426 BIRKELAND. THE NORWEGIAN AURORA POLARIS EXPEDITION, 19021903. 



Po = iL sin \L sin cos . Ii Z. sin cos 6 . L (g) 



L J /' = ,"i 



P0 = /Z. sin (Z, cos cos (>) . I t + Z, sin sin 9 . L (10) 



L J /' = ,"i 



. .. L cos o cos 9 

 Pw = iL sin C . - - 5 - 



o cos 9 I" 



5 -. 



sin [ 



s--. , i i 



e Z. sin sin [ y/,2 _|_ p a __ 2 ? Z, (cos cos + sin sin cos (w /) J ,, = ,,, 



where <<i and ^ represent the values of ft at the ends of the arc. 



We may, then, say 



Po = Po * (to //.) Z J P (w //j) j 



da) 



If, therefore, we calculate the quantities /-"' (w ,11) [i. e. P ( , n (to ft), Pg (to /(), and P,,, n (w u)\ 

 for all values of 6 and ft, we can afterwards determine the length of the piece of current. 



These formulae cannot, however, be employed for 6=0 and 6 = 180, as in these cases P u be- 

 comes, as will easily be seen, infinitely great. Here, therefore, special formulas must be developed for 

 the forces. The following formulae are found for these special cases. 



- i T Z. sin A 



Po = + iL sin C - si A, 



(L- + p 2 + iLq cos ;f/* ' 



-T r 2(i,cos $) . A/' 



o = iL sin L - 5^ sin =5 



' (Z. 2 + e a + 2Z, ? cos ) 3 /a 



cos to 



/'a + /'A 

 - 



2 / 



, Y 2 (Z. COS '+ P) . A/' / /'2 + j" A 



,,, = //, sin L - 57 sin - sin w - 



' Z. 8 >"- 2L cos 3/ 2 2 / 



+ Q"' + 2LQ cos ) /a 



where A," = ,"2 /'i> ar >d where the upper signs will be employed for = 0, and the lower for = 180. 



While P,,i is expressed in algebraic form, the other two components, as we may easily convince 

 ourselves, are expressed as elliptic integrals. 



We have, then, to get these put into a practical form for the numerical calculation. This may be 

 accomplished by using Legendre's normal forms, by means of which we can make a direct use of his 

 tables of elliptic integrals. 



We put 



10 ft = ic 2r, i. e., cos (w ft) = cos 2r = I -(- 2 sin-r (13) 



7,2 _|_ g-2 _ 27, ? (cos cos sin sin 0) = L- + Q- 2Z.0 cos ( -)- 0) = ; (14! 



4Z.p sin sin 



~^T" ~ = ' l ' 5 



Hence the expression for d becomes 



d k-> y 1 AJ sin 2 r (16) 



If we introduce this, we have 



_ a [" l-2sin% , 4 | 1 -* sin'r - 1 + -^L. 



or, if we assume // so that T<> = 0, 



rr 



ch 



