490 
Proceedings of Royal Society of Edinburgh. [sess. 
The first integral is evidently a complete elliptic function of the 
first order, and therefore not expressible in terms of elementary 
transcendents. For convenience this function will he referred to, 
as is usual, by F 1 and its value taken from tables or determined 
by quadrature for any value of e (or e 2 preferably). 
/: 
2 cos 0 d0 = 
o 
Sill 
T =1 
0 
.*. the sum of all the sections = 26(Fj + 1) 
25(F, + 1) u 
and the mean projection width = ^ = — (F T + 1) 
Instead of using tables of elliptic functions, it is instructive to 
use approximate methods of solution. 
(1) Expanding the radical 
1 
by the Binomial 
J 1 - e 2 sin 2 0 
Theorem, the series 1 + \e 2 sin 2 0 4- f e 4 sin 4 0 + .... is obtained 
which is uniformly convergent from 0 = 0 to 0 = ^ radians, since 
e 2 < 1. Integrating this series term by term and using the formula 
* sin" 0 d6 = fr- 1 ) (n-3) . 1 ^ 
o n (n - 2) .... 2 2 
the value of the function is obtained as 
I; 
/, 
do 
“ / 1 I I 
o ^1 - e 2 sin 2 0 2 \ 4 + 64 
9e 4 
which can be easily evaluated for all values of e 2 and to any 
degree of approximation by taking sufficient terms of the series. 
From the nature of the problem, it is, however, not only unneces- 
sary but misleading to. use more than three or four significant 
figures. 
(2) The graphical solution . — Calculate the value of the expres- 
sion 2 b( -- — ^ + cos o') for 10 values of 0 , viz., 0°, 10°, 
20° .... 90°, keeping e 2 constant, say T. Plot these values as 
ordinates and 0 as abscissae. Draw a curve through the plotted 
points. The mean height of the diagram gives, as before, the 
mean projection width for e 2 —'l. Plot out the results on the 
same sheet for e 2 = ’2, *3 . . . . and the different curves on the 
same diagram will render evident to the eye at a glance how the 
