Prof. D. N. Mallik on a Potential Problem. 63 
Now it can easily be shown that a helix on a cone cuts all 
the generators at a constant angle on, where cos en = -_. 
1 sin 2 a cos 2 a, cos 2 /3 — cos 2 a 
p l r 2 ' cos 2 a cos' 
sin 2 a sin 2 00 
r* 
p = r cosec a cosec to. 
Similarly <r = r sec a cosec co. 
This, again, is a known result. 
Apparently the above method would apply to a helix on 
any form of surface, bat the results would be too complicated 
to be of any use, except in very special cases. The only 
difficulty is to find doo. The tangent line at any point P 
makes an angle a with OZ and lies in a known plane, i. e. 
the tangent plane at P. It might be possible to calculate 
from this the angle made by the vertical plane containing 
the tangent at P with (say) the plane x = 0. Say this angle 
was found to be -\jr. Then doo = d-^ ; but unless the surface 
was of a very simple nature, the results would be useless 
because of their length and want of simplicity. - This is not 
of much importance, as the most interesting helices are those 
described on surfaces of revolution. These may very con- 
veniently be dealt with by the method given above. 
IV. A Potential Problem. By D. N. Mallik*. 
IP V is the potential of an ellipsoid and X Y Z the com- 
ponent forces, then of course 
X=+Aa'&c, and V=§Axdx+..\ 
At an internal point, ABC) are constants and then 
V=V + i(A^ 2 +B/+G^). 
(Minchin's Statics, vol. ii. p. 326.) 
Ihe same method is also applicable for an external point, 
although I have not seen it given in the text-books. 
* Communicated by Prof. A. W. Porter. 
