428 
MR. J. H. MICHELL ON THE THEORY OF FREE STREAM LINES. 
and exactly as in the treatment of Schwarz’s formula we have m = ayjiT — 1, where 
a,, is the exterior angle of the polygon corresponding to 
So that, finally,-'' 
£ = Bn|sin(*.-,^,)|}"' ‘ 
As an example, take the case of a rectangle. 
The four singular points may be taken to be 
4> =■ — ct, <f) = a, (j) = I — a, ff) = I a, 
and we have, taking for simplicity I = ^ v, 
~ = A \/ — sin [w — a) sin [iv + ci) cos {iv — a) cos {iv + a), 
= -I A v^(sin^ 2a — siiA 2u’). 
Hence 
2 = -I A I v''(shA 2a — sin^ 2io) cliv + B, 
and 2 is an elliptic integral of w. 
Problem IV. 
Suppose now there are two polygonal prismatic conductors, one at potential 
= — k, the other at xjj = k, and at first (A) suppose that one of the conductors is 
within the other. 
Let (f) increase by 2l in going round either polygon. 
The area in the w plane is now a fiinte rectangle bounded by <^ = — I, (j) = l, 
xp = — k, ^ = k. 
y = ic 
(p = — l 
0 = l 
y = — A; 
In order to satisfy the condition clYlclxp = 0 over xp = — k and \p = k we must 
have a double system of images of the singular points, viz., at 
fpQ “h 2ml, -|- ^')ik, 
(pQ “h 2ml, xpQ “h ( 2?2 -j- 1) 2k, 
* Mr. Brill lias already given this formula (‘Messenger of Math.,’ August, 1889), and I only Insert 
it here for the sake of completeness. 
Note. —Ajiril 29 . Since the above paper was read, Mr. Brill has given the next transformation in 
the same journal. 
