2. How much timber in a tree 18 feet long, and its mean girt 5 feet 8 inches?

Ans. Common rule, 36 feet 1 inch; true content, 46 ft. 2 inches 10" 6"".

LEVELLING.

Art. 319.-PROB. I. To find the difference in the height of two places, by levelling rods.

RULE.

Set up the levelling rods perpendicular to the horizon, and at equal distances from the spirit level; observe the points where the line of level strikes the rods before and behind, and measure the heights of these points above the ground; level in the same manner, from the second station to the third, from the third to the fourth, etc. The difference between the sum of the heights at the back stations, and at the forward stations, will be the difference between the height of the first station and the last.

If the stations are numerous, it will be expedient to place the back and forward heights in separate columns in a table, as in the following example.

Back heights.

Fore heights.

7

6

5

3

1

24

2d

66

3d

286571422

539272

If the sum of the forward heights is less than the sum of the back heights, it is evident that the last station must be higher than the first.

Art. 320.-PROB. II. To find the difference between the

true and apparent level, for any given distance.

OBS.-1. The true level is a curve, which either coincides with, or is parallel to, the surface of water at rest.

2. The apparent level is a straight line, which is a tangent to the true level, at the point where the observation is made.

3. The difference between the true and the apparent level is nearly equal to the square of the distance, divided by the diameter of the earth.

1. What is the difference between the true and apparent level, for a distance of one English mile, supposing the earth to be 7940 miles in diameter ?

Ans. 7.98 inches, or 8 inches, nearly. 2. A tangent to a certain point on the ocean strikes the top of a mountain 23 miles distant. What is the height of the mountain? Ans. 352 feet.

PHILOSOPHICAL PROBLEMS.

Art. 321.-PROB. I. To find the time in which pendulums of different lengths would vibrate, that which vibrates seconds being 39.2 inches.

The time of the vibrations of pendulums are to each other, as the square roots of their lengths; or, their lengths are as the squares of their times of vibrations.

RULE.

As the square of one second is to the square of the time in seconds in which a pendulum would vibrate, so is 39.2 inches to the length of the required pendulum.

EXAMPLES.

1. Required the length of a pendulum that vibrates once in 8 seconds. 12: 82: 39.2 in. : 2508.8 in.=209 ft. Ans. 2. How often will a pendulum vibrate, whose length is 100 feet? Ans. 5.53 seconds. Art. 322.-PROB. II. By having the height of a tide on the earth given, to find the height of one at the moon.

RULE.

As the cube of the moon's diameter, multiplied by its density, is to the cube of the earth's diameter, multiplied by its density, so is the height of a tide on the earth, to the height of one at the

moon.

EXAMPLE.

The moon's diameter is 2180 miles, and its density 494; the earth's diameter is 7964 miles, and its density 400. If, then, by the attraction of the moon, a tide of 6 feet is raised at the earth, what will be the height of a tide raised by the attraction of the earth at the moon? Ans. 236.8+ feet.

I. If the diameters of two globes be equal, and their densities different, the weight of a body on their surfaces will be as their densities.

II. If their densities be equal, and their diameters different, the weight of a body will be as of their circumferences.

III. If their diameters and densities be both different, the weight will be as of their semidiameters multiplied by their densities.

Art. 323.-PROB. III. To find how far a heavy body will fall in a given time, near the surface of the earth.

OBS.-Heavy bodies, near the surface of the earth, fall 16 feet in 1 second of time; and the velocities they acquire in falling are as the squares of the times; therefore, to find the distance any body will fall in a given time, we adopt the following

RULE.

As 1 second is to the square of the time in seconds that the body is falling, so is 16 feet to the distance in feet, that the body will fall in the given time.

How far will a leaden bullet fall in 8 seconds?

1282 16 ft.: 1024 ft.:= Ans.

Art. 324.-PROB. IV.

The velocity given, to find the space

fallen through, to acquire that velocity.

RULE.

Divide the velocity by 8, and the square of the quotient will

be the distance fallen through to acquire that velocity.

1. The velocity of a cannon-ball is 424 feet per second. From what height must it fall to acquire that velocity?

Ans. 2809 feet. 2. At what distance must a body have fallen to acquire the velocity of 1024 feet per second? Ans. 3 miles, 544 feet. Art. 325.-PROB. V. The velocity given per second, to find the time.

RULE.

Divide the velocity by 8, and a fourth part of the quotient will be the time in seconds.

1. How long must a body be falling to acquire a velocity of 304 feet per second? Ans. 9 seconds. 2. How long must a body be falling to acquire a velocity of 864 feet per second? Ans. 27 seconds.

Art. 326.-PROB. VI. The space through which a body has fallen, given, to find the time it has been falling.

RULE.

Divide the square root of the space fallen through by 4, and the quotient will be the time in which it was falling.

How long would a ball be falling from the top of a tower, 900 feet high, to the earth? Ans. 7 seconds.

Art. 327.-PROB. VII. The time given, to find the space fallen through.

RULE.

Multiply the time by 4, and the square of the product will be the space fallen through in the given time.

1. What is the difference between the depth of two wells, into each of which, should a stone be dropped at the same instant, one would reach the bottom in 5 seconds, and the other in 3?

5×4=20, and 20 × 20-400: then 3×4=12, and 12 × 12 144: then 400-144-256 ft. Ans.

2. A ball was seen to fall half the way from the top of a tower in the last second of time. How long was it in descending, and what was its height before its descent?

Ans. 186.486+ feet.

Art. 328.-PROB. VIII. To find the velocity, per second, with which a heavy body will begin to descend, at any distance from the earth's surface.

RULE.

As the square of the earth's semidiameter is to 16 feet, so is the square of any other distance from the earth's centre, inversely, to the velocity with which it begins to descend per second.

With what velocity per second will a ball begin to descend, if raised 3000 miles above the earth's surface?

As 4000 x 4000:16:: 4000+3000 × 4000 × 3000: 5.22449 feet, Ans.

And if the height is required, and the velocity given, thus, as 16: 4000 x 4000 5.22449: 49000000, and √/490000004000 3000 miles, Ans.

Art. 329.-PROB. IX. fallen through, given, to strike.

The weight of a body, and the space find the force with which it will

RULE.

Multiply the space fallen through by 64; then multiply the square root of this product by the weight, and the product is the momentum, or force with which it will strike.

There is a monument 64 feet high. weighing 4 tons, should fall from its top would be its force, or momentum ?

Supposing a stone,

to the earth, what Ans. 573440 lbs.

That is, it would strike the earth with more force than the weight of two hundred and fifty tons.

Art. 330.-PROB. X. To find the magnitude of any thing, when the weight is known.

RULE.

Divide the weight by the specific gravity found in the table, and the quotient will be the magnitude sought.

What is the magnitude of several fragments of clear glass, whose weight is 13 ounces?

13-2600.005 of a cubic foot; and .005 × 1728 8.640 cubic inches, Ans.

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