high. First of all a diagonal is carefully measured along ac and found to be 30 feet; perpendiculars are next measured from e and f to the opposite angles, and these are found to be 5 and 7 feet; required the cubic capacity.

Here the sum of the perpendiculars is 7 + 5 = 12; the area is 12 x 30

2

= =

180; and the cube is 180 x 12 = 2,160. Or, by the method of triangles, the base of the triangle acd multiplied by the perpendicular df gives twice area of triangle ade, and the base ac of the triangle abc multiplied by perpendicular be gives twice triangle area of abc, or

30 x 7 = 210

30 x 5 = 150

2) 360

180

12

2160,

and generally the area of any irregular room may be obtained by dividing it into triangles, and then the cube obtained by multiplying the sum of these areas by the height.

(68) The Cube of Cylinders, such as pillars.

It may occasionally happen that in cubing a room the cube of pillars supporting the ceiling will have to be subtracted; the cubic contents, also, of water mains, of wells and cylindrical cisterns are all calculated on the same principles.

To cube cylindrical bodies it is necessary to remember a few elementary properties of a circle, and to know how to get the areas of circles.

The relation of the circumference to the diameter is as 1 is to 3-1416, a number which may be perhaps committed to memory better by making the nonsense word tofos—the t standing for the three, the o for the one, the ƒ four, and so on.

It hence follows that the diameter may be known if the circumference is given, and vice versa; for the diameter multiplied by 3-1416 gives the circumference, and the circumference divided by 3.1416 equals the diameter.

To obtain the area of a circle the usual way is to square the diameter, and multiply by 7854; this number can always be remembered if 31416 is remembered, for 7854 is one-fourth of 3.1416; another method is to divide both circumference and diameter by 2, multiply them together, and the result is the area. The cube of a cylinder is of course the area multiplied by its height.

For instance, take the case of a pillar of two feet diameter and 15 feet high. d2 ×·7854 × h = cube of pillar, or taking the figures 4× 7854 × 15 = 47.124 cubic feet. Supposing the circumference only of the pillar be given and its height, the process is as in the following example. A pillar is 52 feet in circumference, and 15 feet high; required its cube:—

1.6552× 7854×15 = 32.27, which is the cubic content in feet. The area of a sector of a circle is found by getting the area of the whole circle, multiplying it by the number of degrees in the arc, and dividing by 360; this will not often be required, save as a preliminary step to obtain area of segment. On the other hand the areas of segments are frequently found; many bow windows, for instance, are segments of circles, and in order to obtain the cube of the space occupied by a bow window, the ground plan of which is

a

C

less or greater than a semicircle, the first step is to get the area, and then to multiply by the height.

The area of the sector abce is first obtained, next the area of the triangle abe, the two areas are then subtracted one from the other, if the segment is less than a semicircle; or added if greater than a semicircle, this area multiplied by the height gives the cube. Should only the chord (Fig. 11) ab and the height or length cd be given the diameter and the number of degrees must be calculated out as follows:

f FIG. 11.

Having got the diameter we know from trigonometry that the sine

ad

aec= and from a table of sines the angle aec, and therefore

ae

aeb is obtained from this number, the sum of the two angles equalling the number of degrees in the arc. For instance, given the chord ab = 8 and cd = 2, ad is

therefore the diameter is 10 and radius

4 5

aec =

of 8 = 4, and

42

+ 2 = 10,

2

ae equals 5. The sine of

8, which on reference to a table of sines is equal to 53° 8',

and the whole angle in the sector is therefore twice that or 106° 16', or the area may be found by the following rule:-to of product of chord and height, add the cube of the height divided by cd3 Applying these principles to 2ab

a

g

b

twice the chord (ab×gf×3)+ practice. In the room of which Fig. 12 is a ground plan the length from a to d is 36 feet, the breadth de is 21 feet, the distance from f to g is 14 feet, and the height of the room is 12 feet, required the cubic capacity of the whole. First of all the room is cubed, neglecting the bow 36 × 21 × 12 = 9,072, the next step is to obtain the number of degrees in the sector agb, to get which we want to know the diameter of the circle of which it is an arc, the chord 10.52 ab = 1.4

10.5, and therefore

+1.480.15.

The diameter is then 80-15, and hence the

af

that is

ac

FIG. 12.

radius is 40.075, the sine acg = 10.5 = 15° 11' nearly. The number of degrees in the arc is there40.075 fore 2 x 15° 11', or 30° 22′. The area of the sector can now be 80.152 x 7854 × 30° 22′ obtained

360

=

425.6, from this area subtract the area of the triangle abc which is cf x ab, or 18 x 21 = 378, and 425-6-378 gives 476 as the area of the sector agb, this number multiplied by the height gives the cube 47.6 x 12=571.2, and 571-2+9,072 gives the total cube of the room as 9,643 cubic feet.

(69) Area of Ellipses.

The bow windows in some rooms may be in the form of an elliptical curve, and the rule to find the area of an ellipse is, multiply the product of the two diameters by 7854. If, for instance, in Fig. 12, the bow window is to be considered as half an ellipse, then the major diameter is 21 feet, the minor 14 x 2, or 2·8 and 28 × 21 388; 588 x 7854-46-18 this must be halved = 23:09, and 21-388; the cube of the elliptical bow window is 12 x 2309, or 277.08. The area of hexagons may be obtained by the general rule, multiply the sum of the sides of the polygon by the perpendicular let fall from its centre upon one of the sides and half the product will be the area.

For example, in Fig. 13 let ab 73 feet and op. = 6.5 feet, the sum of the sides is then 6 x 73 = 441, and the area equals 441 × 65 2

side

143:32. If the cube of a poly

gonal pillar or other solid be required, it is of course obtained by multiplying the area by the height. Should the side of a polygon and the number of its sides only be given, its area may be obtained by the trig180° onometrical formula, area = xxx cot

n

when equals the number of sides, and z = the length of one For instance, supposing the area of a regular hexagon be required the side of which is 8 The number of sides is of course 6, therefore #-6

In the case a still easier way may be adopted, for a hexagon with son of No may be considered as omposed of 6 equilateral tena on the arva of one of which may be calculated by the rule sivach ghen page too), 13^x, 3, and area of the polygon = 6 ×

mocadory of a cone of pyramid as found as follows:-Mul

tiply the area of the base by the perpendicular height, and

d

g

a

FIG. 14.

of the product will be the solidity or cubic content. By the aid of this rule the cubic contents of some peculiar shaped roofs may be solved. For instance, let Fig. abcde represent a roof, required the cubic content;-ab = 80 feet breadth, ae = 20 feet, 50 feet, and perpendicular height=10 feet, let dgs be a plane section parallel to the end cbp, then dpgs will be a prism, asdg will be a pyramid. Cube of prism × 50 × 20 × 105,000, cube of pyramid × 30×20×10 = 2,000, total 7,000 cubic feet.

dc

=

(70) The Cube of Rectangular Prismoids and Frustums of Cones or Pyramids.

h

A frustum is a cone or pyramid with its top cut off. In modern built houses there are not infrequently rooms in ornamental towers, which may be treated as frustums. The rule is to the sum of the areas of the two ends, add four times the area of the middle or mean section parallel to the ends, multiply this sum by the height, and one-sixth will be the solidity. For example, let the figure abche represent a small room the length of the floor along ab is 12 feet, the breadth ae or bp is 7, and the height is 7 feet, the length of the ceiling is de or fh is 8, and the breadth is 4. Here the area of the floor is 7x12=84, the area of the ceiling is 8 x 4 = 32, the mean section is the half of the two extreme lengths added together § (12+8) = 10, multiplied by the half of the top and bottom breadths (7+4)=55, then by the rule the cube is

a

FIG. 15.

Cubic Capacity of a Dome.-This may be found by multiplying two-thirds of the product of the area of the base by the height (area of base ×h× }).