridian is the Centre thereof.

But for the three Azimuth Circles, they fall in several Points of the Horizon; and the three Hour-Circles in certain Points in the Aequinoctial. How to finde which Points shall be shewed af∣terwards in due place.

• A Is the Pole of the Meridian, Z H N O.
• T Is the Pole of the Azimuth Circle, Z F N.
• G Is the Pole of the Azimuth Circle,Z ☉ N.
• ☉ Is the Pole of the Azimuth Circle,Z G N.
• X Is the Pole of the Hour-Circle P B S.
• Y Is the Pole of the Hour-Circle P D S.
• V Is the Pole of the Hour-Circle P C S.

The Poles of the World P and S are also the Poles of the Tro∣picks and of all the Parallels of Declination. And

The Zenith and Nadir, Z and N, are the Poles of all the Par∣allels of Altitude.

Having sufficiently acquainted the Reader with the several Circles, Lines, Points and Poles, belonging to every Circle, I will now proceed to my intended purpose; namely, to project (or lay down in Plano) all these Circles, Lines, Points and Poles, in their true Positions.

How to project the Sphere upon the Plain of the Meridian.

FIrst, take 60 degr. of your Line of Chords, and with that distance upon the Point A. (as a Centre) describe the Circle Z H N O, representing the Meridian, (within which Circle all the rest are to be projected) and cross it with the two Diameters H A O the Horizon, and Z A N the Prime Ver∣ticall.

Secondly, (because the Latitude of the place for which you draw your Projection, viz. London, is 51 degr. 30 min.) take 51 degr. 30 min. from your Line of Chords, and set them upon the Meridian from Z to AE, and from N to ae, and draw

the Line AE A ae for the Aequinoctial. Also set 51 degr. 30 m. from O to P, and from H to S, and draw the Line P A S, re∣presenting the Axis of the World and the Hour-Circle of 6 a clock.

Thirdly, take 23 degr. 30 min. the quantity of the Sun's greatest Declination, and also of the distance of the two Tro∣picks from the Aequinoctial, and set them upon the Meridian from AE to ♋, above the Aequinoctial, and also from AE to ♑, below the Aequinoctial. In like manner set the same distance of 23 degr. 30 min. from ae to ♋ above the Aequinoctial, and from ae to ♑ below it. This done, lay a Ruler upon the Points AE and ♋, and it will cut the Axis of the World P A S in the Point I. So a Circle drawn which shall pass through these three Points, ♋ I ♋, shall be the Tropick of Cancer. Again, lay a Ruler to AE and ♑, and it will cut the Axis in the Point K. So a Circle drawn through ♑ K ♑ shall be the Tropick of Capricorn. But to shew how you may find the Centres upon which these Tropical Circles are to be described, I must make this

Diversion.

Ʋpon a long piece of stiff Paper, or rather fine Pastboard, with 60 Degrees of your Line of Chords describe a Quadrant, as A B C, and upon the Point C erect the Perpendicular C D; in doing whereof you must be very carefull, for a small errour committed in that will produce a great one in finding of the true Centres.—Be∣ing thus prepared, you may readily find the Centres of the two Tropicks, and of any other Parallels of Declination or Altitude. But (1.) for the Tropicks: Being the Tropicks are distant from the Aequinoctial 23 degr. 30 min. on either side, take 23 degr. 30 min. out of your Line of Chords, and set them upon the Qua∣drant from B to f, and through the Point f draw the Line A f g, cutting the Perpendicular C D in the Point g. This done, take

[illustration] geometrical diagram

in your Compasses the di∣stance C g, and setting one foot in the Point I in your Projection, where the other Point falleth upon the Line I P, it being sufficiently ex∣tended, will be the Centre of the Tropick of Cancer; and a Circle described with this distance of the Compas∣ses, C g, will pass exactly through the Points ♋ I ♋, and so describe your Tro∣pick of Cancer. The like you must doe for the Tro∣pick of Capricorn, by taking the distance C g in your Compasses, and setting one foot in the Point K of your Projection, and where the other reacheth upon the Line K. S, being extended, there is the Centre of the Tropick of Capricorn. But (2.) to finde the Centres of any of the other Parallels of Decli∣nation, as of the two in the Projection, namely, ♊ a ♌ and ♒ c ♐, either of which are 20 d. distant from the Ae∣quinoctial; the Centres of them are also found in the same manner as the Centres of the Tropicks were. For take 20 degr. out of your

Line of Chords, and set them upon the Quadrant from B to e, drawing the Line A c h, the distance C h shall be the Semidia∣meter of the Parallel of 20 degr. of Declination; which, being set upon your Projection from a, upon the Line a P, (being ex∣tended) will there give you the Centre of the Parallel; and the Compasses at the distance of C h, being there set, will describe the Parallel of 20 degr. of Declination ♊ a ♌. And in the same manner that the Centre of this Parallel was found, so may you find the Centre of that on the other side of the Aequinoctial, ♒ c ♐. But (3.) you have in your Projection a Parallel or Circle of Altitude, namely, M E L, which is 12 degr. from or above the Horizon, the Centre whereof is found in the same man∣ner as were the Parallels of Declination. For if you set 12 degr. of your Chord upon the Quadrant A B C, they will reach to k: draw a Line therefore from A through k, extending it till it meet with the Line C D being also extended; so shall the distance between this Point, where the two Lines meet, and the Point C, be the Semidiameter of the Parallel of Altitude of 12 degr. Where∣fore that distance set from the Point m of your Projection, up∣on the Line m Z, being extended, will be the Centre of the Par∣allel, and the Compasses, at that distance, will describe the Par∣allel of Altitude M m L.

But for those Parallels of Altitude which fall near the Hori∣zon, those Circles or Parallels of Declination which fall near to the Aequinoctial, those Hour-Circles which fall near to the Axis of the World or Hour of Six, and those Azimuth Circles which are near to the Prime Verticall or Azimuth of East and West; those that make Mathematicall Instruments have an In∣strument called a Bow, which, by the help of one or more Screws, (according to the length of the Bow) may be extended to touch any three Points which lie near in a straight Line; by the edge of which Bow you may draw your Hour-Circles, Azimuths, Par∣allels of Declination and Altitude, as easily as you may draw a right Line by the edge of a Ruler.—But to return again to our Projection.

Fourthly, draw a right Line ♋ A ♑ between the two Tro∣picks, touching the Tropick of Cancer above the Horizon at ♋, and the Tropick of Capricorn below the Horizon at the Point ♑.—This Circle hath upon it the Characters of the 12 Signs of the Zodiac, which are to be put on in this manner.—Take 23 d. 30 min. out of your Line of Chords, and set them from P to Q, and from S to R: which Points, Q and R, are the two Poles of the Ecliptick. Then take 60 degr. from your Line of Chords, and set them from Q to 1, and from Q to 3. Also set the same distance from ♋ to 2, and from ♑ to 4.—This done, lay a Ruler to the Pole R and the figure 1, it will cut the Ecliptick in the Point ♊ and ♌: the Ruler laid to R and 2 will cut it in the Point ♉ and ♍; and laid to R and 4, in ♏ and ♓; and laid to R and 3, in ♐ and ♒. So have you the true Points for the Sun's entrance into every Sign. And if you would have every tenth degree of each Sign, divide every of the Spaces ♋ 1, 12, 2 Q, Q 4, 4 3, and 3 ♑, into three equal parts; so will each part contain 10 d. and a Ruler laid to each of them and the Point R shall give you the Points upon the Ecliptick answering to the 10. degr. of every Sign. And in the same manner may you (if your Projection be large) put on every Degree.

Fifthly, for the putting on of the Hour-Circles; consider how far the Circle you are to put on is distant from the Meri∣dian, and set so many degrees upon the Meridian from the Aequinoctial: a Ruler laid from Z to those degrees will cross the Aequinoctial, and through that Point in the Aequinoctial where the Ruler so crosseth, the Hour-Circle will pass.—Ex∣ample: The Hour-Circle P B S, in this Projection, is distant from the Meridian 62 d. 46 m. wherefore take 62 d. 46 m. from your Chords, and set them from a to b; then laying a Ruler from Z to b, it will cut the Aequinoctial in B, through which Point the Hour-Circle of 62 d. 46 m. must pass.—To find the Centre of this Hour-Circle, (and so of any other) repair to the former Scheme for finding of the Centres of the Parallels of Altitude and De∣clination;

and (because this Hour-Circle is distant from the Me∣ridian 62 degr. 46 min.) take 62 degr. 46 min. from your Line of Chords, and set them upon the Quadrant A B C, from C to l, and draw the Line A l m. So shall the Line A l m be the Semidiameter of the Hour-Circle P B S; which being taken in your Compasses, and set upon your Projection from B, upon the Line B AE, (being extended,) shall there give you the Cen∣tre of that Hour-Circle. And in the same manner may the Centres of all the rest be found.

Sixthly, the Azimuth Circles are to be drawn upon the Pro∣jection, and the Centres of them found in all respects as the Hour-Circles were.—So the Azimuth Circle Z ☉ N, being 56 degr. 40 min. from the Meridian, take 56 degr. 41 min. out of your Line of Chords, and set them upon the Meridian of your Projection from O to d; then laying a Ruler unto Z and d, it will cut the Horizon in the Point ☉ through which the Azimuth of 56 degr. 41 min. Z ☉ N, must pass.—Then, to find its Centre, repair to the former Scheme for finding of Centres, and upon the Quadrant A B C set 56 degr. 41 min. of your Chords, from C to n, and draw the Line A u o: so shall the Line A u o be the Semidiameter of the Azimuth Circle Z ☉ N; which being taken in your Compasses, and set upon your Projection from ☉, upon the Line ☉ H, (being extended,) shall there give you the Centre of the Azimuth Circle Z ☉ N. And in this manner may the Centre of any other Azimuth Circle be found.

And here note (I.) That the Centres of all Azimuth Circles fall in the Horizon H A D, being extended where need is. The Centres of all the Hour-Circles fall in the Aequinoctial Line AE A ae, being extended. The Centres of the Tropicks and Par∣allels of Declination fall in the Axis of the World P A S, ex∣tended. And the Centres of the Circles of Altitude fall in the Prime Verticall Circle Z A N.

Note (II.) That if the middle Point of any Hour-Circle do not fall just in the Aequinoctial, or any Azimuth Circle just in

the Horizon, but on either side of them; then you may find the Centres by the Geometricall Propositions at the beginning of this Book; though there be other waies to find the Centres upon the Projection it self, which I omit, for that I would not cumber the Scheme with unnecessary Lines.

Seventhly, Every Circle in the Projection hath its proper Pole, as was before intimated. Now for the finding of them, you are to note, that the Pole of every great Circle is 90 degr. or a Quadrant of a Circle, distant from the Circle it self, upon that Line which cutteth the Circle at right Angles. Thus the Poles of all the Hour-Circles are upon the Aequinoctial, and the Poles of all the Azimuths upon the Horizon.—Now if you would find the Pole of the Hour-Circle P D S, lay a Ruler up∣on P and D, and it will cut the Meridian Circle in e: then take 90 degr. of your Line of Chords, and set them from e to f, a Ruler laid from P to f will cut the Aequinoctial in Y: so is Y the Pole of the Hour-Circle P D S.

Lastly, The finding of the Poles of the Azimuth Circles is the same with the Hour-Circles. So if you would find the Pole of the Azimuth Circle Z G N, lay a Ruler upon Z and G, it will cut the Meridian Circle in g; then set 90 degr. of your Chord from g to d, so a Ruler laid from Z to d will cut the Horizon H A O in the Point ☉, which Point ☉ is the Pole of the Azimuth Circle Z G N. And thus have you found the Poles of one of the Hour and one of the Azimuth Circles. And by the same manner of Work you may find the Poles of all the rest. As

• The Pole of the Hour-Circle P D S will be found at Y
• The Pole of the Hour-Circle P C S will be found at V
• The Pole of the Hour-Circle P A S will be found at AE or ae
• The Pole of the Hour-Circle P B S will be found at X
• The Pole of the Azimuth Circle Z G N will be found at ☉
• The Pole of the Azimuth Circle Z A N will be found at H or O
• The Pole of the Azimuth Circle Z F N will be found at T
• The Pole of the Azimuth Circle Z ☉ N will be found at G

• The Poles of the Horizon H A O are Z and N, the Zenith and Nadir.
• The Poles of the Aequinoctial AE A ae are P and S, the Poles of the World.
• The Poles of the Ecliptick ♋ A ♑ are Q and R.

Thus have I given you at large a plain and easie method how to project the Sphere upon the Plain of the Meridian Circle, by help of the Line of Chords onely: Ʋpon which Projection, by the intersection or crossing of the severall Circles thereof, are constituted divers Sphericall Triangles; some Right-angled, and others Oblique-angled. By the resolving of which Triangles variety of Questions appertaining to Astronomie, Geographie and Navigation, may (with speed and exactness) be resolved. But before I come to shew the manner of working particular Que∣stions of any kind, it will be expedient that I shew you, (1.) how to measure or find the quantity of the Sides and Angles of a Sphericall Triangle, as they are here projected; and (2.) how to project or lay down an Angle or Side of any quantity that shall be required.

I. A Sphericall Triangle being projected, how to find the quantity of any Angle thereof.

LAY a Ruler to the angular Point, and the extremity of the Sides containing the Angle, they being continued to Quadrants; and note where the Ruler cuts the Meri∣dian or outward Circle; at both which places make marks up∣on the Meridian: the distance between those two marks, being measured upon your Line of Chords, shall give you the quan∣tity of the Angle required.

Example I.

IN the Triangle P ☉ O, in the Projection, let it be requi∣red to find the quantity of the Angle ☉ P O.—First, lay a Ruler upon the angular Point P, and to the extreme ends of the Sides P ☉ and P O, they being extended to Quadrants, which is, to that Circle which measures that Angle: (as the Aequinoctial measures all the Angles at P, the Pole of the World; the Horizon all the Angles at Z, the Zenith, &c.) So the Ru∣ler laid from P to ae, will cut the Meridian in ae; and being laid from P to B, it will cut the Meridian in the Point b. The distance b ae, being taken in your Compasses and measured upon your Line of Chords, will be found to contain 62 degr. 46 min. which is the quantity of the Angle ☉ P O.—But if upon the Point P you were to project an Angle to contain 62 degr. 46 min. then take 90 degr. of your Chords, and set them from P to ae, and through the Centre A draw the Line AE A ae; then take 62 degr. 46 m. out of your Line of Chords, and set them from ae to b; and laying a Ruler from P to b, it will cut AE A ae in the Point B: the Circle P B S being drawn, the Angle at P will contain 62 degr. 46 min.

Example II.

LET it be required to find the quantity of the Angle Z E P.—Lay a Ruler to ☉, the Pole of the Circle Z E N, and the Point E, it will cut the Meridian Circle in M; from M set 90 degr. to z; a Ruler laid from ☉ to z will cut the Circle Z E N (it being extended beyond the Zenith Z) at the Point δ.

Again, Lay a Ruler upon Y, the Pole of the Circle P E S, and it will cut the Meridian Circle in v; set 90 degr. from v

to x upon the Meridian; a Ruler laid from Y to x will cut the Circle P E S in y.

This done, lay a Ruler from E to δ, and it will cut the Me∣ridian in θ; also lay the Ruler from E to y, it will cut the Me∣ridian in λ: the distance θ λ, being taken in your Compasses and applied to your Line of Chords, will be found to contain 21 degr. 45 min. And such is the quantity of the Angle Z E P.

These two sorts of Angles are the most troublesome to find their quantities, and therefore I have instanced in them. There are other Angles in the Projection which render their measures to the eye, without farther Instructions for finding their quan∣tities.

II. A Sphericall Triangle being projected, to find the quantity of any Side thereof.

A Ruler laid upon the Pole of the Circle which is to be measured, and to the extreme ends of the Side of the Triangle; note where the Ruler, so laid, cuts the Me∣ridian at both ends of the Side: that distance, taken in your Compasses and measured upon the Line of Chords, will give you the quantity of the Side of the Triangle.

Example I.

LET it be required to find the Side E Z of the Triangle Z E P.—Lay a Ruler to ☉ (the Pole of the Circle Z E N) and the angular Point E, it will cut the Meridian in M; and a Ruler laid to Z will cut the Meridian in Z. So the distance M Z, taken in the Compasses and measured upon the Line of Chords, will be found to contain 78 degr. And such is the quantity of the Side Z E.

Example II.

LET it be required to find the Side ☉ B of the Sphericall Triangle A ☉ B.—Lay a Ruler upon X, the Pole of the Circle P B S, and the Point B, it will cut the Meridian Cir∣cle in ae.—Also lay a Ruler from X to ☉, it will cut the Meri∣dian in the Point ♌. The distance between ae and ♌, being taken and measured on the Line of Chords, will contain 20 d. And such is the quantity of the Side ☉ B.

I could instance in divers other Examples concerning the Mea∣suring of the Sides and Angles of Triangles upon the Projection; but I here omit them, because in the resolving of the following Propositions they will come in practice, and the Manner of the performance is there plainly expressed: onely I deemed it conve∣nient here to give some taste thereof, as a Preparative to that which followeth.—But before I come to shew the Manner of resolving of particular Questions in Astronomie, Geogra∣phy, &c. I will declare the Variety of Sphericall Problems that will naturally arise out of every Sphericall Triangle, being pro∣jected.

THE VARIETY OF SPHERICALL PROBLEMS Naturally arising out of every Sphericall Triangle, both Right and Oblique-angled, and that are resolvable thereby, described as they are perspicuous to the Eye in the Projection. The Fifth EXERCISE.

IN the foregoing Part of this Book you have the Doctrine of Plain and Sphericall Tri∣angles Geometrically performed. And in the Solution of Right-angled Sphericall Triangles there were 16 Cases; and in Ob∣lique-angled there were 12 Cases: but the 16 Cases of Right-angled Triangles will by this projective way be reduced to 5 Ca∣ses, and the 12 of Oblique-angled will be reduced to 6; so that in both there will be but 11 Cases, whereas before there were 28. That this may appear plain to the Reader, I will make use of two

Triangles in the Projection; one whereof shall be Right-angled, as the Triangle P O ☉, Right-angled at O; and the other shall be the Oblique-angled Triangle Z E P.—The Right-angled Triangle is constituted by the Intersection of three great Circles of the Sphere; namely, of P O, an Arch of the Meridian, ☉ O, an Arch of the Horizon, and P ☉, the Arch of an Hour-Circle. —The Oblique-angled Triangle, Z E P, is constituted also of three Arches of great Circles of the Sphere, (as all Sphericall Trian∣gles whatsoever are;) namely, of Z P, an Arch of the Meri∣dian, P E, an Arch of an Hour-Circle, and Z E, an Arch of an Azimuth Circle.

In the Right-angled Sphericall Triangle, P ☉ O,

• The Side P O is the Latitude of the Place.
• The Side ☉ O is the Sun's Amplitude from the North.
• The Side P ☉ is the Sun's distance from the Pole, or the Complement of his Declination.
• The Angle ☉ P O is the Hour from Midnight, or from the North part of the Meridian.
• The Angle P ☉ O is the Angle of the Sun's Position at the time of the Question.
• The Angle P O ☉ is the Right Angle.

The Parts of the Triangle being declared, and of what Cir∣cles of the Sphere the Sides do consist; I will now come to the Cases, which (as I said before) are 5 in every Right-angled Triangle. So that any two parts of the Triangle (besides the Right Angle) being given, I will shew in every of the 5 Cases what parts may be found.

I. In a Right-angled Sphericall Triangle.

CASE I. The Base and Perpendicular being given, to finde the other parts of the Triangle.

IN the Triangle P ☉ O here is given P O, the Latitude, and ☉ O, the Amplitude from the North part of the Meri∣dian; by which you may find 3.

• 1. P ☉, the Complement of the Sun's Declination.
• 2. ☉ P O, the Hour from Midnight.
• 3. P ☉ O, the Angle of the Sun's Position.

CASE II. The Hypotenuse and Perpendicular being given, to find the other parts.

IN the Triangle P ☉ O here is given P O, the Latitude, and P ☉, the Complement of the Sun's Declination; by which may be found 6.

• 1. ☉ O, the Amplitude from the North.
• 2. O P ☉, the Hour from Midnight.
• 3. P ☉ O, the Angle of the Sun's Position.
  • And if in stead of the Perpendicular there had been given the Base ☉ O, you might then find
• 4. P O, the Latitude.
• 5. ☉ P O, the Hour from Midnight.
• 6. P ☉ O, the Angle of the Sun's Position.

CASE III. The Hypotenuse and an Angle being given, to find the other Parts.

HERE is given in the Triangle P ☉ O, the Complement of the Sun's Declination, P ☉ and ☉ P O, the Hour from Midnight; by which may be found 6.

• 1. ☉ O, the Amplitude from the North.
• 2. P O, the Latitude.
• 3. C ☉ O, the Angle of the Sun's Position.
  • And if in lieu of the Angle at P, the Angle at ☉ P had been taken, then you might have found
• 4. ☉ O, the Amplitude from the North.
• 5. P O, the Latitude.
• 6. ☉ P A, the Hour from Midnight.

CASE IV. The Perpendicular, or Base, and either of the Angles given, to find the other Parts.

IN the Triangle P ☉ O let there be given the Amplitude, ☉ O, and the Angle of the Sun's Position, P ☉ O; by which you may find 12.

• 1. P O, the Latitude.
• 2. ☉ P O, the Hour from Midnight.
• 3. P ☉, the distance of the Sun from the Pole.
  • But if the Side given had been ☉ O, and the Angle given ☉ P O, then you might have found
• 4. P ☉, the Complement of the Sun's Declination.
• 5. P O, the Latitude.
• 6. P ☉ O, the Angle of the Sun's Position.
  • But again, if ☉ O and ☉ P O had been given, then might be found
• 7. ☉ P, the Complement of the Sun's Declination.
• 8. ☉ O, the Amplitude from the North.

• 9. P ☉ O, the Angle of the Sun's Position.
  • And again, if P O and O ☉ P had been given, we might then have also found
• 10. ☉ P, the distance of the Sun from the Pole.
• 11. ☉ O, the Amplitude from the North.
• 12. ☉ P O, the Hour from Midnight.

CASE V. The Angles being given, to find the other Parts.

IF the two Angles P ☉ O and O P ☉ be given, there may be found 3.

• 1. P O, the Latitude.
• 2. ☉ O, the Amplitude from the North.
• 3. P ☉, the Complement of the Sun's Declination.

Thus you see, that in this one Right-angled Sphericall Tri∣angle, by the several Parts given in these five Cases, there are 30 Sphericall Problems resolved; and so many are resolvable in every Right-angled Triangle.

II. In an Oblique Sphericall Triangle.

In the Oblique-angled Sphericall Triangle, Z P E,

• The Side Z P is the Complement of the Latitude.
• The Side P E is the Complement of the Sun's Declination, or his distance from the North Pole.
• The Side Z E is the Complement of the Sun's Altitude.
• The Angle E Z P is the Sun's Azimuth from the North part of the Meridian.
• The Angle Z P E is the Hour from Noon.
• The Angle Z E P is the Angle of the Sun's Position.

CASE I. The three Sides being given, to find an Angle.

IN the Triangle Z E P, if there be given the Side E Z, the Complement of the Sun's Altitude, Z P, the Complement of the Latitude, and E P, the Sun's distance from the Pole, or the Complement of his Declination, you may find 3.

• 1. E Z P, the Sun's Azimuth from the North.
• 2. Z E P, the Angle of the Sun's Position.
• 3. Z P E, the Hour from Noon.

CASE II. Two Sides and the Angle comprehended by them being given, to find the other Parts of the Triangle.

IF in the Triangle Z E P there be given the Side E Z and Z P, and the Angle between them E Z P, you may find 9.

• 1. Z E P, the Angle of the Sun's Position.
• 2. Z P E, the Hour from the South, or Noon.
• 3. E P, the Sun's distance from the Pole.
  • But if the Sides Z P and P E, and the Angle Z P E between them, had been given, then might have been found
• 4. P E Z, the Angle of the Sun's Position.
• 5. E Z, the Complement of the Sun's Altitude.
• 6. E Z P, the Azimuth of the Sun from the North.
  • And if the Sides Z E and P E, with the Angle Z E P contained by them, had been given, there might be found
• 7. E Z P, the Sun's Azimuth from the North.
• 8. Z P, the Complement of the Latitude.
• 9. Z P E, the Hour from Noon.

CASE III. Two Angles, and a Side contained by them, being given, to find the other Parts.

IF in the Triangle Z E P the Angle E Z P and the Angle Z P E, with the Side contained between them, Z P, be given, we may find 9.

• 1. Z E, the Complement of the Sun's Altitude.
• 2. Z E P, the Angle of the Sun's Position.
• 3. E P, the Complement of the Sun's Declination.
  • But if the Side E P, and the Angles Z E P & Z P O, had been given, then might be found
• 4. E Z, the Complement of the Sun's Altitude.
• 5. E Z P, the Azimuth from the North.
• 6. Z P, the Complement of the Latitude.
  • And if the Side Z E, and the Angles P Z E and P E Z, had been given, then you might find
• 7. Z P, the Complement of the Latitude.
• 8. Z P E, the Hour from Noon.
• 9. P E, the Sun's distance from the Pole.

CASE IV. Two Sides, with an Angle opposite to one of them, being given, to find the other Parts.

IF there be given the Side Z P, the Side E P, and the Angle Z E P, there may be found 18.

• 1. E Z, the Complement of the Sun's Altitude.
• 2. E Z P, the Sun's Azimuth from the North.
• 3. Z P E, the Hour from the South.
  • But if the Side Z P, and the Side E P, with the Angle E Z P, had been given, then might be found
• 4. E Z, the Complement of the Sun's Altitude.

• 5. Z E P, the Angle of the Sun's Position.
• 6. Z P E, the Hour from Noon.
  • And if there had been given E P, E Z, and E Z P, then might be found
• 7. Z P, the Complement of the Latitude.
• 8. Z P E, the Hour from Noon.
• 9. Z E P, the Angle of the Sun's Position.
  • In like manner, if the Sides P E and E Z, with the Angle Z P E, had been given, then might be found
• 10. Z P, the Complement of the Latitude.
• 11. E Z P, the Azimuth from the North.
• 12. Z E P, the Angle of the Sun's Position.
  • Again, if the Sides E Z and Z P, with the Angle Z P E, had been given, then would be found
• 13. P E, the Sun's distance from the Pole.
• 14. Z E P, the Angle of the Sun's Position.
• 15. E Z P, the Sun's Azimuth from the North.
  • Lastly, if the Sides E Z and Z P, with the Angle Z E P, had been given, then you might find
• 16. P E, the Complement of the Sun's Declination.
• 17. Z P E, the Hour from Noon.
• 18. E Z P, the Azimuth from the North.

CASE V. Two Angles, and a Side opposite to one of them, being given, to find the other Parts of the Triangle.

IN the Triangle Z E P, if there be given the Angles E Z P and Z P E, with the Side P E, there may be found 18.

• 1. Z P, the Complement of the Latitude.
• 2. Z E, the Complement of the Sun's Altitude.
• 3. Z E P, the Angle of the Sun's Position.
  • But if there were given E Z P and Z P E, with the Side Z E, then might be found

• 4. Z P, the Complement of the Latitude.
• 5. P E, the Sun's distance from the Pole.
• 6. Z E P, the Angle of the Sun's Position.
  • And if the Angles Z P E and Z E P, with the Side E P, were given, then might be found
• 7. Z P, the Complement of the Latitude.
• 8. P E, the Hour from Noon.
• 9. E Z P, the Sun's Azimuth from the North.
  • Again, if there were given the Angles Z P E and Z E P, with the Side Z P, you might then find
• 10. Z E, the Complement of the Sun's Altitude.
• 11. P E, the Complement of the Sun's Declination.
• 12. E Z P, the Sun's Azimuth from the North.
  • Also if there were given the Angles Z E P and E Z P, with the Side Z P, you might find
• 13. Z E, the Complement of the Sun's Altitude.
• 14. P E, the Complement of the Sun's Declination.
• 15. Z P E, the Hour from Noon.
  • And lastly, if there were given the Angles Z E P and E Z P, with the Side P E, then might be found
• 16. Z E, the Complement of the Sun's Altitude.
• 17. Z P, the Complement of the Latitude.
• 18. Z P E, the Hour from Noon.

CASE VI. The three Angles being given, to find the other Parts.

IN the Triangle Z E P, if the three Angles E Z P, Z P E, and P E Z, be given, there may be found 3.

• 1. Z P, the Complement of the Latitude.
• 2. P E, the Sun's distance from the Pole.
• 3. Z E, the Complement of the Sun's Altitude.

Thus have you in these six Cases all the Varieties that will arise out of an Oblique-angled Sphericall Triangle, in the Conversion of

which Cases you may observe 60 Questions of the Sphere to be re∣solved; and so many are resolvable in every Oblique-angled Sphe∣ricall Triangle, and 30 in every Right-angled: So that in these two Triangles 90 Questions are resolved. For,

In a Right-angled Sphericall Triangle,

By the	First	Case are resolved	3	Sphericall Questions.
Second	6
Third	6
Fourth	12
Fifth	3
		In all 30.

In an Oblique-angled Sphericall Triangle,

By the	First	Case are resolved	3	Sphericall Problems.
Second	9
Third	9
Fourth	18
Fifth	18
Sixth	3
		In all 60.

PROPOSITIONS ASTRONOMICALL, Usefull in the Practice of NAVIGATION: Performed by the resolving of severall Sphericall Tri∣angles upon the Projection. The Sixth EXERCISE.

OF the vast number of Sphericall Questions that are resolvable by the severall Sphericall Triangles that are (by the Intersections of the great Circles of the Sphere) constitu∣ted upon the Projection, seeing there may be 30 in every Right-angled, and 60 in every Oblique-angled Sphericall Triangle resolved; I have made choice onely of such as I conceived most necessary for Sea-mens use. Of which, some of them will be assistent to them in finding the Latitude; others for finding the Sun's Altitude upon some certain Points

of the Compass, and at some particular Hours: others will help them to find the Variation of their Compass; and some of them will help them to the Hour of the Day and Night.

By the manner of Working of these Propositions upon the Pro∣jection, the ingenious Practitioner may propose divers other Questions to himself, and project the Sphere in Plano sutable unto them; in the Practice and true performance whereof he will accumulate to himself both Pleasure and Profit.

Besides the working of these Propositions upon the Projection, I have to every of them added the Analogie or Proportion, by which they may be resolved by the Canons of Artificial Sines and Tangents; so that such as require more exactness then can be ex∣pected from Instrumental performance, may by those Canons work and resolve their Questions more accurately. Many more Propositions I could have added, but these (for the Astrono∣micall part) I deemed sufficient: such as concern Geographie and Navigation shall follow in convenient place.

PROP. I. The distance of the Sun from the nearest Aequinoctial Point (either Aries or Libra) given, to find his Declination.

IN the Projection this Proposition is to be resolved upon the Right-angled Sphericall Triangle A k ♒, right-angled at ♒: which Triangle is constituted by the Intersection of the Arches of three great Circles of the Sphere; namely, of A ♒, an Arch of the Ecliptick; A k, an Arch of the Aequinoctial; and of ♒ k, an Arch of a great Circle passing through Q and R the Poles of the Ecliptick, and cutting the Ecliptick at right Angles in ♒, which is the place of the Sun at the time of the Que∣stion.

In which Triangle you have given (besides the right Angle

at ♒) (1.) the Side A ♒, 59 degr. which is the distance of the Sun from the nearest Aequinoctial Point Libra; (2.) the Angle ♒ A k, 23 degr. 30 min. which is the Angle that the Ecliptick makes with the Aequinoctial, and is alwaies equal to the greatest Declination of the Sun. Wherefore in this Tri∣angle you having given the Base A ♒, and the Angle at the Base k A ♒, you may find the Perpendicular k ♒, the Sun's Declination, by the 11. Case of Right-angled Sphericall Tri∣angles in the first Part of this Book. For which this is

The Analogie or Proportion.

As the Radius 90 degr. is to the Sine of the Sun's greatest De∣clination 23 degr. 30 min.

So is the Sine of the Sun's distance from the next Aequinoctial Point Libra 59 degr. to the Sine of the Sun's present Decli∣nation 20 degr.

To resolve the Triangle upon the Projection,

Lay a Ruler upon the Pole of the Circle R ♒ Q, (which is at ♉,) and the Point k, it will cut the Meridian in the Point l: Also lay a Ruler from ♉ to ♒, it will cut the Meridian in the Point ♑. So the distance l ♑, being measured upon your Line of Chords, will contain 20 degr. the Sun's Declination, he being in the 29. degree of ♒.

The like Declination the Sun hath when he is in 29 degr. of Taurus, in 1 degr. of Leo, or 29 degr. of Scorpio, every of which Points are distant from one of the Aequinoctial Points Aries or Libra 59 degr.

PROP. II. The Latitude of the Place, and the Declination of the Sun, being given, to find the Ascensional Diffe∣rence.

UPON the Projection this Proposition is to be resolved by finding the Side A B of the Right-angled Sphericall Triangle A ☉ B, Right-angled at B; which Triangle is com∣pounded of three Arches of great Circles, namely, of A ☉, an Arch of the Horizon, A B, an Arch of the Aequinoctial, and ☉ B, an Arch of an Hour-Circle.

In this Triangle you have given (1.) the Side ☉ B, the Sun's Declination 20 d. (2.) the Angle ☉ A B, the Complement of the Latitude 38 d. 30 m. and the right Angle at B. In this Triangle therefore you have given ☉ B, the Perpendicular, and ☉ A B, the Angle at the Base, to find the Base A B, which you may doe by the 14. Case of Right-angled Sphericall Triangles. For which this is

The Analogie or Proportion.

As the Co-tangent of the Latitude 38 degr. 30 min. is to the Tangent of the Sun's Declination 20 degr.

So is the Radius 90 degr. to the Sine of the Ascensional Dif∣ference 27 degr. 14 min.

To resolve the Triangle upon the Projection,

Lay a Ruler to P, the Pole of the World, (and also of the Aequinoctial,) and the Point B, it will cut the Meridian Circle in the Point b; the distance b S, being taken in your Compasses and measured upon your Line of Chords, will reach from the beginning thereof to 27 degr. 14 min. the Ascensional Dif∣ference; which is so much as the Sun riseth or setteth before or after Six a Clock. So these 27 degr. 14 min. being turned

into Time (by allowing 15 deg. for one Hour, and one Degree for 4 minutes of Time) is 1 Hour and 49 min. and so much doth the Sun rise or set before or after the hour of Six, ac∣cording to the time or season of the Year: for if the Sun hath North Declination, then he riseth before Six, and sets after; but if the Sun have South Declination, then doth he rise after, and set before Six.

This Ascensional Difference being added to 6 Hours will give you the Semidiurnall Arch or Half-length of the Day; and being taken from six Hours, will leave the Seminocturnall Arch or Half-length of the Night.

The Semidiurnall Arch, when the Sun hath 20 degr. of North Declination, is represented in the Projection by the Arch ♊ a ☉, and the Seminocturnall Arch by ☉ ♌. The Semidiurnall Arch, when the Sun hath 20 degr. of South Declination, is represented by the Arch ♒ E r, and the Seminocturnall by the Arch r c ♐.

Though I have shewed how these may be found by adding and subtracting the Ascensionall Difference; yet they may be found by the Projection, for the Arches are measured upon the Aequino∣ctial. Wherefore lay a Ruler to P, the Pole of the World, and the Point B, it will cut the Meridian Circle in b: So the distance b AE, being measured by your Chord, will be 117 degr. 14 min. the Semidiurnall Arch, and b ae measured will be 62 degr. 11 min. for the Seminocturnall Arch.

PROP. III. The Latitude of the Place, and the Declination of the Sun, being given, to find his Amplitude.

THERE are two Triangles upon the Projection, by the resolving of either the Proposition may be resolved, and both of them Right-angled. The one is the Triangle P☉O,

Right-angled at O. The other is the Triangle made use of in the last Proposition, A ☉ B. The first Triangle is constituted of these three Arches, viz. P O, an Arch of the Meridian, P ☉, an Arch of an Hour-Circle, and ☉ O, an Arch of the Horizon.—The second Triangle consisteth of A ☉, an Arch of the Horizon, ☉ B, an Arch of an Hour-Circle, and A B, an Arch of the Aequinoctial. In the first Triangle you have given the Perpendicular P O, the Latitude of the Place, and the Hypotenuse P ☉, the Complement of the Sun's Decli∣nation 70 degr. by which you are to find the Base ☉ O, the Sun's Amplitude from the North part of the Meridian, which may be found by the 7. Case of Right-angled Sphericall Tri∣angles.—In the second Triangle B A ☉ (which is that I will first exemplifie in) you have given (1.) the Perpendicular ☉ B, the Sun's Declination 20 degr. (2.) the Angle at the Base ☉ A B, the Complement of the Latitude 38 degr. 30 min. to find the Hypotenuse ☉ A, the Sun's Amplitude from the East or West. So having the Perpendicular ☉ B, and the Angle at the Base ☉ A B, you may find the Hypotenuse ☉ A by the 4. Case of Right-angled Sphericall Triangles. And for it this is

The Analogie or Proportion.

As the Co-sine of the Latitude 38 degr. 30 min. is to the Ra∣dius 90 degr.

So is the Sine of the Sun's Declination 20 degr. to the Sine of the Amplitude from the East or West Points of the Horizon 33 degr. 20 min.

To work the Proposition upon the Projection,

Lay a Ruler upon Z the Zenith, (which is also one of the Poles of the Horizon,) and to the Point ☉, it will cut the Meridian Circle in d; and laid from Z to A, it will cut the Nadir Point in N: So the distance N d, being taken in your Compasses and measured upon your Line of Chords, will con∣tain

33 degr. 20 min. the quantity of the Hypotenuse, which is the Amplitude of the Sun's rising or setting from the true East or West Points of the Horizon.

To perform the same Work by the other Triangle P ☉ O, Lay a Ruler to Z the Zenith, and the Point ☉, and it will cut the Meridian Circle in the Point d, as before: So the distance d O, measured upon your Line of Chords, will con∣tain 56 degr. 40 min. which is the Amplitude of the Sun's ri∣sing or setting from the North Point of the Horizon.

PROP. IV. The Latitude of the Place, and the Declination of the Sun, being given, to find the Angle of the Sun's Position at the time of his rising.

THIS Proposition is resolvable upon the Triangle P ☉ O: In which there is given (1.) the Hypotenuse P ☉, the Complement of the Sun's Declination; (2.) the Perpendi∣cular P O, the Latitude: and it is required to find the Angle P ☉ O: which may be found by the 15. Case of Right-angled Sphericall Triangles, and by the following

Analogie or Proportion.

As the Co-sine of the Declination 70 degr. is to the Radius 90 degr.

So is the Sine of the Latitude 51 degr. 30 min. to the Sine of the Angle of the Sun's Position at the time of his rising.

By the Projection.

Lay a Ruler to X, (the Pole of the Hour-Circle P ☉ S,) and the Point ☉; the Ruler so laid will cut the Meridian Circle near the Point ♌: then set 90 degr. from ♌ to x upon the Me∣ridian, and lay the Ruler from X to x; so shall it cut the Hour-Circle

S ☉ P (being continued without the Meridian Circle) in the Point Φ. Again, the Ruler laid from ☉ to Φ will cut the Meridian Circle in the Point 2. So the distance O 2, being ta∣ken in the Compasses and applied to the Line of Chords, will be found to contain 56 degr. 29 min. which is the quantity of the Angle P ☉ O.

PROP. V. The Sun's Declination, and his Amplitude from the North part of the Horizon, being given, to find the Latitude.

IN the same Triangle as in the last Proposition P ☉ O, you have given (1.) the Base ☉ O, the Sun's Amplitude from the North part of the Horizon; (2.) the Hypotenuse P ☉, to find the Perpendicular P O. So there are two Sides given to find the third, which you may doe by the 8. Case of Right-an∣gled Sphericall Triangles.

The Analogie or Proportion.

As the Co-sine of the Amplitude from the North 33 degr. 20 min. is to the Radius 90 degr.

So is the Sine of the Declination 20 degr. to the Co-sine of the Latitude 38 degr. 30 min.

By the Projection.

This is easie; for if you take in your Compasses the distance from O to P upon the Meridian, and measure it upon your Line of Chords, it will be found to contain 51 degr. 30 min. the Latitude required.

PROP. VI. The Sun's greatest Declination, with his Distance from the next Aequinoctial Point (Aries or Libra,) being given, to find his right Ascension.

THE Triangle A k ♒ in the Projection resolves this Proposition, in which you have given (1.) the Side A ♒, the distance of the ☉ from ♎; (2.) the Angle ♒ A k, the Sun's greatest Declination, and the right Angle at ♒: So you have given the Base, and the Angle at the Base to find the Hypotenuse, which you may doe by the Case of Right-an∣gled Sphericall Triangles, and this

Analogie or Proportion.

As the Radius 90 degr. is to the Co-sine of the greatest De∣clination 66 degr. 30 min.

So is the Tangent of the Sun's distance from the next Aequi∣noctial point Libra 59 d. to the Tangent of the right Ascen∣sion 56 degr. 50 min.

By the Projection.

Lay a Ruler to P and k, it will cut the Meridian Cir∣cle in s: the distance S s, taken and measured upon your Line of Chords, will contain 56 degr. 50 min. the Sun's right Ascension.

I should here shew how the right Ascension and Declination of a Star might be found; but, the Calculation Trigonometricall being very laborious, I have therefore in this place omitted it, be∣cause I have in another Treatise, now speedily to be published, fra∣med Tables of the Longitude, Latitude, Right Ascension, De∣clination,

and Semidiurnall Arches, of the most eminent Fixed Stars in the Heavens, exactly calculated for the Year of our Lord 1670; and also other Tables of the Sun's right Ascension for every Degree of the Ecliptick and Day of the Year: by help of which Tables, the Hour of the Night, the Rising, Setting, and the time of their coming to the South, may be obtained by the Rules and Directions prescribed in the forementioned Treatise, which will contain (besides the Tables here mentioned) all such others as at any time the Sea-man shall have occasion for in his Practice, and divers other things too tedious here to enumerate. But in the mean time I shall request the Reader to be satisfied with these two Tables following: These Exercises being things onely Practicall.

A Table of the Sun's right Ascension for every De∣gree of the Ecliptick.

Degr.	♈	♉	♊	♋	♌	♍
0	0 0	27 54	57 48	90 0	122 12	152 6
1	0 55	28 51	58 51	91 5	123 14	153 3
2	1 50	29 49	59 53	92 11	124 16	154 1
3	2 45	30 46	60 56	93 16	125 19	154 58
4	3 40	31 44	61 59	94 22	126 20	155 54
5	4 35	32 42	63 2	95 27	127 22	156 51
6	5 30	33 40	64 6	96 32	128 24	157 48
7	6 25	34 38	65 9	97 38	129 25	158 44
8	7 21	35 37	66 13	98 43	130 26	159 40
9	8 16	36 36	67 17	99 48	131 27	160 37
10	9 11	37 34	68 21	100 53	132 28	161 33
11	10 6	38 33	69 25	101 58	133 28	162 29
12	11 1	39 33	70 29	103 3	134 29	163 25
13	11 57	40 32	71 34	104 8	135 29	164 20
14	12 52	41 31	72 38	105 13	136 29	165 16
15	13 48	42 31	73 43	106 17	137 29	166 12
16	14 44	43 31	74 47	107 22	138 29	167 7
17	15 40	44 31	75 52	108 26	139 28	168 3
18	16 35	45 31	76 57	109 31	140 27	168 58
19	17 31	46 32	78 1	110 35	141 27	169 54
20	18 27	47 32	79 7	111 39	142 26	170 49
21	19 23	48 33	80 12	112 43	143 24	171 44
22	20 20	49 34	81 17	113 47	144 23	172 39
23	21 16	50 35	82 22	114 51	145 22	173 35
24	22 12	51 36	83 28	115 54	146 21	174 30
25	23 9	52 38	84 33	116 57	147 18	175 25
26	24 6	53 40	85 38	118 1	148 16	176 20
27	25 2	54 41	86 44	119 4	149 14	177 15
28	25 59	55 44	87 49	120 7	150 11	178 10
29	26 56	56 46	88 55	121 9	151 9	179 5
30	27 54	57 48	90 0	122 12	152 6	180 1

A Table of the Sun's right Ascension for every De∣gree of the Ecliptick.

Degr.	♎	♏	♐	♑	♒	♓
0	180 0	207 54	237 48	270 0	302 12	332 6
1	180 55	208 51	238 51	271 5	303 14	333 3
2	181 50	209 49	239 53	272 11	304 16	334 1
3	182 45	210 46	240 56	273 16	305 19	334 58
4	183 40	211 44	241 59	274 22	306 20	335 54
5	184 35	212 42	243 2	275 27	307 22	336 51
6	185 30	213 40	244 6	276 32	308 24	337 48
7	186 25	214 38	245 9	277 38	309 25	338 44
8	187 21	215 37	246 13	278 43	310 26	339 40
9	188 16	216 36	247 17	279 48	311 27	340 37
10	189 11	217 34	248 21	280 53	312 28	341 33
11	190 6	218 33	249 25	281 58	313 28	342 29
12	191 2	219 33	250 29	283 3	314 29	343 25
13	191 57	220 32	251 34	285 8	315 29	344 20
14	192 52	221 31	252 38	286 13	316 29	345 16
15	193 48	222 31	253 43	287 17	317 29	346 12
16	194 44	223 31	254 47	288 22	318 29	347 7
17	195 40	224 31	255 52	289 26	319 28	348 3
18	196 35	225 31	256 57	290 31	320 27	348 58
19	197 31	226 32	258 1	291 35	321 27	349 54
20	198 27	227 32	259 7	292 39	322 26	350 49
21	199 23	228 33	260 12	293 43	323 24	351 44
22	200 20	229 34	261 17	294 47	324 23	352 39
23	201 16	230 35	262 22	295 51	325 22	353 35
24	202 12	231 36	263 28	296 54	326 21	354 30
25	203 9	232 38	264 33	297 57	327 18	355 25
26	204 6	233 40	265 38	298 1	328 16	356 20
27	205 2	234 41	266 44	299 4	329 14	357 15
28	205 59	235 44	267 49	300 7	330 11	358 10
29	206 56	236 46	268 55	301 9	331 9	359 5
30	207 54	237 48	270 0	302 12	332 6	360 0

The right Ascension, Declination, and Ma∣gnitude of some principal Fixed Stars.

The Stars Names.	Right A∣scension.	Declina∣tion.		Magni∣tude.
	D. M.	D. M.
The Pole-Star	7 47	87 27	N	2
The Girdle of Andromeda	12 32	33 48	N	2
The former Horn of the Ram	23 38	17 37	N	4
Bright Star in the Ram's Head	26 56	21 48	N	3
The Whale's Jaw	41 3	2 42	N	2
The Head of Medusa	41 27	39 35	N	3
The Bull's Eye	64 0	15 46	N	1
The Goat	72 44	45 35	N	1
The former Shoulder of O∣rion	76 38	4 59	N	2
The latter Shoulder of O∣rion	84 7	7 18	N	2
The great Dog	97 27	16 13	S	1
The uppermost Head of the Twins	108 1	32 35	N	2
The little Dog	110 17	6 6	N	2
The lower Head of the Twins	111 0	28 49	N	2
The Crib	125 4	20 52	N	Neb.
Hydra's Heart	137 39	7 10	S	2
Lion's Heart	147 27	13 39	N	1
Lion's Loins	163 54	22 26	N	2
Lion's Tail	172 49	16 32	N	1
The Virgin's Girdle	189 32	5 20	N	3

The right Ascension, Declination, and Ma∣gnitude of some principal Fixed Stars.

The Stars Names.	Right A∣scension.	Declina∣tion.		Magni∣tude.
	D. M.	D. M.
Aliot	189 36	57 36	N	2
Vindemiatrix	191 15	12 51	N	3
The Virgin's Spike	196 44	9 17	S	1
Arcturus	209 56	21 4	N	1
The Southern Balance	217 56	14 32	S	2
The Northern Balance	224 31	8 2	S	2
In the Serpent's Neck	231 49	7 35	N	3
The Scorpion's Heart	242 4	25 34	S	1
Hercules Head	254 40	14 51	N	3
Ophiuchus Head	259 41	12 52	N	3
The Harp	276 17	38 30	N	1
The Vulture	293 27	8 1	N	2
The upper Horn of the Goat	299 30	13 32	S	3
Left Hand of Aquarius	307 10	10 43	S	4
Left Shoulder of Aquarius	318 18	7 2	S	3
Pegasus Mouth	321 49	8 18	N	3
Right Shoulder of Aquarius	326 59	1 58	S	3
Fomahant	339 29	31 23	S	1
Upper Wing of Pegasus	341 53	13 21	N	2
In the tip of Pegasus Wing	358 52	13 15	N	2

PROP. VII. The Latitude of the Place and the Sun's Declination being given, to find at what Hour the Sun will be upon the true East or West Points.

UPON the Projection there are two Right-angled Sphe∣ricall Triangles, by either of which this Proposition may be solved. The one is the Triangle Z P o, made by the Intersections of Z o, an Arch of the Prime Verticall, P o, an Arch of an Hour-Circle, and Z P, an Arch of the Meridian. In which Triangle there is given Z P, the Perpendicular, the Complement of the Latitude of the Place 38 degr. 30 min. and the Hypotenuse P o, the Complement of the Sun's De∣clination 70 degr. to find the Angle at the Perpendicular Z P o, which you may doe by the 14. Case of Right-angled Sphericall Triangles.

The other Triangle is o C A, right-angled at C, and is con∣stituted of o C, an Arch of an Hour-Circle, C A, an Arch of the Aequinoctial, and o A, an Arch of the Prime Verticall. In which Triangle you have given, (1.) the Perpendicular, O C, the Sun's Declination; (2.) the Angle at the Base, C A o, the Latitude 51 degr. 30 min. to find the Base C A. Thus having the Perpendicular and the Angle at the Base, you may find the Base C A as followeth, this being

The Analogie or Proportion.

As the Tangent of the Latitude 51 degr. 30 min. is to the Tangent of the Sun's Declination 20 degr.

So is the Radius 90 degr. to the Co-sine of the Hour from Noon.

To resolve the Proposition by the Projection,

Lay a Ruler upon P, the Pole of the World, and the An∣gle C of your Triangle, the Ruler will cut the Meridian Cir∣cle in the Point g: So g Ae, being taken in your Compasses and measured upon your Line of Chords, will be found to contain 73 degr. 10 min. which converted into Hours and Mi∣nutes will be 4 hours and about 53 min. So that the Sun, when he hath 20 degr. of Declination, will come to the East Point at 7 min. past 7 in the Morning, and will be due West 53 min. after 4 in the Afternoon.

PROP. VIII. Having the Latitude of the Place, and the Sun's Decli∣nation, given, to find what Altitude the Sun shall have when he is upon the true East or West Points.

THIS Proposition may be resolved by either or both of the Triangles mentioned in the last Proposition. For in the Triangle Z P o you have given P Z, the Perpendicular, and P O, the Hypotenuse, to find Z o, the Base, by the 8. Case of Right-angled Sphericall Triangles.

But in the Triangle o C A, you have given (1.) the Per∣pendicular o C, the Sun's Declination 20 degr. (2.) the An∣gle at the Base o A C, the Latitude of the Place 51 degr. 30 m. to find the Hypotenuse o A, for which this is

The Analogie or Proportion.

As the Sine of the Latitude 51 degr. 30 min. is to the Sine of the Declination 20 degr.

So is the Radius 90 degr. to the Sine of the Sun's Altitude be∣ing due East or West 25 degr. 55 min.

To resolve the Proposition by the Projection,

Lay a Ruler upon O, (one of the Poles of the Prime Ver∣ticall) and to the Angle o of the Triangle; a Ruler thus laid will cut the Meridian Circle in the Point p: So the distance H p, being taken and measured upon the Line of Chords, will be 25 degr. 55 min. and such height will the Sun have when he is either East or West.

PROP. IX. The Latitude of the Place, and the Declination of the Sun, being given, to find what Altitude the Sun shall have at Six of the Clock.

FOR finding of the Triangles upon the Projection, which will resolve this and the following Propositions, you must suppose another Azimuth Circle to be drawn in the Projecti∣on from Z to N, and through that Point where the Parallel of Declination ♊ ☉ ♌, and the Axis of the World, or Hour-Cir∣cle of Six, P A S, do cross each other. The drawing of which Azimuth Circle I purposely omitted, chiefly because the Scheme in that place is more cumbred with Lines and Letters then any other part thereof: But you may well enough, for the solving of these two Propositions, imagine it to be drawn, the Pole whereof is at *. This Azimuth Circle being suppo∣sed to be drawn, you have upon the Projection two Triangles like-angled, which will perform the Work of resolving this Proposition. In one of which you have given the Base, which is the Complement of the Declination, and the Perpendicular, which is the Complement of the Latitude, to find the Hypo∣tenuse, which is the Complement of the Sun's Altitude requi∣red. This Triangle may be resolved by the first Case afore∣going.—In the other Triangle there will be given the

Hypotenuse, which is the Sun's Declination, and the Angle at the Base, which is the Latitude, to find the Perpendicular, which is the Sun's Altitude at Six a Clock: To find which this is

The Analogie or Proportion.

As the Radius 90 degr. is to the Sine of the Sun's Declinati∣on 20 degr.

So is the Sine of the Latitude 51 degr. 30 min. to the Sine of the Sun's Altitude at Six 15 degr. 30 min.

To resolve the Proposition by the Projection,

Lay a Ruler upon the Point *, and that Point where the Parallel of Declination ♊ ☉ ♌ crosseth the Axis or Hour of Six; the Ruler thus laid will cut the Meridian Circle in the Point g. So O g, being measured upon the Chords, will give you 15 degr. 30 min. And such Altitude will the Sun have at the Hour of Six in the Latitude of 51 degr. 30 min. when he hath 20 degr. of Declination.

PROP. X. The Latitude of the Place and the Declination of the Sun being given, to find what Azimuth the Sun shall have at Six a Clock.

THE two Triangles that were supposed in the last Pro∣position to be drawn upon the Projection will resolve this Proposition also; but seeing the Triangles are not drawn, but supposed, I will onely give you the Analogie, and then the way of working it upon the Projection.

The Analogie or Proportion.

As the Co-sine of the Latitude 38 degr. 30 min. is to the Ra∣dius 90 degr.

So is the Co-tangent of the Sun's Declination 70 degr. to the Tangent of the Sun's Azimuth counted from the North part of the Meridian 77 degr. 14 min.

To resolve the Proposition upon the Projection,

Lay a Ruler to the Zenith-point Z, and upon the Point where the Parallel of Declination cuts the Hour-Circle of Six; the Ruler thus laid will cut the Meridian Circle in r: So the distance r O, being measured upon the Line of Chords, will contain 77 degr. 14 min. which is the Azimuth from the North part of the Meridian.—The distance N r, measured upon the Chords, will give you 12 degr. 46 min. which is the Azimuth from East or West.—And r H, measured upon your Chord, will contain 102 degr. 46 min. his Azimuth from the South.

PROP. XI. The Latitude of the Place, the Declination of the Sun, and the Sun's Altitude, being given, to find the Sun's Azimuth either from the East, North or South Points of the Horizon.

ALL the foregoing Propositions have been performed by the resolving of a Right-angled Sphericall Triangle: This and some others following require the resolving of an Oblique-angled Triangle for their Solution. So this Pro∣position is performed upon the Oblique-angled Triangle Z E P, which in the Projection is constituted by the Inter∣section of P E S, an Hour-Circle, Z E N, an Azimuth Circle; and Z H N O, the Meridian: and the Arches of these Cir∣cles intersecting each other in the Points Z, E, and P, do make the Triangle Z E P; in which you have given the three Sides, (1.) Z P, the Complement of the Latitude 38 degr.

30 min. of the Place, (2.) P E, the Complement of the Sun's Declination South 110 degr. (3.) Z E, the Complement of the Sun's Altitude 78 d. to find the Angle E Z P, which may be re∣solved by the 11. Case of Oblique-angled Sphericall Triangles. But this being to resolve a particular Proposition, I shall give another Analogie or Proportion whereby to work it by the Canon.

Adde all the three Sides together into one Sum, and take the half thereof, from which half Sum subtract the Side P E, noting the difference, as here you see done.

		d.	m.
The Side	Z P	38	30
P E	110	00
Z E	78	00
The Sum	226	30
The half Sum	113	15
The difference between the half Sum and P E	3	15

Having found the Sum, the half Sum, and the difference, you may work by the following.

Analogie or Proportion.

(1.) As the Radius 90 degr. is to the Co-sine of the Altitude 78 degr.

So is the Co-sine of the Latitude 38 degr. 30 min. to the Sine of a fourth number, which is 37 degr. 30 min.

(2.) As the Sine of the fourth number 37 degr. 30 min. is to the Sine of the half Sum 113 degr. 15 min.

So is the Sine of the difference 3 degr. 15 min. to another Sine, viz. 4 degr. 54 min. Unto which seventh Sine if you adde the Sine of 90 degr. half that Sum shall be the Sine of an Arch, whose Complement being doubled is the Azimuth from the North.

To resolve this Probleme by the Projection.

That which is most intricate and difficult to perform by Numbers, is by Projection effected with the same ease as any of the rest. As in this Proposition, it is the Angle E Z P which is required.—Lay a Ruler upon the Zenith-point Z, and to the Point G, upon the Horizon; the Ruler thus laid will cut the Meridian Circle in the Point g. So the di∣stance g O, being taken in your Compasses and measured up∣on your Line of Chords, will be found to contain 146 degr. which is the Sun's Azimuth from O, the North part of the Me∣ridian.—But if you measure the distance between the g and H, it will contain 34 degr. which is the Azimuth from H, the South part of the Meridian.—And if you measure the distance g N upon your Chord-Line, you shall find that to contain 56 degr. and so much is the Sun's Azimuth from A, the East and West Points of the Horizon.

This Example of finding the Azimuth was taken when the Sun had 20 degr. of South Declination. I will now farther exemplifie this Proposition by finding the Azimuth when the Sun hath North Declination.—As let the Latitude be as before 51 d. 30 min. the Sun's Altitude 12 degr. and the Declination 20 d. North.

To work this by the Canon of Sines differeth nothing from the former, for the Analogie or Proportion is general in all Cases.

Upon the Projection it is resolved (though the same way, yet) upon another Triangle, namely, the Triangle Z P a, in which is given (1.) Z P, the Complement of the Latitude 38 d. 30 min. (2.) Z a, the Complement of the Altitude 78 degr. (3.) the Complement of the Sun's Declination North 70 degr. and you are to find the Angle P Z a, the Sun's Azimuth from the North.

Lay a Ruler upon Z unto the Point a, it will cut the Meri∣dian Circle in the Point s; the distance s O, being taken in your Compasses and applied to your Line of Chords, will there give you 72 degr. 52 m. And such is the Sun's Azimuth from the North.

If you subtract this Azimuth from the North 72 degr. 52 m. from 180 degr. the remainer 107 degr. 8 min. will give you the Azimuth from the South, which upon the Projection is the di∣stance s H.—And if from this Azimuth from the South 107 d. 8 min. you take 90 degr. the remainer 17 degr. 8 min. is the Azimuth from the East or West, which in the Projection is the di∣stance N s.

PROP. XII. The Latitude of the Place, the Sun's Declination, and the Sun's Altitude, being given, to find the Hour of the Day.

THIS Proposition is performed by the resolving of the Oblique-angled Sphericall Triangle Z P a, composed of Z P, an Arch of the Meridian, Z a, an Arch of an Azi∣muth Circle, and of P a, the Arch of an Hour-Circle: In which you have given (as in the last Proposition) the three Sides, to find the Angle Z P a, which you may doe by the 11. Case of Oblique Sphericall Triangles.

To resolve this Proposition by the Canon; Adde the three Sides together, and from the half Sum of them subtract the Complement of the Sun's Altitude, and note the difference, as you see here done.

		d.	m.
The Side	Z P, the Complement of the Latitude	38	30
Z a, the Complement of the Altitude	78	00
P a, the Complement of the Declination	70	00
	The Sum	186	30
	The half Sum	93	15
The difference between the half Sum and Z a, the Complement of the Altitude	15	15

Being thus prepared, you may resolve the Proposition by the Canon of Sines, by this

Analogie or Proportion.

(1.) As the Radius 90 degr. is to the Co-sine of the Sun's Al∣titude 78 degr.

So is the Co-sine of the Latitude 38 degr. 30 min. to a fourth. Sine, viz. 35 degr. 48 min.

(2.) As this fourth Sine of 35 degr. 48 min. is to the Sine of the half Sum 93 degr. 15 min.

So is the Sine of the Difference 15 degr. 15 min. to another Sine, viz. to the Sine of 26 degr. 40 min. Unto which Sine if you adde the Sine of 90 degr. (or Radius,) half that Sum shall be the Sine of an Arch, whose Complement being doubled is the Hour from the Meridian 95 degr. 52 min.

To resolve the Proposition by the Projection.

In the Triangle Z P a, it is the Angle at P that is to be found. Wherefore lay a Ruler from the Point P to the Point a, and it will cut the Meridian Circle in t: So the Arch t AE, being mea∣sured upon your Line of Chords, will be found to contain 95 d. 52 min. which is the Hour from the Meridian; and the Arch t ae, being measured, will contain 84 degr. 8 min. which is the

Hour from Midnight. Also the Arch t S, being measured upon the Chord, will contain 5 degr. 52 min. the Hour from Six.

		d.	m.		hours	m.
The Arch	t AE	95	52	converted in∣to Time is	6	23
t ae	84	08	5	36
t S	05	52	0	23

To convert Degrees and Minutes of the Aequinoctial into Hours and Minutes of Time: Note that 15 Degrees of the Aequi∣noctial make one Hour of Time, and one Degree 4 Minutes of Time. Therefore divide the Degrees of the Aequinoctial by 15, the Quotient is Hours; and multiply the Degrees by 4, and the Product will be Minutes of Time.—So the Hour from the Me∣ridian being 95 degr. 52 min. divide 95 by 15, the Quotient is 6 Hours, and 5 remaining, which 5 multiply by 4, and it makes 20 Minutes of Time, and the 52 min. make 3 minutes of Time and more, almost 4 minutes. So that 95 degr. 52 min. of the Aequinoctial do make in Time 6 hours and almost 24 minutes.

PROP. XIII. The Declination, Altitude, and Azimuth of the Sun, be∣ing given, to find the Hour of the Day.

THE Triangle Z E P in the Projection resolves this Problem: in which there is given (1.) E P, the Com∣plement of the Sun's Declination 70 degr. South; (2.) the Side E Z, the Complement of the Sun's Altitude 78 degr. and (3.) the Angle E Z P, the Sun's Azimuth from the North. So that you have two Sides and an Angle opposite to one of them given, to find the Angle opposite to the other, which you may doe by the 8. Case of Oblique-angled Sphericall Trian∣gles, and by the following

Analogie or Proportion.

As the Co-sine of the Declination 70 degr. is to the Sine of the Azimuth 146 degr. or 34 degr.

So is the Co-sine of the Altitude 78 degr. to the Sine of the Hour from Noon 35 degr. 36 min.

By the Projection.

Lay a Ruler to the Pole P, and upon the Point D in the Ae∣quinoctial; a Ruler thus laid will cut the Meridian Circle in e; the distance e Ae, being measured upon the Line of Chords, will give you 35 degr. 36 min. the Hour from Noon, which in Time is 2 hours and 22 minutes.

PROP. XIV. The Sun's Declination, his Altitude, and the Hour from Noon, being given, to find the Sun's Azimuth from the North part of the Meridian.

IN the same Triangle Z E P you have given (1.) the Side E P, the Complement of the Sun's Declination South 70 d. (2.) the Side Z E, the Complement of the Altitude 78 degr. and (3.) the Angle Z P E, the Hour from Noon 36 degr. 35 m. That is, you have given (as before) two Sides and an Angle opposite to one of them, to find the Angle opposite to the other, which you may doe by the 8. Case of Oblique Spheri∣call Triangles; or by this

Analogie or Proportion.

As the Co-sine of the Altitude 78 degr. is to the Sine of the Hour from Noon 36 degr. 35 min.

So is the Co-sine of the Sun's Declination 70 d. to the Sine of the Azimuth from the North part of the Meridian 146 degr. or 34 degr. from the South.

By the Projection.

Lay a Ruler to Z, and upon the Point G, which will cut the Meridian Circle in g: So g H, being measured upon your Line of Chords, will be found to contain 34 degr. the Azimuth from the South part of the Meridian, which being taken from 180 degr. the remainer will be 146 degr. equal to the Arch g O, the Azimuth from the North part of the Meridian.

PROP. XV. The Hour from Noon, the Latitude of the Place, and the Altitude of the Sun, being given, to find the An∣gle of the Sun's Position.

IN the Oblique-angled Sphericall Triangle Z P E you have given the Side Z P, the Latitude, Z E, the Complement of the Sun's Altitude, and Z P E, the Hour from Noon, to find the Angle Z E P, which is the Angle of the Sun's Position at the time of the Question: So in the Triangle Z P E you have two Sides, with an Angle opposite to one of them, given, to find the Angle opposite to the other Side, which you may find by the 2. Case of Oblique-angled Sphericall Triangles: For which this is

The Analogie or Proportion.

As the Co-sine of the Sun's Altitude 78 degr. is to the Sine of the Hour from Noon 35 degr. 36 min.

So is the Co-sine of the Latitude 38 degr. 30 min. to the Sine of the Angle of the Sun's Position at the time of the Que∣stion 21 degr. 45 min.

By the Projection.

This is the most troublesome Proposition that we have yet

met withall to be resolved by the Projection; and yet it is also thereby easily resolved in this manner.

Take in your Compasses 90 degr. of your Chords; then lay a Ruler upon Y, (the Pole of the Hour-Circle P E S,) and the angular Point E; it being so laid will cut the Meridian Circle in v. Then set 90 degr. of your Line of Chords from v to x upon the Meridian Circle, and the Ruler laid from Y to x will cut the Hour-Circle P E S in the Point y.

Again, lay a Ruler to ☉, (the Pole of the Azimuth Circle Z E N,) and to the angular Point E; it being so laid will cut the Meridian Circle in the Point M. Set 90 degr. from M to z upon the Meridian Circle, and lay a Ruler upon ☉ and z; it will cut the Azimuth Circle (it being continued without the Meridian Circle) in the Point δ.

Lastly, Lay a Ruler to the angular Point E, and this Point δ, it will cut the Meridian Circle in λ; also lay a Ruler from E to y, it will cut the Meridian in λ. The distance θ λ, being taken in the Compasses and measured upon your Line of Chords, will contain 21 degr. 45 min. and that is the quanti∣ty of the enquired Angle Z E P, which is the Angle of the Sun's Position at the time of the Question.

PROP. XVI. The Sun's Altitude, his Declination, and Azimuth from the North, being given, to find the Latitude.

IN the former Triangle Z E P, let there be given (1.) E Z, the Complement of the Sun's Altitude 78 degr. (2.) E P, the Complement of the Sun's Declination (or his distance from the North-pole) 110 degr. (3.) the Angle E Z P, the Azimuth from the North.—By the last Proposition find the Angle of the Sun's Position, Z E P, which had, the Side Z E, the Complement of the Latitude, may be found by this

Analogie or Proportion.

As the Sine of the Sun's Azimuth 146 degr. (or 34 degr.) is to the Sine of the Sun's distance from the North-pole 110 d. (or 70 degr.)

So is the Sine of the Angle of the Sun's Position 21 degr. 45 m. to the Complement of the Latitude 38 degr. 30 min.

By the Projection.

It being the Side Z P that is required, forasmuch as that is an Arch of the Meridian Circle, you have no more to doe but to take the distance Z P in your Compasses, and measure it up∣on your Line of Chords; which being done, you will find it will contain 38 degr. 30 m. the Complement of the Latitude, which taken from 90 degr. leaves 51 degr. 30 m. the Latitude it self.

By the Solution of the foregoing Propositions, both by Trigono∣metricall Calculation by the Canons of Sines and Tangents, and also by the Projecting of the Sphere, you may easily discern the facility of the one above the other, namely, that of Projecti∣on; forasmuch as those Propositions which by Calculation are most troublesome, are by the Projective way most easie.—For, here is no need of letting fall Perpendiculars to reduce the Ob∣lique Triangle into two Right-angled Triangles, and so making of two works for solving of one Problem. Besides, the Projection renders every Triangle so naturally to the eye, that they are re∣solved (as it were) by looking on them.—Many more Exam∣ples I might have given upon this one Projection, but I see the Scheme grow too full of Lines and Letters, that makes me here break off. And now I will shew how the ingenious young Sea∣man may apply these Propositions to his use at Sea.

[illustration] geometrical diagram

The Sphere Projected upon the Plain of the Meridian.

page 122.

The foregoing PROPOSITIONS applied to Practice: By which the Ingenious young Sea-man may make them serviceable to him at Sea, to severall good and usefull Purposes. The Seventh EXERCISE.

OF the foregoing Astronomicall Propositions the ingenious Sea-man may make good and profitable use at Sea. For some of them will be assistent to him in the finding of the Lati∣tude of the Place he is in.—Some of them will help him to find the time of the Sun's rising and setting in any place, and at any time of the Year.—Some will help him to the Hour of the Day,—Some to the Hour of the Night, at any time and in any place.—And divers of them to find the Varia∣tion of the Compass. Examples of all which I will instance in, so that he may put them in practice at Sea.

I. Propositions assistent to find the Latitude.

THE Propositions which may be applied to the finding of the Latitude are the First and the Sixteenth.

The first Proposition is to find The Sun's Declination, which being obtained, and the Sun's Meridian Altitude observed at Sea or Land in any part of the World, the Latitude of that Place, by help of them, may be known; in which there are severall Cases, according as the Sun hath either North or South Declination, and as the Sun is situate, he being either upon the North or South-side of the Meridian.—The severall Varie∣ties are these which follow.

When the Sun is in the Aequinoctial, ha∣ving no Declination, and the Meridian Altitude is observed on the

• SOUTH-side of the Meridian,
  • The Meridian Alti∣tude taken from 90 degr. leaves the Elevation of the North-pole.
• NORTH-side of the Meridian,
  • The Meridian Alti∣tude taken from 90 degr. leaves the Elevation of the South-pole.

When the Sun's Declination is

• NORTH,
  • If the Meridian Altitude be less then 90 d. and the Sun upon the South-side of the Meridian; the Sun's Decli∣nation, being taken from the Meridian Altitude, leaves the height of the Aequinoctial, which taken from 90 d. gives the Latitude North.
• SOUTH,
  • If the Meridian Altitude be less then 90 d. and the Sun upon the South-side of the Meridian, adde the Meridi∣an Altitude and Declination together; their Sum is the height of the Aequinoctial, which taken from 90 degr. leaves the Latitude North. But if the Sum of the De∣clination and Altitude exceed 90 degr. take 90 degr. therefrom, the remainer is the Latitude South.

When the Sun's Declination is

• NORTH,
  • If the Meridian Altitude be less then 90 d. and the Sun upon the North-side of the Meridian, adde the Alti∣tude and Declination together; their Sum is the height of the Aequinoctial, which taken from 90 d. leaves the Latitude South.—But if the Sum be above 90 d. take 90 d. therefrom, the remainer is the Latitude North.
• SOUTH,
  • If the Meridian Altitude be less then 90 d. and the Sun upon the North-side of the Meridian, subtract the De∣clination from the Meridian Altitude; the remainer is the height of the Aequinoctial, which taken from 90 degr. leaves the Latitude South.

When the Sun's Decli∣nation is

• NORTH, If the Meridian Altitude be just 90 degr. the Sun's Declination is the Latitude North.
• SOUTH, If the Meridian Altitude be just 90 degr. the Sun's Declination is the Latitude South.

If the Meridian Altitude be observed under the Pole, within the bounds of the Polar Circles, in such case the Sun's Decli∣nation must be taken from 90 degr. and what remains is his distance from the Pole; which being added to the Meridian Altitude, the Sum is the Latitude of the Place.

The other Proposition which will be assistent to find the La∣titude is the Sixteenth. If you set your Compass to the given Azimuth, and when the Sun is upon that Azimuth, if you take his height, you may find your Latitude either by Trigonometri∣call Calculation or by Projection. As in the Sixteenth Propo∣sition.

II. Propositions assistent to find the time of the Sun's rising and setting.

THE principal Proposition for this purpose is the Se∣cond, which is to find the Ascensionall Difference, from which the time of the Sun's rising and setting, the Semidiurnall and Seminocturnall Arches, may be gathered; and from thence the length of the Day and Night: all which are plainly shew∣ed in the Proposition it self.

III. Propositions assistent to find the Hour of the Day and Night.

THE Twefth and Thirteenth Propositions will be ser∣viceable to find the Hour of the Day. The Twelfth giving the Hour at any time of the Day by the Work of that Proposition it self.—The Thirteenth findeth the Hour upon a given Azimuth and Altitude. Wherefore set your Compass to the given Azimuth, and observe his Altitude when he cometh upon that Azimuth; the Sun's Declination (or time of the year) being known, you may then find the Hour by the Work of the Thirteenth Proposition.

The Proposition that will be assistent to you in finding the Hour of the Night is chiefly the Sixth, it shewing how to find the Sun's right Ascension, which, with the assistence of those other Tables which follow that Proposition, will help you to the Hour of the Night, and also to find at what time any of the Stars there inserted in the Table will be upon the Meridian. The manner how to effect either shall be shewed in these two following Problems.

PROBL. I. How to find at what time any of the Stars in the Table of the Sixth Proposition will be upon the Meridian.

SUbstract the Right Ascension of the Sun from the Right As∣cension of the Star, the remainer is the time of the Star's coming to the Meridian after noon. But if the right Ascension of the Star be less then the Right Ascension of the Sun, adde 360 degr. thereto, and substract the Right Ascension of the Sun from the Sum, and the Remainer is the time of the Star's coming to the Meridian.

Example. Upon the fourth of October 1667, the Sun being in 21. degr. of Libra, I would know at what time Sirius (or the Great Dog) will be upon the Meridian.

	d.	m.
The Right Ascension of Sirius is	97	27
The Right Ascension of the Sun that day is	199	23
Because Substraction cannot be made, adde 360 d. to the Right Ascension of the Star	360	00
The Sum is	457	27
The Sun's Right Ascension substracted from 458 degr. 4 min. leaves the time of the Star's coming to the Meridian	258	04

Which 258 degr. 4 min. being converted into Time make 17 hours almost, that is, at 5 of the Clock the next Morning Sirius will be upon the Meridian.

PROBL. II. To find the Hour of the Night by any of the Stars that are in the Table of the Sixth Proposition.

TAKE the Altitude of the Star, and by his Declination and Altitude find the Hour (by the Twelfth Proposition) as if it were by the Sun, which I call the Star's Hour. Then comparing the Right Ascension of the Sun with the Right Ascen∣sion of the Star, you may come to find the Hour of the Night.

Example. Upon the 16. day of November, in the Morning, I took the Altitude of Arcturus, finding it to be 27 degr. 12 m. and his Declination (by the Table) I find to be 20 degr. 58 m. By help of these two and the Latitude I find the Star's Hour to be 72 degr. Then compare the Sun's Right Ascension with the Star's Right Ascension, and find his time of coming to the Meridian, as in the former Probl. the difference between the Star's Hour and his coming to the Meridian is the Hour of the Night. See the manner of the Operation.

	d.	m.
The Right Ascension of Arcturus	210	13
The Right Ascension of the Sun	242	00
Adde to make Subt.	360	00
The Sum is	570	13
The Sun's Right Ascension substracted, rests	328	13
From which take 180 degr. or 12 hours	180	00
Rests	148	13
The Star's Hour substracted	72	00
Leaves the Hour of the Night	76	13

Which converted into Time is 5 h. 5 m. and that is the Hour in the Morning.

IV. Propositions assistent to the finding of the Variation of the Compass.

THE Propositions that will be serviceable herein are the 3. 7. 8. 9. 10. 11. and 14. but more especially the Third and the Eleventh: and those I shall here illustrate by Exam∣ple, though all the rest (as occasion may fall out) will be also usefull thereunto. By the Third Proposition you may find the Amplitude of the Sun's Rising and Setting.—By the Eleventh you may find the Azimuth at any time of the day.— By either of which the Variation of the Compass may be found, and also which way it varieth.

I. To find the Variation by the Amplitude.

BY the Third Proposition you found the Sun's Amplitude at his rising or setting to be 33 degr. 20 min. from the true East or West Points of the Horizon towards the North. Having thus before-hand found the Amplitude, in the Morning I set my Compass to the Sun at his Rising; and if I find that the Sun by my Compass do rise 33 d. 20 m. from the West-point thereof towards the North, then may I be ascertain'd that my Compass hath no Variation, but that the Fly or Wires do point directly North and South.—But finding before-hand the Amplitude to be 33 d. 20 m. and I should find the Sun to rise but 28 degr. from the East-point of my Compass, then substracting 28 degr. from 33 degr. 20 min. the difference is 5 degr. 20 min. and so much doth my Compass varie from the true East-point, and con∣sequently all the other Points of the Compass as much.

Now to find which way the Compass varieth, you must ob∣serve whether your Amplitude, found by your Calculation, be to the Right or Left-hand of the Sun's rising or setting. And

if it be on the Right-hand, you may conclude the Variation to be Easterly; but if on the Left-hand, Westerly.

As for Example; Finding by the Amplitude that the Sun should rise 33 d. 20 min. from the East Northerly, when I come to set my Compass to the Sun at his rising, I find that the Sun riseth but 28 degr. from the East Northerly; wherefore the Am∣plitude found is on the Left-hand, and so I conclude the Varia∣tion to be 5 d. 20 min. Westerly.

II. To find the Variation by the Azimuth.

SUppose the Sun's Azimuth found by the Eleventh Propositi∣on to be 107 degr. 30 min. from the North, and when I set the Compass, I find the Magneticall Azimuth to be 102, the dif∣ference between the true and the Magneticall Azimuth being 5 d. 30 m. which is the Variation.

Now to know whether this Variation be towards the East or towards the West: seeing by the Azimuth found the Sun should have been 107 d. 30 min. from the North, which is 17 degr. 30 min. from the East; but setting of the Sun with my Compass, I find that it was from the East to the Southward onely 12 degr. so that the Degree upon which the Sun should have been was more towards the Right-hand then the Degree on which it was; therefore I conclude the Variation to be 5 degr. 30 min. Easterly.

PROPOSITIONS GEOGRAPHICALL, Shewing how the Distance of any two Places upon the Terrestriall Globe may be found, both by Trigono∣metricall Calculation and Geometricall Projection. The Eighth EXERCISE.

BEfore I come to shew how the Distances of Places are to be found upon the Terrestri∣all Globe, it will be necessary, first, to de∣scribe unto you the manner how any two Pla∣ces, whose Longitudes and Latitudes are given, may be laid down upon the Projecti∣on. Wherefore in the Scheme, the outward Circle thereof, N AE S ae, represents the first Meridian, passing over the Islands of Azores, from whence the Ancients did begin their account of Longitude, because (say some) the Compass hath there no Variation.—The Line AE ae is the Aequinoctial, upon which the Longitude is counted from the first Meridian.—The Circles N E S, N R S, N C S, and N D S, are Circles of Longitude passing over severall Places.—The lesser Circles E, C, D, and V, are Circles or Parallels of Latitude.

—The Points *, *, *, &c. are severall Places whose distance we are to find by the following Propositions.—And the Points Sun;, ☉, ☉, &c. are the Poles of the Arches of great Circles which pass through the respective Places whose distance is to be found.

And here note, That the Circles of Longitude in this Pro∣jection are the same as the Azimuth Circles in the former Pro∣jection; and the Centres and Poles of them are found in the same manner.—Likewise, the Parallels of Latitude in this Scheme are the same with the Parallels or Circles of Altitude in the former Projection, and their Centres are found in the same manner as is before, in the Description of that Projection, prescribed; and therefore it shall here need no more Precepts for its Delineation, but we will proceed to the Propositions which shew how to find the Distance of Places.

PROP. I. Two Places which differ onely in Latitude, to find their Distance.

IN this Proposition there are two Varieties.

1. If both the Places lie under one and the same Meridian, and on one and the same Side of the Aequinoctial, either on the North or South Side thereof, then substract the lesser Lati∣tude from the greater, and the Difference converred into Miles (by allowing 60 Miles to one Degree) shall give you the Distance.

Example. London and Ribadio lie both under one Meridian, namely of 20 degr. of Longitude; but they differ in Latitude, for London hath 51 d. 30 min. and Ribadio hath Latitude 43 d. both North; the difference of Latitude is 8 degr. 30 m. which being turned into Miles makes 510 miles.

2. If the two Places lie under one and the same Meridian,

but one on the North, and the other on the South-side of the Aequinoctial, adde both the Latitudes together, the Sum is the Distance.

[illustration] geometrical diagram

Example. London and the Island Tristan Dacunhu lie both under one Meridian; but London hath 51 degr. 30 min. North Latitude, and the Island hath 34 d. South Latitude: their Sum is 85 degr. 30 min. which converted into Miles (by dividing

the Degrees by 60. and allowing for every Minute one Mile) makes 5130 miles. And such is the distance of London and the Island Tristan Dacunhu.

To find the distance of these Places upon the Projection.

Seeing that they all lie under one Meridian, namely, N E G H S, find the Pole thereof at K; then lay a Ruler to K and E, it will cut the first Meridian in a; also a Ruler laid from K to G will cut the Meridian in b: the distance a b, measured upon the Line of Chords, will give 8 degr. 30 min. the Distance of London and Ribadio. Again, to find the Distance between London and the Island Tristan Dacunhu, lay a Ruler from K to E, it will cut the first Meridian in a, (as before) and laid from K to H, it will cut the first Meridian in c: the Distance a c, being mea∣sured upon the Line of Chords, will contain 85 degr. 30 m. the Distance between London and the Island, which in Miles is 5130.

PROP. II. Two Places which differ onely in Longitude, to find their Distance.

IN this Proposition there are two Varieties also. For 1. The two Places may lie both under the Aequinoctial, and have no Latitude: in this Case the difference of their Lon∣gitudes (if it be less then 180 degr.) reduced into Miles is their Distance; but if their difference exceed 180 degr. take it out of 360 degr. the remaining Degrees turned into Miles will be the Distance of the two Places.

Example. The Island Sumatra and the Island of S. Thoma lie both under the Aequinoctial, the Island of S. Thoma having 33 d. 10 m. of Longitude, and the Island Sumatra 137 d. 10 m. The lesser Longitude taken from the greater leaves 104 d. 0 m.

which converted into Miles is 6240. And that is the Distance of the two Islands.

[illustration] geometrical diagram

2. But if the two Places differ onely in Longitude, and lie not under the Aequinoctial, but under some other intermediate Parallel of Latitude, between the Aequinoctial and one of the Poles, then to find their Distance, this is

The Analogie or Proportion.

As the Radius 90 degr. is to the Co-sine of the common Lati∣tude 47 degr.

So is the Sine of half the difference of Longitude 21 d. 37 m. to the Sine of half their Distance 15 degr. 38 m.

So let the two Places be the Cities Constantinople and Com∣postella, both lying in the Latitude of 43 degr. but they differ in Longitude 43 degr. 15 min. So that their Distance accor∣ding to the former Analogie will be found to be 31 degr. 16 m. which, converted into Miles, is 1876 miles, which is the Di∣stance between the two Cities Constantinople and Compostella.

To find the Distance of these two Places upon the Projection.

The two Places upon the Projection are noted with the Letters C and D, both lying in the Latitude of 43 degr. but Constantinople in the Longitude of 63 degr. and Compo∣stella in the Longitude of 106 d. 15 min. So that their diffe∣rence of Longitude is 43 degr. 15 min. Wherefore through the two Places C and D draw the Arch of a great Circle, and find the Pole thereof; (which to effect is already taught at the beginning of this Book:) which Pole will be at the Point M. Then laying a Ruler upon M and C, it will cut the first Meridian in the Point d; and laid from M to D, it will cut the first Meridian in e: the Distance between d and e, mea∣sured upon the Line of Chords, will be found to contain 31 degr. 16 min. which, converted into Miles, giveth 1876, the Distance as before.

PROP. III. Two Places differing both in Longitude and Latitude being proposed, to find their Distance.

THERE are three Varieties contained in this Propositi∣on. For

1. One of the Places may lie under the Aequinoctial, and have no Latitude; and the other under some Parallel of La∣titude, between the Aequinoctial and one of the Poles. For finding the Distance of Places that are so situate, this is

The Analogie or Proportion.

As the Radius 90 degr. is to the Co-sine of the difference of Longitude 76 degr. 50 min.

So is the Co-sine of the Latitude given 38 degr. 30 min. to the Co-sine of the Distance required 52 degr. 41 min.

Thus suppose the two Places to be the Island of S. Thoma, ly∣ing under the Aequinoctial in the Longitude of 33 d. 10 m. and London under the Parallel of 51 d. 30 m. of North Latitude, having 20 degr. of Longitude, their Distance by the former Proportion will be found to be 52 degr. 41 min. which, con∣verted into Miles, gives 3161 miles for their Distance.

By the Projection.

The two Places upon the Projection are represented by the Letters A and E. The Point A lying under the Aequinoctial, and in 33 degr. 10 min. of Longitude, represents the Island of S. Thoma; and the Point E, under the Parallel of 51 d. 30 m. North, and in Longitude 20 degr. represents London.— Through the two Places A and E (according to former Dire∣ctions) draw an Arch of a great Circle, and find the Pole thereof, which will be at the Point L. A Ruler laid to L and

the Point E will cut the first Meridian in n; and the Ruler being laid from L to A will cut the first Meridian in f: the Di∣stance n f, being measured upon the Line of Chords, will be found to contain 52 degr. 41 min. as before, which in Miles is 3161.

[illustration] geometrical diagram

2. If both the Places proposed shall be without the Aequi∣noctial, but on one Side, either both towards the North, or

both towards the South, the finding of their Distance is by this

Analogie or Proportion.

(1.) As the Radius 90 degr. is to the Co-sine of the diffe∣rence of Longitude 44 degr.

So is the Tangent of 38 degr. 30 min. to a fourth Tangent 28 degr. 55 min. which taken from the Complement of the Lat. of Jerusalem 58 degr. 20 min. leaves 29 d. 25 m.

(2.) As the Co-sine of the fourth Tangent 61 degr. 35 min. is to the Co-sine of 60 degr. 35 min.

So is the Co-sine of the Latitude of London 38 degr. 30 min. to the Co-sine of the Distance 51. degr. 9 min.

So the two Places propounded being London, lying in North Latitude 51 degr. 30 min. and Longitude 20 degr. and the other, Jerusalem, lying in North Latitude also 31 degr. 40 min. and Longitude 66 degr. you may find their Distance by the foregoing Analogie to be 38 degr. 51 min. which in Miles makes 2331.

By the Projection.

The two Places in the Projection are represented by the Letters E and F, E being London, F Jerusalem; through which Points draw the Arch of a great Circle, and find its Pole: the Circle (in this Example) comes so near a right Line, that I have so drawn it; and therefore his Pole is but little within the outward Circle, viz. at P. Wherefore lay a Ruler to P and E, it will cut the first Meridian in g; and being laid from P to F, it will cut the Meridian in h: the Distance g h, being measured upon the Line of Chords, will be found to con∣tain 38 degr. 51 min. and in Miles 2331, as before.

3. The two Places propounded may be so situate, that one of them may lie on the North, and the other on the South-side of the Aequinoctial. For finding the Distance of such Places follow this

Analogie or Proportion.

(1.) As the Radius 90 degr. is to the Co-sine of the difference of Longitude 40 degr.

So is the Co-tangent of the greater Latitude 50 d. to the Tan∣gent of a fourth Arch 37 d. 10 m. which being substracted out of the other Latitude, and 90 d. added thereto, say,

[illustration] geometrical diagram

(2.) As the Co-sine of the Arch found 52 degr. 50 min. is to the Co-sine of the Arch remaining 52 degr. 50 min.

So is the Co-sine of the Latitude first taken 50 degr. 00 min. to the Co-sine of the Distance 40 degr. which taken from 180 degr. there remains 140 degr. for the Distance of the two Places.

So the two Places propounded being the Cape of Good hope, lying in the Latitude of 40 degr. South, and Longitude 50 d. and the other Place Malibrigo, lying in 26 degr. of North La∣titude, and in 180 degr. of Longitude, you may find their Di∣stance by the foregoing Analogie to be 140 degr. which, con∣verted into Miles, make 8400. And such is the Distance of the two Places.

By the Projection.

In the Projection the two Places are represented by the Let∣ters T and V; the Letter V representing the Cape of Good hope, and T Malibrigo. Now Malibrigo having 180 degr. of Longi∣tude, (which is just half the Circumference of the Aequinoctial, and is as far remote as any Place can be from the first Meridi∣an; for if you were to project any Place having above 180 d. Longitude, (as suppose 230 degr.) you must substract such Longitude from 360 degr. and project the remainer; so 230 degr. being taken from 360 degr. leaves 130 degr. which must be projected in stead of 230 degr. and by this means it is that Malibrigo is projected upon the outermost Circle or first Meridian.)

Through these two Points T and V draw the Arch of a great Circle, T V X, and find its Pole at R: then a Ruler laid at R and the Point V will cut the first Meridian in k, and T k, being measured upon your Line of Chords, will be found to contain 140 d. and that is their Distance, which in Miles maketh 8400.

These are all the Varieties of Positions of Places upon the Terrestriall Globe; for no two Places (whose Distance can be

required) can be proposed, but they must fall under one or other of the Varieties contained in some of these three Propositions. And note that this way of finding the Distance of Places is the most absolute and exact of any other.—And what is here said con∣cerning finding these Distances the ingenious may apply to Cir∣cular Sailing, of all other waies the most perfect: which I shall leave to the industrious Sea-man to find out of himself, till I pre∣sent him with something of that kind: in the mean time let him make use of the foregoing EXERCISES, and this which follows.

The Doctrine of RIGHT-LINED TRIANGLES applied to Practice in NAVIGATION: Whereby Sundry Nauticall Questions are resolved; and many Problems of Sailing, both by the Plain and Merca∣tor's Chart, performed by Protraction, by Calculati∣on, and also wrought upon the Chart it self. The Ninth EXERCISE.

SECTION I.

BEfore I come to the Resolving of such Problems as principally appertain to Navigation, which are such as concern Longitude, Latitude, Rhumb and Distance; I shall shew how the Solution of plain Triangles may be made applicable to the ta∣king of Heights and Distances, and so (in the first place) pro∣pose and work severall Nauticall Questions, which to the indu∣strious

Mariner will be both delightfull and profitable, and give occasion to him to invent and put in Practice others of his own contrivance.

QUESTION I. There are two Ships set sail from the Port A, the one saileth di∣rectly North 24 Centesms, (or 4 Leagues and ⅘ parts of a League,) and the other directly East 32 Centesms, (or 6 ⅖ Leagues;) I demand how the two Ships bear one from the other, and also how far they are asunder.

[illustration] geometrical diagram

DRAW a right Line A B, and upon A raise the Perpen∣dicular A C: let the Point A represent the Port from whence the two Ships set sail: then, because the first Ship sailed 24 Centesms North, from a Scale of equal parts take 24 Cent. and set them from A to C; so shall C be the place of the first Ship. Then, because the other Ship sailed directly

East, which is a Quadrant or Quarter of the Compass distant from the North, therefore the Angle at A must be a right Angle: And because the second Ship sailed East 32 Cent. take 32 Cent. from the same Line of equal parts, and set them upon the Line A B, from A unto B; so shall B be the place of the second Ship.

Now first, To know how these two Ships bear one from a∣nother, Draw the Line C B, and measure the Quantity of the Angle at B, which you shall find to be 33 degr. 45 min. which is three Points from the West Northerly, that is the N. W. by West Point of the Compass; and so doth the second Ship B bear from the first Ship C.—Again, find the quantity of the Angle at C, which you shall find to be 56 degr. 15 m. which is five Points from the South Easterly, that is the S. E. by East Point of the Compass; and so do the two Ships bear one from the other.—Then for the Distance that the two Ships are from each other, Take in your Compasses the di∣stance between B and C, which measure upon your Scale of equal parts, and you shall find it to contain 40 Centesms or 8 Leagues; and so far asunder are the two Ships B and C.

The Bearing of the Ships one from the other is found by the first Case of Right-angled plain Triangles by this Ana∣logie.

As the Distance that the first Ship sailed is to the Distance that the second Ship sailed;

So is the Radius to the Tangent of the Angle that the first Ship bears to the second. The Complement whereof is the bearing of the second Ship to the first.

The Distance of the Ships from each other is found by the seventh Case, by the following Analogie. Having found the Angle C, the bearing of the first Ship from the second, say,

As the Bearing of the first Ship is to the Distance that the se∣cond Ship sailed;

So is the Radius to the Distance of the two Ships.

QUEST. II. A Ship at A discovers an Island at C, lying from her directly East, but she sails from A towards B 32 Cent. or 6 ⅖ Leagues directly South; but her Compass coming to some mischance, that use cannot be made of it, she again at B discovers the same Island, and sails upon an unknown Point of the Compass directly upon the Island, and touches upon it, having sailed 8 Leagues.—I demand upon what Point of the Compass the Ship sailed from B to C, and also how far off the Island was from A, where it was first discovered.

DRAW a Line C A, representing a Line of East and West, and upon A erect a Perpendicular A B, and from A to B set off 32 Cent. or 6 ⅖ Leagues, the distance that the Ship sailed from A to B. Then take out of your Scale of equal

[illustration] geometrical diagram

parts 40 Cent. or 8 Leagues, the distance that the Ship sailed from B to the Island; and setting one foot of the Compasses in B, with the other describe an obscure Arch of a Circle m m, crossing the East and West Line in C: so is C the place of the Island.

Now first, to find upon what Point of the Compass the Ship sailed from B to the Island, you must find the quantity of the Angle at B, (either by your Line of Chords, or Pro∣tracting Quadrant,) and you shall find it to contain 33 degr. 45 min. which is three Points from the North Easterly, that is N. E. by N. and upon that Point did the Ship sail from B to the Island at C.—Then, to know how far the Island C was from A, where it was first discovered, Take in your Com∣passes the length of the Line A C, and measure it upon your Scale; so shall you find that to contain 24 Cent. or ⅘ Leagues: and so far distant was the Island from A.

The Point of the Compass that the Ship sailed upon from B to C may be found by the second Case of Right-angled plain Triangles, by this Analogie.

As the Distance which the Ship sailed from B to C is to the Radius;

So is the Distance sailed between A and B to the Co-sine of the Point that the Ship sailed upon from B to C.

The Distance that the Ship was from the Island, when first discovered, may be found by the fifth Case of Right-angled plain Triangles, by the following Analogie.

(1.) As the Distance that the Ship sailed from B to C is to the Radius;

So is the Distance that the Ship sailed from A to B to the bear∣ing of the Island from B.

(2.) As the Radius is to the Distance that the Ship sailed from C to B;

So is the Sine of the Rhumb that the Ship sailed upon from B to C to the Distance of the Island from A.

QUEST. III. There are two Ports at A and B which are distant 6 ⅖ Leagues, and lie directly North and South of each other; from whence two Ships set sail, both for the Port C: the Ship at B sails away upon a South-W. by South Point; and the Ship at A sails directly West.—I demand how many Leagues either of the Ships had sailed when they met at the Port C, and also how the Port C did bear from that at B.

[illustration] geometrical diagram

DRAW a right Line A B, and upon it set off 32 Cen∣tesms, or 6 ⅖ Leagues. Now because the Ship at B steered a S. W. by S. Course, which is three Points from the South-Westerly, therefore upon the Point B protract an An∣gle of 33 degr. 45 min. and draw the Line B C.—Then, because the Ship at A steered a Westerly Course, which is a Quarter from the North, upon the Point A protract an Angle

of 90 degr. and draw the Line A C, cutting the former Line B C in C.—Now to know how many Leagues each Ship sailed, take in your Compasses the length of the Line B C, and measuring it upon your Scale, you shall find it to contain eight Leagues; and so many did the Ship that came from B sail. Also take the length of the Line A C in your Compasses, and measuring that upon your Scale, it will be found to contain 24 Centesm. or 4 ⅘ Leagues; and so much did the Ship that came from A sail. Now to know how the Port at C did bear from that at B, find the quantity of the Angle at C, which you shall find to be 56 degr. 15 min. that is, five Points from the East Northerly, namely, N. E. by N. and so did the Port C bear from B.

The finding of the Distance that each Ship sailed may be done by the third Case of Right-angled plain Triangles by this Analogie.

As the Distance of the two Ports A and B is to the bearing of the Port C from B;

So is the Sine of the Rhumb that the Ship sailed upon from B to C to the Distance that the Ship sailed from A to C;

And so is the Radius to the number of Leagues that the Ship sailed from B to C.

QUEST. IV. A Ship at C discovers a Point of Land at A bearing from her S. S. E. but she shapes a Course E. by S. and sails away 8 Leagues to B, and at B she discovers the same Point of Land bearing from her W. S. W.—I demand how far the Ship was from Land being at C and B.

DRAW a Line C B containing 40 Cent. or 8 Leagues, and upon C protract an Angle of 56 degr. 15 min. or five Points, which is the difference the Point of Land did bear

from the Ship being at C, and the Point upon which he sailed from C to B; and draw a right Line C A.—Then upon the Point B protract an Angle of 33 degr. 45 min. which is the difference of the Ship's bearing from C and A, she being at B, namely, W. S. W. and draw the Line B A, cutting the Line C A, before drawn, in A.

Now to find how far the Ship was from Land being at C, measure the Line CA upon your Scale of equal parts, and you shall find it to contain 24 Centesms, or 4 ⅘ Leagues: and so far was the Ship from the Land, when she was at C. Also measure the length of the Line B A, and you shall find that to contain 32 Cent. or 6 ⅖ Leagues: and so far from Land was the Ship being at B.

To find these Distances by the Canon of Sines and Table of Logar. you may doe it by the fourth Case of Right-angled Triangles, by this Analogie.

As the Radius is to the Distance that the Ship sailed from C to B;

So is the Bearing of the Ship, being at C, to her Distance from Land, being at B.

Or,

The Bearing of the Sip, she being at B, to her Distance from Land at C.

QUEST. V. A Ship being at A discovers two other Ships at C and B; the Ship at C bears from her directly East, and the other Ship at B bears from her directly South. The Ship at A sails directly South 32 Cent. to B, and being at B, steers away upon an un∣known Course to C 40 Cent. or 8 Leagues.—I demand upon what Point the Ship failed from B to C,—and also how far C is distant from A.

DRAW a right Line A B, for the bearing of the Ship B from the Ship A, which was direct South. Also from A draw another Line A C, for the bearing of the Ship C from the Ship A, which was directly-East. Now because between the South and the East is 90 degr. or one Quarter of the Compass, therefore upon the Point A protract an Angle of 90 degr. drawing the Lines A C and A B at right Angles. This done, take 32 Cent. out of your Scale of equal parts, which is the distance that the Ship sailed South from A to B. Then take from the same Scale 40 Cent. which is the distance that the Ship sailed from B to C upon an unknown Point. And with this distance, setting one foot of the Compasses in B, with the other describe an obscure Arch of a Circle m m, cutting the Line A C in the Point C, and draw the Line C B.—Now to find upon what Point of the Compass the Ship sailed from B to C, find the quantity of the Angle at B, which you shall find to contain 33 degr. 45 min. that is three Points from the North Easterly, namely, N. E. by N. and upon that Point did the Ship sail from B to C.—Then to find how far C is distant from A, Take the Line C A in your Compasses, and measuring it upon your Scale, you shall find it to contain 24 Cent. or 4 ⅘ Leagues: and so far is C distant from A.

The Point upon which the Ship sailed from B to C may be found by the second Case of Right-angled Triangles, by this Analogie.

As the Distance that the Ship sailed from B to C is to the Ra∣dius;

So is the Distance that the Ship sailed from A to B to the Co-sine of the Rhumb from the Meridian.

Then for the Distance of C from A.

As the Radius is to the Distance that the Ship sailed from B to C;

So is the Rhumb from the Meridian that the Ship sailed upon from B to C to the Distance of C A.

QUEST. VI. Two Islands at A and C are discovered by a Ship at B, the Island A bears from the Ship at B N. N. W. and the Island at C bears N. by E. from B; the Ship being at B sails away N. N. W. to the Island A, and having sailed 32 Cent. touches upon the Island, and being there findes that the Island C bears from the Island A E. N. E.—I demand how far the Ship at B was from the Island C, and also how far the two Islands were asunder.

DRAW a Line A B, and upon it set 32 Cent. which is the Distance that the Ship sailed from B to the Island at A. And because the Island A did bear from B N. N. W. and the Island at C N. by E. which are three Points, or 33 degr. 45 min. asunder, upon the Point B protract an Angle of 33 d. 45 min. and draw the Line B C.—Then because the Island at C bears from the Island at A E. N. E. which is eight Points, or 90 degr. from N. N. W. upon the Point A pro∣tract an Angle of 90 degr. and draw the Line A C, cutting the Line B C in C.

Now to find the Distance of the Ship being at B from the Island C, take the Line C B in your Compasses, and applying it to your Scale, you shall find it to contain 40 Cent. or 8 Leagues; and so far was the Ship at B from the Island at C. And to find the Distance of the Islands one from the other, take C A in your Compasses, and measure it upon your Scale, you shall find it to contain 24 Cent. or 4 ⅘ Leagues; and so far distant were the Islands one from the other.

The Distance from A to C may be found by the sixth Case of Right-angled plain Triangles, by this Analogie.

As the Co-sine of the Rhumb that the Ship sailed upon from B to A is to the Distance that the Ship sailed from B to A;

So is the Radius to the Distance of the Ship at B from the Island at C.

Then for the Distance of the two Islands, by the fourth Case say,

As the Radius is to the Distance C B;

So is the Sine of the Difference between the bearing of the two Islands from B to the Distance of the two Islands C and A.

QUEST. VII. Two Ships set out from one and the same Port A; the Ship C sails 24 Cent. or 4 ⅘ Leagues directly East, and the Ship B sails away 32 Cent. or 6 ⅖ Leagues directly South.—When they have thus sailed, I demand how far the two Ships are from each other.

DRAW a right Line A B, and set off upon it 32 Cent. the Distance that the Ship sailed from A to B South. —Then because the other Ship sailed directly East, which is 90 degr. from the South, upon the Point A erect the Perpen∣dicular A C, and upon it set off 24 Cent. or 4 ⅘ Leagues from A to C, which was the Distance the other Ship sailed East. —Then draw the Line C B, which being taken in your Compasses, and measured upon your Scale, will be found to contain 40 Cent. or 8 Leagues. And so far are the two Ships from each other.

This Distance, by the seventh Case of Right-angled plain Triangles, may be found by this Analogie.

(1.) As the Distance that the Ship sailed from A to B is to the Distance that the Ship sailed from A to C;

So is the Radius to the Tangent of the Angle at B.

(2.) As the Sine of the Angle at B is to the Distance C A; So is the Radius to the Distance C B.

QUEST. VIII. Two Ships set sail from the Port at K; the one sails 3 77/100 Leagues upon the S. W. Point towards M, the other sails 8 Leagues upon the West Point towards L.—I demand how many Leagues the Ships at M and L are asunder, and also how the Ship at M bears from the Port K, and the other Ship at L.

DRAW a right Line K L, and by help of your Scale set off upon it 8 Leagues, the Distance that the Ship sailed from K to L upon the West-Point. Then because the other Ship sailed 3 77/100 Leagues from K towards M upon the S. W. Point, which is 45 degr. or 4. Points from the West, therefore upon the Point K protract an Angle of 45 degr. and draw the Line K M, setting off upon it from K to M 3 77/100 Leagues, the Distance that the Ship sailed from K to M, and draw the Line M L.

[illustration] geometrical diagram

Now to know, First, how far distant the Ships at M and L are from each other, take in your Compasses the length of the Line M L, which applie to your Scale, and you shall find it to contain 6 62/100 Leagues.—And, Secondly, to find how the Ship at M bears from the Port K and the other Ship at L, you must find the quantity of the Angle at M, which you will find to be 112 degr. 30 min. that is, eleven Points. Now because the Course from K to M was S. W. therefore the Ship at M bears from the Port K N. E. And seeing that the Angle at M is 112 degr. 30 min. or eleven Points; therefore eleven Points counted from the N. E. Point is the Bearing of the Ship at M from that at L, which is W. N. W.

The Distance of the Ships M and L may be found by tho fifth Case of Oblique-angled plain Triangles; and the Bear∣ings by the second Case.

QUEST. IX. There are three Ships, K, L, and M: the Ship K is distant from the Ship L 8 Leagues; the Ship at L is distant from that at M 6 62/100 Leagues; and the Ship at M is distant from that at K 3 77/100 Leagues; and they lie directly North and South.—I demand how the Ship at M bears to that at L, and how that at L bears to that at K.

DRAW a right Line, and out of your Scale take 8 Leagues, and set them thereon from K to L, for the Di∣stance of the Ships at K and L. Then take 3 77/100 Leagues, the Distance of the Ships K and M, out of your Scale; and setting one foot of the Compasses in K, with the other describe the obscure Arch of a Circle o o. Again, take 6 62/100 Leagues from your Scale, which is the Distance that the Ship L was from the Ship M; and setting one foot of the Compasses in L, with the other describe the obscure Arch of a Circle n n, crossing

the former Arch in the Point M. Then draw the Lines M K and M L; so have you their true Positions.

Now to find their Bearing one from another; forasmuch as the Ships M and K did lie North and South of each other, find the quantity of the Angle at M, which is 112 degr. 30 min. that is, eleven Points from the South Eastward, (or 3 Points from the East Northward,) either of which will be the N.E. by E. Point: and so doth the Ship M bear from that at K. And for the Bearing of that at K from that at L, finde the quantity of the Angle at L, which will be 22 degr. 30 min. or two Points; so two Points from the S. W. by W. Point Southward is S. W. by S. and so doth the Ship L bear to that at K.

[illustration] geometrical diagram

The Bearings of the Ships from each other may be found by the third Case of Oblique-angled plain Triangles, by the Analogie in that Case set down.

PROBLEMS Of Sailing by the Plain Sea-Chart. SECTION II.

AMONG Sea-men there are Three principal waies of Sailing most in Use and Practice: Two whereof are Rectilineal, performed by Right Lines, the Third is Sphericall or Cir∣cular, performed by Arches of great Cir∣cles of the Sphere.

Of the Two first, the one is called Plain Sai∣ling, or, Sailing by the Plain Sea-Chart.

The other is called Mercator's Sailing, or, Sailing by Merca∣tor's Chart.

These two Charts are both of them composed of Right Lines, yet differ both in their Construction and Use, though not so much in their Use as in their making or composition.

The Plain Sea-Chart consisteth of Meridians and Parallels, which are drawn in all parts equal from the Aequinoctial towards either of the Poles; which is erroneous, as hereafter shall be dis∣covered.

Mercator's Chart hath the Degrees of Longitude in every Parallel of Latitude equal to those in the Aequinoctial, as the Plain Chart hath: But the Degrees of Latitude do increase more and more (as they grow nearer the Poles) in such a Pro∣portion as every Parallel of Longitude doth decrease.

The way of Sailing by the Plain Sea-Chart is much in use, nay too much, considering the Errours that it leads Sea-men into; though they are not so easily discovered in short as in long Voi∣ages, nor in places near the Aequinoctial as those nearer the Poles. But, I suppose, it is more used, for the ease there is in Projecting of this Chart, more then in that of Mercator's: other∣wise I know not why that should be so as it is embraced, and the other (I mean that of Mercator's) so much neglected; which comes so near to the Spherical way of Sailing, that there is an insensible difference between them. But I shall bring them face to face, that the ingenious Sea-man may see their Difference, and thereby abandon Errour, and embrace the Truth. For in the fol∣lowing Problems I shall perform the same thing by both Charts, by which the Errours may more palpably be discovered. And to retain the Method which I have observed in all the foregoing EXERCISES, I shall shew how these Nauticall Problems may be Trigonometrically performed by the Tables of Loga∣rithms, and Canons of Artificial Sines and Tangents. And to be∣gin this Exercise, the first thing that I shall propose unto you is

The Making of the Plain Sea-Chart.

A Sea-Chart may be made either general, or particular. A General Sea-Chart is that whose Degrees of Latitude pro∣ceed from the Aequinoctial to either Pole, which in the com∣mon Sea-Chart may be done; but it will be egregiously false, as the Degrees of Latitude grow nearer the Pole, as I have al∣ready declared.—A Particular Sea-Chart is such a one as is made properly for one particular Navigation: as if your

whole Navigation were not to exceed the Latitudes of 48 and 60 degr. of Latitude, and not to differ in Longitude above 8 degrees.

Now to project or make such a Chart; First, draw a right Line A B, representing the Meridian, and cross it at right Angles in the Point A with another right Line A D, representing the Parallel of your least Latitude, namely, of 48 degr.— Secondly, consider what Distance you will have your Parallels of Longitude and Latitude to be, (for in this Chart they are both equal,) whether an Inch, 2, 3, or 4 Inches, (for the lar∣ger the better.) But in this Example I have made them one∣ly half an Inch. I take therefore half an Inch out of an exact Scale, and run it up upon the Meridian Line A B, from A to 49, from 49 to 50, from 50 to 51, &c. till I come to my greatest Latitude, which is here supposed to be 60 degr.—Thirdly, run the same Distance of half an Inch from A towards D, upon the Line A D, eight times, because the Difference of Longitude in your whole Navigation will not exceed 8 degrees.— Fourthly, draw the Line C D, parallel to A B, and B C, par∣allel to A D, and run the same Distances upon the Line B C as are upon the Line A D, and the same upon C D as are upon the Line A B.—Fifthly, from each Degree of Latitude in the Line A B draw to the like Degree of Latitude in the Line C D a right Line, as 49, 49; 50, 50; 51, 51; 52, 52; &c. till you have drawn all your Parallels of Latitude.—Sixthly, for your Meridians, they are to be drawn in like manner as were the Parallels of Latitude, all of them equidistant, and par∣allel to your first Meridian A B, as the Lines 1, 1; 2, 2; 3, 3; &c. And by this means have you the Meridians and Parallels drawn.

The grand Divisions, or whole Degrees, being thus set upon your Chart, we now come to sub-divide them. And for the dividing of the Degrees of the Aequinoctial at the top and bottom of your Chart, let each of them be divided into 5 or 10 parts, and each of those parts sub-divided into 5 or 10

more less parts, according as Quantity will permit; for eve∣ry one of them is supposed to be divided into 100 or 1000 parts.

For the dividing of the Degrees of Latitude; they may be di∣vided as those of Longitude were, into 100 parts. But some∣times each Degree is subdivided into 60 Minutes, or English Miles, or into 20 Leagues.—Now I have divided the Degrees of Latitude in this Chart each of them into 5 parts, by which means it is capable of the Numeration either by Miles, Leagues, Centesms, or 100 parts.—For if you count by 60 minutes, or miles, then every of those Divisions will be 12 minutes, or miles; if by 20 Leagues, then every Division will contain 4 Leagues; and if by Centesms or 100 parts, then every of them is 20 Centesms. And thus much concerning the Making or Projecting of this Chart. I now come to shew

Some Uses of the Plain Sea-Chart.

THE Problems that are to be resolved by (or upon) the Sea-Chart are chiefly such as concern Longitude, Lati∣tude, Rhumb or Course, and Distance.

Longitude is the Distance of a Place from some known Meri∣dian to that Place, and is alwaies counted upon the Aequinoctial.

Latitude is the Distance of any Place from the Aequinoctial, counted upon that Meridian Circle which passeth over that Place.

Rhumb or Course is the Angle that a Ship in his Sailing makes with the Meridian, and is discovered in the general by the Magneticall Needle, which alwaies respecteth the North; and (though not directly, yet) its Variation being often observed, and the Chart rectified thereby, (as I have before shewed how it may be done by severall means) is the best help that Navigators yet have to steer their Course by.

Distance is the number of Leagues, Miles, or Centesms, that any Ship hath sailed.

Raising of the Pole is when a Ship sails from a lesser to a greater Latitude.

Depressing of the Pole is sailing from a greater to a lesser Latitude.

These Terms thus explained, I will proceed to Practice, as followeth.

PROBL. I. How to set any Place upon your Chart according to its Longitude and Latitude.

IF the two Places lie under one and the same Parallel, dif∣fering not at all in Latitude, but onely in Longitude, then the Course leading from the one to the other is directly East or West. As E and F are two Places lying under the Par∣allel of 50 degr. of Latitude, and differ in Longitude 5 ½ deg. Lay a Ruler to 5 ½ degr. both at the top and bottom of the Chord, and where the Ruler crosseth the Parallel of 50 degr. as at F, there is your other Place upon the Chart. So E and F lie in 50 degr. of Latitude, and differ in Longitude 5 ½ degr.

But if the two Places to be set upon the Chart differ onely in Latitude, and lie under the same Meridian as G F, then the Course leading from the one to the other is directly North or South, and the Difference of Latitude of F and G is 2 degr. G lying in the Latitude of 48 degr. and F in the Latitude of 50 degr.

But if the Places to be set upon the Chart differ both in Longitude and Latitude, as A and F, then the Course lea∣ding from the one to the other is upon some other Point of the Compass, so far distant from the Meridian as is the quanti∣ty of the Angle E A F, which here is 70 degr. 1 min. that is upon the E. N. E. Point 2 degr. 31 min. Easterly.

This Angle may be found either by Protraction by your Line of Chords, or it may be protracted by your Protracting

Quadrant, which in all these Operations upon the Chart is best, for that it avoids the drawing of Arches of Circles upon your Chart or Blank. So then if you were to protract the Angle E A F by your Protracting Quadrant; Lay the Centre A of your Quadrant upon the Point A in your Chart, and the Meridian Line of the Quadrant A B upon the Meridian Line of your Chart; then will the Line A C of the Quadrant lie upon the Parallel A D of your Chart: and the Angle that you are to protract being 70 degr. 1 min. by the edge of your Quadrant make a small Mark or Prick with your Needle, and from A through that Point draw a right Line, which will be the Line A F.

[illustration] geometrical diagram

And in the same manner as you set any Place upon your Chart, you may find in what Latitudes and difference of Lon∣gitude any Places already set upon your Chart are in.

PROBL. II. Any Places being set upon the Chart, to find in what La∣titudes they are, and also how they differ in Longitude.

LET the Points Q and R upon the Chart be two Places, and I would know in what Latitudes they lie. First, through the Point Q draw a Line parallel to the Line B C of your Chart; and also through the Point R draw another Line parallel to A C. The Line that is drawn through Q

shoots upon the Latitude of 58 degr. 36 min. and the Line passing through R cuts the Meridian of the Chart on either side at the Latitude of 57 degr. 16 min. And under those two Latitudes are the two Places Q and R.

Then to find their Difference of Longitude; Take in your Compasses the Distance between R and S, and measuring it upon the bottom of the Chart, it will reach from A to 4 degr. 24 min. And such is the Difference of Longitude of the two Places Q and R.

PROBL. III. Having the Rhumb, and the Distance that the Ship hath run upon that Rhumb, to find the Difference of Lon∣gitude and Latitude.

The Analogie or Proportion.

As the Radius is to the Distance run;

So is the Sine of the Rhumb to the Difference of Longitude:

And

So is the Co-sine of the Rhumb to the Difference of Latitude.

So the Rhumb being 70 degr. 1 min. that is E. N. E. 2 degr. 31 min. Easterly, and the Distance run 117 Leagues, the Dif∣ference of Longitude will be found to be 5 ½ degr. and the Difference of Latitude 2 degr.

Ʋpon the Chart.

UPON the Point A protract an Angle of 70 d. 1 m. as the Angle E A F, and draw the Line A F, which is the Rhumb upon which the Ship sailed. Upon this Line set 117, the number of Leagues that the Ship sailed from A to F. Then through the Point F draw the Line F E parallel to A D. So shall E F be the Difference of Longitude, 5 d. and an half, and A E the Difference of Latitude, 2 degr.

PROBL. IV. The Difference of Latitude and the Rhumb being given, to find the Distance run and the Difference of Lon∣gitude.

The Analogie or Proportion.

As the Co-sine of the Rhumb is to the Difference of Latitude; So is the Radius to the Distance run:

And

So is the Sine of the Rhumb to the Difference of Longitude.

So the one Latitude being 48 degr. and the other 50 degr. the Difference is 2 degr. and the Rhumb being E. N. E. 2 deg. 31 min. Easterly, the Distance run will be found to be 117 Leagues, and the Difference of Longitude 5 ½ degr.

Ʋpon the Chart.

SET the Difference of Latitude 2 degr. from A to E, and draw the Line E F parallel to A D. Then upon the Point A protract the Angle of the Rhumb 70 degr. 1 min. E. N. E. 2 degr. 31 min. Easterly, and draw the Line A. F, cutting the other Line E F in F. Then taking in your Compasses the length of the Line A F, and measuring it upon the side of the Chart, you shall find it to contain 117; which is the number of Leagues the Ship sailed: And the Line E F, being so measured, will contain 5 ½ degr. the Difference of Longitude.

PROBL. V. Having the Difference of Longitude and the Rhumb gi∣ven, to find the Distance run and Difference of Latitude.

The Analogie or Proportion.

As the Sine of the Rhumb is to the Difference of Longitude; So is the Radius to the Distance run:

And

So is the Co-sine of the Rhumb to the Difference of Latitude.

So the Rhumb being E. N. E. 2 degr. 31 min. Easterly, and the Difference of Longitude 5 ½ degr. the Distance run will be found to be 117 Leagues, and the Difference of La∣titude 2 degr.

Ʋpon the Chart.

UPON the Point A protract an Angle of the Rhumb 70 degr. 1 min. and draw the Line A F. Then the Difference of Longitude being 5 ½ degr. count 5 ½ degr. upon the bottom of your Chart from A to G, and upon the Point G raise a Perpendicular G F, cutting the Line A F before drawn in F. Then the Line A F, being measured upon the Side of your Chart, will be found to contain 117 Leagues, the Distance run: And F G, there also measured, will be found to be 2 degr. the Difference of Latitude.

PROBL. VI. The Distance that the Ship hath run, and the Difference of Latitude, given, to find the Rhumb and Difference of Longitude.

The Analogie or Proportion.

As the Distance run is to the Radius;

So is the Difference of Latitude to the Co-sine of the Rhumb:

And

So is the Sine of the Rhumb to the Difference of Longitude.

So the distance run being 117 Leagues, and the Difference of Latitude being 2 degr. the Rhumb will be found to be E. N. E. 2 degr. 31 min. Easterly, and the Difference of Lon∣gitude 5 ½ degrees.

Ʋpon the Chart.

SET the Difference of Latitude 2 degr. upon your Chart from A to E, and draw the Line E F parallel to A B. Then out of the Side of your Chart take the Distance run, 117 Leagues; and setting one foot of the Compasses in A, turn the other about till it cross the Line E F, which it will doe in F. Then F E, being measured upon the bottome of your Chart, will contain 5 ½ degr. the Difference of Longitude. And by your Line of Chords or Protracting Quadrant find the Quantity of the Angle E A F, which will be 70 degr. 1 min. the E. N. E. Point 2 degr. 31 min. Easterly.

PROBL. VII. The Distance that the Ship hath run, and the Difference of Longitude, being given, to find the Rhumb and Difference of Latitude.

The Analogie or Proportion.

As the Distance run is to the Radius;

So is the Difference of Longitude to the Rhumb:

And

So is the Co-sine of the Rhumb to the Difference of Latitude.

So the Difference of Longitude being 5 ½ degr. and the Di∣stance that the Ship hath run 117 Leagues; the Rhumb will be found to be E. N. E. 2 degr. 31 min. Easterly, and the Difference of Latitude 2 degr.

Ʋpon the Chart.

COUNT the Difference of Longitude upon the bot∣tome of the Chart from A to G, and upon the Point G raise the Perpendicular G F. Then take out of the Side of your Chart the Distance run, 117 Leagues, and setting one foot of the Compasses in A, with the other cross the Perpen∣dicular F G in the Point F. Now if you take F G in your Compasses, and measure it on the Side of your Chart, you shall find it to contain 2 degr. for the Difference of Latitude; and the Angle E A F, being measured by your Chord or Quadrant, will be 70 degr. 1 min. that is the E. N. E. Point 2 d. 31 m. Easterly for the Rhumb.

PROBL. VIII. The Difference of Longitude and Difference of Latitude being given, to find the Rhumb and the Distance run.

The Analogie or Proportion.

As the Difference of Latitude is to the Radius;

So is the Difference of Longitude to the Tangent of the Rhumb:

And

As the Sine of the Rhumb is to the Difference of Longitude;

So is the Radius to the Distance run.

So the Difference of Longitude being 5 ½ degr. and the Difference of Latitude 2 degr. the Rhumb will be found to be E. N. E. 2 degr. 31 min. Easterly, and the Distance upon the Rhumb 117 Leagues.

Ʋpon the Chart.

COUNT the Difference of Latitude from A to E, and draw the Line E F parallel to A D. Also count the Dif∣ference of Longitude from A to G, and upon the Point G raise the Perpendicular G F, cutting the Line E F in the Point F. Then take in your Compasses the length of the Line A F, and measuring it upon the Side of the Chart, you shall find it to contain 117 Leagues, the Distance that the Ship hath run. And if by your Line of Chords, or Quadrant, you find the Quantity of the Angle E A F, it will be the Rhumb, which you may find to be E. N. E. 2 degr. 31 min. Easterly, or 70 degr. 1 min.

PROBL. IX. The Rhumb that a Ship hath sailed upon, and the num∣ber of Leagues she hath sailed upon that Rhumb, being given, to know how much she hath raised or depressed the Pole.

The Analogie or Proportion.

As the Radius is to the Distance run;

So is the Co-sine of the Rhumb from the Meridian to the Difference of both Latitudes.

So the Rhumb being E. N. E. 2 degr. 31 min. Easterly, that is, 70 degr. 1 min. and the Distance that the Ship hath sailed upon that Rhumb 117 Leagues, the Pole will be found to be raised 2 degr.

Ʋpon the Chart.

UPON the Point A, the lesser Latitude, protract an Angle of 70 degr. 1 min. and draw the Line of the Rhumb A F, and out of the Side of your Chart take 117 Leagues, (the Distance the Ship sailed) and set them upon the Rhumb from A to F. Then through the Point F draw the Line E F parallel to A D, cutting the Meridian of your Chart in E, which is 2 degr. from A: so that the Ship hath raised the Pole 2 degrees.

PROBL. X. The Longitude and Latitude of the Place from whence you came, with the Rhumb and Distance sailed, be∣ing given, to find the Longitude and Latitude of the Place to which you are come.

The Analogie or Proportion.

As the Radius is to the Distance run;

So is the Sine of the Rhumb from the Meridian to the Diffe∣rence of Longitude:

And

So is the Co-sine of the Rhumb to the Difference of Latitude.

So the Latitude of the Place from whence you came being 52 degr. and the Longitude 35 degr. the Rhumb upon which you have sailed N. E. by N. 33 degr. 45 min. and the Distance which you have sailed upon that Rhumb 96 2/10 Leagues; you shall find the Difference of Longitude to be 2 degr. 40 min. and the Difference of Latitude 4 degr. So that the Place to which you are come is in the Latitude of 56 degr. and in the Longitude of 37 degr. 40 min.

Ʋpon the Chart.

THE Place from whence you came being in the Latitude of 52 degr. and in the Longitude of 35 degr. is repre∣sented by H. The Rhumb you have sailed upon being N. E. by N. 33 degr. 45 min. upon the Point H protract an Angle of 33 degr. 45 min. and draw the Line H K for the Rhumb. Then out of the Side of your Chart take 96 2/10 Leagues, which is so much as the Ship sailed, and set that upon the Rhumb-Line from H to K, and through the Point K draw the Line K L

parallel to B C, (or perpendicular to A B,) and it will cut the Line A B in L. So K L, being measured on the bottom of your Chart, will be found to contain 2 degr. 40 min. the Dif∣ference of Longitude; which added to 35 degr. the Longitude you came from, gives 37 degr. 40 min. for the Latitude you are in. Also the Line H L, being measured on the Side of your Chart, will be found to contain 4 degr. And such is the Difference of Latitude, which added to 52 degr. the Latitude from whence you came, gives 56 degr. the Latitude in which you are.

PROBL. XI. The Longitude and Latitude of the Place from whence you came, the Rhumb upon which you sailed, and the Latitude of the Place to which you are come, being given, to find the Distance and Difference of Lon∣gitude.

The Analogie or Proportion.

As the Difference of Latitude is to the Radius;

So is the Tangent of the Rhumb from the Meridian to the Difference of Longitude:

And

As the Sine of the Rhumb is to the Difference of Longitude;

So is the Radius to the Distance run.

So the Latitude of the Place from whence you came being 52 degr. and the Longitude 35 degr. and the Rhumb upon which you sailed the third from the Meridian N. E. by N. 33 degr. 45 min. you shall find the Distance run to be 96 Leagues 2/10, and the Difference of Longitude 2 degr. 40 min.

Ʋpon the Chart.

UPON your Chart assign H for your Place from whence you came, in the Latitude of 52 degr. and Longitude 35 degr. Upon this Point H protract the Angle of the Rhumb 33 d. 45 m. N. E. by N. and draw the Rhumb-Line H K. Then the Latitude of the Place where you are being found by ob∣servation (or being otherwise given) to be 56 degr. draw a Line quite cross your Chart at the 56^th degree of Latitude, as the Line 56. 56 in the Chart crossing the Rhumb-Line in the Point K. So K L, being measured at the bottome of your Chart, will be found to contain 2 degr. 45 min. which added to 35 degr. the Longitude you came from, makes 37 degr. 40 min. And that is the Longitude in which you are. In like manner measure H K upon the Side of your Chart, and you shall find it to contain 96 2/10 Leagues. And so much hath the Ship run upon that Point N. E. by N.

PROBL. XII. The Latitude of two Places, and the Difference of Longi∣tude between them, being known, to find what Rhumb leadeth from one to the other, and how many Leagues distant they are asunder.

The Analogie or Proportion.

As the Difference of Latitude is to the Radius;

So is the Difference of Longitude to the Tangent of the Rhumb:

And

As the Sine of the Rhumb is to the Difference of Longitude;

So is the Radius to the Distance of the two Places.

[illustration] geometrical diagram

The Plaine See Chart.

page 172.

So the Latitude of one of the Places being 50 degr. and the other 52 degr. 30 min. and the Difference of Longitude 6 ½ degrees; the Rhumb will be found to be 67 degr. 23 min. and the Distance upon the Rhumb 6 ½ degr. or 120 Leagues.

Ʋpon the Chart.

UPON the Point of the greater Latitude at N 52 deg. 30 min. draw a Line N M, parallel to A D, upon which Line set 6 degr. the Difference of Longitude of the two Places (being taken from the bottom of the Chart) from N to M. Then from the Point M draw the Line to E, the lesser Latitude, 52 degr. which Line, taken in the Compasses and measured upon the Side of the Chart, will be found to contain 6 ½ degr. or 130 Leagues. Also the Angle N E M, being measured by your Chord, or Protracting Quadrant, will be found to contain 67 degr. 23 min. which is the Rhumb leading from one to the other, namely, short of the E. N. E. Point 7 degr. or, N. E. by E. 11 degr. 8 min. Easterly.

PROBLEMS Of Sailing by Mercator's Chart. SECTION III.

PROBL. I. How to make a Sea-Chart according to MERCATOR's Projection, by your Line of Chords.

UPON a piece of thick and smooth Paper (or ra∣ther Past-board) draw a right Line, as A B, and upon the Point A, with 60 degr. of your Line of Chords, describe the Quadrant A C D, which divide into 90 equal Parts or Degrees, as here in this Figure there is onely every fifth Degree. —This done, upon the Point D erect the Perpendicular D F, (in which you must be very exact.) And from the Point A (through each Degree of the Quadrant) draw right Lines, as A 10, 10; A 20, 20; A 30, 30; A 40, 40; &c. to∣wards 90. till they touch the Line D F.—Then with your

Compasses, one foot being placed in A, extend the other to 60 degr. in the Line D F, and with that Distance describe the Arch F H G B. Again, extend the Compasses from A to 55 in the Line F D, and keeping one foot in A, with the other describe 55 K; so shall the Point K be the Point of 55 degr. in the Line A B.—Also, extend the Compasses from A to 50 in the Line F D, and draw the Arch 50, 50. Doe so with 45, 40, 35, &c. till you come to the beginning of the Degrees at D. So shall these Arch-lines, by meeting with the Line A B, divide that part of it D B into unequal parts, at 10, 20, 30, 40, 50, 60, and so forward. But this Figure is sufficient for Example.

[illustration] geometrical diagram

Now from this Line A B, being thus unequally divided, you may divide the Meridian-line of a Sea-Chart according to Mercator's Projection of any bigness, so that the Distance be∣tween Degree and Degree in the Aequinoctial be less then the Distance A D, which is here two Inches. And if a Chart were made that the Aequinoctial Degrees were two Inches di∣stant, and it passed upon a smooth Board, many Nauticall Con∣clusions might be wrought upon it very exactly. Being thus far prepared, I will now shew you how, from the Line A B,

To divide the Meridian Line of a Sea-Chart.

A Sea-Chart, according to this Projection, may be made either General, or Particular. I call that a General Sea-Chart, whose Line E H, in the following Figure, represents the Aequinoctial, as the Line E H there doth the Parallel of 49 degr. and so I will make the Chart following to contain all Latitudes between 49 degr. and 57 degr. whose Difference of Longitude exceedeth not 8 degr.

Now to project such a Chart, having drawn the Line E F for the Meridian, and crossed it at right Angles with another Line representing the Parallel of 49 d. parallel thereto draw another Line F G, representing the Parallel of 57 degr. and another Meridian G H, parallel to F E. So shall you have made the Parallelogram E F G H.

This done, consider how far distant you would have your Degrees of Longitude upon the Aequinoctial each from other, as suppose (and as in this Chart I have made them to be) half an Inch. Take half an Inch out of a Line of Inches, and run that Distance along the Line E H from E to 1, from 1 to 2, from 2 to 3, &c. And also doe the like upon the Line F G, at the top of the Chart, drawing the Lines 1, 1; 2, 2; 3, 3; &c.

Now for the Dividing of the Meridians E F and H G, re∣pair to the foregoing Figure, taking in your Compasses the Distance that is between Degree and Degree of the Aequino∣ctial,

which in our Example is half an Inch. With this Di∣stance, set one foot of the Compasses in the Point D, and with the other describe the Arch m m; by the very Edge whereof draw the Line A G: so is your Figure prepared to divide the Meridian-line of a Sea-Chart whose Degrees of Longitude are half an Inch distant.

Now in respect that your first Parallel of Latitude E H in your Chart is drawn for 49 degr. your next Parallel must be 50 degr. Wherefore set one foot of your Compasses upon 50 degr. in the Line A B, and with the other take the nearest Distance to the Line A G: that is done by turning the Com∣passes about till the moveable foot do onely touch the Line A G; which when it so doth, that Distance at which your Com∣passes then are, being set upon the Meridian of your Chart, will reach from 49 degr. to 50, which being set upon your Chart, on both sides thereof, from 49 draw the Line 50. 50 will give you the Parallel of 50 d. of Latitude. In like man∣ner for the Parallel of 51 degr. Set one foot of the Compasses in 51 degr. upon the Line A B of the former Figure, and with the other take the least Distance to the Line A G: this Distance set upon the Meridian of your Sea-Chart, on both sides thereof, will reach from 50 to 51; and there draw the Parallel 51, 51.—Likewise for the Parallel of 52 degr. Set one foot of the Compasses in 52 degr. in the Line A B, taking the nearest Distance to the Line A G: that Distance set upon the Meridian of your Sea-Chart, on both sides thereof, will reach from 51 to 52; and there draw the Parallel of 52, 52. Doe thus with all the Degrees, as 53, 54, 55, 56, and 57. So shall the Meridians of your Chart E F and H G be divided into whole Degrees.

For the Sub-divisions of these Degrees, they may be divi∣ded each of them into equal parts, as the Divisions at the top and bottome of the Chart ought to be; but the Degrees of the Meridian, as they grow higher, they ought still to grow greater. But the Difference is so small, that it cannot produce

any considerable Errour, though the Sub-divisions be all made equal between Degree and Degree. You may therefore di∣vide them either into 60 Minutes or English Miles, or into 20 Leagues, or into 100 parts of Degrees, as you shall best like of.

But if you would make a Chart that the Distance between De∣gree and Degree upon the Aequinoctial should be an Inch, or any other Distance less then A D in the foregoing Figure; take that Di∣stance (as suppose an Inch) in your Compasses, and setting one foot in D, with the other describe the Arch o o, and draw the Line A H onely to touch the Arch o o. The least Distance taken from each Degree to this Line A H shall give you the Distance of the Degrees upon the Meridian of a Sea-Chart, whose Distance of Degrees up∣on the Aequinoctial are an Inch from each other.

Your Chart being thus prepared, I will now come to shew you how to resolve severall Problems upon it.

PROBL. II. To find how many Leagues do answer to one Degree of Longitude in every severall Latitude.

UPON the two edges of your Protracting Quadrant there are two Lines, the one divided into 20, the other into 60 equal parts.

Take therefore the least Distance from the Complement of the Parallel's distance from the Aequator, (or the Complement of the given Latitude:) this Distance, being measured upon the edge that is divided into 20, shall shew you what number of Leagues make one Degree of Longitude in that Parallel of Latitude. And the same Distance, being measured upon the other edge that is divided into 60, will give so many of our Miles, or so many Minutes of the Aequinoctial, or any other

great Circle, as are answerable to one Degree of Longitude in that Latitude.

Example. Let it be required to find how many Leagues do answer to one Degree of Longitude in the Latitude of 18 d. 12 min.

Set one foot of your Compasses in 71 degr. 48 min. the Complement of the given Latitude, and with the other take the nearest Distance to the side of the Quadrant which is di∣vided into 20: that Distance, measured upon the Line 20, will reach from the beginning thereof to 19: and so many Leagues do answer to one Degree of Longitude in the Lati∣tude of 18 degr. 12 min.

Or, If you take the least Distance from 18 degr. 12 m. the Latitude it self, in the Limb of the Quadrant, to that edge which is divided into 60, that Distance will also reach to 19 upon the Line 20, as before.

And the same Distance, being measured upon the Line 60 of the Quadrant, will give you 57 parts: and so many Mi∣nutes of the Aequator are answerable to one Degree of Lon∣gitude in the Parallel of 18 degr. 12 min. of Latitude.

So likewise in the Latitude of 25 degr. 15 min. if you take the least Distance from the Complement thereof, or from the Latitude it self, to the edges of the Quadrant, you shall find that Distance to reach 18 in the Line of 20: and so many Leagues do answer to one Degree of Longitude in the Lati∣tude of 25 degr. 15 min. or unto 54 in the Line of 60: and so many Minutes of the Aequator do answer to one Degree of Longitude in that Parallel of Latitude.

The Analogie or Proportion.

As the Radius is to the Co-sine of the Latitude;

So is

• 20 Leagues to the num∣ber of Leagues
• 60 Minutes to the num∣ber of Minutes

which answer to one Degree of Longitude in that Lati∣tude.

PROBL. III. By the Latitude of two Places and their Distance, to find the Rhumb.

The Analogie or Proportion.

As the Distance upon the Rhumb is to the Radius;

So is the Difference of Latitudes to the Co-sine of the Rhumb from the Meridian.

Thus if the Places given were one in the Latitude of 50 d. and the other in the Latitude of 55 degr. and the Distance up∣on the Rhumb 6 degr. or 120 Leagues; the Rhumb leading from one to the other will be found to be the third from the Meridian, namely, N. E. by N. 33 degr. 45 min.

Ʋpon the Chart.

LET A represent the Place in the Latitude of 50 degr. and C that in 55 degr. whose Distance from A to C is 6 degr. Take 6 degr. out of the Meridian-line, by setting one foot as much below the lesser Latitude as above the grea∣ter, which will be from K in the Latitude of 49 ½ degr. to L in the Latitude of 55 ½; either of which are half a Degree above and under the two given Latitudes. Take this Distance K L in your Compasses, and setting one foot in A, (the lesser Latitude) with the other cross the Parallel of the greater La∣titude 55 degr. in the Point C, and draw a right Line from A to C. So shall the quantity of the Angle B A C, being found (either by your Chord or Quadrant,) shew you the Inclina∣tion of the Rhumb to the Meridian to be 33 degr. 45 min. the N. E. by N. Point.

Note, That in the Propositions following, the Difference of Lon∣gitude must always be taken out of the Aequator, and measured

thereupon also. But the Difference of Longitude and Distance upon the Rhumb must alwaies be measured upon, and taken out of, the Meridian Line of your Chart. And hereafter I shall call them the proper Difference, and proper Distance.

PROBL. IV. The Longitude and Latitude of two Places being gi∣ven, to find the Rhumb.

The Analogie or Proportion.

As the proper Difference of Latitude is to the Radius;

So is the Difference of Longitude to the Tangent of the Rhumb from the Meridian.

Thus if the Places should lie one in the Latitude of 50 deg. and the other in the Latitude of 55 degr. and the Difference of Longitude between them were 5 degr. 30 min. the Rhumb leading from one Place to the other will be found to be the third from the Meridian N. E. by N. 33 degr. 45 min.

Ʋpon the Chart.

THE Meridians and Parallels being drawn through the two Places at A and C, and a straight Line from A to C, for the Rhumb, by your Chord or Quadrant find the quanti∣ty of the Angle B A C, which you will find to be 33 d. 45 m. or the third Rhumb from the Meridian N. E. by N.

But if this Rhumb were to be found by the Common Sea-Chart, it would be found to be above 47 degr. that is, N. E. 2 degr. Easterly, that is, one whole Point and 2 degr. more Easterly then it should be.

PROBL. V. The Latitude of two Places and the Rhumb being given, to find the Difference of Longitude.

The Analogie or Proportion.

As the Radius is to the Tangent of the Rhumb from the Meri∣dian;

So is the proper Difference of Latitudes to the Difference of Longitude.

Thus the Latitude of one Place being 50 degr. and the other 55 degr. and the Rhumb leading from one to the other being the third from the Meridian, the Difference of Longi∣tude will be found to be 5 ½ degr.

Ʋpon the Chart.

LET a Meridian be drawn through A, and a Parallel of Latitude through C. Then upon the Angle A protract the Angle of the Rhumb 33 degr. 45 min. So the Distance B C upon the Parallel, being measured upon the bottome of the Chart, will be found to contain 6 degr. 30 min.

But if this Difference of Longitude were to be found by the Plain Sea-Chart, the Difference of Longitude would be found to be but 3 degr. 20 min. which is more then 3 degr. less then the truth; a vast Difference. And yet this Errour would be yet greater, if either the Latitude be greater, or the Rhumb farther from the Me∣ridian.

PROBL. VI. The Difference of Longitude of two Places, the Latitude of one of them, and the Rhumb leading from one to the other, given, to find the Latitude of the other Place.

The Analogie or Proportion.

As the Radius is to the Co-tangent of the Rhumb from the Meridian;

So is the Difference of Longitude to the proper Difference of Latitude.

Thus if the Latitude of one of the Places were 50 degr. the Rhumb leading from that to the other N. E. by N. 33 d. 45 min. and the Difference of Longitude between the two Places were 5 degr. 30 min. the Latitude of the other Place will be found to be in 55 degr.

Ʋpon the Chart.

LET A B and D C be two Meridians drawn through A and C, at 5 ½ d. the Difference of Longitude, and a Par∣allel of Latitude through A, crossing the Meridian C D in D. Then upon the Point A protract an Angle equal to the Rhumb from the Meridian given 33 degr. 45 min. So the Line C D, being measured upon the Meridian from A, the given Latitude, 50 degr. will reach to 56 degr. the proper Difference of Latitude. So that the other Place lies in the Latitude of 56 degr.

But if this Difference of Latitude were to be found by the Plain Sea-Chart, this Difference of Latitude would be found to be 8 d. 13 min. and the Latitude sought would be found to be 58 degr.

13 min. above three Degrees more then the truth. As by the Tri∣angle for that purpose drawn upon the Plain Sea-Chart, marked with T V E, may appear.

PROBL. VII. Having the Latitude of one Place, the Rhumb leading from that Place to another unknown, and the Distance upon the Rhumb from the first to the second Place, to find the Difference of Longitude of the two Places.

The Analogie or Proportion.

As the Radius is to the Sine of the Rhumb from the Meridian;

So is the proper Distance upon the Rhumb to the Difference of Longitude.

Thus if the two Places were one in the Latitude of 50 degr. and the other in a greater Latitude, but unknown; the proper Distance upon the Rhumb leading from one place to the other being 6 degr. and the Rhumb N. E. by N. 33 degr. 45 min. the Difference of Longitude will be found to be 5 ½ degr.

Ʋpon the Chart.

THrough the Point A in the Latitude of 50 degr. let be drawn a Meridian A B, and a Parallell A D; and upon the Point A protract an Angle equal to the Rhumb from the Meridian 33 degr. 45 min. Then take with the Compasses 6 degres, the proper Distance upon the Rhumb, out of the Meridian-line, (having respect to the Latitude of the Places) as from K to L, and set that Distance upon the Rhumb from A to C. Then through C draw another Meridian C D, cros∣sing the Parallel drawn through A in the Point D. So the Line A D, being measured at the bottom of the Chart, will be found to contain 5 ½ d. the Difference of Longitude sought.

But if this Difference of Longitude had been to be found by the Common Sea-Chart, it would be found to have been onely 3 d. 20 min. which is 2 degr. 10 min. less then the truth; as in the Plain Chart may be seen, where the third Rhumb from the Me∣ridian cuts the Parallel of 55 degr. of Latitude in 3 degr. 20 m. of Longitude at the Point X.

PROBL. VIII. The Difference of Longitude between two Places, the Rhumb leading from one Place to the other, and the Latitude of one of the Places, being given, to find their Distance.

The Analogie or Proportion.

As the Sine of the Rhumb from the Meridian is to the Diffe∣rence of Longitudes;

So is the Radius to the proper Distance of the two Places up∣on the Rhumb.

Thus, if the Latitude of one Place were in 50 degr. the other in a greater Latitude unknown, the Difference of Lon∣gitude between the two Places 5 ½ degr. and the Rhumb N. E. by N. 33 degr. 45 min. from the Meridian; the proper Distance upon the Rhumb will be found to be 6 degrees.

Ʋpon the Chart.

LET two Meridians, A B and C D, be drawn through A and C, according to the Difference of Longitude, and a Parallel of Latitude through A, crossing the Meridian C D in the Point D. Then upon the Point A protract an Angle of 33 degr. 45 min. the quantity of the Rhumb from the Meri∣dian, and draw the Line A C crossing the Meridian C D in C. So the Distance C D, being taken in the Compasses, and

measured upon the Meridian-line of the Chart, (respect be∣ing had to the Latitude of the Places) that is, so much above the greater Latitude as below the lesser Latitude, you will find it to contain 6 degr.

But if this settting of the Compasses so much above one La∣titude as below another seem difficult, it may be thus other∣wise done.—For, the Rhumb Line being drawn, it will cut the Meridian C D in C: so a Parallel drawn through C will cut the Meridian A B in B: so is B the Latitude of the second Place, viz. 55 degr. Then divide the Distance be∣tween the two Latitudes A and B in two equal parts in the Point M; also divide the Rhumb-Line A C in two equal parts in N: then take the Distance N C or N A, and setting one foot of the Compasses in M, the other will reach to L above the greater Latitude, and from M to K as much below the lesser Latitude, namely, 30 min. or half a Degree on either side; so that between K and L are contained 6 degr. and that is the proper Distance upon the Rhumb.

But if this Distance were to be found by the Plain Chart, it would be almost 10 degr. or 197 Leagues, which is 77 Leagues more then in truth it should be. As may appear, if you measure the Line A L in the Plain Chart, upon the Side thereof.

PROBL. IX. The Difference of Longitude, and Distance of two Places, with the Latitude of one of the Places, being given, to find the Rhumb that leads from one to the other.

The Analogie or Proportion.

As the proper Distance upon the Rhumb is to the Difference of Longitude;

So is the Radius to the Sine of the Rhumb from the Meridian.

Thus, if one of the Places lay in the Latitude of 50 degr. and the other in a greater Latitude, but unknown; the Dif∣ference of Longitude between them 5 ½ degr. and their pro∣per Distance upon the Rhumb 6 degr. the Inclination of the Rhumb to the Meridian which leadeth from one Place to the other will be found to be 33 degr. 45 min. that is the N. E. by N. Point.

Ʋpon the Chart.

LET the Meridians A B and D C be drawn through A and C, and through A a Parallel of Latitude A D. Then open the Compasses (having respect to the Latitudes) from K to L, the quantity of 6 degr. in the Meridian; and setting one foot of that Extent in A, with the other foot cross the Meridian C D in C, and draw the right Line A C for the Rhumb. Lastly, by your Chord or Quadrant find the quantity of the Angle B A C, 33 degr. 45 min. and that is the Rhumb required N. E. by N.

But if you were to find this Rhumb by the Plain Sea-Chart, it would be found almost the E. N. E. Point within 1 degr. 30 min. differing from truth very near 3 whole Points to the Eastward.

PROBL. X. The Longitude and Latitude of two Places being given, to find the Distance upon the Rhumb.

The Analogie or Proportion.

As the proper Difference of Latitudes is to the Radius;

So is the Difference of Longitudes to the Tangent of the Rhumb from the Meridian:

And

As the Sine of the Rhumb from the Meridian is to the Diffe∣rence of Longitude;

So is the Radius to the proper Distance upon the Rhumb.

Thus, the two Places being one in the Latitude of 50 degr. the other in the Latitude of 55 degr. and the Difference of Longitude between them being 5 ½ degr. the proper Distance upon the Rhumb will be found to be 6 degr.

Ʋpon the Chart.

DRAW the Meridians A B and C D, the Difference of Longitude between them being 5 ½ degr. and through A and B draw two Parallels B C and A D, and then the Line for the Rhumb leading from the one to the other A C. So A C, being taken in the Compasses, and measured upon the Meridian-line of the Chart, with this Condition, that at the resting of the Compasses upon the Meridian-line, one foot be so many Degrees above the greater Latitude as the other foot is below the lesser Latitude; so will the feet of the Compasses rest in the Points K and L, one being 30 min. below the lesser Latitude, and the other 30 min. above the greater.

But if this Distance upon the Rhumb were to be found by the Plain Chart, it would be found to be almost 7 degr. 15 min. or 245 Leagues, which is 25 Leagues more then it should be.

PROBL. XI. The Latitude of two Places and their Distance upon the Rhumb being given, to find their Difference of Lon∣gitude.

The Analogie or Proportion.

As the proper Distance upon the Rhumb is to the Radius;

So is the proper Difference of Latitudes to the Co-sine of the Rhumb from the Meridian:

And

So is the Sine of the Rhumb from the Meridian to the Diffe∣rence of Longitude.

Thus, if one of the Places be in the Latitude of 50 degr. and the other in 55 degr. and their proper Distance upon the Rhumb 6 degr. or 120 Leagues; their Difference of Longi∣tude will be found to be 5 degr. 30 min.

Ʋpon the Chart.

DRaw A D and B C, two Parallels of Latitude, through 50 degr. and 55 degr. which were the two given Lati∣tudes. Then out of the Meridian Line take the proper Distance upon the Rhumb (having respect to both Lati∣tudes) from K to L: the Compasses being opened to this Distance, one foot being set in A, the lesser Latitude, the other will cross the Parallel of the greater Latitude in C. So the Distance B C, being measured at the bottome of the Chart from E, will reach to 5 degr. 30 min. And such is the Diffe∣rence of Longitude between the two Places.

But if this Difference of Longitude were to be found by the Plain Chart, it would be but 3 degr. 20 min. which is no less then 2 d. 10 min. less then the truth; as by the Triangle T V. E drawn upon the Plain Chart may appear.

PROBL. XII. The Difference of Longitude of two Places, their Distance upon the Rhumb, and the Latitude of one of the Pla∣ces, being given, to find the Difference of Latitudes.

The Analogie or Proportion.

As the proper Distance of the two Places upon the Rhumb is to the Radius;

So is the Difference of Longitudes to the Inclination of the Rhumb to the Meridian:

And

So is the Co-sine of the Rhumb from the Meridian to the Dif∣ference of Latitudes.

Thus, the Difference of Longitudes being 5 ½ degr. their pro∣per Distance upon the Rhumb 6 degr. and the Latitude of one of the Places 50 d. the Difference of Latitudes will be found to be 5 d.

Ʋpon the Chart.

THrough the given Latitude A draw a Meridian AB, and a Parallel A D, and upon the Parallel set the Difference of Longitude 5 ½ d. taken from the bottom of the Chart, from A to D, and through D draw the Meridian D C. Then out of the Meridian-line take the proper Distance upon the Rhumb, 6 d. from K to L, and setting one foot of the Compasses in A, with the other cross the Meridian C D in C: so a Parallel of Latitude drawn through C will be the Parallel of 55 d. So is 55 d. the Latitude of the other Place, and 50 being taken from 55, leaves 5 d. for the Difference of Latitudes required.

Which Difference, had it been to be found by the Plain Chart, would have been but 2 d. 25 m. that is, 2 d. 35 m. less then the truth; as by the Triangle T V E upon the Plain Chart may appear.

PROBL. XIII. The Latitude of two Places and their Difference of Lon∣gitudes being given, to find the Rhumb leading from one to the other, and also how many Degrees distant they are asunder.

THIS Proposition is already performed in the Example of the two Places A and B; but for Variety I will take two other Places, and onely shew the manner of working upon the Chart.

Suppose then two Places, one (as before) in the Latitude of 50 d. the other in the Latitude of 52 degr. 30 min. whose Dif∣ference of Longitudes is 6 degr.

Ʋpon the Chart.

Through the two given Latitudes 50 d. and 52 ½, at A and O draw two Parallels, O P and A D, upon which set the Dif∣ference of Longitudes from O to P, and from A to Q, 6 degr. Then draw the Line A P, which shall be the Line of the Rhumb leading from one Place to the other: wherefore, by your Chord or Protracting Quadrant find the quantity of the Angle O A P, which shall be the Inclination of the Rhumb to the Me∣ridian, and will be found to be 56 d. 15 m. that is the N. E. by E. Point; which was the First thing that was required.

Then to find the proper Distance upon the Rhumb; Take the Line A P in your Compasses, and measure it upon the Me∣ridian-line, so that one foot may be above the greater Latitude so much as the other is below the lesser; and you will find the Compass-points to rest in E and S, E being one whole Degree below the lesser Latitude, and S one Degree above the grea∣ter. So that there is intercepted between E and S 4 ½ degr. And that is the proper Distance upon the Rhumb; which was the Second thing required.

But if this Problem had been wrought upon the Plain Chart, the Rhumb from the Meridian would be found to be 67 d. 23 m. that is, within 7 m. of the 6^th Rhumb; which is more then the truth by 11 d. 8 m.

PROBL. XIV. A Ship set sail from the Latitude of 50 degr. upon the fifth Rhumb N. E. by E. after that she had made 36 Leagues of way upon that Rhumb, the wind changing, she was constrained to sail 50 Leagues upon the 7^th Rhumb E. by N. I would know in what Longitude and Latitude the Ship is.

Ʋpon the Chart.

THE Rhumb-Line A P being drawn, set off thereupon 36 Leagues (which was the way that the Ship made upon the fifth Rhumb before the Wind changed) from A to T, (which Distance must be taken out of the Meridian-line by opening the Compasses from 50 d. to 51, 48. or better, to as much below 50 d. as above 51 d.) So shall the Point T be the Place that the Ship was in when the Wind altered. So a Paral∣lel drawn through T upon the Chart will cut the Meridian at V in 51 d. and in that Latitude the Ship was. Now to find in what Longitude she was; Take in your Compasses the Line T V, and measure it at the bottom of the Chart, you shall find it will reach from E to 2 d. 21 m. And in that Longitude the Ship then was.

This done, upon the Point T (where the Wind changed, and drove the Ship 2 Points more Eastwardly, namely, upon the E. by N. Point) protract an Angle of 22 d. 30 m. namely, the An∣gle P T X, which is the Rhumb upon which the Ship sailed 50 Leagues after the Wind changed. Therefore take 50 Leagues out of the Meridian-line, and set them from T to X. So shall X be the Place that the Ship was in after she had sailed 50 Leagues upon the E. by N. Point; which, by drawing a Parallel through K, will be found in the Latitude of 51 d. 30 m. and by drawing of a Meridian through K also, it will be found to be in the Lon∣gitude of 6 degr. 16 min.

But if these Courses had been protracted according to the Plain Sea-Chart, the Point T would fall in the Latitude of 51 degr. and the Point X in the Latitude of 51 degr. 30 m. But the Longitude of T would be onely 1 d. 30 m. and the Longitude of X in 3 d. 57 min. Both these Longitudes being added, make but 5 d. 27 m. for the Difference of Longitude between X and the first Meridian; whereas by the other Chart it is 6 d. 16 m. So that the Ship at X is 33 m. Westward of the Place to which she was bound.

These Differences, which I have observed to be between the Plain and Mercator's Chart, may be seen by comparing the Scheme of the two Charts together.

FINIS.