From (143), to first order in f, find 



sin d'= sin d - (f/4) Y sin 2d 

 From (141), (144), and (145) find 



X'=X + 2fX-fX' 



Y'= Y + 2fY - fXY + (f/2) (X^ - Y^) cos d . 

 Hence the transformations (140) are from (143), (146), and (147) the following: 



d'= d - (f/2) Y sin d + (fVl6) [4Y (X - 3) sin d + (2Y'-X') sin 2d] 



sin d'= sin d - (f/4) Y sin 2d 



X'=X + 2fX-fX' 



Y ' = Y + 2fY - fXY + (f/2) (X' - Y') cos d 

 Substution of the relations (148) into (138) produces equations (13?^ providing a second 

 check of Sodano's formula for the inverse solution 



The inverse of the transformations (148) which will carry (139) into (138) are: 



d = d ' + (f/2) Y ' sin d ' + (fVl6) [4Y ' (X '- 1) sin d '+ (2Y'' - X ") sin 2d '] 



sin d = sin d'+ (f/4) Y'sin 2d' 



X = X'-2fX'+fX'' 



Y = Y'-2fY'+fX'Y'+(f/2) (Y'^-X'^) cos d'. 



(146) 



(147) 



(148) 



(149) 



DIFFERENCE FORMULAE FOR THE TWO SECOND ORDER FORMS 

 From equation (142) to second order in f, find 



2 sin '<;6o = 2 sin % (1 + 2f - 2f sin '6, + 3P - 7f sin '6^ + 4f sin "do), 

 and extending (144) to second order in f 



cos (d/ + dj') = cos (d, + dj) + f sin ^do cos d sin ^ {i^ + dj) 



- (fV2) sin \ sin ^(d. + d,) 



M sin ^do cos (d, + dj) 



+ sin ^6o cos d - (3/2) cos d 



+ (3/2) sin ^do cos 2d cos (d, + dj) 



(150) 

 (151) 



From equations (148), by factoring sin d out of every term of the expression for d ', we 

 can write: 



d'= sin d {T - (f/2) Y + (fV8) [2Y(X-3) + (2Y' - X') cos d]! 



Since we can write X'= 2 sin ^(f>o, X = 2 sin ^^o, Y'= 2 sin ^cf)^ cos (d,'+ dj'), 

 Y = 2 sin ^00 cos (d, + dj) we have from (150) and then combining (150) and (151) 

 (multiplying respective members together) 



(152) 



91 



