240 Proceedings of the Royal Society of Edinburgh. [Sess. 
0 = 3-^ = (32 + 2 V + 20g 4 ) 
a cos 0 
+ (80^ +11 6# 3 ) cos 0 
- (96 — 12^ 2 - 177# 4 ) cos 2 6 
- (160g + 232 q 3 ) cos 3 
- (HO7 2 + 560g 4 ) cos 4 6 
+ 231<p cos 6 6 — 0 ..... (8) 
The solution of this equation is 
cos. 0 = 0-57778 + 0T3457? .... (9) 
Greatest Numerical Error. 
It appears from Table II that the minimum value of e is in every case 
numerically greater than its maximum value ; hence, confining our atten- 
tion to the former, we have for the greatest numerical error 
e= -0-0miq 2 (l + '7222q+l-21q 2 +V18q 3 ). . . . (10) 
Table III. 
p- 
2- 
cos 0. 
| 
e. 
e. 
a. 
4 
•242424 
•6104 
52-38 
- -00357 
1-4031 
5 
•196078 
•6042 
52-83 
- -00221 
1-4071 
7 
•141414 
•5968 
53-36 
- *00108 
1-4106 
10 
•099502 
•5912 
53-76 
- -00051 7 
1-4124 
16 
•062378 
•5862 
54-11 
- -000197 
1-4135 
QO 
0 
•5774 
54-736 
0 
1-4142 
It will be observed from fig. 8 that as p increases, the positive and 
negative parts of the curve became more and more alike ; but that complete 
similarity is only attained in the limiting case where p — go , and the curves 
have become coincident with the base line. 
Amplitude of Harmonic. 
The column giving the amplitude, headed a in Table III, is calculated 
thus : — 
From equation 3 
a = V( 2p 2 +l + 2p)- A /(V+i-2p) 
.*. ia 2 = (2^ 2 +l)-V(^ 4 +l); 
and sufficiently nearly 
la 2 = 1 - — + 1 
2 (2p) 2 (2p)« 
( 11 ) 
For all practical purposes the value of a within the range from p = 4 to 
p = oo may be taken as constant and equal to 1-41. 
