276 Mr. W. D. Niven on the Calculation 



?i=3, aud suppose that we are integrating over an arc of 5°, in which one 

 fourth of the horizontal velocity is lost. Then 



«£X j. T 1 



— = -tan0 . gyV^; 

 _ 



-^=cot 2(p y-Vo" approximately. 



It will therefore be observed that the error committed is trifling. As 

 (p will be small over the greater part of any trajectory, this error will 

 chiefly affect T ; but by good fortune the sign of the error is opposite 

 to the error we shall presently find, and therefore helps to neutralize that 

 error. 



Eeturning to the integral for X, we are now to take account of the 

 square of ty. Let the right-hand side of equation (6) be put equal to 



' ' ^?isec" +1 mi/. 

 9 



Then 



V =\jj - ^±1 tan en// 2 , 



Substituting this value of \p in the expansions of cos n_1 and cos' l " J ^sin<p, 

 we get 



cos n _1 a + (n - 1) cos n ~ 2 a sin av// 



+ — i— { 2n sin2 « — 1} cos" -3 a\p' 2 

 + etc., 



and 



cos" -2 a sina+ {(>i — 2) sin 2 a — cos 2 a} cos n_3 a^ 



+ {2(n -l)(n-2) sin 2 a - 4(n- 1) cos 2 a} cos"" 4 a sin a ^ 

 + etc. 



When the expressions just found are substituted in the (C) and (D) 

 integrals, it is obvious that we have to determine these two integrals — 



