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with the square of EC=EB, is equal to the squares EB, BD; take away the common square EB, and the remaining rectangle AD, DC, is equal to the square of BD the visual ray. And because the rectangle AD, DC, is equal to the square of BD, (Euclid. 17th. 6.) DC : DB : : DB : AD.: Therefore, DB2 =AD and AD-DC=CA the diameter.

:

DC.

To find FD.

To find CF.

As rad. 90°
10.00000 As rad. 90°

10.00000 is to DC 4 0.60206

is to DC
4

0.60206 So sec. 87° 25' 55" 11.34866 So is tan. 87° 25' 55" 11.34822 To FD 89.27 1.95072 / To CF 89.18

1.95028

Here it must be observed, that if from any point without a circle, two straight lines be drawn to touch the circle, they are equal to one another, (Eucl. 37. 3.); therefore, FC is equal to FB, but BF and FD make up BD the visual ray; consequently,

; it will be 89.18+89.27=178.45=BD, and 178.45'=7951

4

=AD, and 7961–4=7957, the earth's diameter nearly.

To find BE the semidiameter.
As rad. 90°

10.00000
is to BD 178.4

2.25139 So is tang. 87° 25' 55" '- 11.34822

TO BE the semidiameter, 3978 3.59961

The diameter of the earth

7956

Several

6

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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 s if the earth be an exact sphere, the circumference of a great circle of it, will be 123.249,600 Paris feet, and the femidiameter of the earth, 19.615,800 feet: but the French mathematicians, who, of late, examined Mr Picarts observations,

aflure 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; 6 and the axis or diameter that pafles 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 likewise been measured at the equator, and found to be considerably 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 meridian would be elliptical, and the equatorial would exceed the polar diameter, by about 44 miles.'

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PROBLEM JII. Plate 5. fig. 2.

To find the height of an object, by means of one fiaff:

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

nient

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.

5 : 4 :: 80

4

5)320

65=AB.

PROBLEM IV.

a

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; fo is the length of the object's shadow: to its height.

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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 41 feet, and of the thadow, 6 feet, required the height of the steeple.

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To measure the height of an object, by a plane mirror, or by a bucket

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.

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Distances may also be menfured by loud sounds, 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 elapsed, be multiplied by 1142, the product will be the anfwer in feet.

EXAM

EXAMPLE I.

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

1142

8

5280)913611

5280

3)3856

1285; Anf. i mile 12851 yards.

EXAMPLE II.

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

EXAMPLE UI.

After observing a man striking a bell with a hammer, 5 feconds elapsed 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 fay, As the number of seconds elapsed

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