PRISM.

A solid whose bases or ends are similar, equal, and parallel plane figures; and whose sides are parallelograms. To compute volume, multiply area of base by height. A prism may

be triangular, rectangular or square; a brick is a rectangular prism; a piece of square scantling, with both ends squared, is a square prism.

PRISMOID.

A solid resembling a prism in form, but whose sides are not parallel; the base is parallel with the top end, though they are of different areas. To compute volume, add together the areas of the two ends, to this sum add four times the area of middle section parallel to them, and multiply this sum by one-sixth of perpendicular height.

Example: What is the volume of a rectangular prismoid with base, EG, and, GH, "x6"; top end, AB, and, BD, 3" x 2", and height 15 feet?

×6=42, area of base; 3x2=6, area of top.

42+6=48, sum of areas of two ends.

Length and breadth of middle section are respectively equal to half the sum of lengths and breadths of base and top.

7+3

=5, length of middle section;

6+2
2

=4, breadth of

2 middle section. 4x5=20, area of middle section. 20x4=80. four times area of middle section. 80+48=128, sum of areas of two ends plus area of middle section.

15' x 12"180", height in inches. 180÷6=30. Then, 180 x 30 3840 cu. in. volume of prismoid.

If this prismoid is a piece of lumber and it is desired to find the number of board feet it contains, divide its volume in cubic inches by 144; there are 144 cubic inches in a board foot. 3840144-26.6; therefore, in a piece of lumber 7"x 6" at one end, 3" x 2" at the other end, and 15' long there are 26.6 board feet.

CYLINDER.

To compute volume, multiply area of base by height.

CONE.

To compute volume, multiply area of base by perpendicular height and take one-third of this product.

Example: A cone with diameter of base AB, 6" and height, EC, 12" what is volume?

C

FRUSTUM OF CONE.

To compute volume, add together the square of the diameters of the greater and lesser ends and the product of the two diameters; multiply this sum by .7854, and this product by the height, then divide this last product by three.

FIG. 17.-Cone.

FIG. 18.-Frustum

of Cone.

Example: What is volume of frustum of cone having diameter 5" at greater end, BD, diameter 3" at lesser end, AC, and height, EO, 9"?

52+32=34, sum of squares of diameters of ends.

5x3=15, product of diameters.

15+34=49; 49 x.7854-38.4846.

38.4846 x 9" (height) = 346.36143 115.4538 cu. in.

volume.

SPHERE.

To find volume, multiply cube of diameter by .5236.

B

FIG. 19.-Sphere.

Example: Find volume of sphere having diameter AB, 5". 53-125; 125x.5236 65.45 cu. in. volume. To compute hollow sphere, subtract volume of internal space from sphere.

=

HOW TO FIND THE QUANTITY OF WATER, TONS OR GALLONS, A TANK WILL HOLD.

The tank is 10' long, 6' wide, and 6' deep; multiply, the length by the breadth, 10×6=60; multiply this by the depth, 60×6=360, the number of cubic feet in the tank. A cubic foot of fresh water weighs 62.5 lbs., and of sea water, 64 lbs. ; in this example we will use fresh water. Multiply cubic feet in tank by weight of 1 cu. ft. of water, 360 × 62.5=22500 lbs. of fresh water in the tank; divide this by 2240, the number of pounds in a ton (long), and the result will be the number of tons in the tank, when it is filled with fresh water.

There are about 7 gals. of water in 1 cu. ft.; therefore, to find the number of gallons the above tank will hold, multiply the number of cubic feet (360) by 71⁄2, which will give 2700 gals., the amount the tank will hold.

CAPACITY OF ROUND TANK.

How much sea water, tons and gallons, will a cylindrical tank hold which is 4' in diameter and 12' high?

First find area of circle 4'; the formula is

πα

4

area.

4216; 16×3.1416-50.2656; 50.2656÷4=12.5664, area. of section of tank in square feet. 12.5664 × 12=150.79+, capacity of tank in cubic feet.

Sea water weighs 64 lbs. per cu. ft. 150.79 × 64=9650.56 lbs. 2240=4.30, capacity of tank in tons.

Or as sea water is taken at 35 cu. ft. per ton, then 150.79÷ 35=4.30, capacity of tank in tons of sea water.

There are about 7.5 gals. in 1 cu. ft. 150.79 x 7.5=1130.92, capacity of tank in gallons.

TO FIND AMOUNT OF COAL TAKEN BY CHANGE OF DRAFT.

A ship before coaling has a draft of 24' 06" ford., and 25' 08" aft. After coaling, the draft is 26′ 08′′ ford. and 27′ 04′′ aft. The mean tons per inch immersion of these drafts are 62 tons. How much coal has been taken on board?

Rule. Subtract mean draft before coaling from mean draft after coaling; the product of this remainder in inches, multiplied by the tons per inch immersion, will give the number of tons of coal taken.

24' 06"

25' 08"

50′ 02′′÷2=25′ 01′′, mean draft before coaling.

26' 08"

27' 04"

54′ 00′′÷2=27' 00", mean draft after coaling.

=

27′ 00′′-25′ 01′′-1′ 11′′ 23"; then 23 × 62=1426 tons of coal taken.

The mean tons per inch immersion is taken at the mean of the mean draft before coaling, and the mean draft after coaling. In this case it is found as follows:

25′ 01′′, mean draft before coaling.
27′ 00′′, mean draft after coaling.

52′ 01′′÷2 26' 0",

the draft at which the tons per inch immersion should be taken in the above problem.