88 
MR. DAVIES’S GEOMETRICAL INVESTIGATIONS 
But by the triangle TNU we have 
r u sin 0 n = r, sin 0 t , 
which substituted in (84.) gives 
d y sin 3 — sin 3 6 tl 
dx sin 2 0 t cos + sin 2 0 /; cos d t ’ ••••••• (85.) 
First. That the tangent may be parallel to the magnetic axis, we must have 
sin 3 0 t — sin 3 0 U = 0, 
which resolves itself into 
and 
sin 0, — sin 0 U == 0 
(86.) 
sin 2 0 t + sin 0 t sin 0 n + sin 2 0 n = 0 (87.) 
The latter of these equations (87.) being imaginary, (for sin 0 n = — — - sin 0 P ) 
the only points that exist where the property holds good are to be determined from 
the former (86.). 
This equation may be fulfilled by the four following systems of values j 
1 . + 0 t and + 0 U ; 2. 0 t and cr + 0 n ; 3. *r ip 0 t and 0 U ; and 4. <z Ip and % ^ 6 n . 
But whichever of these we employ, it must be consistent with the equation of the 
curve itself ; viz. with 
cos d t + cos 0 lt = 2 cos /3. 
1. The first of these obviously gives the points P and P', and is consistent with the 
equation of the curve. 
2. The fourth is virtually the same as the first, if we consider that in taking one 
supplement we should take all the supplements. 
3. The second and third are incompatible with the equation of the curve itself. 
There are hence only the two points P and P' at which the needle can be parallel 
to the axis. 
Secondly. That the tangent may be perpendicular to the axis, we must have 
sin 2 0 t cos 6, + sin 2 0 U cos 0 U 0 (88.) 
Expressing sin 2 sin 2 0„, and cos 0 n in terms of (3 and 0 t from the equation of the 
curve, and inserting the results in (88.), we shall obtain, after slight reductions, 
- . sin /3 1 
cos 0 t = cos (3 + -^j=, 
, . n sin /3 
and cos 0 n — cos (3 -}- ^=. 
> 
(89.) 
7 r 
From these equations it appears that if (3 be less than -j, there can be no point or 
the system at which the tangent is perpendicular to the axis, as in that case either 
cos 0, or cos 0 n would be greater than unity. 
