Page:Scientific Memoirs, Vol. 2 (1841).djvu/212

Resolving the horizontal magnetic force into two portions, one of which, $X$, acts in the direction of the geographical meridian, and the other, $Y$, perpendicularly to that meridian,—and considering $X$ as positive when directed towards the north, and $Y$ as positive when directed towards the west,—then

${\begin{aligned}&X=-{\frac {(1-(2\epsilon -\epsilon ^{2})\cos u^{2}){\frac {3}{2}}}{(1-\epsilon )^{2}}}\centerdot {\frac {d\,V}{R\,d\,u}}\\&Y=-{\sqrt {(1-(2\epsilon -\epsilon ^{2})\cos u^{2})}}\centerdot {\frac {d\,V}{R\sin u.d\,\lambda }}\end{aligned}}$.

The total horizontal force is then

$={\sqrt {(X^{2}+Y^{2})}}$,

and the tangent of the declination

$={\frac {Y}{X}}$.

Neglecting the square of the ellipticity, $\epsilon$, the expressions become

${\begin{aligned}&X=-(1+(2-3\cos u^{2})\epsilon )\centerdot {\frac {d\,V}{R\,d\,u}}\\&Y=-(1-\epsilon \cos u^{2})\centerdot {\frac {d\,V}{R\sin u.d\,\lambda }},\end{aligned}}$

or, setting the ellipticity quite aside,

$X=-{\frac {d\,V}{R\,d\,u}}$

$Y=-{\frac {d\,V}{R\sin u.d\,\lambda }}$.

The data furnished by the observations which we possess are much too scanty, and most of them much too rude, to make it advisable at present to take into account the spheroidal form of the earth. It would not be difficult to do so; but it would complicate the calculations without affording any corresponding advantage. We will therefore adhere to the last-mentioned formula, in a which the earth is considered as a sphere, whose semi-diameter $=R$.

15.

If $X$ be expressed by a given function of $u$ and $\lambda$, $Y$ can be be deduced from it a priori.

Let the integral $\int _{0}^{u}X\,d\,u=T$, considering $\lambda$ as constant in the integration: it is then clear that if we differentiate in a similar