Cosmographia, or, A view of the terrestrial and cœlestial globes in a brief explanation of the principles of plain and solid geometry applied to surveying and gauging of cask : the doctrine of primum mobile : with an account of the Juilan & Gregorian calendars, and the computation of the places of the sun, moon, and fixed stars ... : to which is added an introduction unto geography / by John Newton ...

Practical Geometry; OR, THE ART of SURVEYING.

CHAP. I.

Of the Definition and Division of Geometry.

GEometry is a Science explaining the kinds and properties of continued quantity or magnitude.

2. There are three Kinds or Spe∣cies of Magnitude or continued Quantity, Lines, Superficies and Solids.

3. A Line is a Magnitude consisting only of length without either breadth or thickness.

4. In a Line two things are to be considered, the Terms or Limits, and the several Kinds.

5. The term or limit of a Line is a Point.

6. A Point is an indivisible Sign in Magni∣tude which cannot be comprehended by sense, but must be conceived by the Mind.

7. The kinds of Lines are two, Right and Ob∣lique.

8. A Right Line is that which lieth between his Points, without any going up or going down on either side. As the Line AB lieth streight and equally between the Points A and B. Fig. 1.

9. An Oblique Line is that which doth not lie equally between its Points, but goeth up and down sometimes on the one side and sometimes on the other. And this is either simple or various.

10. A simple Oblique Line, is that which is ex∣actly Oblique, as the Arch of a Circle; of Various Oblique Lines there is but little use in Geometry.

11. Thus are Lines to be considered in them∣selves, they may be also considered as compared to one another, and that either in respect of their distances, or in respect of their meetings.

12. In respect of their distances, they may be either equally distant, or unequally.

13. Lines equally distant are two or more, which by an equal space are distant from one ano∣ther, and these are called Parallels; and these though infinitely extended will never concur.

14. Lines unequally distant, are such as do more or less incline to one another, and these be∣ing extended will at last concur.

15. Concurring Lines are either perpendicu∣lar or not perpendicular.

16. A Perpendicular Line, is a Right Line falling directly upon another Right Line, not de∣clining or inclining to one side more than ano∣ther; as the Line AB in Fig. 1.

17. A Perpendicular Line is twofold, to wit, ei∣ther falling exactly in the middle of another Line, or upon some other Point which is not the middle.

18. A line exactly Perpendicular, may be drawn in the same manner, as any Right Line

may be divided into two equal Parts; the which may thus be done. If from the two Terms or Points of the Right Line given, there shall be described two Arches crossing one another above and below, a Line drawn through the Intersecti∣ons of those Arches, shall be exactly Perpendicu∣lar, and also divide the Right Line given into her equal Parts. Fig. 1.

For Example; Let CD be the Right Line given, and let it be required, to bisect this Line, and to erect a Perpendicular in the middle thereof. 1. Then set∣ting one of your Compasses in the Points C, draw the Arches E and F. 2. Setting one Foot of your Com∣passes in D, draw the Arches G and H, and from the Intersections of these Arches draw the Right Line KL, so shall the Right Line KL be Perpendicular to the Right-Line CD, and the Right Line CD also di∣vided into two equal Parts, in the Point A.

19. A Line Perpendicular to any other Point than the middle is twofold: for it is either drawn from some Point given in the Line; or from some Point given without the Line.

20. From a Point given in the Line, at Per∣pendicular may thus be drawn. In Fig. 2. Let the given Line be CD, and let it be required to draw a Perpendicular Line to the Point C, your Compasses being opened to any reasonable distance, set one Foot in the Point C, and the other in any place on either side the Line CD, suppose at A, then describe the Arch ECF, this done draw the Line EA, and where that Line being extended shall cut the Arch ECF, a Right Line drawn from C to that Intersection shall be Perpendi∣cular to the Point C in the Line CD, as was required.

21. From a Point given without the Line, a Perpendicular may be drawn in this manner.

In Fig. 2. Let the given Line be CD, and let it be re∣quired to draw another Line Perpendicular thereunto, from the Point F without the Line. From the Point F draw a streight Line to some part of the Line CD at pleasure, as FE, which being bisected, the Point of Bi∣section will be A, if therefore at the distance of AF, you draw the Arch ECF, the Right Line CF shall be Perpendicular to the Line CD, as was required.

22. Hitherto concerning a Perpendicular Line. A Right Line not Perpendicular, is a Right Line falling indirectly upon another Right Line, in∣clining thereto on the one side more, and on the other less.

23. Lines unequally distant, and at last con∣curring, do by their meeting make an Angle.

24. An Angle therefore is nothing else, then the place, where two Lines do meet or touch one another, and the two Lines which constitute the Angle, are in Geometry called the sides of the Angle.

25. Every Angle is either Heterogeneous, or Ho∣mogeneous: that is called an Hetorogeneous Angle, which is made by the meeting of one Right Line, and another that is Oblique and Crooked; and that is called an Homogeneous Angle, which is made by the meeting of two Lines of the same kind, that is, of two Right Lines, or of two curved or Circular Lines.

26. An Homogeneous Angle made of two curved or Circular Lines, is to be considered in Geome∣try as in Spherical Triangles, but the other which is made of Right Lines, is in all the Parts of Geometry of more frequent use.

27. Right lined Angles are either Right or Oblique.

28. A Right Angle is that whose legs or sides are Perpendicular to one another, making the comprehended space on both sides equal. Thus in Fig. 1. the Line AK is Perpendicular to the Line CD, and the Angles KAC and KAD, are right and equal to one another.

29. An Oblique Angle is that, whose sides are not Perpendicular to one another.

30. An Oblique Angle is either acute or ob∣tuse.

31. An Acute Angle is that which is less than a Right.

32. An Obtuse Angle, is that which is greater than a Right. Thus in Fig. 1. The Angle BAC is an Acute Angle because less than the Right Angle CAK. And the Angle BAD is an Obtuse Angle being greater than the Right Angle DAK.

The Geometrical Propositions concerning Lines and Angles are very many, but these follow∣ing we think sufficient for our present purpose.

Proposition I.

To divide a Right Line given into any Number of equal Parts.

Let it be required to divide the Right Line AB into five equal Parts. From the extream Points of the given Line A and B, let there be drawn two Parallel Lines, then from the Point A at any di∣stance of the Compasses, set off as many equal Parts wanting one, as the given Line is to be di∣vided into, which in our Example is four, and are noted thus, 1. 2. 3. 4. and from the Point B set off the like Parts in the Line BC, and let them be

noted likewise thus, 1. 2. 3. 4. then shall the Pa∣rallel Lines, 14. 23. 32, and 41. divide the Right Line AB into 5 equal Parts, as was required.

Proposition II.

Two Right Lines being given, to find a Mean pro∣pertional between them.

Let the two Right Lines given be DB and CB, which let be made into one Line as CD, which being besected the Point of bisection is A, from which as from a Centre describe the Arch CED, and from the Point B erect the Perpendi∣cular BE, so shall BE, be the Mean proportional required; for, BC. BE∷BE. BD.

Proposition III.

Three Right Lines being given, to find a fourth proportional.

Let the three given Lines be AB. BC. and AD. Fig. 5. to which a fourth proportional is required: draw AE at any Acute Angle, to the Line AD in the Point A; and make DE paral∣lel to BC, so shall AE be the fourth proportio∣nal required; for, AB. BC∷AD. AE.

Proposition IV.

Vpon a Right Line given, to make a right-lined. Angle, equal to an Angle given.

Let it be required upon the Line CD in Fig. 6.

to make an Angle, equal to the Angle DAE in Fig. 5. From the Point A as a Center, at any ex∣tent of the Compasses describe the Arch BG, between the sides of the Angle given, and with the same extent describe the Arch HL from the Point D, and then make HL equal to BG, then draw the Line DL, so shall the Angle CDL be equal to the Angle DAE given, as was re∣quired.

CHAP. II.

Of Figures in the general, more particularly of a Circle and the affections thereof.

HItherto we have spoken of the first kind of Magnitude, that is, of Lines, as they are considered of themselves, or amongst them∣selves.

2. The second kind of Magnitude is that which is made of Lines, that is, a Figure con∣sisting of breadth as well as length, and this is otherwise called a Superficies.

3. And in a Superficies there are three things to be considered. 1. The Term or Limit. 2. The middle of the Term. 3. The Thing or Figure made by the Term or Limit.

4. The Term or Limit is that which compre∣hendeth and boundeth the Figure, it is common∣ly called the Perimeter or Circumference.

5. The Term of a Figure is either Simple or various.

6. A Simple Term is that which doth consist of a Simple Line, and is properly called a Cir∣cumference

or Periphery: A Periphery therefore is the Term of a Circle or most Simple Figure.

7. A various Term is that which hath bending or crooked Lines, making Angles, and may there∣fore be called Angular.

8. The middle of Term is that which is the Center of the Figure; for every Figure, whe∣ther Triangular, Quadrangular, or Multangular, hath a Center as well as the Circular, differing in in this, that the Lines in a Circle drawn from the Center to the Circumference are all equal, but in other Figures they are not equal.

9. The Thing or Figure made by the Term or Limit, is all that Area or space which is inclu∣ded by the Term or Terms. And here it is to be observed, that the Term of a Figure is one thing, and the Figure it self another; for Example, A Periphe∣ry is the Term of a Circle, but the Circle it self is not properly the Periphery, but all that Area or space which is included by the Periphery, for a Periphery is nothing but a Line, but the Circle is that which is in∣cluded by that Line.

10. As the Term of a Figure is either Simple or Various; so the Figure it self is either Simple and Round, or Various and Angular.

11. A Simple Figure is that which is contained by a Simple or Round Line, and is either a Circle or an Ellipsis.

12. A Circle therefore is such a Figure which is made by a Line so drawn into it self, as that it is every where equally distant from the middle or Center.

13. An Ellipsis is an oblong Circle.

14. In a Circle we are to consider the affecti∣ons which are as it were the Parts or Sections

thereof, as they are made by the various applica∣tions of Right Lines.

15. And Right Lines may be applied unto a Circle, either by drawing them within, or with∣out the Circle.

16. Right Lines inscribed within a Circle, are either such as do cut the Circle into two equal or unequal Parts, as the Diameter and lesser Chords, or such as do cut the Diameter and lesser Chords into two equal or unequal Parts, as the Right and versed Sines.

17. A Diameter is a Right Line drawn through the Center from one side of the Cir∣cumference to the other, and divideth the Circle into two equal Parts, As in Fig. 7. The Right Line GD drawn through the Center B is the Diameter of the Circle GEDL dividing the same into the two e∣qual Parts GED, and GLD: and this is also called the greatest Chord or Subtense.

18. A Chord or Subtense is a Right Line in∣scribed in a Circle, dividing the same into two equal or unequal Parts; if it divide the Circle into two equal Parts, it is the same with the Diameter, but if it divide the Circle into two unequal Parts it is less than the Diameter, and is the Chord or Sub∣tense of an Arch less than a Semi-circle, and also of an Arch greater than a Semi-circle. As in the former Figure, the Right Line CAK divideth the Circle into two unequal Parts, and is the Chord or Subtense of the Arch CDK, less than a Semi-circle, and of the Arch CGK greater than a Semi-circle: and these are the Lines which divide the Circle into two equal or unequal Parts. And as they divide the Circle into two equal Parts, so do they also divide one another; The lesser Chords when they are divided by

the Diameter into two equal Parts, those Parts are called Right Sines, and the two Parts of the Diame∣ter made by the intersection of the Chords are called versed Sines.

19. Sines are right or versed.

20. Right Sines are made by being besected, by the Diameter, and are twofold, Sinus totus, the whole Sine or Radius, and this is the one half of the Diameter, as the Lines BE or BD, and all Lines drawn from the Center to the Circumference.

21. Sinus simpliter, or the lesser Sines, are the one half of any Chord less than the Diameter, as in the former Figure CA or AK, which are the equal Parts of the Chord CAK, are the Sines of the Arch∣es CD. and DK less than a Quadrant, and also the Sines of CEG and KLG greater than a Quadrant.

22. Versed Sines are the Segments of the Dia∣meter, made by the Chords intersecting it, at Right Angles, as AD is the versed Sine of CD or DG and the other Segment AG is the versed Sine of the Arch CEG or KLG.

23. The Right Lines drawn without the Cir∣cle are two, the one touching the Circle, and is called a Tangent, and the other cutting the Cir∣cle, and is called a Secant.

24. A Tangent is a Right Line touching the Circle, and drawn perpendicular to the Diame∣ter, and extended to the Secant.

25. A Secant is a Right Line drawn from the Center through the Circumference, and extended to the Tangent. As in the former Figure, the Right Line DF is the Tangent of the Arch CD, and the Right Line BF is the Secant of the same Arch CD.

Proposition I.

The Arch of a Circle being given to describe the whole Periphery.

Let ABC be an Arch given, and let the Cir∣cumference of that Circle be required. Let there be three Points taken in the given Arch at plea∣sure, as A, B, C; open your Compasses to more than half the distance of A, B, and setting one Foot in A describe the Arch of a Circle, and the Compasses remaining at the same distance, set∣ting one Foot in B, describe another Arch so as it may cut the former in two Points, suppose G, and H, and draw the Line HG towards that Part on which you suppose the Center of the Center of the Circle will fall.

In like manner, opening your Compasses to more than half your distance of B, C, describe two other Arches from the Points E and C, cut∣ting each other in E and F, then draw the Line EF till it intersect the former Line HG, so shall the Point of Intersection be the Center of the Circumference or Circle required, as in Fig. may be seen.

Proposition II.

The Conjugate Diameters of an Ellipsis being given, to draw the Ellipsis.

Let the given Diameter in Fig. 24. be LB and ED, the greatest Diameter. LB being bisected in the Point of Bisection, erect the Perpendicular

AD. which let be half of the lesser Diameter ED, then open your Compasses to the extent of AB, and setting one Foot in D, with the other make a mark at M and N in the Diameter BL, then cutting a thred to the length of BL, fasten the thred with your Compasses in the Points NM, and with your Pen in the inside of the thred de∣scribe the Arch BFKL, so shall you describe the one half of the Ellipsis required, and turning the Thred on the other side of the Compasses, you may with your Pen in the like manner describe the other half of the Ellipsis GBHL.

CHAP. III.

Of Triangles.

HItherto we have spoken of the most Simple Figure, a Circle. Come we now to those Figures that are Various or Angular.

2. And an Angular Figure is that which doth consist of three or more Angles.

3. An angular Figure consisting of three Angles, otherwise called a Triangle, is a Super∣ficies or Figure comprehended by three Right Lines including three Angles.

4. A Triangle may be considered either in re∣spect of its Sides, or of its Angles.

5. A Triangle in respect of its Sides, is either Isopleuron, Isosceles, or Scalenum.

6. An Isopleuron Triangle, is that which hath three equal sides. An Isoscecles hath two equal Sides. And a Scalenum hath all the three Sides unequal.

7. A Triangle in respect of its Angles is Right or Oblique.

8. A Right angled Triangle is that which hath one Right Angle and two Acute.

9. An Oblique angled Triangle, is either Acute or Obtuse.

10. An Oblique acute angled Triangle, is that which hath all the three Angles Acute.

11. An Oblique obtuse angled Triangle, is that which hath one Angle Obtuse, and the other two Acute.

Proposition I.

Vpon a Right Line given to make an Isopleuron or an Equilateral Triangle.

In Fig. 8. let it be required to make an Equila∣teral Triangle upon the Right Line AB. Open your Compasses to the extent of the Line given, and setting one Foot of your Compasses in A, make an Arch of a Circle above or beneath the Line given, then setting one Foot of your Com∣passes in B, they being full opened to the same extent, with the other foot draw another Arch of a Circle crossing the former, and from the In∣tersection of those Arches draw the Lines AC and AB, so shall the Triangle ACB be Equilate∣ral as was desired.

Proposition II.

Vpon a Right Line given to make an Isosceles Tri∣angle, or a Triangle having two Sides equal.

In Fig. 8. let AB be the Right Line given, from the Points A and B as from two Centers, but at a lesser extent of the Compasses than AB;

if you would have AB the greatest Side, at a greater extent; if you would have it to be the least Side, describe two Arches cutting one ano∣ther, as at F, and from the Intersection draw the Lines AF, and FB, so shall the Triangle AFB have two equal Sides, as was required.

Proposition 3.

To make a Scalenum Triangle, or a Triangle, whose three Sides are unequal.

In Fig. 9. let the three unequal Sides be EFG make AB equal to one of the given Lines, sup∣pose G, and from A as a Center, at the extent of E describe the Arch of a Circle; in like manner from B at the extent of F describe another Arch intersecting the former, then shall the Right Lines AC. CB and BA comprehend a Triangle, whose three sides shall be unequal, as was required.

CHAP. IV.

Of Quadrangular and Multangular Figures.

WE have spoken of Triangles or Figures con∣sisting of three Angles, come we now to those that have more Angles than three, as the Quadrangle, Quinquangle, Sexangle, &c.

2. A Quadrangle is a Figure or Superficies, which is bounded with four Right Lines.

3. A Quadrangle is either a Parallelogram or a Trapezium.

4. A Parallelogram is a Quadrangle whose oppo∣site

Sides are parallel having equal distances from one another in all Places.

5. A Parallelogram is either Right angled or Oblique.

6. A Right angled Parallelogram, is a Quadran∣gle whose four Angles are all Right, and is either Square or Oblong.

7. A Square Parallelogram doth consist of four equal Lines. The Parts of a Square are, the Sides of which the Square is made, and the Dia∣gonal or Line drawn from one opposite Angle to another through the middle of the Square.

8. An Oblong is a Right angled Parallelogram, having two longer and two shorter Sides.

9. An Oblique angled Parallelogram, is that whose Angles are all Oblique, and is either a Rhombus or a Rhomboides.

10. A Rhombus is an Oblique angled and equi∣lateral Parallelogram.

11. A Rhomboides is an Oblique angled and inequilateral Parallelogram.

12. A Trapezium is a Quadrangular Figure whose Sides are not all parallel; it is either Right angled or Oblique.

13. A Right angled Trapezium hath two op∣posite Sides parallel, but unequal, and the Side between them perpendicular.

14. An Oblique angled Trapezium is a Qua∣drangle, but not a Parallelogram, having at least two Angles Oblique, and none of the Sides pa∣rallel.

15. Thus much concerning Quadrangles or four sided Figures. Figures consisting of more than four Angles are almost infinite, but are re∣ducible unto two sorts, Ordinate and Regular, or Inordinate and Irregular.

16. Ordinate and Regular Polygons are such, as are contained by equal Sides and Angles, as the Pentagon, Hexagon, and such like.

17. Inordinate or irregular Polygons, are such as are contained by unequal Sides and Angles. The construction of these Quadrangular and Multangular Figures is explained in the Proposi∣tions following.

Proposition. I.

Vpon a Right Line given to describe a Right an∣gled Parallelogram, whether Square or Oblong.

In Fig. 10. let the given Line be AB, upon the Point A erect the Perpendic••lar AD equal to AB if you intend to make a Square, but long∣er or shorter, if you intend an oblong, and upon the Points D and B at the distance of AB and AD describe two Arches intersecting one ano∣ther, and from the Intersection draw the Lines ED and EB, so shall the Right angled Figure AE be a Square, if AB and AD be equal, o∣therwise an Oblong, as was desired.

Proposition II.

To describe a Rhombus or Rhomboides.

In Fig. 11. To the Right Line AB draw the Line AD at any Acute Angle at pleasure, equal to AB if you intend a Rhombus, longer or short∣er if you intend a Rhomboides, then upon your Compasses to the extent of AD and upon B as a Center describe an Arch; in like manner, at the extent of AB upon D as a Center describe an∣other

Arch intersecting the former, then draw the Lines ED and EB, so shall AE be the Rhom∣bus or Rhomboides, as was required.

Proposition III.

Vpon a Right Line given to make a Regular Pen∣tagon, or five sided Figure.

In Fig. 12. Let the given Line be AB, upon A and B as two Centers describe the Circles EBGH and CAGK, then open your Compasses to the extent of BC, and making G the Cen∣ter, describe the Arch HAFK, then draw the Lines KFE and HFC: so shall AE and BC be two sides of the Pentagon desired, and opening your Compasses to the extent of AB, upon E and C as two Centers describe two Arches inter∣secting one another, and from the Point of Inter∣section draw the Lines ED and DC, so shall the Figure AB and DE be the Pentagon required.

Proposition IV.

To make a Regular Pentagon and Decagon in a given Circle.

In Fig. 13. upon the Diameter CAB describe the Circle CDBL, from the Center AErect the Perpendicular AD, and let the Semidiameter AC be bisected, the Point of Bisection is E, set the distance ED from E to G, and draw the Line GD, which is the side of a Pentagon, and AG the side of a Decagon inscribed in the same Circle.

Proposition V.

In a Circle given to describe a Regular Hexagon.

The side of a Hexagon is equal to the Radius of a Circle, the Radius of a Circle therefore being six times applied to the Circumference, will give you six Points, to which Lines being drawn from Point to Point, will constitute a Regular Hexagon, as was desired.

Proposition VI.

In a Circle given to describe a Regular Hepta∣gon or Figure consisting of seven equal sides.

The side of a Heptagon is equal to half the side of a Triangle inscribed in a Circle, having therefore drawn an Hexagon in a Circle, the Chord Line subtending two sides of the Hexagon lying together, is the side of a Triangle inscrib∣ed in that Circle, and half that Chord applied seven times to the Circumference, will give se∣ven Points, to which Lines being drawn from that Point, will constitute a Regular Heptagon, as in Fig. 14. is plainly shewed.

CHAP. V.

Of Solid Bodies.

HAving spoken of the two first kinds of Magnitude, Lines and Superficies, come we now to the third, a Body or Solid.

2. A Body or Solid is a Magnitude consisting of length, breadth and thickness.

3. A Solid is either regular or irregular.

4. That is called a regular Solid, whose Bases, Sides and Angles are equal and like.

5. And this either Simple or Compound.

6. A simple regular Solid, is that whith doth consist of one only kind of Superficies.

7. And this is either a Sphere or Globe, or a plain Body.

8. A Globe is a Solid included by one round and convex Superficies, in the middle whereof there is a Point, from whence all Lines drawn to the Circumference are equal.

9. A simple plain Solid, is that which doth consist of plain Superficies.

10. A plain Solid is either a Pyramid, a Prism, or a mixt Solid.

11. A Pyramid is a Solid, Figure or Body, contained by several Plains set upon one right lin'd Base, and meeting in one Point.

12. Of all the several sorts of Pyramids, there is but one that is Regular, to wit a Tetrahe∣dron, or a Pyramid consisting of four regular or equilateral Triangles; the form whereof (as it may be cut in Pastboard) may be conceived by Figure 15.

13. A Prism is a Solid contained by several Plains, of which those two which are opposite, are equal, like and parallel, and all others are Paralellogram.

14. A Prism is either a Pentahedron, a Hexahe∣dron, or a Polyhedron.

15. A Pentahedron Prism, is a Solid compre∣hended of five Sides, and the Base a Triangle, as Fig. 16.

16. An Hexahedron Prism, is a Solid com∣prehended of six Sides, and the Base a Quadran∣gle, as Fig. 17.

17. An Hexahedron Prism, is distinguished into a Parallelipipedon and a Trapezium.

18. An Hexahedron Prism called a Trapezium is a Solid, whose opposites Plains or Sides, are neither opposite nor equal.

19. A Parallelipipedon is either right angled or oblique.

20. A right angled Parallelipipedon is an Hexa∣hedron Prism, comprehended of right angled Plains or Sides; and it is either a Cube or an Ob∣long.

21. A Cube is a right angled Parallelipipedon comprehended of six equal Plains or Sides.

22. An Oblong Parallelipipedon, is an Hexahe∣dron Prism, comprehended by unequal Plains or Sides.

23. An Oblique angled Parallelipipedon, is an Hexahedron Prism, comprehended of Oblique Sides.

24. A Polyhedron Prism, is a Solid compre∣hended by more than six Sides, and hath a mul∣tangled Base, as a Quincangle, Sexangle, &c.

25. A regular compound or mixt Solid, is such

a Solid as hath its Vertex in the Center, and the several Sides exposed to view, and of this sort there are only three; the Octohedron, the Icosahe∣dron, of both which the Base is a Triangle; and the Dodecahedron, whose Base is a Quincangle.

26. An Octohedron is a Solid Figure which is contained by eight equal and equilateral Trian∣gles, as in Fig. 18.

27. An Icosahedron is a Solid, which is con∣tained by twenty equal and equilateral Trian∣gles, as Fig. 19.

28. A Dodecahedron is a Solid, which is con∣tained by twelve equal Pentagons, equilateral and equiangled, as in Fig. 20.

29. A regular compound Solid, is such a So∣lid as is Comprehended both by plain and circu∣lar Superficies, and this is either a Cone or a Cylinder.

30. A Cone is a Pyramidical Body, whose Base is a Circle, or it may be called a round Py∣ramis, as Fig. 21.

31. A Cylinder is a round Column every where comprehended by equal Circles, as Fig. 22.

32. Irregular Solids are such, which come not within these defined varieties, as Ovals, Fru∣stums of Cones, Pyramids, and such like.

And thus much concerning the description of the several sorts of continued Quantity, Lines, Plains and Solids; we will in the next place consider the wayes and means by which the Di∣mentions of them may be taken and determined, and first we will shew the measuring of Lines.

CHAP. VI.

Of the Measuring of Lines both Right and Circular.

EVery Magnitude must be measured by some known kind of Measure; as Lines by Lines, Superficies by Superficies, and Solids by Solids, as if I were to measure the breadth of a River, or height of a Turret, this must be done by a Right Line, which being applied to the breadth or height desired to be measured, shall shew the Perches, Feet or Inches, or by some other known measure the breadth or height desired: but if the quantity of some Field or Meadow, or any other Plain be desired, the number of square Perches must be enquired; and lastly, in measuring of Solids, we must use the Cube of the measure used, that we discover the number of those Cubes that are contained in the Body or Solid to be measured. First, therefore we will speak of the several kinds of measure, and the making of such Instruments, by which the quantity of any Magni∣tude may be known.

2. Now for the measuring of Lines and Su∣perficies, the Measures in use with us, are Inches, Feet, Yards, Ells and Perches.

3. An Inch is three Barley Corns in length, and is either divided into halves and quarters, which is amongst Artificers most usual, or into ten equal Parts, which is in measuring the most useful way of Division.

4. A Foot containeth twelve Inches in length, and is commonly so divided; but as for such things as are to be measured by the Foot, it is far

better for use, when divided into ten equal Parts, and each tenth into ten more.

5. A Yard containeth three Foot, and is com∣monly divided into halves and quarters, the which for the measuring of such things as are usually sold in Shops doth well enough, but in the mea∣suring of any Superficies, it were much better to be divided into 10 or 100 equal Parts.

6. An Ell containeth three Foot nine Inches, aud is usually divided into halves and quarters, and needs not be otherwise divided, because we have no use for this Measure, but in Shop Com∣modities.

7. A Pole or Perch cotaineth five Yards and an half, and hath been commonly divided into Feet and half Feet. Forty Poles in length do make one Furlong, and eight Furlongs in length do make an English Mile, and for these kinds of of lengths, a Chain containing four Pole, divided by Links of a Foot long, or a Chain of fifty Foot, or what other length you please, is well enough, but in the measuring of Land, in which the number of square Perches is required; the Chain called Mr. Gunters, being four Pole in length divided into 100 Links, is not without just reason reputed the most useful.

8. The making of these several Measures is not difficult, a Foot may be made, by repeating an Inch upon a Ruler twelve times, a Yard is eight Foot, and so of the rest; the Subdivision of a Foot or Inch into halves and quarters, may be performed by the seventeenth of the first, and into ten or any other Parts by the first Proposition of the first Chapter, and all Scales of equal Parts, of what scantling you do desire. And this I

think is as much as needs to be said concerning the dividing of such Instruments as are useful in the measuring Right Lines.

9. The next thing to be considered is the mea∣suring of Circular Lines, or Perfect Circles.

10. And every Circle is supposed to be divi∣ded into 360 Parts called Degrees, every Degree into 60 Minutes, every Minute into 60 Seconds, and so forward this division of the Circle into 360 Parts is generally retained, but the Sub∣division of those Parts, some would have be thus and 100, but as to our present purpose either may be used, most Instruments not exceeding the fourth part of a Degree.

11. Now then a Circle may be divided into 360 Parts in this manner, Having drawn a Dia∣meter through the Center of the Circle dividing the Circle into two equal Parts, cross that Dia∣meter with another at Right Angles through the Center of the Circle also, so shall the Circle be divided into four equal Parts or Quadrants, each Quadrant containing 90 Degrees, as in Fig. 7. GE. ED. DL and LG, are each of them 90 Degrees; and the Radius of a Circle being equal to the Chord of the sixth Part thereof, that is to the Chord of 60 Degrees, as in Fig. 14. if you set the Radius GB from L towards G, and also from G towards L, the Quadrant GL will be sub∣divided into three equal Parts, each Part con∣taining 30 Degrees, GM. 30. MH 30 and HL 30, the like may be done in the other Quadrants also; so will the whole Circle be divided into twelve Parts, each Part containing 30 Degrees.

And because the side of a Pentagon inscribed in a Circle is equal to the Chord of 72 Degrees, or

the first Part of 360, as in Fig. 13. therefore if you set the Chord of the first Part of the Circle given from G to L or L to G, in Fig. 7. you will have the Chord of 72 Degrees, and the difference between GP 72 and GH 60 is HP 12, which be∣ing bisected, will give the Arch of 6 Degrees, and the half of six will give three, and so the Circle will be divided into 120 Parts, each Part con∣taining three Degrees, to which the Chord Line being divided into three Parts, the Arch by those equal Divisions may be also divided, and so the whole Circle will be divided into 360, as was de∣sired.

12. A Circle being thus divided into 360 Parts, the Lines of Chords, Sines, Tangents and Secants, are so easily made (if what hath been said of them in the Second Chapter be but considered) that I think it needless to say any more concerning their Construction, but shall rather proceed unto their Use.

13. And the use of these Lines and other Lines of equal Parts we will now shew in circular and right lined Figures; and first in the measuring of a Circle and Circular Figures.

CHAP. VII.

Of the Measuring of a Circle.

THe squaring of a Circle, or the finding of a Square exactly equal to a Circle given, is that which many have endeavoured, but none as yet have attained: Yet Archimedes that Famous Mathe∣matician hath sufficiently proved, That the Area of a Circle is equal to a Rectangle made of the Ro∣dius and half the Circumference: Or thus, The Area of a Circle is equal to a Rectangle made of the Diameter and the fourth part of the Circum∣ference. For Example, let the Diameter of a Circle be 14 and the Circumference 44; if you multiply half the Circumference 22 by 7 half the Diameter, the Product is 154; or if you multiply 11 the fourth part of the Circumference, by 14 the whole Diameter, the Product will still be 154. And hence the Superficies of any Circle may be found though not exactly, yet near enough for any use.

2. But Ludolphus Van Culen finds the Circum∣ference of a Circle whose Diameter is 1.00 to be 3.14159 the half whereof 1.57095 being mul∣tiplied by half the Diameter 50, &c. the Product is 7.85395 which is the Area of that Circle, and from these given Numbers, the Area, Circumfe∣rence and Diameter of any other Circle may be found by the Proportions in the Propositions fol∣lowing.

Proposition I.

The Diameter of a Circle being given to find the Circumference.

As 1. to 3.14159: so is the Diameter to the Circumference. Example. In Fig. 13. Let the Diameter IB be 13. 25. I say as 1. to 3. 14159. so IB. 13.25 to 41.626 the Circumference of that Circle.

Proposition II.

The Diameter of a Circle being given to find the Superficial Content.

As 1. to 78539; so is the Square of the Dia∣meter given, to the Superficial Content required. Example, Let the Diameter given be as before IB 13.25 the Square thereof is 175.5625 therefore.

As 1. to 78539: so 175.5625 to 137.88 the Superficial Content of that Circle.

Proposition III.

The Circumference of a Circle being given, to find the Diameter.

This is but the Converse of the first Propositi∣on: Therefore as 3.14159 is to 1: so is the Circumference to the Diameter; and making the Circumference an Unite, it is. 3. 14159. 1∷ 1. 318308, and so an Unite may be brought into the first place. Example, Let the given Cir∣cumference

be 41. 626. I say,

As 1. to 318308: so 41.626 to 13. 25. the Diameter required.

Proposition IV.

The Circumference of a Circle being given to find the Superficial Content.

As the Square of the Circumference of a Cir∣cle given is to the Superficial Content of that Circle: so is the Square of the Circumference of another Circle given to the Superficial Con∣tent required. Example, As the Square of 3.14159 is to 7853938: so is 1. the Square of another Circle to 079578 the Superficial Content required, and so an Unite for the most easie work∣ing may be brought into the first place: Thus the given Circumference being 41. 626. I say,

As 1. to 0.79578: so is the Square of 41.626 to 137.88 the Superficial Content required.

Proposition V.

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

This is the Converse of the second Propositi∣on, therefore as 78539 is to 1. so is the Superfici∣al Content given, to the Square of the Diameter required. And to bring an Unite in the first place: I say.

As 7853978. 1∷1. 1. 27324, and there∣fore if the Superficial Content given be 137.88, to find the Diameter: I say,

As 1. to 1.27324: so 137.88 to 175.5625 whose Square Root is 13.25, the Diameter sought.

Proposition VI.

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

This is the Converse of the Fourth Propositi∣on, and therefore as 079578 is to 1 : so is the Su∣perficial Content given, to the Square of the Cir∣cumference required, and to bring an Unite in the first place: I say,

As 079578. 1 :: 1. 12.5664, and therefore if the Superficial Content given be 137.88, to find that Circumference: I say,

As 1. to 12.5664: so is the 137.88 to 1732.7 whose Square Root is 626 the Circumference.

Proposition VII.

The Diameter of a Circle being given to find the Side of the Square, which may be inscribed within the same Circle.

The Chord or Subtense of the Fourth Part of a Circle, whose Diameter is an Unite, is 7071067, and therefore, as 1. to 7071067: so is the Dia∣meter of another Circle, to the Side required. Example, let the Diameter given be 13.25 to find the side of a Square which may be inscribed in that Circle: I say,

As 1. to 7071067: so is 13.25 to 9.3691 the side required.

Proposition VIII.

The Circumference of a Circle being given, to find the Side of the Square which may be inscribed in the same Circle.

As the Circumference of a Circle whose Dia∣meter is an Unite, is to the side inscribed in that Circle; so is the Circumference of any other Circle, to the side of the Square that may be in∣scribed therein. Therefore an Unite being made the Circumference of a Circle.

As 3.14159 to 7071067: so 1. to 225078.

And therefore the Circumference of a Circle being as before 41.626, to find the side of the Square that may be inscribed: I say,

As 1. to 225078. so is 41.626 to 9.3691 the side inquired.

Proposition IX.

The Axis of a Sphere or Globe being given, to find the Superficial Content.

As the Square of the Diameter of a Circle, which is Unity, is to 3.14159 the Superficial Content, so is the Square of any other Axis given, to the Superficial Content required. Ex∣ample, Let 13.25 be the Diameter given, to find the Content of such a Globe: I say,

As 1. to 3.14159: so is the Square of 13.25 to 551.54 the Superficial Content required.

Proposition X.

To find the Area of an Ellipsis.

As the Square of the Diameter of a Circle, is to the Superficial Content of that Circle; so is the Rectangle made of the Conjugate Diame∣ters in an Ellipsis, to the Area of that Ellipsis; And the Diameter of a Circle being one, the Area is 7853975, therefore in Fig. 26. the Diameters AC8 and BD5 being given, the Area of the Ellipsis ABCD may thus be found.

As 1. to 7853975: so is the Rectangle AC in BD40 to 3.1415900, the Area of the Ellipsis required.

CHAP. VIII.

Of the Measuring of Plain Triangles.

HAving shewed the measuring of a Circle, and Ellipsis, we come now to Right lined Figures, as the Triangle, Quadrangle, and Mul∣tangled Figures, and first of the measuring of the plain Triangles.

2. And the measuring of Plain Triangles is either in the measuring of the Sides and Angles, or of their Area and Superficial Content.

3. Plain Triangles in respect of their Sides and Angles are to be measured by two sorts of Lines, the one is a Line of equal Parts, and by that the Sides must be measured, the other is a Line of Chords, the Construction whereof hath

been shewed in the sixth Chapter, and by that the Angles must be measured, the Angles may in∣deed be measured by the Lines of Sines, Tan∣gents or Secants, but the Line of Chords being not only sufficient, but most ready, it shall suffice to shew how any Angle may be protracted by a Line of Chords, or the Quantity of any Angle found, which is protracted.

4. And first to protract or lay down an Angle to the Quantity or Number of Degrees propo∣sed, do thus, draw a Line at pleasure as AD in Figure 5, then open your Compasses to the Num∣ber of 60 Degrees in the Line of Chords, and setting one Foot in A, with the other describe the Arch BG, and from the Point A let it be requi∣red to make an Angle of 36 Degrees: open your Compasses to that extent in the Line of Chords, and setting one Foot in B, with the other make a mark at G, and draw the Line AG, so shall the Angle BAG contain 36 Degrees, as was re∣quired.

5. If the Quantity of an Angle were re∣quired, as suppose the Angle BAG, open your Compasses in the Line of Chords to the extent of 60 Degrees, and setting one Foot in A, with the other draw the Arch BG, then take in your Com∣passes the distance of BG, and apply that extent to the Line of Chords, and it will shew the Num∣ber of Degrees contained in that Angle, which in our Example is 36 Degrees.

6. In every Plain Triangle, the three Angles are equal to two right or 180 Degrees, there∣fore one Angle being given, the sum of the other two is also given, and two Angles being given, the third is given also.

7. Plain Triangles are either Right Angled or Oblique.

8. In a Right Angled Plain Triangle, one of the Acute Angles is the Complement of the other to a Quadrant or 90 Degrees.

9. In Right Angled Plain Triangles, the Side subtending the Right Angle we call the Hypotenuse, and the other two Sides the Legs, thus in Fig. 5▪ AE is the Hypotenuse, and AD and ED are the Legs; these things premised, the several cases in Right Angled and Oblique Angled Plain Tri∣angles may be resolved, by the Propositions fol∣lowing.

Proposition I.

In a Right Angled Plain Triangle, the Angles of one Leg being given, to find the Hypotenuse and the other Leg.

In the Right Angled Plain Triangle ADE in Fig. 5. Let the given Angles be DAE 36, and DEA 54, and let the given Leg be AD 476; to find the Hypotenuse AE, and the other Leg ED.

Draw a Line at pleasure, as AD, and by your Scale of equal Parts set from A to D 476 the Quantity of the Leg given, then erect a Perpen∣dicular upon the Point D, and upon the Point A lay down your given Angle DAE 36 by the fourth hereof, and draw the Line AE till it cut the Perpendicular DE, then measure the Lines AE and DE upon your Scale of Equal Parts, so shall AE 588.3 be the Hypotenuse, and DE 345.8 the other Leg.

Proposition II.

The Hypotenuse and Oblique Angles given, to find the Legs.

Let the given Hypotenuse be 588, and one of the Angles 36 degrees, the other will then be 54 degrees, Draw a Line at pleasure, as AD, and upon the Point A by the fourth▪ hereof lay down one of the given Angles suppose the less, and draw the Line AC, and from your Scale of equal Parts, set off your Hypotenuse 588 from A to E, and from the Point E to the Line AD let fall the Perpendicular ED, then shall AD be∣ing measured upon the Scale be 476 for one Leg, and ED 345.8 the other.

Proposition III.

The Hypotenuse and one Leg given to find the An∣gles and the other Leg.

Let the given Hypotenuse be 588. and the given Leg 476. Draw a Line at pleasure as AD, upon which set the given Leg from A to D. 476, and upon the Point D, erect the Perpendicular DE, then open your Compasses in the Scale of Equal Parts to the Extent of your given Hypotenuse 588, and setting one Foot of that Extent in A, move the other till it touch the Perpendicular DE, then and there draw AE, so shall ED be 345.8 the Leg inquired, and the Angle DAE, will be found by the Line of Chords to be 36▪ whose Comple∣ment is the Angle DEA. 54.

Proposition IV.

The Legs given to find the Hypotenuse, and the Oblique Angles.

Let one of the given Legs be 476, and the o∣ther 345.8, Draw the Line AD to the extent of 476, and upon the Point D, erect the Perpendi∣cular DE to the extent of 345.8, and draw the Line AE, so shall AE be the Hypotenuse 588, and the Angle DAE will by the Line of Chords be found to be 36 Degrees, and the Angle DEA 54, as before.

Hitherto we have spoken of Right angled plain Triangles: the Propositions following concern such as are Oblique.

Proposition V.

Two Angles in an Oblique angled plain Triangle, being given, with any one of the three Sides, to find the other two Sides.

In any Oblique angled plain Triangle, let one of the given Angles be 26.50 and the other 38. and let the given Side be 632, the Sum of the two given Angles being deducted from a Semi-circle, leaveth for the third Angle 115.50 De∣grees, then draw the Line BC 632. and upon the Points B and C protract the given Angles, and draw the Lines BD and CD, which being mea∣sured upon your Scale of equal Parts BD will be fou••d to be 312.43, and BD 431.09,

Proposition VI.

Two Sides in an Oblique Angled Triangle being given, with an Angle opposite to one of them, to find the other Angles and the third Side, if it be known whe∣ther the Angle Opposite to the other Side given be Acute or Obtuse.

In an Oblique Angled Plain Triangle, let the given Angle be 38 Degrees, and let the Side ad∣jacent to that Angle be 632, and the Side oppo∣site 431. 1. upon the Line BC in Fig. 25. protract the given Angle 38 Degrees upon the Point C, and draw the Line DC, then open your Compasses to the Extent of the other Side given 431. 1. and setting one Foot in B, turn the other about till it touch the Line DC, which will be in two pla∣ces, in the Points D and E; if therefore the Angle at B be Acute the third Side of the Triangle will he CE, according therefore to the Species of that Angle you must draw the Line BD or BE to compleat the Triangle, and then you may measure the other Angles, and the third Side as hath been shewed.

Proposition VII.

Two Sides of an Oblique Angled Plain Triangle be∣ing given, with the Angle comprehended by them to find the other Angles and the third Side.

Let one of the given Sides be 632, and the o∣ther 431.1, and let the Angle comprehended by them be Deg. 26.50, draw a Line at pleasure,

as BC, and by help of your Scale of Equal Parts, set off one of your given Sides from B to C 632. then upon the Point B protract the given Angle 26. 50. and draw the Line BD, and from B to D, set off your other given Side 431. 1. and draw the Line DC, so have you constituted the Triangle BDC, in which you may measure the Angles and the third Side, as hath been shewed.

Proposition VIII.

The three Sides of an Oblique Angled Triangle being given, to find the Angles.

Let the length of one of the given Sides be 632, the length of another 431.1, and the length of the third 312.4, and Draw a Line at pleasure, as BC in Fig. 25, and by help of your Scale of E∣qual Parts, set off the greatest Side given 632 from B to C. then open your Compasses in the same Scale to the extent of either of the other Sides, and setting one Foot of your Compasses in B, with the other describe an occult Arch, then extend your Compasses in the same Scale to the length of the third Side, and setting one Foot in C with the other describe another Arch cutting the former, and from the Point of Intersection draw the Lines BD and DC. to constitute the Triangle BDC, whose Angles may be measured, as hath heen shewed.

And thus may all the Cases of Plain Triangles be resolved by Scale and Compass, he that desires to resolve them Arithmetically, by my Trigome∣tria Britannica, or my little Geometrical, Trigo∣nometry; only one Case of Right Angled Plain

Triangles which I shall have occasion to use, in the finding of the Area of the Segment of a Cir∣cle I will here shew how, to resolve by Numbers.

Proposition IX.

In a Right Angled Plain Angle the Hypote∣nuse and one Leg being given to find the other Leg.

Take the Sums and difference of the Hypotenuse and Leg given, then multiply the Sum by the Dif∣ference, and of the Product extract the Square Root, which Square Root shall be the Leg inqui∣red.

Example. In Fig. 5. Let the given Hypotenuse be AE 588.3, and the given Leg AD 476, and let DE be the Leg inquired. The Sum of AE and AD is 1064.3, and their Difference is 112.3, now then if you multiply 1064.3 by 112.3, the Product will be 119520.89, whose Square Root is the Leg DE. 345. 8.

Proposition X.

The Legs of a Right Angled Plain Triangle being gived, to find the Area or Superficial Content thereof.

Multiply one Leg by the other, half the Product shall be the Content. Example, In the Right angled plain Triangle ADE, let the given Legs be AD 476, and DE 345, and let the Area of that Triangle be required, if you multiply 476 by 345 the Product will be 164220, and the half thereof 82110 is the Area or Superficial Con∣tent required.

Proposition XI.

The Sides of an Oblique angled plain Triangle be∣ing given to find the Area or Superficial Content there∣of.

Add the three Sides together, and from the half Sum subtract each Side, and note their Dif∣ference; then multiply the half Sum by the said Differences continually, the Square Root of the last Product, shall be the Content required.

Example. In Fig. 9. Let the Sides of the Tri∣angle ABC be AB 20. AC 13, and BC 11 the Sum of these three Sides is 44, the half Sum is 22, from whence subtracting AB 20, the Dif∣ference is 2, from whence also if you substract AC 13, the Difference is 9, and lastly, if you subtract BC 11 from the half Sum 22, the Diffe∣rence will be 11. And the half Sum 22 being multiplied by the first Difference 2, the Product is 44, and 44 being multiplied by the Second Dif∣ference 9, the Product is 396, and 396 being mul∣tiplied by the third Difference 11, the Product is 4356, whose Square Root 66, is the Content re∣quired.

Or thus, from the Angle C let fall the Perpen∣dicular DC, so is the Oblique angled Triangle ABC, turned into two right, now then if you measure DC upon your Scale of Equal Parts, the length thereof will be found to be 6.6, by which if you multiply the Base AB 20, the Product will be 132.0, whose half 66, is the Area of the Tri∣angle, as before.

Proposition XII.

The Sides of any Oblique angled Quadrangle be∣ing given, to find the Area or Superficial Content thereof.

Let the Sides of the Oblique angled Quadran∣gle ABED in Fig. 11. be given, draw the Diago∣nal AE, and also the Perpendiculars DC and BF, then measuring AE upon the same Scale by which the Quadrangular Figure was protracted, suppose you find the length to be 632, the length of DC 112, and the length of BF 136, if you multiply AE 632 by the Half of DC 56, the Product will be 35392 the Area of ACED. In like manner if you multiply AE 632, by the half of BF 68, the Product will be 42976 the Area of ACEB, and the Sum of these two Products is the Area of ABED as was required.

Or thus, take the Sum of DC 112, and BF 136; the which is 248, and multiply AE 632 by half that Sum, that is by 124, the Product will be 78368 the Area of the Quadrangular Figure ABED, as before.

Proposition XIII.

The Sides of a plain irregular multangled Figure being given, to find the Content.

In Fig. 26. Let the Sides of the multangled Fi∣gure. A. B. C. D. E. F. G. H. be given, and let the Area thereof be required, by Diagonals drawn from the opposite Angles reduce the Figure given,

into Oblique angled plain Triangles, and those Oblique angled Triangles, into right by letting fall of Perpendiculars, then measure the Diagonals and Perpendiculars by the same Scale, by which the Figure it self was protracted, the Content of those Triangles being computed, as hath been shewed, shall be AF the Content required: thus by the Diagonals AG. BE and EC the mul∣tangled Figure propounded is converted into three Oblique angled quadrangular Figures, AFGH. AFEB and BEDC, and each of these are divided into four Right angled Triangles, whose several Contents may be thus computed. Let GA 94 be multiplied by half HL 27 more Half of KF 29, that is by 23, the Product will be 21, be the Area of AHGF. Secondly, OB is 11, and FN 13, their half Sum 12, by which if you multiply AE 132, the Product will be 1584 the Area of AFEB. Thirdly, let Bp be 18 m D 32, the half Sum is 25, by which if you multiply AEC 125 the Product will be 3125 the Area of BEDC, and the Sum of these Products is 6871 the Area of the whole irregu∣lar Figure. ABCDEFGH, as was required.

Proposition XIV.

The Number of Degrees in the Sector of a Circle being given, to find the Area thereof.

In Fig. 27. ADEG is the Sector of a Circle, in which the Arch DEG, is Degrees. 23.50, and by 1. Prop. of Archimed. de Dimensione Circuli, the length of half the Arch is equal to the Area of the Sector of the double Arch, there the length

of DE or EG is equal to the Area of the Sector ADEG: and the length or circumference of the whole Circle whose Diameter is 1 according to Van Culen, is 3.14159265358979, therefore the length of one Centesme of a Degree, is. 0. 01745329259. Now then to find the length of any Number of Degrees and Decimal Parts, you must multiply the aforesaid length of one Cen∣tesme by the Degrees and Parts given, and the Product shall be the length of those Degrees and Parts required, and the Area of a Sector containing twice those Degrees and Parts. Example, the half of DEG 23.50 is DE or EG 11.75, by which if you multiply 0.01745329259, the Pro∣duct will be 2050761879325, the length of the Arch DE, and the Area of the Sector ADEG.

Proposition XV.

The Number of Degrees in the Segment of a Circle being given, to find the Area of the Segment.

In Fig. 27. Let the Area of the Segment DEGK be required, in which let the Arch DEG be Degrees 23.50, then is the Area of the Sector ADEG 2050761879325 by the last a∣foregoing, from which if you deduct the Area of the Triangle ADG, the remainer will be the Area of the Segment DEGK. And the Area of the Tri∣angle ADG may thus be found. DK is the Sine of DE 11.75, which being sought in Gellibrand's De∣cimal Canon is. 2036417511, and AK is the Sine of DH 78.25, or the Cosine of DE. 9790454724, which being multiplied by the Sine of DE, the Pro∣duct will be 1993745344, or if you multiply AG

[illustration]

the Radius by half DF the Sine of the double Arch DEG, the Product will be 19937453445 as be∣fore, and this Product being deducted from the Area of the Sector ADEG 2050761879325, the remainer will be 57016434875 the Area of the Segment DEGL, as was desired.

Proposition XVI.

The Diameter of a Circle being cut into any Num∣ber of Equal Parts, to find the Area of any Segment made by the Chord Line drawn at Right Angles through any of those equal Parts of the Diameter.

In Fig. 28. The Radius AD is cut into five E∣qual Parts, and the Segment EDFL is made by the Chord Line ELF at Right Angles to AD in the fourth Equal Part, or at eight tenths there∣of: now then to find the Area of this Segment we have given AE Radius, and AL 8, and there∣fore by the ninth hereof EL will be found to be 606000, the Sine of ED 36.87, by which if you multiply 0.0174532, the Product is the Area of the Sector AEDF 64350286, and the Area of the Triangle AEF is 48, which being deducted from the Area of the Sector, the Remainer 16350286 is the Area of the Sector EDFL, as was required. And in this manner was that Ta∣ble of Segments made by the Chord Lines cutting the Radius into 100 Equal Parts.

Another way.

In Fig. 28. Let the Radius AD be cut into 10.100 or 1000 Equal Parts, and let the Area of

the Segments made by the Chord Lines drawn at Right Angles through all those Parts be required: first find the Ordinates GK and M. PN. EL, the double of each Ordinate, will be the Chords of the several Arches, and the Sum of these Chords beginning with the least Ordinate, will orderly give you the Area of the several Segments made by those Chord Lines, but the Diameter must be be divided into 100000 Equal Parts, because of the unequal differences at the beginning of the Diameter: but taking the Area of the Circle to be 3. 1415926535, &c. as before, the Area of the Semicircle will be 1.57079632, from which if you deduct the Chord GH1999999, the Chord answering to 999 Parts of the Radius, the remainer is. 1.56879632 the Area of the Seg∣ment GDH. And in this manner by a conti∣nual deduction of the Chord Lines from the Area of the Segment of the Circle given, was made that Table shewing the Area of the Segments of a Circle to the thousandth part of the Radius.

And because a Table shewing the Area of the Segments of a Circle to the thousandth part of the Radius, whose whole Area is Unity, is yet more useful in Common Practice, therefore from this Table, was that Table also made by this Pro∣portion.

As the Area of the Circle whose Diameter is. Unity, to wit 3.14149 is to the Area of any part of that Diameter, so is Unity the supposed Area of another Circle, to the like part of that Diameter. Example, the Area answering to 665 parts of the Radius of a Circlewhose Area is 3.14159 is 0.91354794 therefore,

As 3.14159265 is to 0.91354794: So is ••.

[illustration]

[illustration]

to 290791, the Area required; and the Table being thus computed to the 1000 parts of the Radius, we have enlarged it by the difference to the 5000 parts of the Radius, and consequently to the ten thousandth part of the Diameter: The use of which Table shall be shewed when we come to the measuring of Solid Bodies.

CHAP. IX.

Of the Measuring of Heights and Distances.

HAving shewed in the former Chapter, how all plain Triangles may be measured, not only in respect of their Sides and Angles, but in respect of their Area, and the finding of the A∣rea of all other plain Figures also, that which is next to be considered, is the practical use of those Instructions, in the measuring of Board, Glass, Wainscot, Pavement, and such like, as also the measuring or surveying of Land; and first we will shew the measuring of Heights and Distances.

2. And in the measuring of Heights and Di∣stances, besides a Chain of 50 or 100 Links, each Link being a Foot, it is necessary to have a Qua∣drant of four or five Inches Radius, and the larger the Quadrant is, the more exactly may the Angles: be taken, though for ordinary Practice, four or five Inches Radius will be sufficient.

Let such a Quadrant therefore be divided in the Limb into 90 Equal Parts or Degrees, and numbred from the left hand to the right, at every tenth Degree, in this manner 10. 20. 30. 40. 50. 60. 70. 80. 90. and within the Limb of the Qua∣drant

draw another Arch, which being divided by help of the Limb into two Equal Parts, in the Point of Interfection set the Figure 1. represent∣ing the Radius or Tangent of 45 Degrees, and from thence both ways the Tangents of 63.44 Deg. 71.57 Deg. 75.97 Deg.78.70 Deg. 80.54 Deg. that is, 2. 3. 4. 5 and 6 being set also, your Quadrant will be fitted for the taking of Heights several ways, as shall be explained in the Propositions following.

Proposition I.

To find the Height of a Tower, Tree, or other Object at one Station.

At any convenient distance from the Foot of the Object to be measured, as suppose at C in Fig. 30. and there looking through the Sights of your Quadrant till you espie the top of the Object at A, observe what Degrees in the Limb are cut by the Thread, those Degrees from the left Side or Edge of the Quadrant to the Right, is the Quan∣tity of the Angle ACB, which suppose 35 De∣grees; then is the Angle BAC 55 Degrees, be∣ing the Complement of the former to 90 Degrees. This done with your Chain or otherwise mea∣sure the distance from B the Foot of the Object, to your Station at C, which suppose to be 125 Foot. Then as hath been shewed in the 1. Prop. Chap. 8. draw a Line at pleasure as BC, and by your Scale of Equal Parts, set off the distance measured from B to C 125 Foot, and upon the Point C lay down your Angle taken by observa∣tion 35 Degrees, then erect a Perpendicular upon

the Point B, and let it be extended till it cut the Hypothenusal Line AC, so shall AB measured on your Scale of Equal Parts, be 87.5 Foot for the Height of the Object above the Eye; to which the Height of the Eye from the Ground being added, their Sum is the Height required.

Another way.

Let AB represent a Tower whose Altitude you would take, go so far back from it, that looking through the Sights of your Quadrant, to the top of the Tower at A the Thread may cut just 45 Degrees in the Limb, then shall the distance from the Foot of the Tower, to your Station, be the Height of the Tower above the Eye.

Or if you remove your Station nearer and near∣er to the Object, till your Thread hang over the Figures 2. 3. 4 or 5 in the Quadrant, the Height of the Tower at 2. will be twice as much as the distance from the Tower to the Station, at 3. it will be thrice as much, &c. As if removing my Station from C to D, the Thread should hang o∣ver 2 in the Quadrant, and the distance BD 62 Foot, then will 124 Foot be the Height of the Tower, above the Eye.

In like manner if you remove your Station backward till your Thread fall upon one of those Figures in the Quadrant; between 45 and 90 De∣grees, the distance between the Foot of the Tower, and your Station will at 2. be twice as much as the Height, at 3. thrice as much, at 4. four times so much, and so of the rest.

A Third way by a Station at Random.

Take any Station at pleasure suppose at C, and looking through the Sights of your Quadrant, observe what Parts of the Quadrant the Thread falls upon, and then measure the distance be∣tween the Station, and the Foot of the Object, that distance being multiplied by the parts cut in the Quadrant, cutting off two Figures from the Product shall be the Height of the Object above the Eye?

Example, Suppose I standing at C, that the Thread hangs upon 36 Degrees, as also upon 72 in the Quadrant which is the Tangent of the said Arch, and let the measured distance be CB 125 Foot, which being multiplied by 72, the Product is 9000, from which cutting off his Figures be∣cause the Radius is supposed to be 100, the Height inquired will be 90 Foot, he that desires to per∣form this work with more exactness, must make use of the Table of Sines and Tangents Natural or Artificial, this we think sufficient for our pre∣sent purpose.

Proposition II.

To find an inaccessible Height at two Stations.

Take any Station at pleasure as at D, and there looking through the Sights of your Quadrant to the top of the Object, observe what Degrees are cut by the Thread in the Limb, which admit to be 68 Degrees, then remove backward, till the An∣gle taken by the Quadrant, be but half so much

as the former, that is 34 Degrees, then is the di∣stance between your two Stations equal to the Hypothenusal Line at your first Station, viz. AD. if the distance between your two Stations were 326 foot, then draw a Line at pleasure as BD, upon the Point D protract, the Angle ADB 68 Degrees, according to your first Observation, and from your Line of equal parts set off the Hy∣pothenusal 326 Foot from D to A, and from the Point A let fall the Perpendicular AB which be∣ing measured in your Scale of Equal Parts, shall be the Altitude of the Object inquired.

Or working by the Table of Sines and Tan∣gents, the Proportion is.

As the Radius, is to the measured distance or Hypothenusal Line AD; so is the Sine of the Angle ADE, to the height AB inquired.

Another more General way, by any two Stations taken at pleasure.

Admit the first Station to be as before at D, and the Angle by observation to be 68 Degrees, and from thence at pleasure I remove to C, where observing aim I find the Angle at C to be 32 De∣grees, and the distance between the Stations 150 Foot. Draw a Line at pleasure as BC, and upon Clay down your last observed Angle 32 Degrees, and by help of your Scale of Equal Parrs, set off your measured distance from C to D 150 Foot, then upon D lay down your Angle of 68 De∣grees, according to your first Observation, and where the Lines AD and AC meet, let fall the Perpendicular AB, which being measured in your Scale of Equal Parts, shall be the height of the Object as before.

Or working by the Tables of Sines and Tan∣gents, the Proportions.

1. As the Sine of DAC to the Distance DC. So the Sine of ACD, to the Side AD.

2. As the Radius, to the Side AD; so the Sine ADB, to the Perpendicular height AB inquired.

The taking of Distances is much after the same manner, but because there is required either some alteration in the sights of your Quadrant or some other kind of Instrument for the taking of Angles, we will particularly shew, how that may be also done several ways, in the next Chapter.

CHAP. X.

Of the taking of Distances.

FOr the taking of Distances some make use of a Semicircle, others of a whole Circle, with Ruler and Sights rather than a Quadrant, and al∣though the matter is not much by which of these Instruments the Angles be taken, yet in all Cases the whole Circle is somewhat more ready, than either a Semicircle or Quadrant, the which with its Furniture is called the Theodolite.

2. A piece of Board or Brass then about twelve or fourteen Inches Diameter, being made Circular like a round Trencher, must be divided into four Quadrants, and each Quadrant divided into 90 Degrees, or the whole Circle into 360, and each Degree into as many other Equal Parts, as the largeness of the Degrees will well permit: let your Circle be numbred both ways to 360, that is from the right hand to the left, and from the left to the right.

3. Upon the backside of the Circle there must be a Socket made fast, that it may be set upon a three legged Staff, to bear it up in the Field.

4. You must also have a Ruler with Sights fixed at each end, for making of Observation, either fixed upon the Center of your Circle, or loose, as you shall think best; your Instrument being thus made, any distance whether accessible or inacces∣sible may thus be taken.

5. When you are in the Field, and see any Church, Tower, or other Object, whose Distance from you, you desire to know, choose out some other Station in the same Field, from whence you may also see the Object, and measure the distance between your Stations; then setting your Ruler upon the Diameter of your Circle, set your In∣strument so, as that by the Sights on your Ruler, you may look to the other Station, this done turn your Ruler to that Object whose distance you de∣sire to know, and observe how many Degrees of the Circle are cut by the Ruler, as suppose 36 De∣grees, as the Angle ACD in Fig. 30. Then re∣moving your Instrument to D, lay the Ruler on the Diameter thereof, and then turn the whole In∣strument about till through your Sights you can espy the mark set up at your first Station at C, and there fix your Instrument, and then upon the Cen∣tre of your Circle turn your Ruler till through the Sights you can espy the Object whose distance is inquired, suppose at A; and observe the De∣grees in the Circle cut by the Ruler, which let be 112, which is the Angle ADC, and let the di∣stance between your two Stations be DC 326 Foot; so have you two Angles and the side be∣tween them, in a plain Triangle given, by which

to find the other sides, the which by protraction may be done as hath been shewed, in the fifth Pro∣position of Chapter 8. but by the Table of Sines and Tangents, the Proportion is.

As the Sine of DAC, is to DC; so is the Sine of ACD to the Side AD.

Or, as the Sine of DAC, is to the given Side DC.

So is the Sine of ADC to the Side AC.

6. There is another Instrument called the plain Table, which is nothing else, but a piece of Board, in the fashion and bigness of an ordinary sheet of paper, with a little frame, to fasten a sheet of pa∣per upon it, which being also set upon a Staff, you may by help of your Ruler, take a distance there∣with in this manner.

Having measured the distance between your two Stations at D and C, draw upon your paper a Line, on which having set off your distance place your Instrument at your first Station C, and laying your Ruler upon the Line so drawn there∣on, turn your Instrument till through the Sights you can espy the Station at D, then laying your Ruler upon the Point C, turn the same about till through the Sights you can espy the Object at A, and there draw a Line by the side of your Ruler, and remove your Instrument to D, and laying your Ruler upon the Line DC, turn the Instru∣ment about, till through the Sight you can espy the Mark at C, and then laying your Ruler upon the Point D, turn the same, till through the Sights you can espy the Object at A, and by the side of your Ruler draw a Line, which must be extended till it meet with the Line AC, so shall the Line AD being measured upon your Scale of Equal

Parts, be the distance of the Object from D, and the Line AC shall be the distance thereof from C.

7. And in this manner may the distance of two, three or more Objects be taken, from any two Stations from whence the several Objects may be seen, and that either by the plain Table, or The∣odolite.

CHAP. XI.

How to take the Plot of a Field at one Station, from whence the several Angles may be seen.

ALthough there are several Instruments by which the Plat of a Field may be taken, yet do I think it sufficient to shew the use of these two, the plain Table and Theodolite.

2. In the use of either of which the same chain which is used in taking of heights and distances, is not so proper. I rather commend that which is known by the Name of Gunter's Chain, which is four Pole divided into 100 Links; being as I con∣ceive much better for the casting up the Content of a Piece of Ground, than any other Chain that I have yet heard of, whose easie use shall be explain∣ed in its proper place.

3. When you are therefore entered the Field with your Instrument, whether plain Table, or Theodolite, having chosen out your Station, let visible Marks be set up in all the Corners thereof, and then if you use the plain Table, make a mark upon your paper, representing your Stati∣on, and laying your Ruler to this Point, direct

your Sights to the several Corners of the Field, where you have caused Marks to be set up, and draw Lines by the side of the Ruler upon the paper to the point representing your station, then measure the distance of every of these Marks from your Instrument, and by your Scale set those distances upon the Lines drawn upon the paper, making small marks at the end of every such distance, Lines drawn from Point to Point, shall give you upon your paper, the Plot of the Field, by which Plot so taken the content of the Field may easily be computed.

Example. Let Fig. 31. represent a Field whose Plot is required; your Table being placed with a sheet of paper thereupon, make a Mark about the middle of your Table, as at A. apply your Ruler from this Mark to B and draw the Line AB, then with your Chain measure the distance there∣of which suppose to be 11 Chains 36 Links, then take 11 Chains 36 Links from your Scale, and set that distance from A to B, and at B make a mark.

Then directing the Sights to C, draw a Line by the side of your Ruler as before, and measure the distance AC, which suppose to be 7 Chains and 44 Links, this distance must be taken from your Scale, and set from A to C upon your paper.

And in this manner you must direct your Sights from Mark to Mark, until you have drawn the Lines and set down the distances, between all the Angles in the Field and your station, which being done, you must draw the Lines from one Point to another, till you conclude where you first began, so will those Lines BC. CD. DE. FG. and GB, give you the exact Figure of the Field.

4. To do this by the Theodolite, in stead of

drawing Lines upon your paper in the Field, you must have a little Book, in which the Pages must be divided into five Columns, in the first Column whereof you must set several Letters to signifie the several Angles in the Field, from which Lines are to be drawn to your place of standing, in the second and third Columns the degrees and parts taken by your Instrument, and the fourth and fifth, to set down your distances Chains and Links, this being in readiness, and have placed your Instrument direct your Sights to the first mark at B, and observe how many De∣grees are comprehended between the Diameter of your Instrument, and the Ruler, and set them in the second and third Columns of your Book against the Letter B, which stands for your first Mark, then measure the distance AB as before, and set that down, in the fourth and fifth Co∣lumns, and so proceed from Mark to Mark, until you have taken all the Angles and Distances in the Field, which suppose to be, as they are ex∣pressed in the following Table.

	Degr.	Part	Chains	Links
B	39	75	11	56
C	40	75	7	44
D	96	00	7	48
E	43	25	8	92
F	80	00	6	08
G	59	25	9	73

5. Having thus taken the Angles and Distances in the Field, to protract the same on Paper or

Parchment, cannot be difficult; for if you draw a Line at pleasure as EB representing the Dia∣meter of your Instrument about the middle there∣of, as at A, mark a Mark, and opening your Com∣passes to 60 Degrees in your Line of Chords, up∣on A as a Center describe a Circle, then lay your Field book before you seeing that your first Ob∣servattion cut no Degrees, there are no Degrees to be marked out in the Circle, but the Degrees at C are 40.75 which being taken from your Line of Chords, you must set them from H to I, and draw the Line AI. the Degrees at D are 96 which must in like manner be set from I to K, and so the rest in order.

This done observe by your Field-book the length of every Line, as the Line AB at your first Observation was 11 Chains and 36 Links, which being by your Scale set from A will give the Point B in the Paper, the second distance be∣ing set upon AI will give the Point C, and so proceeding with the rest, you will have the Points BCDEF and G, by which draw the Lines BC. CD. DE; EF. FG and GB, and so at last you have the Figure of the Field upon your Paper, as was required.

And what is here done at one station, may be done at two or more, by measuring one or two distances from your first station, taking at every station, the Degrees and distances to as many An∣gles, as are visible at each station.

And as for taking the the Plot of a-Field by In∣tersection of Lines, he that doth but consider how the distances of several Objects may be taken at two stations, will be able to do the other also, and therefore I think it needless, to make any il∣lustration by example.

CHAP. XII.

How to take the Plot of a Wood, Park or other Cham∣pion Plain, by going round the same, and making Observation at every Angle.

BY these Directions which have been already given, may the Plot of any Field or Fields be taken, when the Angles may be seen alone or more stations within the Field, which though it is the case of some Grounds, it is not the case of all; now where observation of the Angles cannot be observed within, they must be observed without, and although this may be done by the plain Ta∣ble, yet as I judge it may be more conveniently done by the Theodolite, in these cases thereof I chiefly commend that Instrument, I know some use a Mariners Compass, but the working with a Needle is not only troublesom, but many times uncertain, yet if a Needle be joyned with the Theodolite the joynt Observations of the Angles may serve to confirm one another.

2. Suppose the Fig. 32. to be a large Wood whose Plot you desire to take; Having placed your Instrument at the Angle A, lay your Ruler on the Diameter thereof, turning the whole Instru∣ment till through the Sights you espy the Angle at K, then fasten it there, and turn your Ruler up∣on the Center, till through the Sights you espy your second Mark at B, the Degrees cut by the Ruler do give the quantity of that Angle BAK, suppose 125 Degrees, and the Line AB 6 Chains, 45 Links, which you must note in your Field-book, as was shewed before.

3. Then remove your Instrument to B, and laying your Ruler upon the Diameter thereof, turn it about, till through the Sights you can espy your third mark at C, and there fasten your In∣strument, then turn the Ruler backward till through the Sights you see the Angle at A, the Degrees cut by the Ruler being 106.25 the quan∣tity of the Angle ABC, and the Line BC contain∣ing 8 Chains and 30 Links, which note in your Field-book, as before.

4. Remove your Instrument unto C, and laying the Ruler on the Diameter thereof, turn the In∣strument about till through the Sights you see the Angle at D, and fixing of it there, turn the Ruler upon the Center till you see your last station at B, and observe the Degrees cut thereby, which sup∣pose to be 134 Degrees, and the Line CD 6 Chains 65 Links, which must be entered into your Field-book also, and because the Angle BCD is an in∣ward Angle, note it with the Mark 〈☐〉〈☐〉 for your better remembrance.

5. Remove your Instrument unto D, and Iaying the Ruler on the Diameter, turn the Instrument a∣bout, till through the Sights, you see the Angle at E, and there fixing your Instrument, turn your Ruler backward till you espy the Mark at C, where the Degrees cut are, suppose 68.0 and the Line DE 8 Chains and 23 Links.

6. Remove your Instrument unto E, and laying the Ruler on the Diameter, turn the Instrument about, till through the Sights you see the Angle at F, and there fix it, then turn the Ruler back∣ward till you see the Angle at D, where the De∣grees cut by the Ruler suppose to be 125 and the Line EF 7 Chains and 45 Links.

7. Remove your Instrument unto F, and laying your Ruler upon the Diameter, turn the Instru∣ment about, till through the Sights, you see the Angle at G, where fix the same, and turn the Ruler backward till you see the Angle at E, where the Degrees cut by the Ruler are 70, and the Line FG 4 Chains 15 Links, which must be set down with this 〈☐〉〈☐〉 or the like Mark at the Angle.

8. Remove your Instrument unto G, and lay∣ing your Ruler upon the Diameter, turn the In∣strument about, till through the Sights you see the Angle at H, where fix the same, and turn the Ru∣ler backward till you see the Angle at F, where the Degrees cut by the Ruler are 65.25, and the Line GH 5 Chains 50 Links.

9. Remove your Instrument in like manner to H and K, and take thereby the Angles and Di∣stances as before, and having thus made observa∣tion at every Angle in the Field, set them down in your Field-book, as was before directed, the which in our present Example will be as follow∣eth.

A	151.00	6.45
B	106.25	8.30
C 〈☐〉〈☐〉	134.00	6.65
D	68.00	8.23
E	125.00	7.45
F 〈☐〉〈☐〉	70.25	4.15
G	65.25	5.50
H	130.00	6.50
K	140.00	11.00

The taking of the inward Angles BCD and EFG was more for Conformity sake than any

necessity, you might have removed your Instru∣ment from B to D, from E to G, the Length of the Lines BC. CD. EF and G, would have given by protraction the Plot of the Field without taking these Angles by observation; many other com∣pendious ways of working there are, which I shall leave to the discretion of the Ingenious Pra∣ctitioner.

10. The Angles and Sides of the Field being thus taken, to lay down the same upon Paper, Parchment, another Instrument called a Protractor is convenient, the which is so well known to In∣strument-makers, that I shall not need here to describe it, the chief use is to lay down Angles, and is much more ready for that purpose than a Line of Chords, though in effect it be the same.

11. Having then this Instrument in a readiness draw upon your Paper or Parchment upon which you mean to lay down the Plot of that Field, a Line at pleasure as AB. Then place the Center of your Protractor upon the Point A, and be∣cause the Angle of your first observation at A was 115 Degrees 00 Parts, turn your Protractor a∣bout till the Line AK lie directly under the 115 Degree; and then at the beginning of your Pro∣tractor make a Mark, ând draw the Line AB, set∣ting off 6 Chains 45 Links from A to B.

12. Then lay the Center of your Protractor upon the Point B, and here turn your Protractor about, till the line AB lie under 106 Degrees 25 Parts, and draw the Line BC, setting off the Distance 8 Chains, 30 Links from B to C.

13. Then lay the Center of your Protractor upon the Point C, and turn the same about till the Line BC lie under 134 Degrees, but remember

to make it an inward Angle, as it is marked in your Field-Book, and there make a Mark, and draw the Line CD, setting off 6 Chains, 65 Links from C to D.

And thus must you do with the rest of the Sides and Angles, till you come to protract your last Angle at H, which being laid down accord∣ing to the former Directions the Line HK will cut the Line AK making AK 11 Chains and HK 6 Chains, 50 Links. This work may be also per∣formed by protracting your last observation first; for having drawn the Line AK, you may lay the Center of your Protractor upon the Point K, and the Diameter upon the Line AK; and because your Angle at K by observation was 140 Degrees, you must make a Mark by the Side of your Protractor at 140 Degrees; and draw the Line KH, setting off 6 Chains, 50 Links from K to H. And thus proceeding with the rest of the Lines and Angles, you shall find the Plot of your Field at last to close at A, as before it did at K.

CHAP. XIII.

The Plot of the Field being taken by any Instrument, how to compute the Content thereof in Acres, Roods, and Perches.

THe measuring of many sided plain Figures hath been already shewed in the 13 Propositi∣on of the 8 Chapter, which being but well conside∣red, to compute the Content of a Field cannot be difficult; It must be remembred indeed that 40 square Pearches do make an Acre.

2. Now then if the Plot be taken by a four Pole Chain divided into 100 Links, as 16 square Poles are the tenth part of an Acre; so 10.000 square Links of such a Chain are equal to 16 square Pole, or Perches; and by consequence 100.000 square Links are equal to an Acre, or the square Pearches.

3. Having then converted your Plot into Tri∣angles, you must cast up the Content of each Triangle as hath been shewed, and then add the several Contents into one Sum, and from the aggre∣gate cut off five Figures towards the right hand; the remainer of the Figures towards the left hand are Acres, and the five Figures so cut off towards the right hand are parts of an Acre, which being multiplied by four, if you cut off five Figures from the Product, the Figures remaining towards the left hand are Roods, and the five Figures cut off are the parts of a Rood, which being multi∣plied by forty, if you cut off five Figures from the Product, the Figures remaining towards the left hand are Perches, and the Figures cut off are the Parts of a Pearch.

Example. Let 258.94726 be the Sum of seve∣ral Triangles, or the Content of a Field ready cast up, the three Figures towards the left hand 258 are the Acres, and the other Figures towards the right hand 94726 are the Decimal Parts of an Acre, which being multiplied by 4, the Pro∣duct is 3.78904, that is three Roods and 78904 Decimal Parts of a Rood, which being multiplied by 40, the product is 31.56160, that is 31 Perch∣es and 56160 Decimal Parts of a Perch; and therefore in such a Field there are Acres 258, Roods 3, Pearches 31, and 56160 Decimal Parts of a Perch.

CHAP. XIV.

How to take the Plot of Mountainous and uneven Grounds, and how to find the Content.

VVHen you are to take the Plot of any Mountainous or uneven piece of Ground, such as is that in Figure 33, you must first place your Instrument at A, and direct your Sights to B, measuring the Line AB, observing the Angle GAB, as was shewed before, and so proceed from B to C, and because there is an as∣cent from C to D, you must measure the true length thereof with your Chain, and set that down in your Book, but your Plot must he drawn according to the length of the Horizontal Line, which must be taken by computing the Base of a right angled Plain Triangle, as hath been shewed before, and so proceed from Angle to Angle until you have gone round the Field, and having drawn the Figure thereof upon your Paper, reduce into Triangles and Trapezias, as ABC. CDE. ACEF and AFG. then from the Angles B. C. D. F and G; let fall the Perpendiculars, BK. CN. DL. FM. and GH. This done you must measure the Field again from Angle to Angle, setting down the Distance taken in a straight Line over Hill and Dale, and so likewise the several Perpendiculars, which will be much longer than the streight Lines measured on your Scale, and by these Lines thus measured with your Chain cast up the Content; which will be much more than the Horizontal Content of that Field according to the Plot, but if it should be otherwise plotted than by the Ho∣rizontal

Lines, the Figure thereof could not be contained within its proper limits, but being laid down among other Grounds, would force some of them out of their places, and therefore such Fields as these must be shadowed off with Hills, if it be but to shew that the Content thereof is computed according to the true length of the Lines from Corner to Corner, and not according to their Distance measured by Scale in the Plot.

CHAP. XV.

How to reduce Statute Measure into Customary, and the contrary.

VVHereas an Acre of Ground by Statute Measure is to contain 160 square Perches, measured by the Pole or Perch of sixteen foot and a half: In many places of this Nation, the Pole or Perch doth by custom contain 18 foot, in some 20. 24. 28 Foot; it will be therefore re∣quired to give the Content of a Field according to such several quantities of the Pole or Perch.

2. To do this you must consider how many square Feet there is in a Pole according to these several Quantities.

• In 16.5 to the Pole, there are 272.25 sq. feet.
• In 18 to the Pole there are 324 square feet.
• In 20 to the Pole there are 400 square feet.
• In 24 to the Pole there are 576 square feet.
• In 28 to the Pole there are 784 square feet.

Now then if it were desired to reduce 7 Acres, 3 Roods, 27 Perches, according to Statute Measure, into Perches of 18 Foot to the Perch; first re∣duce

your given quantity, 7 Acres. 3 Rods, 27 Poles into Perches, and they make 1267 Perches.

Then say, as 324. to 272. 25. so is 1267 to 1065. 6. that is 1065 Perches, and 6 tenths of a Perch. But to reduce customary Measure into sta∣tute measure, say as 272. 25. is to 324 so is 1267 Perches in customary measure, to 1507. 8 that is 1507 Perches and 8 tenths of a Perch in statute measure, the like may be done, with the custo∣mary measures of 20.24 and 28 or any other mea∣sure that shall be propounded.

CHAP. XVI.

Of the Measuring of solid Bodies.

HAving shewed how the content of all plains may be computed, we are now come to the measuring of solid Bodies, as Prisms, Pyramids and Spheres, the which shall be explained in the Propositions following.

Proposition. I.

The base of a Prism or Cylinder being given, to find the solid content.

The base of a Prism is either Triangular, as the Pentahedron; Quadrangular, as the Hexahedron, or Multangular, or the Polyhedron Prism, all which must be computed as hath been shewed, which done if you multiply the base given by the altitude, the product shall be the solid content required.

Example. In an Hexahedron Prism, whose base

is quadrangular, one side of the Base being 65 foot and the other 43, the Superficies or Base will be 27. 95. Which being multiplyed by the Alti∣tude, suppose 12. 5. the product. 359. 375. is the so∣lid content required.

In like manner the Base of a Cylinder being 45. 6. and the altitude 15. 4. the content will be 702. 24.

And in this manner may Timber be measured whether round or squared, be the sides of the squa∣red Timber equal or unequal.

Example. Let the Diameter of a round piece of Timber be 2. 75 foot. Then, As 1 it to 785397. so is the square of the Diameter 2. 75. to 5.9395 the Superficial content of that Circle.

Or if the circumference had been given 8. 64. then, As 1 is to 079578, so is the square of 8. 64. to 5.9404 the superficial content.

Now then if you multiply this Base 5. 94. by the length, suppose 21 foot, the content will be 124. 74.

If the side of a piece of Timber perfectly square be 1.15 this side being multiplyed by it self, the product will be 1.3225 the superficial content, or content of the Base, which being multiplyed by 21 the length, the content will be 27. 7745.

Or if a piece of Timber were in breadth 1. 15. in depth 1.5 the content of the Base would be 1.725 which being multiplied by 21 the length, the content will be. 36. 225.

Proposition. II.

The Base and Altitude of a Pyramid or Cone being given, to find the solid content.

Multiply the Altitude by a third part of the Base, or the whole Base by a third part of the Al∣titude, the Product shall be the solid content re∣quired.

Example. In a Pyramid having a Quadrangu∣lar Base as in Fig. 22. The side CF 17. CD 9. 5. the Product is the Base CDEF. 161. 5, which being multiplyed by 10.5 the third of the Alti∣tude AB 31.5 the Product is 1695.75 the con∣tent. Or the third of the Base. viz. 53. & 3 being multiplied by the whole Altitude AB 31.5 the Product will be the content as before.

2. Example. In Fig. 21. Let there be given the Diameter of the Cone AB 3. 5. The Base will be 96. 25. whose Altitude let be CD 16.92 the third part thereof is 5.64 & 96.25 being multipli∣ed by 5.64, the Product 542.85 is the solid con∣tent required.

Proposition. III.

The Axis of a Sphere being given, to find the so∣lid content.

If you multiply the Cube of the Axis given by 523598 the solid content of a Sphere whose Ax∣is is an unite, the Product shall be the solid content required.

Example. Let the Axis given be ••, the Cube thereof is 27, by which if you multiply. 523598, the Product 14.137166 is the solid content re∣quired.

Proposition. IV.

The Basis and Altitude of the Frustum of a Pyra∣mid or Cone being given, to find the content.

If the aggregate of both the Bases of the Frust∣an and the mean proportional betwe••n them, shall be multiplied by the third part of the Altitude, the Product shall be the solid content of the Frustum.

Example. In Fig. 22. Let CDEF represent the greater Base of a Pyramid, whose superfici∣al content let be 1. 92, and let the lesser Base be HGLKO. 85 the mean proportional between them is. 1. 2775 and the aggregate of these three numbers is. 4. 0475. Let the given Altitude be 15. the third part thereof is. 5 by which if you mul∣tiply 4.0475 the Product 20. 2375 is the content of the Frustum Pyramid.

And to find the content of the Frustum Cone. I say.

As. 1. ro 78539. so 20.23 to 15. 884397, the content of the Cone required.

But if the Bases of the Frustum Pyramid shall be square, you may find the content in this man∣ner.

Multiply each Diameter by it self and by one another, and the aggregate of these Products, by the third part of the altitude, the last Product shall be the content of the Frustum Pyramid.

Example. Let the Diameter of the greater Base be 144, the Diameter of the lesser Base 108, and the altitude 60.

The Square of 144 is

20736

The Square of 108 is

11664

The Product of 1444108 is

15552

The Sum of these 3 Products is

47952

Which being multiplyed by 20 the third part of the Altitude, the Product 959040 is the con∣tent of the Frustum Pyramid.

And this content being multiplied by .785 39 the content of the Frustum Cone will be .753 .228.

Another way.

Find the content of the whole Pyramid of the greater and lesser Diameter, the lesser content de∣ducted from the greater, the remain shall be the content of the Frustum. To find the content of the whole Pyramid, you must first find their se∣veral Altitudes in this manner.

As the difference between the Diameters,

Is to the lesser Diameter.

So is the Altitude given, to the Altitude cut off.

Example. The difference between the former Diameter. 144. and 108 is 36, the Altitude 60. now then As 36. 108∷60. 108. the altitude cut off.

Now then if you mnltiply the lesser Base 1 1664 by 60 the third part of 180 the Product 699840 is the content of that Pyramid.

And adding 60 to 180 the Altitude of the great∣er Pyramid is 240, the third part whereof is 80, by which if you multiply the greater Base before found, 70736, the Product is the content of the

greater Pyramid. 1658880, from which if you deduct the lesser 699840 the remainer 959040 is the content of the Frustum Pyramid as before.

And upon these grounds may the content of Taper Timber, whether round or square, and of Brewers Tuns, whether Circular or Elliptical, be computed, as by the following Propositions shall be explained.

Proposition. V.

The breadth and depth of a Taper piece of Squared Timber, both ends being given together with the length, to find the content.

Let the given Dimensions.

• At the Bottom be A. 5.75 and B 2.34
• At the Top. C. 2.16 and D. 1.83.
• And let the given length be 24 Foot.

According to the last Proposition, find the A∣rea or Superficial content of the Tree at both ends thus.

Multiply the breadth	3.75	0.574031
By the depth	2.34	0.369215
The Product	8.7750	0.943246
2. Multiply the breadth	2.16	0.334453
By the depth	1.82	0.262451
The Product is	3.9528	0.596904

3. Multiply the 1. Content.	87750	0.943246
by the second content.	3.9528	0.596904
And find the square root	5.8986	1.540150
		0.770075

The Sum of these 18.6264 being multiplyed by 8 one third of the length, the content will be found to be 149. 0112. Thus by the Table of Logarithms the mean proportional between the two Bases is easily found, and without extracting the square Root, may by natural Arithmetick be found thus.

A 4 2/2 CX A half C multiplyed by B: And C more half A multiplyed by D being added toge∣ther and multiplyed by 30, the length shall give the content. Example.

A. 3.75	C 2.16
1/2 C. 1.08	1/2 A 1.875
Sum 4.83	Sum. 4.035
B- 2.34	D. 1.83
1932	12105
1449	32280
966	4035
11.3022	7.38405
	11.30220
The sum of the Products	18.68625

Being multiplyed by 8 the third of the length, the content will be. 149. 49000. The like may be done for any other.

Proposition VI.

The Diameters of a piece of Timber being given at the Top and and Bottom, together with the length, to find the content.

The Proposition may be resolved either by the Squares of the Diameters, or by the Areas of the Circles answering to the Diameters given, for which purpose I have here annexed not only a Ta∣ble of the Squares of all numbers under a thou∣sand, but a Table sharing the third part of the Areas of Circles in full measure, to any Diame∣ter given under 3 foot.

And therefore putting S = The Sum of the Tabular numbers answering to the Diameters at each end.

X = The difference between these Diameters.

L = the length of the Timber, C = The content.

Then 1 ½ S = ½ - XX. + L. = C.

If you work by the Table of the squares of Numbers. you must multiply the less side of the Equation, by 0.26179 the third part of 0.78539 the Product being multiplyed by the length, will give the content.

But if you work by the Table of the third parts of the Areas of Circles in full measure, the ta∣bular Numbers being multiplyed by the length will give the content. Only instead of the square of the difference of the Diameter, you must take half the Tabular number answering to that Dif∣ference, and you shall have the content as be∣fore. Example.

Let the greatest Diameter by 2.75, and the less 1. 93.

Their difference is 0.83

The square of 2.75 is

7.5625

The square of 1.93 is

3.7249.

The Sum of the Squares

11.2874

The half Sum

5.6437

The Sum of them is

16.9311

Half the square of 0.82 deduct.

0.3362

The Difference is

16.5949

Which being multiplyed by

26179

 

1493541

1161643

165949

995694

331898

 

1493541

1161643

165949

995694

331898

 

1493541

1161643

165949

995694

331898

 

1493541

1161643

165949

995694

331898

 

1493541

1161643

165949

995694

331898

The Product will be.

4.344378871

Or by the Table of Areas.

The Area of 2.75 is

1.979857

The Area of 1.93 is

0.975176

The Sum

2.955033

The half Sum

1.477516

The Sum of them

4.432549

Half the Area of 0.82 deduct

0.088016

The former Product

4.344533

Which being multiplyed by

24

 

17378132

8689066

 

17378132

8689066

The content is

104268792

But because that in measuring of round Tim∣ber the circumference is usually given and not the Diameter, I have added another Table by which the circumference being given, the Diame∣ter may be found.

Example. Let the circumference of a piece of Timber be 8325220 looking this Number in the second column of that Table, I find the next less to be 8.168140 and thence proceeding in a streight Line, I find that in the seventh Column the Num∣ber given, and the Diameter answering thereun∣to to be 2. 65. and thus may any other Diameter be found not exceeding the three foot. The Proportion by which the Table was made, is thus. As 1. to 3.14159 so is the Diameter given, to the circumference required.

[illustration]

Or the Circumference being given, to find the Diameter, say: As. 1. to 0.3183, so is the Circum∣ference given to the Diameter required.

And although by these two Tables all round Timber may be easily measured, yet it being more usual to take the Circumference of a Tree, then the Diameter, I have here added a third Table, shewing the third part of the Areas of Circles answering to any circumference under 10 foot, and that in Natural and Artificial numbers, the use of which Table shall be explained in the Proposition following.

Proposition. VII.

The Circumference of a piece of round Timber at both ends, with the length being given, to find the con∣tent.

The Circumference of a Circle being given, the Area thereof may be found as hath been shewed, in the 7 Chapter, Proposition 4. and by the first Pro∣position of this; and to find the third part of the Area, which is more convenient for our purpose I took a third part of the number given by which to find the whole, that is a third part of 07957747 that is 0.02652582 and having by the multiplying this number by the square of the Circumference computed three or four of the first numbers, the rest were found by the first and second differences.

The Artificial numbers were computed by ad∣ding the Logarithms of the Squares of the cir∣cumference, to 8.42966891 the Logarithm of 0. 02652582.

And by these Natural and Artificial numbers

the content of round Timber may be found two ways

By the Natural numbers in the same manner as the content was computed, the Diameters being given, and by the Natural and Artificial numbers both, by finding a mean proportional between the two Areas at the top and bottom of the Tree, as by Example shall be explained.

Let the given Diamensions, or Circumferences be At the Bottom 9.95 Their difference is 6.20 At the Top 3.75

The tabular Numbers.

	Natural	Artificial.
Answering to 9.95	2.626162	0.418931
And to 3.75	0.373019	9.571731

The Sum of the Logarith.

9.990662

The half Sum or Logarith. 989300

9.995331

The Sum of the Number is

3.988481

The Sum of the Natural Numbers is

2.9••9181

The half Sum

••.499190

The Sum of them

4.498771

Half the number answer. to. 6. 20 is

0.509826

The remainer is

3.988945

Which being multiplyed by the length 24, the content will be 95. 73468.

Mr. Darling in his Carpenters Rule made easie, doth propound a shorter way, but not so exact, which is by the Circumference given in the middle of the piece to find the side of the Square, name∣ly

by multiplying the Circumference given by 28209, or 2821. which side of the Square being computed in Inches, and lookt in his Table of Timber measure, doth give the content of the Tree not exceeding 31 foot in length, the which way of measuring may be as easily performed by this Table. Example.

The circumference at the top and bottom of the Tree being given 9.95 and 3.75 the Sum is 13.70 The half thereof is the mean circumfer. 6.85 Which sought in the Table, the Numbers are.

The Natural number is 1.244657, which being multiplyed by 3 the Product is 3.733971, which multiplyed by the length 24, the content is 89. 615304.

The Artificial number is

0.095049

The Logarithm of 24 is

1.380211

The Absolute Number 29.871

1.475260

Which multiplyed by 3, the Product is

89613

Proposition. VIII.

The Diameters of a Brewers Tun at top and bottom being given with the height thereof, to find the con∣tent.

In Fig. 29. Let the given Diameter.

At the top be AC 136 BD 128

At the bottom. KG 152 HF 144 Altit. 51 Inches.

The which by the 5 Proposition of this Chap. may thus be computed. AC 139 + ½ KG 76 = 212 × BD 128 the Product is 27136.

And KG 1524 ½ AC 68 = 220 × HF 144 the Product is 31680. the Sum of these 2 Products is 58816 which being multiplyed by onethird of 51, that is by 17, and that Product multiplyed by 26179 the third of 78539 will give the content.

The Logarithm of 58816. is

54.76949

The Logarithm of 17 is

1.230449

The Product

1.999944

The Logarithm of. 26179

9.417968

The content is. 261765

5.417912

Thus the content of a Tun may be found in In∣ches, which being divided 282 the number of In∣ches in an Ale Gallon, the quotient will be the con∣tent in Gallons.

Or thus; divide the former. 26179 by 282 the quotient will be 00092836. by which the content may be found in Ale Gallons in this manner.

The former Product

5.999944

The Logarithm of 0.00092836

6.967719

The content in Gallons 928.24

2.967663

Proposition. IX.

The Diameters of a close Cask, at head and bung with the length given, to find the content.

In the resolving of this Proposition, we are to consider the several forms of Casks, as will as the kind of the Liquor, with which it is filled, for one and the same Rule will not find the con∣tent in all Cask.

And a Coopers Cask is commonly taken, ei∣ther for the middle Frustum of a Spheroid, the middle Frustum of a Parabolical Spindle, the mid∣dle Frustum of two Parabolick Conoids, or for the middle Frustum of two Cones abutting upon one common Base.

And the content of these several Casks may be found either by equating the Diameters, or by e∣quating the Circles. for the one, a Table of Squares is necessary, and a Table shewing the third part of the Areas of a Circle to all Diameters. The making of the Table of Squares, every one knows, to be nothing else but the Product of a Number multiplyed, by it self, thus the Square of 3 is 9. the Square of 8 is 64 and so of the rest.

And the Area of a Circle to any given Diame∣ter may be found, as hath been shewed, in Chap. 7 Proposition 2. But here the Area of a Circle in In∣ches, will not suffice, it will be more fit for use, if the third part of the Area be found in Ale and Wine Gallons both, the which may indeed be done by dividing the whole Area in Inches by 3 and the quotient by 282 to make the Table for Ale-measure, and by 231 to make the Table for Wine-measure; but yet these Tables (as I think) may be more readily made in this manner.

The Square of any Diameter in Inches, being divided by 3.81972 will give the Area of the Cir∣cle in Inches: And this Division being multiply∣ed by 282 will give you 1077.161 for a common Division, by which to find the Area in Ale-Gal∣lons, or being multiplyed by 231 the Product, 882.355 will be a commou Division by which to find the Area in Wine-Gallons.

But because it is easier to multiply then divide:

If you multiply the several Squares by 26178 the third part of 78539 the Product will give the Area in Inches, or if you divide. 26179 by 282 the quotient will be. 00092886 for a common Mul∣tiplicator, by which to find the Area in Ale-Gal∣lons, or being divided by 231 the quotient will be 0011333 a common Multiplicator, by which to find the content in Wine-Gallons. An Exam∣ple or two will be sufficient for illustration. Let the Diameter given be 32 Inches, the Square thereof 1024 being divided by 3.81970 the quo∣tient is 268.083, and the same Square 1024 be∣ing multiplyed by 261799, the Product will be 268. 082.

Again if you divide 1024 by 1077.161 the quotient will be 9508, or being multiplied by 00092836, the Product will be 9508.

Lastly if you divide 1024 by 882.755, the quo∣tient will be 1.1605, or being multiplied by 00113333 the Product is 1.1605,

And in this manner may the Tables be made for Wine and Beer-measure, but the second differen∣ces in these Numbers being equal, three or four Numbers in each Table being thus computed, the rest may be found by Addition only.

Thus the Squares of 1. 2. 3. and 4 Inches are. 1. 4. 9 and 16 by which if you multiply 00113333, the several Products will be third part of the Area, of the Circles answering to those Diameters in Wine-Gallons. Or 00092836 being multiplied by those Squares, the several Products, will be the third part of the Areas of the Circles answering to those Diameters in Ale-Gallons; the which with their first and second differences are as fol∣loweth.

The Products or Areas in Wine-Gallons:

1.	00113333
2.	00453332	33999	226666
3.	01019997	566665	226666
4.	01813328	796331

The Products in Ale-Gallons.

1.	00092836
2.	00371344	278508	185672
3.	00835524	464180	185672
4.	01485376	649852

And by the continual addition of the second differences to the first, and the first differences to the products before found, the Table may be con∣tinued as far as you please.

The construction of the Tables being thus shewed: We will now shew their use in finding the content of any Cask.

Let S = the Sum of the Tabular Numbers an∣swering to the Diameters at the Head and Bung. D = their difference X = the difference of the Diameters themselves. L = the length of the Vessel, and C = the content thereof.

1. If a Cask be taken for the middle Frustum of a Spheroid, intercepted between two Planes parallel, cutting the Axis at right Angles: Then 1 ½ S + ½ D × L = C.

2. If a Cask be taken for the middle Frustum of a parabolical Spindle, intercepted between two planes parallel cutting the Axis at right Angles. Then 1 ½ S + ½ D × L = C.

3. If a Cask be taken for the middle Frustum of two Parabolick Conoids, abutting upon one common Base, intercepted between two Planes parallel, cutting the Axis at right Angle: Then 1 ½ S: × L = C.

4. If a Cask be taken for the middle Frustum of two Cones, abutting upon one common bafe, intercepted between two Planes parallel cutting the Axis at Right Angles. Then 1 ½ S—⅓ XX. × L = C.

In all these four Equations, if you work by the Table of Squares of numbers, you must multi∣ply the less side of the Equation by 262, if you would have the content in Cubical Inches; by 001133 if you would have the content in Wine-Gallons; and by 000928, if you would have the content in Ale-Gallons.

But if you work by the Tables of the third parts of the Areas Circle, the Tabular Numbers being multiplyed by the length only will give the content required, only in the fourth Equation instead of half the Square of the Difference of the Diameters, take half the Tabular Number answering to that difference, and you shall have the content required; as by the following Exam∣ples will better appear, then by many words.

Examples in Wine-measure by the Table of the Squares of Numbers.

The Diameter of a Vessel

At the Bung being 32 Inches.

At the Head 22 Inches.

The difference of the Diameters 10 Inches.

And the length of the Vessel 44 Inches.

Spheroid.	Parabolick Spindle.
1024	1024
484	484
1508	1508
754	754
270	540
2532	23160
2532	23160
7596	69480
7596	69480
7596	69480
28695156	262472280
44	44
114780624	104988912
114780624	104988912
126.2586864	115.4878032

Parabolick Conoid	Cone.
1024	1024
484	484
1508	1508
754	754
	50
2262	2212
2262	2212
6786	6636
6786	6636
6786	6636
25635246	25068596
44	44
102540984	200274384
102540984	100274384
112.79508241	110.30182224

This which hath been done by the Table of Squares may be more easily performed, by the Table of the third part of the Areas of Circles, ready reduced to Wine-Gallons.

Spheroid	Parabolick Spindle.
1.16053	1.16053
0.54853	0.54853
1.70906	1.70906
85453	85453
30600	61200
2.86959	2.624790
44	44
1147836	1049916
1147836	10499160
126.26196	115.490760

Parabolick Conoid	Cone.
1.16053	1.16053
0.54853	0.54853
1.70906	1.70906
85453	85453
	56666
2.56359	2.506924
44	44
1025436	10027696
1025436	10027696
112.79796	110.304656

Examples in Ale-measure by the Table of the Squares of Numbers.

Spheroid.	Parabolick Spindle.
1024	1024
484	484
1508	1508
754	754
270	540
2532	2316.0
00092836	00092836
22758	20844
5064	4632
20256	18528
7596	6948
15192	138960
235660752	2.150081760
44	44
948623008	860032704
940643008	860032704
103.22673088	94.60359744

Parabolick Conoid	Cone.
1024	1024
484	484
1508	1508
754	754
	50
2262	2212
20358	19909
4524	4424
18096	17696
6786	6636
13527	13272
2.09995032	2.05423232
44	44
8.39980128	821692928
839980128	821692928
92.39781408	90.38622208

By the Areas of Circles.

Spheroid.	Parabolick Spindle.
0.95052	0.95052
0.44930	0.44930
1.39982	1.39982
.69991	69991
.25061	050122
2.35034	2.149852
34	44
940136	8599408
940136	8599408
103.41496	94.593488

Parabolick Conoid.	Cone.
0.95052	0.95052
0.44930	0.44930
1.39982	1.39982
69991	.69991
	46425
209973	2.053305
44	44
839892	8213220
839892	8213220
90.345420	90.345420

And here for the Singularity of the Example, I will set the Dimensions of a Cask lately made in Herefordshire, for that excellent Liquor of Red streak Cyder, the like whereof either for the largeness of the Cask, or incomparable goodness of that kind of Drink, is not to be found in all England, nay and perhaps not in the World.

The length of the Cask is 104 Inches.

The Diameter at the Bung 92 Inches.

And the Diameter at the Head 74 Inches.

The Numbers in the Table of Ale-Gallons an∣swering to these Dimensions are.

Spheroid	Parabolick Spindle.
Bung. 92 7.859639	7.859639
Head. 74 5.083699	5.083699
12.941338	12.941338
6.470669	6.470669
1.386770	.277394
20.798777	19.689401
104	104
83195108	78.757604
20798777	19689401
Con. 2163.072808	2047.697704

Parabolick Conoid.	Cone.
7.857639	7.857639
5.083699	5.083699
12.941338	12.941338
6.470669	6.470669
	0.150394
19.412007	19.261613
104	104
77648028	77046452
19412007	19261613
201. 8. 848728	2003.207752

And thus you have the content of this Cask by four several Ways of Gauging, but that which doth best agree with the true content, found by these that filled the same is the second way or that which takes a Cask to be the middle Frustum of a Parabolick Spindle, according to which the content is 2047 Gallons. That is allowing 64 Gallons to the Hogshead. 32 Hogsheads very near.

Proposition. X.

If a Cask be not full, to find the quantity of Li∣quor contained in it, the Axis being posited parallel to the Horizon.

To resolve this Proposition, there must be gi∣ven the whole content of the Cask, the Diame∣ter at the Bung, and the wet Portion thereof, then by help of the Table of Segments, whose Area is unity, and the Diameter divided into 10.000 equal parts, the content may thus be found.

As the whole Diameter, is to its wet Por∣tion.

So is the Diameter in the Table. 10.000 to its like Portion, which being sought in the Table of Segments, gives you a Segment, by which if you multiply the whole content of the Cask, the Product is the content of the Liquor remaining in the Cask.

But in the Table of Segments in this Book, you have the Area, to the equal parts of one half of the Diameter only, when the Cask therefore is more then half full, you must make use of the dry part of the Diameter instead of the wet, so shall you find what quantity of Liquor is wanting to fill up the Cask, which being deducted from the whole content of the Cask; the remainer is the quantity of Liquor yet remaining, an Exam∣ple in each will be sufficient, to explane the use of this Table.

1. Example, In a Wine Cask not half full, let the great Diameter be as before 32 Inches, the

content 126.25 Gallons, and let the wet part of the Diameter be 12 Inches, First I say.

As the whole Diameter 32. is to the wet part 12. so is 10.000 to 3750, which being sought in the Table, I find, the Area of that Segment to be. 342518 which being multiplyed by the whole content of the Cask 126.25, the Product is 43.24289750 and therefore there is remaining in the Cask 43 & 1/4 ferè.

2. Example. In the same Cask let the wet part of the Diameter be 18 Inches. I say.

As 32.18 :: 10000.5625 whose Complement to 10000 is 4375 which being sought in the Ta∣ble, I find the Area answering thereto to be 420630; now then I say.

As the whole Area of the Circle 1000000 is to the whole content of the Cask 126. 25.

So is the Area of the Segment sought. 420630, to the content 53.1044375 which is in this case the content of the Liquor that is wanting, this therefore being deducted from the content of the whole Cask, 136. 25. the part remaining in the Vessel is. 73. 1455625.

Thus may Casks be gauged in whole or in part, in which a Table of Squares is sometimes necessary, as being the Foundation, from whom the other Tables are deduced; such a Table therefore is here exhibited, for all Numbers un∣der 1000, by help whereof the Square of any Number under 10.000 may easily be found in this manner.

The Rectangle made of the Sum and Difference of any two Numbers, is equal to the Difference of the Squares of these Numbers.

Example, Let the given Numbers be 36 and 85

their Sum is 121, their difference 49, by which if you multiply 121, the Product will be 5929. The Square of 36 is 1296, and the Square of 85 is 7225, the difference between which Squares is 5929 as before.

And hence the Square of any Number under 10.000 may thus be found, the Squares of all Numbers under 1000 being given.

Example. Let the Square of 5715 be required. The Square of 571 by the Table is 326041, there∣fore the Square of 5710 is 32604100: the Sum of 5710 and 5715 is 11425, and the difference 5, by which if you multiple 11425, the Product is 52125 which being added unto 32604100 the Sum 32656325 is the Square of 5715. The like may be done for any other.