EXAMPLE. Find the solidity of a squared log of timber, of the invariable breadth and thickness of 32 and 20 inches, its length being 40 feet 6 inches.

v = btl = 32 × 2 × 401 180 cubic feet.

Or, find (57) the mean proportional between 32 and 20, which is 25-3; then set 40-5 on C to 12 on D, and opposite to 25.3 on D is 180 on C, the content.

The first rule is that of article 332, Part I.

The method by the sliding rule is derived thus :-Let m = the mean proportional between 6 and t, then m2 = bt, and as m is in inches,

2 L 12;

that is, the distance between 12 and m on D is equal to that between b and v on C.

EXERCISES.

1. Find the solidity of a log of wood 30 inches broad, 18 thick, and 16 feet long.

Ans. 60 feet. 2. What is the content of a plank, the end of which is 30 inches by 20, and length 20 feet?

Ans. 833 feet. 3. What is the content of a square log of wood, the side being 14 inches, and the length 12 feet? Ans. 163 ft. 4. The side of a square block of sandstone is 3 feet, and its length is 6 feet; what is its content? Ans. 54 feet. 5. Find the cubic content of a log of wood 20 feet 3 inches long, its end being 32 by 20 inches. Ans. 90 feet. 6. The side of a square log of wood is 2 feet, and its length 24 feet 1 inch; what is its content? Ans. 963.

63. PROBLEM III.-To find the content of squared tapering timber.

'Find the mean breadth and thickness, and multiply their product by the length.'

As in last problem, v = btl.

By the sliding rule-The method is the same as that of last problem, using the mean breadth and thickness.

EXAMPLE. The breadth of a tapering plank of wood at the two ends is 18 and 12 inches, and its thickness at the

ends is 14 and 10 inches, and its length is 19 feet 10 inches; what is its solidity?

Here b(18+12)=15, and t = (14+10) = 12, and v=btl=1512 × 1952412 feet.

The rule above, though it is generally used, is correct only in one case, namely, when two of the sides are parallel and the other two converge; for the solid is then a prism, having one of the parallel sides for its base. In other cases, this rule gives the content a little less than the real solidity, and the error is greater the more the difference between the breadth and thickness. When all the sides converge to the same point, the log is then the frustum of a pyramid, and its solidity can be found by 342, Part I.

64. In every case when the sides are straight, the solid is a prismoid, and its content can be found accurately by problem 347, Part I. When the lateral surface is convex, but straight, the solid is then a cylindroid, and its solidity is found in the same manner.

65. When the breadth is irregular, it may be measured at several places, and the sum of these breadths, divided by their number, may be taken for the mean breadth. same way the mean thickness may be found.

In the

The preceding example, calculated by the rule for the solidity of a prismoid (347, Part I.), gives for the content 25.067 cubic feet instead of 2410.

EXERCISES.

1. Find the content of a squared tapering log of wood, the breadth and thickness at one end being 34 and 20 inches, and those at the other end 26 and 16 inches, and the length 32 feet. Ans. 100 feet.

2. Find the cubic content of a log, the breadth and thickness at one end being 33 and 22 inches, and those at the other end 27 and 18 inches, and the length 40 feet. Ans. 1663 feet.

3. The breadth and thickness of one end of a piece of timber are 21 and 15 inches, and those at the other end are 18 and 12 inches, and the length is 41 feet; what is its solidity? Ans. 74.95 feet.

4. The breadth and thickness at the greater end of a piece of timber are 1·78 and 1.23 feet, and at the smaller end 1.04 and 0.91 feet; what is its content, its length being 27.36 feet? Ans. 41.278 feet.

66. PROBLEM IV.-To find the content of round or unsquared timber.

I. Find the quarter girt, that is, one-fourth of the mean circumference, and multiply its square by the length.'

By the sliding rule- Set the length on C to 12 on D, and opposite to the quarter girt in inches on D is the content on C.'

II. 'Find one-fifth of the girt, and multiply its square by twice the length.'

By the sliding rule- Set twice the length on. C to 12 on D, and opposite to one-fifth of the girt on D is the content on C."

Let 7 and c denote the length and mean circumference of a piece of round timber, and its volume, then,

by rule I., v=

2

= ( 2 ) * 1 = 1; c2 = ·0625 c2ol,

2

16

by rule II., v = 2 (-)*‍1 — •08 col.

EXAMPLE. The mean circumference of a piece of unsquared timber is 6 feet 8 inches, and its length is 16 feet 4 inches; what is its contents?

When the piece of timber is of a cylindric form, its volume, by 336, Part I. is, v=bh='07958 c2l, if l = h. Therefore the first rule in this case gives the content too small by more than one-fifth part of the true solidity; and the second gives it too much by about the 190th part.

When the tree tapers uniformly, it is then a frustum of a cone, and its true volume can be found by the rule in article 344, Part I. In this case, the first rule gives a result still farther from the truth, for a conic frustum exceeds a cylinder of the same length, whose circumference is the mean girt of the frustum.

The first rule is generally followed in practice, and the deficiency in the content given by it is intended to be a compensation to the purchaser for the loss of timber caused by squaring it.

In the following exercises, the first answer is the result by the first rule, and the other is that by the second rule.

EXERCISES.

1. The mean girt of a tree is 8 feet, and its length 24 feet; required its content. Ans. 96 feet, or 122.88 feet.

2. What is the content of a piece of round timber, the girt at the thicker end being 16 feet, and at the smaller 12 feet, and its length 19 feet? Ans. 2323 ft., or 297.93 ft.

3. Find the content of a tree, whose mean girt is 3.15 feet, and length 14 feet 6 inches. Ans. 8.992 ft., or 11.51 ft. 4. The girts of a piece of round timber at five different places are 9.43, 7·92, 6·15, 4·74, and 3.16 feet, and its length is 17 feet 3 inches; what is its content?

Ans. 42.5195 feet, or 54.4065 feet.

RELATIONS OF WEIGHT AND VOLUME
OF BODIES.

67. The relations of the weights and volumes of bodies are determined by means of their specific gravities.

68. The specific gravity of a body is the ratio of the weight of a given volume of it to the weight of the same volume of some body assumed as a standard, as of water.

Thus, if the weight of a cubic inch of any body is 10 times the weight of a cubic inch of water, its specific gravity is 10. The weight of any volume of water being denoted by 1, the weight of an equal volume of the preceding body would evidently be 10, its specific gravity; that is, if the

weight of any volume of water is taken for a standard or unit of weight, the weight of an equal volume of any other body is its specific gravity.

69. The following table contains the specific gravities of the most common solids and fluids :

TABLE OF SPECIFIC GRAVITIES.

Amethyst, common

I.-Water being the standard.

SOLIDS.

Flint, black

2.582