104 Samuel Karp, Jack Kotik, and Jerome Lurye 
Then (A24) becomes 
1/2 1/2 
R = -877F? | | M(x) MCx') Y((F|x-x'|) dxdx'. (A26) 
-1/2 
-1/2 
Now let the density M(x) =—{c/27)(V/2L) f(x), as in Eq. (25) of the text. Then 
1/2 
R = -20c? iy se 
Gorge VOE 
-1/2 - 
The drag coefficient, C,,, is obtained by dividing R by 
2 
sg 
2 2L 
1/2 
f(x) f(x') Y((F|x-x'|) dxdx'.  (A27) 
1/2 
Thus, the final result is 
j f(x) f(x') Yj(F|x-x'|) dxdx' (A28) 
-1/2 
in agreement with Eq. (29). 
Appendix B 
NUMERICAL SOLUTION OF THE INTEGRAL EQUATION 
The equation to be solved is Eq. (30) of the text, viz., 
1/2 
[69 ¥CFlxx! |) dt = 1. (B1) 
-1/2 
It is easily shown that g,(x) = g,(—x), so that (B1) can be written 
1/2 : 
| 6,(x') {y,(Flx-x'|) oY SIR G@at a) dx" ==) (B2) 
0 
As already noted, MacCamy has proved [10] that g (x) has the form 
