Of the Construction of the Sector.

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.

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.

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.

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.

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.

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.

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.

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.