take off the transverse distance of 4 and 4, and set it off from a and B to 4 and 3 2. Make this extent from A to 4 a transverse distance at 10; then the transverse distance at 5 bisects A 4 and 3 B in 2 and 5; set off from 3, gives 1, and from 4 gives 6; and thus the line A B is divided into seven equal parts as required.

To open the Sector so that the Line of Lines may answer for any required Scale of equal Parts.-Take one inch in the com passes, and open the sector, till this extent becomes a transverse distance at the division indicating the number of parts in an inch of the required scale; or, if there be not an inte gral number of parts in one inch, it will be better to take such a number of inches as will contain an integral number of parts, and make the extent of this number of inches, if it be not too great, a transverse distance at the division indicating the number of parts of the required scale in this extent

Example.-To adjust the Sector as a Scale of One Inch to Four Chains.-Make one inch the transverse distance of 4 and 4; then the transverse distances of the other corresponding divisions and subdivisions will represent the number of chains and links indicated by these divisions: thus, the transverse distance from 3 to 3 will represent three chains; the transverse distance at 47, or the seventh principal subdivision after the primary division marked 4, will represent 4 chains 70 links, and so on.

To construct a Scale of Feet and Inches in such a manner that an extent of Three Inches shall represent Twenty Inches.1. Make three inches a transverse distance between 10 and 10, and the transverse distance of 8 and 8 will represent 16 inches. 2. Set off this extent from A to B, divide it by con tinual bisection into 16 equal parts, and place permanent strokes to mark the first 12 of these divisions, which will represent inches. 3. Place the figure 1 at the twelfth stroke, and set off again the extent of the whole 12 parts, from 1 to 2, 2 to 3, &c., to represent the feet.

As an Example of the Use of the Line of Lines in reducing Lines, let it be required to reduce a drawing in the Proportion of 5 to 8.-Take in the compasses the distance between two points of the drawing, and make it a transverse distance at 8 and 8; then the transverse distance of 5 and 5 will be the distance between the two corresponding points of the copy 2. These two points having been laid down, make the distance between one of them and a third point a transverse distance at 8, and with the transverse distance at 5 describe, fron

that point as center, a small arc. 3. Repeat the operation with the other point, and the intersection of the two small arcs will give the required position of the third point in the copy. In the same manner all the other points of the reduced copy may be set off, each one from two points previously laid down.

LINE OF CHORDS

The double scales of chords upon the sector are more generally useful than the single line of chords described on the plain scale; for, on the sector, the radius with which the arc is to be described may be of any length between the transverse distance of 60 and 60 when the legs are close, and that of the transverse of 60 and 60 when the legs are opened as far as the instrument will admit of but, with the chords on the plain scale, the arc described must be always of the same radius.

To protract or lay down a right-lined Angle в A C, which shall contain a given number of Degrees, suppose 46°.-Case 1. When the angle contains less than 60°, make the transverse distance of 60 and 60 equal to the length of the radius of the circle, and with that opening describe the arc в C. (Fig. at page 40.) Take the transverse distance of the given degrees 46, and lay this distance on the arc from the point B to c.

From

the center A of the arc draw two lines A C, A B, each passing through one extremity of the distance B c laid. on the arc; and these two lines will contain the angle required. Case 2. When the angle contains more than 60°. Suppose, for example, we wish to form an angle containing 148°. Describe the arc B C D, and make the transverse distance of 60 and 60 equal the radius as before. Take the transverse distance of or, &c., of the given number of degrees, and lay this distance on the arc twice or thrice, as from в to a, a to b, and from b to D. Draw two lines connecting в to a, and a to D, and they will form the angle required.

When the required angle contains less than 5°, suppose 3, it will be better to proceed thus. With the given radius, and from the center A, describe the arc D G; and from some point, D, lay off the chord of 60°, which suppose to give the point G, and also from the same point D lay off in the same direction the chord of 56° (= 60° — 34°), which would give the point E Then through these two points E and G, draw

The line of polygons is chiefly useful for the ready division of the circumference of a circle into any number of equal parts from 4 to 12; that is, as a ready means to inscribe regular polygons of any given number of sides, from 4 to 12, within a given circle. To do which, set off the radius of the given circle (which is always equal to the side of an inscribed hexagon) as the transverse distance of 6 and 6, upon the line of polygons. Then the transverse distance of 4 and 4 will be the side of a square; the transverse distance between 5 and 5, the side of a pentagon; between 7 and 7, the side of a heptagon; between 8 and 8, the side of an octagon; between 9 and 9, the side of a nonagon, &c., all of which is too plain to require an example.

If it be required to form a polygon, upon a given right line set off the extent of the given line, as a transverse distance between the points upon the line of polygons, answering to the number of sides of which the polygon is to consist; as for a pentagon between 5 and 5; or for an octagon between 8 and 8; then the transverse distance between 6 and 6 will be the radius of a circle whose circumference would be divided by the given line into the number of sides required.

The line of polygons may likewise be used in describing, upon a given line, an isosceles triangle, whose angles at the base are each double that at the vertex. For, taking the given line between the compasses, open the sector till that extent becomes the transverse distance of 10 and 10, then the transverse distance of 6 and 6 will be the length of each of the two equal sides of the isosceles triangle.

All regular polygons, whose number of sides will exactly divide 360 (the number of degrees into which all circles are supposed to be divided) without a remainder, may likewise be set off upon the circumference of a circle by the line of chords

Thus, take the radius of the circle between the compasses, and open the sector till that extent becomes the transverse distance between 60 and 60 upon the line of chords; then, having divided 360 by the required number of sides, the transverse distance between the numbers of the quotient will be the side of the polygon required. Thus for an octagon, take the distance between 45 and 45; and for a polygon of 36 sides take the distance between 10 and 10, &c.

LINES OF SINES, TANGENTS, AND SECANTS.

Given the Radius of a Circle (suppose equal to two inches); required the Sine and Tangent of 28° 30' to that Radius.— Open the sector, so that the transverse distance of 90 and 90 on the sines, or of 45 and 45 on the tangents, may be equal to the given radius, viz., two inches; then will the transverse distance of 28° 30′, taken from the sines, be the length of that sine to the given radius, or, if taken from the tangents, will be the length of that tangent to the given radius.

But, if the Secant of 28° 30′ is required, make the given radius of two inches a transverse distance of 0 and 0, at the beginning of the line of secants, and then take the transverse distance of the degrees wanted, viz., 28° 30'.

A Tangent greater than 45° (suppose 60°) is thus found:Make the given radius, suppose two inches, a transverse distance to 45 and 45, at the beginning of the line of upper tangents, and then the required degrees (60) may be taken from

the scale.

The tangent, to a given radius, of any number of degrees greater than 45° can also be taken from the line of lower tangents, if the radius can be made a transverse distance to the complement of those degrees on this line (see note, page 35).

Example.-To find the tangent of 78° to a radius of two inches. Make two inches a transverse distance at 12 on the lower tangents, then the transverse distance of 45 will be the tangent of 78°.

In like manner the secant of any number of degrees may be taken from the sines, if the radius of the circle can be made a transverse distance to the complement of those degrees upon this line. Thus making two inches a transverse distance to the sine of 12°, the transverse distance of 90 and 90 will be the secant of 78°.

To find, by means of the lower tangents and sines, the degrees answering to a given line, greater than the radius which expresses the length of a tangent or secant to a given radius.

For a tangent, make the given line a transverse distance at 45 on the lower tangents; then take the extent of the given radius, and apply it to the lower tangents; and the complement of the degrees at which it becomes a transverse distance will be the number of degrees required. For a secant make the given line a transverse distance at 90 on the sines; then the extent of the radius will be a transverse distance at the complement of the number of degrees required.

Given the Length of the Sine, Tangent, or Secant of any Degrees, to find the Length of the Radius to that Sine, Tangent, or Secant.-Make the given length a transverse distance to its given degrees on its respective scale. Then,

To find the Length of a versed Sine, to a given Number of Degrees, and a given Radius.—1. Make the transverse distance of 90 and 90 on the sines equal to the given radius. 2. Take the transverse distance of the sine of the complement of the given number of degrees. 3. If the given number of degrees be less than 90, subtract the distance just taken, viz., the sine of the complement, from the radius, and the remainder will be the versed sine: but, if the given number of degrees are more than 90, add the complement of the sine to the radius, and the sum will be the versed sine.

To open the legs of a Sector, so that the corresponding double Scales of Lines, Chords, Sines, and Tangents may make each a right Angle.-On the line of lines make the lateral distance 10, a transverse distance between 8 on one leg, and 6 on the other leg.

On the line of sines make the lateral distance 90, a transverse distance from 45 to 45; or from 40 to 50; or from 30 to 60; or from the sine of any degree to their complement. On the line of sines make the lateral distance of 45 a transverse distance between 30 and 30

MARQUOIS'S SCALES. (Plate II. Fig. 3.)

These scales consist of a right-angled triangle, of which the hypothenuse or longest side is three times the length of the shortest, and two rectangular rules. Our figure, which is drawn one-third the actual size of the instruments froin which