36

MENSURATION OF

Several methods have been invented to find the earth's diameter. Mr Picart of the Academy of sciences at Paris, has proposed an exact method, by which, not only the equatorial and polar diameters may be known, but also the figure of the earth determined.

According to Mr Picart, a degree of the meridian at the latitude of 49° 21', was 57.06 French toises, each of which con. tains 6 feet of the same measure; from which it follows, that if the earth be an exact sphere, the circumference of a great • circle of it, will be 123.249,600 Paris feet, and the semi• diameter of the earth, 19.615,800 feet: but the French mathematicians, who, of late, examined Mr Picarts observations, affure us, that a degree in that latitude, is 57.183 toises. They measured a degree in Lapland, in the latitude of 66° 20', ' and found it to be 57.438 toises. By comparing these degrees, • as well as by the observations on pendulums, and the theory ' of gravity, it appears, that the earth is an oblate spheriod; and the axis or diameter that paffes through the poles, will be ⚫ to the diameter of the equator, as 177 is to 178, or the earth ' will be 22 miles higher at the equator, than at the poles. A • degree has likewife been measured at the equator, and found

to be confiderably less than in the latitude of Paris, which ' confirms the oblate figure of the earth. Hence it appears, ' that if the earth were of an uniform density from the surface ' to the centre, then according to the theory of gravity, the me'ridian would be elliptical, and the equatorial would exceed ' the polar diameter, by about 44 miles.'

PROBLEM III. Plate 5. fig. 9.

To find the height of an object, by means of one staff.

Suppose the pole AB of an unkown height, BC a horizontal plane, and ED a staff of a known length. At any conve

:

venient distance from the pole, fix your staff perpendicular in the ground, then move backwards or forwards, till you find the point C, whence you may view the top of your staff, E, in a line with A the top of the object, then say, as CD: DE::CB: BA the height of the object. Fig. 67. plate 5.

EXAMPLE.

Let BC be 80 feet, CD 5, and DE 4, required AB.

1

5:4::80

4

5)320

65=AB.

PROBLEM IV.

To measure the height of an object from the length of its shadow.

Place any staff of a known length in the same plane with the object; then say, as the length of the staff's shadow, is to the length of the staff; so is the length of the object's shadow: to its height.

EXAMPLE.

Wanting to know the height of a steeple, whose shadow I found to be 200 feet, I fixed my staff perpendicular to the horizontal plane, the length of the staff, is 4+ feet, and of the shadow, 6 feet, required the height of the steeple.

To measure the height of an object, by a plane mirror, or by a busket full of water. See fig. 69

Place the mirror or bucket between you and the object. So that the top of the object may appear in the middle of the horizontal surface, then say, As the distance between the object, shadow, and your feet, is to the height of the eye; so is the distance between the object's shadow, and the object; to the height of the object.

PROBLEM VI.

Distances may also be measured by loud founds, fuch as, the firing of a cannon, the talling of a bell, thunder, &c.

It has been found, by many exact experiments, that the uniform velocity of found, is 1142 feet, per second of time. If, therefore, the seconds elapfed, be multiplied by 1142, the product will be the answer in feet.

EXAMPLE I.

After seeing a flash of lightning, it was 8 seconds before I heard the thunder, required the distance.

1142
8

5280)9136(1
5280

3)3856

1285 Anf. I mile 1285 yards.

EXAMPLE II.

After obferving the firing of a cannon, 24 seconds elapsed, before I heard the report, required the distance. Anf. 5 miles 336 yards.

EXAMPLE III.

After observing a man striking a bell with a hammer, 5 feconds elapfed before I heard the found. What was the diftance? Anf. I mile 430 feet.

PROBLEM VII.

To find the velocity of the wind.

Observe the shadow of a cloud at any particular place, then count the number of seconds elapsed, before it reach any other particular place; then say, As the number of seconds elapfed bottom of the object 72° 8', and of its top 78° 20′. Required the height and distance of the object.

Here, because the angle ABC is 42° 18' the angle BAC is 47° 42; consequently, its fupplement, the angle BAD will be 132° 18. And fince all the angles of a triangle are equal to two right angles, and that the angle DBA is 29° 50', the remaining angle BDA will be 17° 52' Again, the angle CDE is a right angle, of which the angle BDC is a part; therefore, the angle BDE is 72° 8', and the angle at E 101° 40'; also the angle DBE will be 6° 12'.

As fine ADB 17° 52′ 9.48686 | As fine BDA 17° 52′ 9.48686 is to AB 170

2.23045 So is fine ABD 29°50′ 9.69677 |

is to AB 170 fois fineBAD=132°18′9.86902

2.23045

EXAMPLE III. Plate 5. fig. 1.

Being on a horizontal plane, I took the angle of elevation of the summit of a hill, and of the top of a tower built upon it, and found them to be 48° 20' and 61° 25′. I then measured back 150 yards, and found the angle fubtended by the height of the tower above the plane to be 38° 19'. Required the height of the tower.

The