The Sector is a Mathematical Instrument, whose Use is to find the Proportion between Quantities of the same kind as between one Line and another, between one Superficies and another, between one Solid and another, &c.

This Instrument is made of two equal Rules, or Legs, of Silver, Brass, Ivory, or Wood, joined to each other by a Rivet, so worked, as to render it’s Motion regular and uniform. To do which, first make two Slits with a Saw, about an Inch deep, at one End of one of the Rules, in order to fit therein the Head-Pieces, which must, be well rivetted. Afterwards the Head must be rounded, by filing off the Superfluities, in such manner, that the Middle-Piece and Head-Pieces may be even with each other. Then to find the Center of the Rivet, set one Foot of your Compasses at the Bottom of the Middle-Piece, and mark with the other Foot four Sections in the middle of the Rivet, by opening the Middle-Piece of the Joint to four or more different Angles, and the Middle-Point of those Sections will be the Center of the Rivet, and, consequently, also the Center of the Sector. This being done, a Line must be drawn upon the Rule from the Center, near the inward Edge, by which Line the inward Edge of the Rule must be filed strait; the inward Edge of the other Rule being also made strait, and slit, to receive the Middle-Piece, you must cut away it’s Corner in an Arc, so as it may well fit the Joint, and then rivet, with three or four little Rivets, the Rule to the Middle-Piece; by which means the two Legs may easily open and shut, and keep at any Opening required. But Care must be taken that the Legs are filed very flat, and do not twist; Care must also be taken that the Sector be well centered, that is, that being entirely opened, both Inside and Outside, may make a right Line, and that the Legs be very equal in Length and Breadth; in a word, that it be very strait every way. Note, The Length and Breadth of the aforesaid Rules are not determinate, but they are commonly six Inches long, three quarters of an Inch broad, and about one quarter in Thickness.

There are commonly drawn upon the Faces of this Instrument six kind of Lines; viz. the Line of equal Parts, the Line of Planes, and the Line of Polygons on one Side; the Line of Chords, the Line of Solids, and the Line of Metals on the other.

There is generally placed, near the Edge of the Sector, on one Side, a divided Line, whose Use is to find the Bores of Cannons; and on the other Side, a Line shewing the Diameters and Weights of Iron-Bullets, from one Quarter to 64 Pounds, whose Construction and Uses we shall give, in speaking of the Instruments belonging to the Artillery.

Section I. Of the Line of Equal Parts.

This Line is so called, because it is divided into equal Parts, whose Number is commonly 200, when the Sector is six Inches long.

Having drawn upon one of the Faces of each Leg the equal Lines AB, AB, from the Center of the Joint A: First, divide them into two equal Parts, each of which will consequently be 100; then each of those Parts being again divided into two equal Parts, and each Part arising will be 50; then divide each of these last Parts into five others, and each Part produced will be 10, and finally dividing each of these new Parts into 2, and each of these last into five equal Parts, and by this means the Lines AB, AB, will be each divided into 200 equal Parts, every 5 of which must be distinguished by short Strokes, and every 10 numbered from the Center A to 200, at the other End.

Now because the two other Lines, drawn upon the same Faces of each Leg, must terminate in the Center A, the Extremity B of the Line of equal Parts must be drawn as near as possible the outward Edges of each Leg, that so there may be Space enough left to draw the Line of Planes in the middle of the Breadth of the said Legs, and the Line of Polygons near their inward Edges; but Care must be taken, in drawing of these Lines, that each one, and it’s Fellow, be equally distant from the interior Edges of each Leg, as may be seen in the Figure.

Section II. Of the Line of Planes.

This Line is so called, because it contains the homologous Sides of a certain Number of similar Plants, Multiples of a small one, beginning from the Center A; that is, whose Surfaces are double, triple, quadruple, &c. that small Plane, from Unity, according to the natural Order of Numbers, to 64, which is commonly the greatest Term of the Divisions, denoted upon the Line AC.

This Line may be divided two ways, both of which are founded upon Prop. 20. lib. 6. Eucl. which demonstrates, That similar Plane Figures are to each other, as the Squares of their homologous Sides. The first way of dividing this Line is by Numbers, and the second without Numbers, as follows:

Having drawn the Line AC, from the Center A, upon each Leg of the Sector; first divide it into eight equal Parts, the first of which; next to the Center A, which represents the Side of the least Plane, hath no need of being drawn. The second Division from the Center, which is double the first, is the Side of a similar Plane quadruple the least Plane, (whose Side is supposed one of the eight Parts the Line AC is divided into), because the Square of 2 is 4. The third Division, which is three times the first, is the Side of a similar Plane, nine times greater than the first, because the Square of 3 is 9. The fourth Division, which is four times the first, and consequently half of the whole Scale, is the Side of a similar Plane, sixteen times greater than the first, because the Square of 4 is 16. Lastly, The eighth Division which is eight times the first, is the Side of a similar Plane, sixty-four times greater, because the Square of 8 is 64.

There is something more to do to find the homologous Sides of Planes that are double, triple, quadruple, &c. of the first. For you must have a Scale divided into 1000 equal Parts (as that whose Construction we have already given in Book I.), whose Length must be equal to the Line AC; and because the Side of the least Plane is \(\frac{1}{8}\) of the Line AC, it will consequently be \(\frac{1}{8}\) of 1000, which is 125. Again, to have in Numbers the Side of a Plane double the least, the square Root of a Number twice the Square of 125 must be found. This Square is 15625, which doubled, is 31250, the square Root of which is about 177, the Side of a similar Plane double the least, whose Side is supposed to be 125. Moreover, to have the Side of a Plane three times the first, the square Root of a Number three times the Square of the first must be found. The Number is 46875, and it’s Root, which is about 216, is the Side of a similar Plane three times the least, and so of others; therefore by laying off from the Center A, upon the Line of Planes, 177 Parts of the aforesaid Scale, you will have the Length of the Side of a similar Plane double the least Plane. Again, laying off 216 Parts of the same Scale from the C enter A, the Length of the Side of a similar Plane will be had, which is three times the least Plane.

According to the aforesaid Directions, the following Table is calculated, that shews the Number of equal Parts which are contained in the homologous Sides of all the similar Planes that are double, triple, quadruple, &c. of a Plane whose Side is 125, to the Plane 64, that is, which contains it 64 times, and whose Side is 1000.

Each of the ten Divisions which the Scale of 1000 Parts contains, is 100; and each of the Subdivisions of the Line AB is 10: therefore if it is to be used for dividing any of the Lines of the Sector; as, for Example, the Line of Planes, take on the Scale a Line denoting the Hundreds, and the Excess above must be taken in the Space between the Points AB: As to denote the first Plane, to which the Number 125 answers, place your Compasses on the fifth Line of the Space marked 100, and open them to the Distance OP; in the same manner, if the Plane 50 is to be denoted, to which the Number 884 answers, for 800 take the 8th Space of the Scale, and for 84 take in the Space AB, the Interaction of the 8th Transversal, with the fourth Parallel, which will be the Distance NL.

The Line of Planes may otherwise be divided in the following manner, without Calculation, founded on Prop 47. lib. 1. Eucl. Make the right-angled Isosceles Triangle KMN, whose Side KM, or KN, let be equal to the Side of the least Plane, and then the Hypothenuse MN will be the Side of a similar Plane double to it; therefore having laid off with your Compasses the Distance MN, on the Side KL produced, from K to 2, the Length K2 will be the Side of a Plane double the least Plane. In like manner lay off the Distance M2, from K to 3, the Line K3 will be the Side of a Plane triple the first. Again, lay off the Distance M3, from K to 4, the Line K4 (twice KM) will be the Side of a Plane four times greater, that is, which will contain the least Plane four times; and so of others, as may be seen in the Figure.

Section III. Of the Line of Polygons.

This Line is so called, because it contains the homologous Sides of the first twelve regular Polygons inscribed in the same Circle, that is, from an equilateral Triangle to a Dodecagon.

The Side of the Triangle being the greatest of all, must be the whole Length of each of the Legs of the Sector; and because the Sides of the other regular Polygons, inscribed in the same Circle, still diminish as the Number of Sides increase, the Side of the Dodecagon is least, and consequently must be nighest the Center of the Sector.

Now supposing the Side of a Triangle to be a thousand Parts, the Length of the Sides of every of the other Polygons must be found; and because the Sides of regular Polygons, inscribed in the same Circle, are in the same Proportion as the Chords of the Angles of the Center of each of the Polygons, it is necessary to shew here how to find the said Angles.

To do which, divide 360 Deg. by the Number of the Sides of any Polygon, and the Quotient will give the Angle of the Center.

If for Example, the Angle of the Center of a Hexagon is required, divide 360 Deg. by 6, and the Quotient will be 60; which shews that the Angle of the Center of a Hexagon is 60 Deg. If likewise the Angle of the Center of a Pentagon be required, divide 360 Deg. by 5, the Number of Sides, and the Quotient will be 72; which shews that the Angle of the Center of a Pentagon is 72 Deg. and so of others.

The Angle of the Center being known, if it be substracted from 180 Degrees, the Remainder will be the Angle of the Polygon: As, for Example, the Angle of the Center of a Pentagon being 72 Degrees, the Angle of the Circumference will be 108 Degrees, and so of others, as may be seen in the following Table.

Regular Polygons.

Angles of the Center.

Angles of the Circumference.

Triangle

120 Deg.

60 Deg.

Square

90 Deg.

90 Deg.

Pentagon

72 Deg.

108 Deg.

Hexagon

60 Deg.

120 Deg.

Heptagon

51 Deg. 26 Min.

128 Deg.

Octagon

45 Deg.

135 Deg.

Nonagon

40 Deg.

140 Deg.

Decagon

36 Deg.

144 Deg.

Undecagon

32 Deg. 44 Min.

147 Deg. 16 Min.

Dodecagon

30 Deg.

150 Deg.

Now to find in Numbers the Sides of the regular Polygons inscribed in the same Circle: Having supposed that the Side of the equilateral Triangle is 1000 equal Parts, instead of the Chords of the Angles of the Center, take their Halves, which are the Sines of half the Angles at their Centers, and make the following Analogy.

For Example, to find the Side of the Square, say, As the Sine of 60 Degrees, half the Angle of the Center of the equilateral Triangle, is to the Side of the same Triangle, supposed 1000; So is the Sine of 45 Degrees half the Angle of the Center of the Square, to the Side of the same Square, which, by calculating, will be found 816.

And in this manner are the following Tables of Polygons constructed.

Polygon.

Sides.

Equal Parts.

The Side of an equilateral Triangle, denoted on the Sector by the Number

3

1000

Of the Square by the Number

4

816

Of the Pentagon by the Number

5

678

Of the Hexagon by the Number

6

577

Of the Heptagon by the Number

7

501

Of the Octagon by the Number

8

442

Of the Nonagon by the Number

9

395

Of the Decagon by the Number

10

357

Of the Undecagon by the Number

11

325

Of the Dodecagon by the Number

12

299

We have neglected the Fractions remaining after the Calculation in this Table, as in all others; as being but thousandth Parts, which are not considerable.

Those that will not denote an equilateral Triangle upon the Sector, because of the Facility of describing it, and which consequently begin at the Square, use the following Table, wherein the Side of the Square is supposed 1000 Parts.

Another Table of Polygons.

Parts.

Square

1000

Pentagon

831

Hexagon

707

Heptagon

613

Octagon

540

Nonagon

484

Decagon

437

Undecagon

398

Dodecagon

366

To make the Line of Polygons upon the Sector (the same Scale of 1000 equal Parts being used, as that for making the Line of Planes), you must lay off from the Center A, upon both the Lines AD, the Number of Parts expressed in the Table, that thereby the Numbers 3, 4, 5, &c. may be graved upon the Sector, signifying the Numbers of the Sides of the regular Polygons.

Section IV. Of the Line of Chords.

This Line is so named, because it contains the Chords of all the Degrees of a Semicircle, whose Diameter is the Length of that Line, which is denoted upon the other Surface of each Leg of the Sector, from the Point A, which is the Center of the Joint, to the End F of each Leg; so that the two Lines AF are exactly equal, and equidistant from the interior Edges of the Sector.

Note, The Line of Chords must be drawn directly under the Line of equal Parts, because of some Operations that require a Correspondence between those two Lines.

It is also proper for the Line of Solids to be drawn under the Line of Planes, and the Line of Metals under the Line of Polygons.

For the Division of the aforesaid Line AF, describe a Semicircle, whose Diameter let be equal to it, which divide into 180 Degrees; afterwards lay off the Lengths of the Chords of all those Degrees upon the Diameter of the Semicircle; then lay the Diameter of the Semicircle upon the Legs of the Sector, and mark upon them Points that represent the Degrees of the Semicircle, every fifth of which, distinguish by short Strokes, and every tenth by Numbers, beginning from the Point A, and going on to F.

The same Degrees may otherwise be denoted, upon the Line of Chords, by help of Numbers, in supposing the Semidiameter of a Circle, or the Chord of 180 Degrees, to be 1000 equal Parts; all of which Numbers may be found ready calculated in the common Tables of Sines: for in (lead of the Chords, there is no more to do but to take their halves, which are the Sines of half their Arcs. As for Example; instead of the Chord of 10 Degrees, the Sine of 5 must, be taken; and because the Calculation in Tables is made for a Radius of 100000 Parts, the two last Numbers must be taken away, as may be seen in the following Table, where the Chords of all the Degrees to 180 are denoted.

Note, This Division is made with a Scale of 1000 Parts.

Section V. Of the Line of Solids.

This Line is so called, because it contains the homologous Sides of a certain Number of similar Solids, Multiples of a lesser from Unity, according to the natural Order of Numbers, to 64, which is commonly the greatest of the Divisions of this Line, which is marked AH, next to the Line of Chords.

To make the Divisions upon it, the Scale of 1000 Parts must be used, and the Side of the 64th and created Solid must be supposed 1000 equal Parts; then because the Cube-Root of 64 is 4, and the Cube Root of 1 is 1, it follows that the Side of the 64th Solid is quadruple the Side of the first and least Solid, which consequently will be 250, because (per Prop. 33. lib. 11. Eucl.) similar Solids are to each other, as the Cubes of their homologous Sides.

The Number 500 (twice 250) is the Side of the eighth Solid, that is, of a Solid eight times as great as the first: because the Cube of 2, which is 8, is eight times the Cube of Unity.

Likewise the Number 750, which is three times 250, is the Side of the 27th Solid; because the Cube of 3, which is 27, is 27 times the Cube of Unity.

There are more Calculations required to find the Sides of Solids double, triple, quadruple, &c. the first, which cannot exactly be expressed in Numbers, because their Roots are incommensurable; nevertheless they may be sufficiently approached for Use, by the following Method.

For Example; To find the Number expressing the Side of a Solid, twice the first and least: it’s Side 250 must be cubed, which is 15625000; then this Number must be doubled, and the Cube-Root of it extracted, which will be almost 315, for the Side of a Solid double the first. To have the Side of a Solid triple the first, the said Cube must be tripled, and it’s Cube-Root, which is 360, will be the Side of a Solid triple the first; and so of others, as may be seen in the following Table.

The Sides of all these Solids being thus found in Numbers, they are denoted on the Line of Solids, by laying off from the Center A the Parts which they contain, taken upon the Scale.

Section VI. Of the Line of Metals.

This Line is so named, because it is used to find the Proportion between the six Metals, or which Solids may be made.

It is placed upon the Legs of the Sector, hard by the Line of Solids, and the Metals are figured thereon by the Characters, which have been appropriated to them by Chymists and Naturalists,

The Division of this Line is founded upon Experiments that have been made of the different Weights of equal Masses of each of these Metals, from whence their Proportions are calculated, as in the following Table.

A Table for the Line of Metals.

Gold

☉︎

730

Lead

♄

863

Silver

☽︎

895

Brass

♀

937

Iron

♂

974

Tin

♃

1000

That of all the six Metals which has the least Weight, which is Tin, is marked at the End of each Leg (as AG) at a Distance from the Center, equal to the Length of the Scale of 1000 Parts; and the other Metals nigher the said Center (each according to the Numbers which correspond with them), taken upon the same Scale.

Because most of the aforementioned Lines, marked on the Sector, are divided by means of the Scale of 1000 equal Parts, it is requisite that they be exactly equal between themselves and to the said Scale; therefore, because they all center in one Point (which is the Center of the Joint), they must all be terminated at the other End by an Arc, made upon the Surface of each of the Legs.

It is not always necessary to divide the Sector by the Methods we have given; for, to make them sooner, prepare a Ruler of the same Length, Breadth, and Thickness as the Sector, and draw upon it the same Lines we have already prescribed: then with a Beam-Compass transfer the same Divisions upon the Sector, having first drawn upon it the Lines to contain them.

Section VII. Containing the Proofs of the Six Lines commonly put upon the Sector.

The Proof of the Line of equal Parts.

The Division of this Line is so easy, that there is no need of any other Proof, but to examine, with your Compasses, whether the two correspondent Lines, drawn upon the Legs of the Sector, are very equal, and equally divided; which may be known by taking between your Compasses (whose Points let be very sharp) any Number at pleasure of those equal Parts, beginning any where: for if the Line of equal Parts be well divided, by carrying that same Opening of your Compasses on the said Line, the two Points will always contain between them the same Number of equal Parts upon either of the Legs, reckoning from the Center, or from any other Point of Division.

The Proof of the Line of Chords.

The Method before explained will not serve to know whether the Line of Chords be well divided, because the Divisions are not equal: the Chord of 10 Degrees, for Example, is greater than half that of 20; likewise the Chord of 20 Degrees is greater than the hall of that of 40 Degrees, and so on: so that the Divisions are greater towards the Center of the Sector, than towards the Ends of it’s Legs, as is manifest from the Nature of the Circle. But because we have given two Methods for dividing the Line of Chords, one by help of Numbers, and the other by means of the Chords of Arcs, one of these Methods will serve to prove the other.

But there is still another Method, which is this: Take at pleasure, on the Line of Chords, two Numbers equally distant from 120 Degrees; as for Example, 110 and 130, which are each 10 Degrees distant from it; the first in Defect, and the last in Excess: Then take in your Compasses the Distance of the two Numbers 110 and 130, which must be equal to the Chord of 10 Degrees, or to the Distance of the Point 10, upon the Line of Chords, from the Center of the Sector.

You will find, by the same Means, that the Distance between 100 and 140 Degrees, is equal to the Chord of 20 Degrees; as likewise the Distance between 90 and 150 is equal to the Chord of 30 Degrees, which is the Number by which 120 exceeds 90, and by which 150 exceeds 120, and so of others, as may easily be noted by the aforegoing Table of Chords, where you may see (for Example) the Number 44, which is the Chord of 5 Degrees, is the Difference between 843, which is the Chord of 115 Degrees; and 887, which is the Chord of 125; as likewise 87, the Chord of 10 Degrees, is the Difference between the Chord of 110 Degrees and 130, &c. which are equally distant from 120 Degrees.

Proof of the Line of Polygons.

You may know whether this Line be well divided, by help of the Line of Chords, in the following manner.

Take in your Compasses, upon the Line of Polygons, the Distance of the Number 6, denoting a Hexagon, from the Center of the Joint; then carry this Distance upon the Line of Chords, putting each Point of your Compasses upon the correspondent Points, from 60 to 60, denoting the Angle of the Center of an Hexagon.

The Sector being thus opened, take upon each Line of Chords the Distance of the two Points marked 72 from the Center, and lay it off upon the Line of Polygons, placing one Foot in the Center of the Joint; then the other loot must reach to the Point 5, which appertains to a Pentagon, whose Angle at the Center is 72 Degrees.

Likewise in taking upon the Line of Chords the Distance of the two Points, denoting go 4 and laying it off upon the Line of Polygons, the Foot of your Compasses must meet the Point 4, appertaining to a Square, whose Angle of the Center is 90 Deg. and so of others.

Proof of the Line of Planes.

Because we have given two Methods for dividing the Line of Planes, one may serve to prove the other; but (fill you may easier know whether the Divisions be well made, in the following manner: Take between your Compasses the Distance of any Point upon this Line from the Center of the Joint, and lay it off from the same Point on the other Side of the same Line of Planes; then the Foot of your Compasses will fall upon the Number of a Plane four times greater than that which was taken towards the Center: and if again your Compasses thus opened should be once more turned over, towards the End of the said Line, the Point would fall upon the Number of a Plane nine times greater. As, for Example; if you take the Distance from the Center to the Plane 2, in placing one Point of your Compasses on 2, the other ought to fall upon 8; and by turning the Compasses once more, one of it’s Points must fall upon 18, which contains 9 times 2. Moreover, in turning the Compasses once more over, the other Point ought to fall upon the Number 32, containing 2, 16 times. If, lastly, you turn over the Compasses again, it must fall upon 50, and so of other similar Planes, because they are to each other as the Squares of their homologous Sides. It is this that facilitates the Division of the Line of Planes; for having the first, these are likewise had, viz. the 4th, the 9th, the 16th, the 20th, the 25th, the 36th, the 49th, and the 64th. Having found the 2d, the 8th, the 18th; the 32d, and the 50th will be had: likewise having found the 3d, the 12th, the 27th, and the 48th will be had, and so of others.

Proof of the Line of Solids.

You may know whether this Line be well divided, in the following manner: Take between your Compasses the Distance of some Point on this Line from the Center of the Joint; then place one of it’s Points, thus opened, upon this Point of Division, and turn the other Point over towards the End of the Line. Now this Point must fall upon the Number of a Solid 8 times greater than that which was taken. Again, if the Compasses be once more turned over, it will fall upon a Solid 27 times greater than that which was first taken. As, for Example; the distance of the first Solid from the Center, will be equal to the Distance from 8 to 27, and from 27 to 64. Likewise, twice the Distance from the Center to 3, will be equal to the Distance from 3 to 24. By the 4th Solid, the 32d will be had. Moreover the 5th Solid will give the 40th; by the 6th the 48th Solid will be had; and, in a word, by help of the 7th, the 56th Solid will be had, because similar Solids are to each other, as the Cubes of their homologous Sides, which facilitates the Division of the Line of Solids.

Proof of the Line of Metals.

We have already mentioned, that the Division of this Line is founded upon Experiments made of the different Weights of a Cubic Foot of each of the six Metals, as they are here denoted.

Metals.

Weights of a Cubic Foot.

Gold

1326 Pounds

4 Ounces

Lead

802

2

Silver

720

12

Brass

627

12

Iron

558

00

Tin

516

2

From these different Weights of the six Metals the beforementioned Table was calculated, by means of which the Line of Metals was divided.

Now because Tin is the lighted of the said six Metals, it is manifest that if, for Example, a Ball of Tin is to be made of the same Weight as a Ball of Iron, or Brass, the Ball of Tin must be greater than either of them; as also the Ball of Iron ought to be greater than that of Brass, and so on to that which will be the lead. Therefore supposing the Diameter of a Ball of Tin to be 1000, the Question is to find the Lengths of the Diameters of Iron and Brass Balls, that may be of the same Weight as the Ball of Tin.

Now to do this, you must make a Rule of Three, whose first Term let always be the heaviest of the two Metals to be compared; the second Term must be the Weight of the Tin, and the third must be the Number 64, which is the greatest Solid of the Table of Solids, to which the Number 1000 answers. As, for Example, to compare Iron, a Cubic Foot of which weighs 558 Pounds, with Tin, a Cubic Foot of which weighs 516 Pounds, 2 Ounces: Having reduced them all into Ounces, the 558 Pounds make 8928 Ounces; and the 516 Pounds, 2 Ounces, make 8258 Ounces: then lay, if 8928 gives 8258, What will 64 give? The Rule being finished, the fourth Term will be 59 and a small Remainder; then look for the Number 59 in the Table of Solids, and the Number answering thereto is 973; instead of which take 974, because of the remaining Fraction: therefore, I say, that the Diameter of the Ball of Iron must be 974. In the same manner, by making four other Rules of Proportion, you may know whether the Numbers, marked against the four other Metals, are well calculated, and consequently, whether the Line of Metals be well divided.