POLYGONS.

Problem 31.

The breadth of a polygon given to find the radius of a circle to contain that polygon, Fig. 26, Plate 2.

Rule. Multiply half the breadth of the polygon by the numbers in the sixth column of the Table No. 14, opposite its name, and the product will be the radius of a circle to contain that polygon.

If the polygon have an uneven number of sides, the half breadth is accounted from its centre to one of its sides.

Problem 32.

Required the radius of a circle to contain an octagon whose breadth A B = 36.4 feet.

Ex. AB 36.4 ÷ 2 = 18.2 x tab. num. 1.08 19.656 feet

Problem 33.

Required the radius of a circle to contain a pentagon whose half breadth A B = 14.4, Fig. 27.

Ex. AB 14.4 × tab. num. 1.238 = 17.827

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Problem 34.

The radius of a circle given to find the length of the side of an octagon, Fig. 26.

Rule. Multiply the radius of any circle by the numbers in the fifth column in the table opposite the polygon required, and the product will be the length of the side nearly, that will divide that circle into the proposed number of sides.

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Given the radius OC 19.656, required the length of the side C D.

Ex. OC 19.656 × tab. num. .76536 15.043916 the length of the side

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Problem 35.

The length of the side C D is 15.036840 of an octagon, given to find the radius D O, Fig. 26.

Rule. Multiply the given length of side by the number on the fourth column in the table opposite the polygon required, and the product will be the radius of a circle to contain the polygon.

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Ex. CD 15.036840 × tab. num. 1.307 19.646 nearly

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Problem 36.

To find the angle at the centre and sides of a regular polygon, Fig. 27.

Rule. Divide 360°, the degrees in a circumference, by the number of sides in the proposed polygon.

Thus: 360° ÷ 5 = 72° for the angle of the pentagon at the centre

Problem 37.

To find the angle formed by the sides.

Subtract the angle at the centre from 180°.

Thus: 180-72° 108° the outside angle of a pentagon

Problem 38.

Methods of describing arcs of circles of large magnitude. Angles in the same segment of a circle are equal to one another.

Let A B C D, Fig. 28, Plate 2, be the segment of a circle; the angles formed by lines drawn from the extremities A and B of the base of the segment to any points C and D of its arc, as the angles AC B, A D B are equal.

Note.-This problem is referred to hereafter in setting out railway curves.-See Euclid, Prob. 31, book iii.

Problem 39.

of finding points in and describing large circles.

If upon the ends A B, Fig. 29, Plate 2, of a right line, A B as an axis, two circles or rollers, CD and E F, be firmly fixed, so that the said line shall pass through the centres, and at right angles to the planes of the circles, and the whole be suffered to roll upon a plane without sliding:

If the rollers CD and E F be equal in diameters, the lines described upon the plane by their circumferences will be parallel right lines, and the axis A B, and every line D F, drawn between contemporary points of contact of the rollers and plane will be parallel among themselves:

If the rollers CD and E F be unequal, these lines formed by their circumferences upon the plane will be concentric circles, and the axis A B, and also the lines D F, will in every situation tend to the centre of those circles:

Then it will be as the diameter of the wheel CD is to the difference of the diameter of CD and EF, so is the radius of the circle proposed to be described by C D to the distance A B, at which the two wheels must be asunder, measured upon the plane on which the circle is to be described.

The wheel will evidently describe simultaneously another circle, whose radius will be less than that of the former by E F.

Note.-This rule is applicable to the wheels of railway carriages rolling over a curve; instead of the outer wheel being greater in diameter than the inner one, the outer rails are gradually raised from their tangents to the proper height, according to the radius of the curve or arc of the circle.

Problem 40.

Through three points to describe an arc of a circle by means of two laths, Fig. 30, Plate 2.

On a given chord A B to describe an arc of a circle that shall contain any number of degrees, performing the operation without compasses, and without finding the centre of the circle.

Place two straight laths forming an angle A C B, equal to the supplement of half the given number of degrees, and fix them at C; fix two pins at the extremities of the given chord A B, hold a pencil at C, then move the two laths against the pins, and the pencil at C will describe the arc required.

Example. It is required to describe an arc of fifty degrees on the given chord A B, subtract 25 degrees (half the given angle) from 180°; the difference will be 155° equal to the supplement;

then form the angle A C B = to 155° with the two rules, and proceed as before described.

If required to extend the arc on either or both sides, as at DE, first mark on the edge of the two legs of the instrument at A and B; bring the centre C to A, and the edge of the right leg over the point at C, point off the distance marked off on the left leg as at D, and fix the pins as before at DAC; in like manner fix the pins at E, and describe the continuation of the arc on both sides.

The whole circle may be described in the same manner by placing the two lower pins below the diameter and describing one half the circle; then fix the lower pins in the same manner on the upper part of the circle (keeping the laths fixed), and describe the other half of the circle in like manner.

Problem 41.

A very neat instrument, called the bevel or centrolinead, for the same purpose as last described, Fig. 31.

It consists of two rules, movable on a common centre, similar to a carpenter's rule; the brass semicircle has a groove and a screw at D to fix the two legs at the required angle.

The instrument is placed against the pins at A D C, as shown in the former example, and the pencil at C; keep the rules close to the pins, holding the pencil at C, and describe the arc required.

Problem 42.

Given three points to draw a line from either of them tending to the centre of the circle, and passing through each, Fig. 32, Plate 2.

Let A B C be the given points; it is required to draw A D, so that if continued it would pass through the centre of the circle containing A B C.

Make the angle BAD equal to the complement of the angle BCA, and AD is the line required; for supposing A E a tangent to the point A, then is E A D a right angle, and EA B =

BCA; whence BAD = right angle less the angle B C A, or the complement of B CA.

Corollary 1. AD being drawn, lines from B and C, or any other points in the same circle, are easily found; thus, make ABG BAD, which gives BG; then make BCFCBG, which gives CF; or CF may be had without the intervention of B G, by making A C F = CAD.

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Corollary 2. A tangent to a circle at any of the points, A for instance, is thus found.

Make BAE = BC A, and the line A E will touch the circle at A.

To perform the same by the bevel.

Set the bevel to the three given points A B C; lay the centre on A and the right leg on C, the other leg will give the tangent HA; from A draw A D perpendicular to HA, for the line required.

CIRCLES.

Problem 43.

To find the centre of a circle, Fig. 33, Plate 2.

Draw any chord A B, and bisect it with the line CD; bisect CD by the chord E F; the intersection at O will be the centre of the circle.

Problem 44.

Through three given points to describe the circumference, Fig. 34.

From the middle point B draw the chords B A and BC; bisect these chords and draw the lines n O, m 0; from O, with the radius O A, O B, O C, describe the circle required.

Problem 45.

To draw a tangent to a given circle, to pass through a given point at A, Fig. 35.

When the point A is in the circumference of the circle.

From the centre O, draw the radius OA; through the point