other square; intersecting the sides A B C D at the points 1, 2, 3, 4, 5, 6, 7, 8, the octagon required.

Problem 61.

To inscribe a nonagon in a circle, Fig. 51.

Draw the diameters, and bisect the radius OD at a; draw the line b a d parallel to A c, intersecting the circumference at b; draw the line Bb; divide the arc B Ab into three equal parts, draw the line B c, will be one side of the nonagon.

Problem 62.

To inscribe an undecagon in a circle, Fig. 52.

Draw in the diameters and bisect the radius O D in a; draw the line b a d parallel to A C, intersecting the circumference at b; the distance from E to O will be one side of the undecagon.

Problem 63.

To find a right line that shall be nearly equal to any given arc of a circle, Fig. 53.

Divide the chord A B into four equal parts; set one part on the circumference from B to D; draw the line from C to the first division on the chord; and twice the length of the line CD will be the length of the arc nearly.

Problem 64.

To find the side of a square nearly equal in area to a given circle, Fig. 54.

Draw the two diameters A B and C D at right angles to each other; bisect the radius OC by a line from one end of the diameter at A, meeting the circumference in E, then will the line A E be the side of a square nearly equal in area to a given circle; and if the line E F be drawn parallel to C D, it will be of the circumference nearly, or three times the diameter, and once the versed sine QH of the angle A O D will be the cir cumference nearly.

E

Problem 65.

To find the length of an arc of a circle, Fig. 55.

Rule 1. Subtract the chord of the whole arc from eight times the chord of half the arc, and the remainder is the length of

the arc nearly.

183

Required the length of the arc AC B, the chord of half the arc A C 1.30, and the chord of the whole arc A B = 2.50.

=

Ex. 1.30 x 8 = 10.40 2.50 = 7.903 2.63 the length of the arc nearly

Rule 2. Multiply the radius by the number of degrees contained in the arc, and the product by .0174524.*

Given radius 250; angle 60°.

Ex. 2.50 × 60° = 150.00 × 0174524 = 2.617 length of arc

Problem 66.

To find the length of an arc by the Table, No. 17.

Rule. Divide the height by the base, and the quotient will be the height of an arc of which the base is unity; multiply the tabular number opposite the corresponding quotient by the base of the arc, and the product equals the length required.

Problem 67.

To find the diameter of a circle by having the chord and versed sine given, Fig. 56, Plate 3.

Rule. Divide the square of half the chord by the versed sine; to the quotient add the versed sine, and the sum will be the diameter, &c.

Or: If the sum of the squares of the semichord and versed sine be divided by the versed sine, the quotient will be the diameter of the circle to which that segment corresponds.

Given the chord A B = 39.2, and the versed sine C D = 13.1; required the diameter C E.

Ex. 39.22

=

=

19.62 384.16 ÷ 131 = 293 + 131 42.4 the diameter required Or: 19.6 + 13.1a = 55577131 42.4 the diameter

*The decimal .0174524 is the natural sine of 1°.

Problem 68.

Having the diameter of a circle given to find the circumference; or the circumference given to find the diameter.

Rule 1. As 7 is to 22, so is the diameter to the circumference.

Or: As 22 is to 7, so is the circumference to the diameter. Rule 2. As 1 is to 3.1416, so is the diameter to the circumference.

Or: As 3.1416 is to 1, so is the circumference to the diameter.

Example 1. Required the circumference of a circle whose diameter is 30.55.

30.55 × 22 67210 ÷ 7 = 96.01 the circumference

Example 2.

96.01 × 7

=

67210 22 = 30.55 the diameter

Example 3. Required the circumference of a circle whose diameter is 30.55.

3.1416 x 30.55 = 95.975880 the circumference

Example 4. Required the diameter of a circle whose circumference is 95.975880.

95.975880 3.141630.55 = the diameter

Problem 69.

To describe an ellipsis, the transverse and conjugate diameters being given, Fig. 57, Plate 3.

Draw the diameters at right angles to each other, and set off the given lengths, as at A B C D; with the difference of the two semidiameters set off O a and Ob, take half the length of ab, and lay it off from a to C; then take the distance O c, and mark O d, O e, and Of, which will be the four centres of the ellipse; place the point of the compasses at c, with the radius c A, describe the arc i Ag; also with the same radius from e describe the arc h B k; then with the radius fc from the point f describe the arc g Ch, and with the same radius from d describe the arc k Di, which will be the ellipse required.

To find the area see problem 93.

Note.-The longest diameter is called the transverse

= А В.

The shortest diameter is called the conjugate = CD.
Focii or focuses are the points on which the periphery of the ellipse is
described as c def.

The most perfect method of drawing ellipses is by the trammel, or elliptic compass.

It consists of a cross with a dovetailed groove at right angles, in which two small pieces are made to run easily up and down each way, to which is a beam (similar to a beam compass) with sliding sockets, one for the pen, pencil, or point, the other two, forming the focii, fitting into the two running slides in the grooves; the instrument, when set to the two diameters, will draw the ellipse by one revolution.

To adjust it: Set the beam over the transverse A B, and slide it backwards and forwards until the pencil or ink point coincide with the point A, and tighten the screw of the slider, which moves on the conjugate axis; now turn the beam, so as to lay over the conjugate axis CD, and make the pencil or point coincide with the point C, then fix the screw, which is over the slider of the transverse axis; the compass being adjusted, move it gently round.

Ellipses may also be drawn by a string or chord; first find the two focii, then fix two pins in the focii, another in one end of the conjugate diameter, tie the string firmly, then with a pencil move round, will describe the ellipse.

Problem 70.

To describe a parabola, any ordinate to the axe and its abscissa being given, Fig. 58, Plate 3.

Let V R and Ro be the given abscissa and ordinate, bisect Ro in m, join V m, and draw mn perpendicular to it, meeting the axe in n, make V C and V F each equal to R n, then will F be the focus of the curve.

Take any number of points, rr, &c., in the axis, and draw the double ordinates of an indefinite length.

From F as a centre, with the radii C F, Cr, &c., describe arcs cutting the corresponding ordinates in the points o, 0, 0, 0, &c., and the curve o V o, drawn through all the points of intersection, will be the parabola required.

Problem 71.

To describe an egg-shape figure, Fig. 59, Plate 3.

Describe a circle A B equal to the given diameter, draw the lines H B and DF; bisect the radius A C or C B, then with half the radius draw the circle ELF K, touching the circle A B at E; then from A and B, with the distance D F, set off GH as centres to the sides A L and B K, draw the lines HIK and GI K, intersecting the circle E L F K at the tangent points K L, will be the figure required.

AREAS;

Otherwise Mensuration of Superfices.

The area or superficial content of any figure is the space contained within length and breadth, without having any regard to

thickness.

As applied to land surveying in finding the areas or contents of plane figures, are calculated by the number of square links in each.

When figures are measured by yards or feet, the calculations are also made by their squares in like manner; in either case the contents are reduced to acres, roods, and perches.

All irregular or crooked fences must first be reduced to straight lines, to form the sides of the regular mathematical figures, which are chiefly confined to the triangle and trapezium.

Other methods are also introduced to obtain the accurate contents of irregular boundaries by mechanical rules, equally correct, and in some instances with greater precision.

The statute acre contains 1.00000 square links, more particu