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EXAMPLE V. Plate 5. fig. 3. From the top of a tree 70 feet high, I took the angle of depreslion of two other trees, lying directly in a straight line from me, and on the same horizontal plane with the tree on which I then stood, viz. that of the nearer 36°, and of the other, 55° 30'. Required their distance from the tree from which the obfervation was taken, and from one another.

To find the dift. of the nearer. To find the dist of the other. As radius 90 1 0.00000

10.00000 to height of tree 70 1.84510 is to heightof tree 70 1.84510 So is tan. dep. 36° 9.86120 | Soistan.2.depr.55°30'10.16287

Assad. 90

To the dist. 50.86

1.70636 | To the dift. 101.9

2.00797

The distance of the farthest
The distance of the nearer

ior 9 feet. 50.86 feet.

51.04 feet.

.

Their distances from one another

EXAMPLE VI. Plate 5. fig. 4. Wanting to know the distance between a house and a tree, the tree being on the other side of a river; I took my first sta- . tion at the house, and marked my second at B; the angle fub

subtended by the distance between my second station, and the tree is 60°. I then measured the distance between my first and fecond stations, 380 yards, and found the angle subtended by the house and tree to be 43o. Required the distance between

° the house and the tree.

As

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12.41356 To the distance 266 2.42484

EXAMPLE VII. Plate 5. fig. 5: I wished to know the distance between a kirk and a mill, which were upon the other side of a river, I choose two ftations, A and B, distant 400 links, and found the angles MAK 40°, KAB 64° 25', and ABM 56°15', MBK 50° 8'. Required the distance between K the Kirk, and M the Mill.

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106° 23' ABK 64 2; KAB

170 48

180 00

9 12 ang. AKB.

In the triangle AKM, to find the angles AMK, MKA. As AK+AM 3405

3.53212 to sum

to fum 70° 0o is to AK---AM 1395

3.14457 | add dif. 48° 23' So is tan. AMK+MKA 70° 10.43893

the greater 118 23

13.58350i the less To tan. AMK-MKA 48° 23'10. 05138

2

21 37

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Neie, The foregoing example may be performed, by ufing MB

and BK as the containing fides.

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If the Peak of Teneriff be four miles above the level of the sta, and the angle of depreflion taken from the farthest visible point, be 87° 25' 55". Required the diameter of the earth, also the farthest visible point that can be seen from the Peak.

If the square of the visual ray, being a tangent to the earth, be divided by the height of the spectator's eye, above the level of the sea, the quotient will give the earth's diameter, and the height of the spectator's eye above the level more.

Demon. Because the straight line AC is equally divided at E, and produced to the point D, the rectangle AD, DC, together with the square of EC, is equal to the square of ED, but the {quare of ED is equal to the squares EB, BD, because DBE is 2 right angle; therefore, the rectangle AD, DC, together 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, DB=AD and AD-DC=CA the diameter.

DC.

To find FD.

To find CE.
As rad. 90°

10.000OO
As rad. 90°

10.00000 is to DC 4 0.60206 is to DC 4

0.60206 So fec. 87° 25' 55" 11.34866 So is tan. 87° 25' 55" 11.34822

To FD 89.27

1.95072 | To CF 89.18

1.95077

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*=7961

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

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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 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, 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 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.'

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PROBLEM III. Plate 5. fig. 9.

5

To find the height of an objec7, by means of one fiaf:

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

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