PROB. VI. To find the true weight of any quantity when weighed in each scale of a balance, whose beam is unequally divided.

RULE. Take the square root of the product of the different weights, for the true weight.

A parcel of sugar weighs in one scale 25 lbs.; in the other 30 lbs. What was its true weight?

/25X30=2734.

PROB. VII. The base and perpendicular given, to find the hypothenuse.

RULE. The square root of the sum of the squares of the base and perpendicular will be the hypothenuse.

This rule is illustrated by the following figure.

If the base of a right angled triangle be 9 feet, and the perpendicular 12, what is the hypothenuse?

PROB. VIII.

B

Given the base and the sum of the perpendicular and hypothenuse of a right angled triangle, to find the perpendicular.

RULE. From the square of the sum subtract the square of the base, and divide the remainder by twice the sum, and the quotient will be the perpendicular.

A tree, 100 feet in height, is broken off-the top of the tree reaches the ground 30 feet from the bottom, while the

How high from the
Answer, 45 feet.

part broken off rests on the stump. ground was it broken off?

difference of the hy

PROB. IX. Given the base and the pothenuse and perpendicular of a right angled triangle, to find the perpendicular.

RULE. From the square of the base subtract the square of the given difference, and divide the remainder by twice the difference.

EXAMPLE.

If the base of a right angled triangle be 30 feet, and the difference of the other two sides 6 feet, what is the length of the perpendicular? Answer, 72 feet.

PROB. X. To find the diameter of the earth, from the known height of a distant mountain, whose summit is just visible in the horizon.

RULE. From the square of the distance, divided by the height, subtract the height.

The highest point of the Andes is about 4 miles above the bed of the ocean. If a straight line from this touch the surface of the water at the distance of 1784 miles, what is the diameter of the earth? Answer, 7940.

PROB. XI. To find the greatest distance at which a given object can be seen on the surface of the earth.

RULE. To the product of the height of the object into the diameter of the earth, add the square of the height; and extract the square root of the sum.

If the diameter of the earth be 7940 miles, and Mount Etna 2 miles high, how far can it be seen at sea?

Answer, 126+miles.

No7940+2+2=126.

NOTE. The actual distance at which an object can be seen is increased by the refraction of the air.

2. A man standing on a level with the ocean, has his eye raised 5 feet above the water. he see the surface?

To what distance can
Answer, 27 miles.

PROB. XII. To find the height of an object at sea, or on the surface of the earth, having only the distance given.

RULE. From the given distance, take the distance which the elevation of the eye above the surface will give,

found by the last problem; then divide the square of the remainder by the diameter of the earth, and the quotient will be the height required.

PROB. XIII. To find the content of squared timber.

RULE. Multiply the mean breadth by the mean thickness the product, multiplied by the length, will give the

content.

Required the content of a log, the length 24 feet 6 inches, mean breadth 1 foot 1 inch, and mean thickness 1 foot 1 inch. Answer, 28 feet, 9 inches, 6"".

PROB. XIV.

To find the content of round timber. COMMON RULE. Take one-fourth of the mean girt, and square it, and multiply it by the length, for the content.

NOTE 1. Tapering timber should be divided into pieces of eight or ten feet long, and these parts should be computed separately, and added.

NOTE 2. In order to reduce the tree to such a circumference as it would have without its bark, a deduction is generally made ofor of an inch for every foot of quarter-girt for young oak, ash, beech, &c.; but 1, or even 14 inch, must be allowed for old oak, for every foot of quarter-girt.

NOTE 3. The common rule gives the content too small, by 3 feet on every 11 feet of content; yet it is universally used in practice, being originally introduced in order to compensate the purchaser of round timber for the waste occasioned by squaring it.

RULE 2d. Take one fifth of the girt, and square it, and multiply by twice the length, for the content.

1. Required the content of a tree, 24 feet long, and its girts at the ends 14 and 2 feet. Answer, 96 feet, by the common rule; the true content is 122.83 feet.

2. How much timber in a tree, 18 feet long, and its mean girt 5 feet 8 inches? Answer, common rule, 36 feet 1 inch; true content, 46 feet, 2 inches, 10" 6.""

PROB. I.

LEVELLING.

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; ob. serve the points where the line of level strikes the rods be

fore 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, &c. 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.

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.

PROB. II. To find the difference between the true and apparent level, for any given distance.

NOTE 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 ?

Answer, 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? Answer, 352 feet.

PHILOSOPHICAL PROBLEMS.

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.

EXAMPLE.

1. Required the length of a pendulum that vibrates once in 8 seconds.

12: 82: 39.2 in.: 2508.8 in.—209 ft., Answer. 2. How often will a pendulum vibrate, whose length is 100 feet? Answer, 5.53+seconds.

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.

1. The moon's diameter is 2180 miles, and its density 494; the earth's diameter is 7064 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? Answer, 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 semi-diameters multiplied by their densities.