Let 2BC=D, the diameter of the earth, (Fig. 12.)
BT=h, the height of the object,
AT=d, the distance required.

Then, (D+h) ×h=d3. And d=vDh+h3.

Ex. If the diameter of the earth be 7940 miles, and Mount Etna 2 miles high; how far can its summit be seen at sea? Ans. 126 miles.

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

24. In this problem, the eye is supposed to be placed at the level of the ocean. But if the observer be elevated above the surface, as on the deck of a ship, he can see to a greater distance. If BT (Fig. 13.) be the height of the object, and B'T' the height of the eye above the level of the ocean; the distance at which the object can be seen, is evidently equal to the sum of the tangents AT and AT'.

Ex. The top of a ship's mast 132 feet high is just visible in the horizon, to an observer whose eye is 33 feet above the surface of the water. What is the distance of the ship? Ans. 21 miles.

25. The distance to which a person can see the smooth surface of the ocean, if no allowance be made for refraction, is equal to a tangent to the earth drawn from his eye, as T'A (Fig. 13.)

Ex. If a man standing on the level of the ocean, has his eye raised 5 feet above the water: to what distance can he see the surface? Ans. 21 miles.

26. If the distance AT, (Fig. 12.) with the diameter of the earth be given, and the height BT be required; the equation in Art. 23 gives

h=viD2+d3 —{D

See Surveying, Section IV, on Leveling.

27. When the diameter of the earth is ascertained, this may be made a base line for determining the distances of the

* See note A.

heavenly bodies. A right angled triangle may be formed, the perpendicular sides of which shall be the distance required, and the semi-diameter of the earth. If then one of the angles be found by observation, the required side may be easily calculated.

Let AC (Fig. 14.) be the semi-diameter of the earth, AH the sensible horizon at A, and CM the rational horizon parallel to AH, passing through the moon M. The angle HAM may be found by astronomical observation. This angle, which is called the Horizontal Parallax, is equal to AMC, the angle at the moon subtended by the semi-diameter of the earth. (Euc. 29. 1.)

PROBLEM X.

To find the distance of any HEAVENLY BODY whose horizontal parallax is known.

28. AS RADIUS, TO THE SEMI-DIAMETER OF THE EARTH; SO IS THE CO-TANGENT OF THE HORIZONTAL PARALLAX, TO THE DISTANCE.

In the right angled triangle ACM, (Fig. 14.) if AC be made radius;

R: AC::Cot AMC : CM.

Ex. If the horizontal parallax of the moon be 0° 57', and the diameter of the earth 7940 miles; what is the distance of the moon from the center of the earth?

Ans. 239,414 miles.

29. The fixed stars are too far distant to have any sensible horizontal parallax. But from late observations it would seem, that some of them are near enough, to suffer a small apparent change of place, from the revolution of the earth round the sun. The distance of the sun, then, which is the semi-diameter of the earth's orbit, may be taken as a base line, for finding the distance of the stars.

We thus proceed by degrees from measuring a line on the surface of the earth, to calculate the distances of the heavenly bodies. From a base line on a plane, is determined the height of a mountain; from the height of a mountain, the diameter

of the earth; from the diameter of the earth, the distance of the sun, and from the distance of the sun the distance of the

stars.

30. After finding the distance of a heavenly body, its magnitude is easily ascertained; if it have an apparent diameter, sufficiently large to be measured by the instruments which are used for taking angles.

Let AEB (Fig. 15.) be the angle which a heavenly body subtends at the eye. Half this angle, if C be the center of the body, is AEC; the line EA is a tangent to the surface, and therefore EAC is a right angle. Then making the distance EC radius,

That is, radius is to the distance, as the sine of half the angle which the body subtends, to its semi-diameter.

Ex. If the sun subtends an angle of 32′ 2", and if his distance from the earth be 95 million miles; what is his diameter ? Ans. 885 thousand miles.

PROMISCUOUS EXAMPLES.

1. On the bank of a river, the angle of elevation of a tree on the opposite side is found to be 469; and at another station 100 feet directly back on the same level, 31°. What is the height of the tree? Ans. 143 feet.

2. On a horizontal plane, observations were taken of a tower standing on the top of a hill. At one station the angle of elevation of the top of the tower was found to be 50°; that at the bottom 39°; and at another station 150 feet directly back, the angle of elevation of the top of the tower was 32°. What are the heights of the hill and the tower?

Ans. The hill is 134 feet high; the tower 63.

3. What is the altitude of the sun, when the shadow of a tree, cast on a horizontal plane, is to the height of the tree as 4 to 37 Ans. 36° 52' 12".

4. If a straight line from the top of the White Mountains in New Hampshire touch the ocean at the distance of 1031 miles; what is the height of the mountains?

Ans. 7100 feet.

5. From the top of a perpendicular rock 55 yards high, the angle of depression of the nearest bank of a river is found to be 55° 54', that of the opposite bank 33° 20'. Required the breadth of the river, and the distance of its nearest bank from the bottom of the rock.

The breadth of the river is 46.4 yards;
Its distance from the rock 37,2,

6. If the moon subtend an angle of 31′ 14′′, when her distance is 240,000 miles; what is her diameter ?

Ans. 2180 miles.

7. Observations are made on the altitude of a balloon, by two persons standing on the same side of the balloon, and in a vertical plane passing through it. The distance of the stations is half a mile. At one, the angle of elevation is 30° 58′, at the other 36° 52'. What is the height of the balloon above the ground? Ans. 11⁄2 mile.

8. The shadow of the top of a mountain, when the altitude of the sun on the meridian is 32°, strikes a certain point on a level plane below; but when the meridian altitude of the sun is 670, the shadow strikes half a mile farther south, on the same plain. What is the height of the mountain above the plain? Ans. 2245 feet.

NAVIGATION.

SECTION I.

PLANE SAILING.

ART. 33. NAVIGATION is the art of conducting a ship on the ocean. The most accurate method of ascertaining the situation of a vessel at sea is to find, by astronomical observations, her latitude and longitude. But this requires a view of the heavenly bodies; and these are often obscured by intervening clouds. The mariner must therefore have recourse to other means for determining the progress which he has made, and the particular part of the ocean through which he is at any time making his way. The common method is to measure the rate of the ship's going by a log-line, and to find the direction in which she sails by a mariner's compass. From these data, the difference of latitude, the departure, and the difference of longitude, may be calculated. The two first may be found by plane sailing; the last by middle latitude sailing, or more correctly by Mercator's sailing. See Sec. II. and III.

34. The log-line is a cord which is wound round a reel, one end being attached to a piece of wood called a log. It is used to determine the distance which a ship runs in an hour, by measuring the distance which she runs in half a minute. The log is commonly a small piece of board, in the form of a quadrant of a circle. The arc is loaded with a quantity of lead sufficient to give the board a perpendicular position, when thrown upon the water. This will prevent it from moving forward toward the vessel, while the line is running off the reel. So that the length of line drawn off by the log