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mountain-ground in the world lies to the north of that continent, and an unbroken expanse of ocean stretches south down to the south pole. Both these causes, by opposite effects, make the plumb-line hang somewhat northerly of the true vertical.

In the following Propositions a method is laid down for calculating the attraction of an irregular superficial stratum of the Earth's surface, and making it depend altogether upon the contour of the surface. The method pursued is this : A law of geometric dissection of the surface is discovered which divides it into a number of four-sided spaces, such that if the height of the attracting mass were the same in them all, they would all attract the given station exactly to the same amount, whether far or near.

In this case it would be necessary only to calculate for one space, then count the number of

spaces the country under consideration, and the final result is easily attained. The country being supposed irregular, the heights in the spaces will not be all alike. The principle, therefore, should be stated thus, that the attractions of the masses on the several compartments are in proportion to their mean heights. These mean heights are known by knowing the contour of the country.

PROP. To discover a Law of Dissection of the surface of the earth into compartments, so that the attractions of the masses of matter standing on them, upon a given station, shall be exactly proportional to the mean heights of the masses, be they far

in

or near.

58. Suppose a number of great circles to be drawn from the station in question to the antipodes, making any angle ß, each with the next, thus dividing the earth's surface (which we may in this calculation suppose to be a sphere, without incurring any sensible error) into a number of Lunes. Then, with the station as centre, describe on the surface a number of circles, at distances the law of which it is our object now to determine, dividing the whole into a number of four-sided compartments.

We will begin by calculating the attraction of a mass of matter, standing on one of these compartments, at a uniform height throughout, upon the station in a horizontal direction.

4

P. A.

Let a and a + $ be the angular distances from the station of the two circles bounding this compartment; h the height of the mass; 0 the angular distance along the surface of an elementary vertical prism of the mass; a the radius of the earth; y the angle which the plane of 0 makes with the plane of the mid-line of the lune, and in which latter plane the resultant attraction evidently acts. The area of the base of the prism = a' sin 0dy do.

Since the height of prism (K) is supposed very small, the distances of its two extremities from the station may be taken to be the same, and = 2a sin 10. Its attraction along the chord of , by Art. 48,

pa’h sin 0 de dy

4a sin' 10 Attraction along the tangent to 6 = ph sin 0 do dys

4 sin 10

cos 10; .. attraction along the tangent to the mid-line of the lune

ph cos' 10 de dys
2 sin je

cos y;

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ato cosa 10 :

dsin jo

tan (a + 0) = 2ph sin B {log.

+cos (a + $) – cos a} tan la

sin (10+£$)+sin 10 2ph sin }B {log. sin (fa+10) – sin $

-2 sin (1 a+10) sin 10} Aph sin (fa+

sin fa+)

neglecting only the cube and higher powers of sin 10,
ph sin .

$coso (fa+10)
sin (1a +10)

neglecting 4* &c. The law of dissection we shall choose will simplify this; for we are to assume such a relation between 0 anda that the expression in $ may be constant, in order to make the attraction the same for all compartments in which h is the same or varying as h where the heights of the masses standing on the compartments are different. As the value of the constant to which we equal the function of a and $ is quite arbitrary, we will assume it such that when a and p are small, o shall

ha 4 = iou In this case it

1a + aa a 21

4 21

$ cos' (1a +10) Hence

sin (1a + 10) defines the Law of Dissection.

(1),

The attraction of the mass standing on the compartment, in consequence,

ph sin ;

}B 21

4

an exceedingly simple expression. We may obtain it in terms of gravity as follows. Let p the density be the same as that of the mountain Schehallien, viz. 2.75; the mean density, according to Mr Baily’s repetition of the Cavendish experiment, being 5.66; g gravity; a=4000 miles.

Now

9

47

a x mean density 3

4п 566

a 3

275

Pi

... attraction of mass on any compartment

4 3 275 the

g=0.000005523h sin 1 B.g, 212 47 566

being expressed in parts of a mile.

Since 0.000005523 = tan (1". 1392); ::: deflection of the plumb-line caused by this attraction =1". 1392h sin .

(2). The method of using this theorem is as follows. When the numerical values of the successive pairs of a and $ are determined by the solution of equation (1) giving the law of dissection, lay them and the lunes down on a map of the country the attraction of which is to be found. It will thus be covered with compartments. After examining the map, write down the average heights of the masses standing on all the several compartments of any one lune; add them together, multiply the sum by 1".1392 sin 1 B, and equation (2) shows that we have the deflection caused by the mass on the whole lune in the vertical plane of its middle line. Multiply by the cosine and then the sine of the azimuth of that middle line, and we have the deflections in the meridian and the primevertical. The same being done for all the lunes, and the results added, we have the effects in meridian and primevertical produced by the whole country under consideration.

59. COR. That the mass on each compartment will attract as if collected at the middle of the mid-line appears from what follows. Let @ be the distance at which the matter may be concentrated so as to produce the same effect as the actual

Now the area of the compartment

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=a*B{cos a - cos (a +$)}=2aß sin dø sin (a + 10).

Then by the last Proposition,

ph sin 1B cos' (fa +20)

= attraction, by hypothesis sin (ža +10)

2a’ß sin £¢ sin (a+) cos 40; =ph

4a sin10

=

sino 10

4B sin 10 sino (1a + 10) cos 10 sin 16 cos (1a + 10) Hence, if ß be not taken larger than 30', and since ® is always small, this gives @ =a + žo, which coincides with the middle of the mid-line of the compartment.

PROP. To calculate the dimensions of the successive compartments from the law of dissection.

60. For this purpose we should solve the equation of last Proposition, viz. o coso (1a + 10) 4

(1). sin (1a +40)

21 But this cannot be done. We must therefore approximate, which will equally well suit our purpose. In order to afford a test of the values we arrive at the equation may be written under the following form. Putting the angle • for the arc ,

4 180 sin (1a +10) 8°

21 7 cos” (1a + 10)

(11.0379639
or = log1{+ log sin (1a +10)

- 2 log cos (1a + )

...

(3).

Equation (1) can be solved by expansion so long as a and $ are not too large. It gives 4 la

1
φ
+

+ +
21 2

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