with the help of a Table of cosines and co-tangents to form the sum of the series in the last Article. The values of u are the first 21 values of a in the Table in Art. 60.

From this Table we gather, that the Deflections caused at the station by the superficial mass one mile thick, when distributed uniformly through the depths 208, 417, 625 miles, are 3".385, 2".088, 1".478. The densities of the matter thus

= 7.101

= 5.028

=2".088

= 1".478

Let a and a + be the angular distances from the station of the two circles bounding this compartment; h the height of the mass; the angular distance along the surface of an elementary vertical prism of the mass; a the radius of the earth; the angle which the plane of @ 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 prisma2 sin @d↓ do.

Since the height of prism (h) 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 0, by Art. 48,

Attraction along the tangent to eph sin Oddy 4 sin20

cos 10;

.. attraction along the tangent to the mid-line of the lune

neglecting only the cube and higher powers of sin 10,

The law of dissection we shall choose will simplify this; for we are to assume such a relation between $ and a that the expression in o 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 4 is quite arbitrary, we will assume it such that when a and are small, & shall

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

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.

a

Since 0.000005523 = tan (1′′. 1392);

.. deflection of the plumb-line caused by this attraction = 1′′.1392h sin ß ........

......

(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 6, 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

mass.

=

a2ß {cos a― cos (a + )} = 2a3ß sin 10 sin (a + 16).

Hence, if ß be not taken larger than 30°, and since is always small, this gives a+, 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.

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,

Equation (1) can be solved by expansion so long as a and are not too large.