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 from which

it is taken, represents the triangle and one of the rules, as being used to draw a series of parallel lines. Either rule is one foot long, and has, parallel to each of its edges, two scales one placed close to the edge and the other immediately within this, the outer being termed the artificial, and the inner, the natural scale. The divisions upon the outer scale are three times the length of those upon the inner scale, so as to bear the same proportion to each other that the longest side of the triangle bears to the shortest. Each inner, or natural scale, is, in fact, a simply divided scale of equal parts (see p. 9), having the primary divisions numbered from the left hand to the right throughout the whole extent of the rule. The first primary division on the left hand is subdivided into ten equal parts, and the number of these subdivisions in an inch is marked underneath the scale, and gives it its name. On one of the pair of Marquois's scales now before us, we have, on one face, scales of 30 and 60, on the obverse scales of 25 and 50, and on the other we have on one face scales of 35 and 45, and on the obverse scales of 20 and 40. In the artificial scales the zero point is placed in the middle of the edge of the rule, and the primary divisions are numbered both ways from this point to the two ends of the rule, and are, every one, subdivided into ten equal parts, each of which is, consequently, three times the length of a subdivision of the corresponding natural scale.

The triangle has a short line drawn perpendicular to the hypothenuse near the middle of it, to serve as an index or pointer; and the longest of the other two sides has a sloped edge.

To draw a Line parallel to a given Line, at a given Distance from it.-1. Having applied the given distance to the one of the natural scales which is found to measure it most conveniently, place the triangle with its sloped edge coincident with the given line, or rather at such small distance from it, that the pen or pencil passes directly over it when drawn along this edge. 2. Set the rule closely against the hypothenuse, making the zero point of the corresponding artificial scale coincide with the index upon the triangle. 3. Move the triangle along the rule, to the left or right, according as the required line is to be above or below the given line, until the index coincides with the division or subdivision corresponding to the number of divisions or subdivisions of the natural scale, which measures the given distance; and the line drawn

along the sloped edge in its new position will be the line required*.

Note. The natural scale may be used advantageously in setting off the distances in a drawing, and the corresponding artificial scale in drawing parallels at required distances.

To draw a Line perpendicular to a given Line from a given Point in it.—1. Make the shortest side of the triangle coincide with the given line, and apply the rule closely against the hypothenuse. 2. Slide the triangle along the rule until a line drawn along the sloped edge passes through the given point; and the line so drawn will be the line required.

The advantages of Marquois's scales are: 1st, that the sight is greatly assisted by the divisions on the artificial scale being so much larger than those of the natural scale to which the drawing is constructed; 2nd, that any error in the setting of the index produces an error of but one-third the amount in the drawing.

If the triangle be accurately constructed, these scales may be advantageously used for dividing lines with accuracy and despatch; our figure, as well as the sliding rule (fig. 4), was drawn by the aid of Marquois's scales alone.

We proceed to explain a method which we have found to answer well for dividing lines with accuracy and despatch, and which is altogether independent of any error in the construction of the triangle.

Let ab represent the line to be divided into any number, n, of equal parts; select one of the natural scales, of which n divisions or some multiple p n of n divisions, are nearly

equal to a b, but less than it; then setting the sloped edge of the triangle perpendicular to a b, so that a line drawn along it passes through a, place the rule closely against the hypothenuse, making a division of the artificial scale, corresponding * If A B C represent the triangle in its new position, and the dotted lines represent its original position, we have,

by similar triangles, A B C, A'A D,

A DAA BC: BA

:

: : 1:3;

and therefore A D contains as many divisions of the natural as A A' contains of the artificial scale.

A

D

B

to the selected natural scale, coincide with the index upon the triangle; and, moving the triangle along to the rule to the right, until the index coincides with the p nth division from that to which it was set, draw a line along the sloped edge intersecting a b in c. Upon ab as hypothenuse describe the right-angled triangle a bd, having the side a d equal to a c, and placing the sloped edge of the triangle perpendicular to ad, so that a line drawn along it passes through a or d; slide the triangle along the rule to the right or left, and drawing a line as the index comes into contact with every pth division of the artificial scale, the line will be divided as required.

THE VERNIER. (Plate II. Figs. 1 and 2.)

The property of this ingenious little subsidiary instrument will be readily comprehended from what has been already said of the construction and use of a vernier scale (p. 11). It is so constructed as to slide evenly along the graduated limb of an instrument, and enables us to measure distances, or read off observations, with remarkable nicety. In the vernier scale before described, the divisions on the lower or subsidiary scale were longer than those on the upper or primary scale; but in the vernier now to be described the divisions are usually shorter than those upon the limb to which it is attached; the length of the graduated scale of the vernier being exactly equal to the length of a certain number (n - 1) of the divisions upon the limb, and the number (n) of divisions upon the vernier being one more than the number upon the same length of the limb.

Let, then, L represent the length of a division upon the limb, and v, the length of a division upon the vernier :

so that (n − 1) L = N V ;

and therefore LVL

or the defect of a division upon the vernier from a division upon the limb is equal to the nth part of a division upon the limb, n being the number of divisions upon the vernier *.

* If n divisions of the vernier were equal to (n + 1) divisions of the limb, or (n+1) Ln v,

then would v-L

n+1

1

L-LL;
n

or the excess of a division upon the vernier above a division upon the limb would be equal to the nth part of a division upon the limb. With this arrangement, however, we should have the inconvenience of reading the vernier backwards.