3 6 



BIRKELAND. THE NORWEGIAN AURORA POLARIS EXPEDITION, 1902 1903. 



p " _ ' 

 1 " s 



5 



(4) 



If we divide (3) by (4), we find that 



<? = + . (5) 



In the cases we shall come to examine here, however, the current is always between Kaafjord 

 and Axeleen, so that <J will always be positive. In equation (5) p and q are known quantities; hence 

 d can be calculated, and from (i) and (2) we obtain 



Dp 



h = ---; sin ip 

 From (5) and (3) / can then be calculated, and we obtain 



sn 



(6) 



(7) 



With regard to the determination of the angle (//, we should remark that if the current-arrows 



for Kaafjord and Axeleen make the same angle with the great circle between these two stations, t// will 



simply equal that angle. The angles will generally be somewhat different. Calling them respectively 

 t^j and tjj. 2> we put 



NUMERICAL VALUES FOR HEIGHT AND STRENGTH OF CURRENT. 



(i) The Perturbation of the ijth December, 1902. 



76. During this perturbation the balance at Kaafjord stuck fast, so the direction can only just be 

 distinguished. As the sensibility of the balance at Axeleen was not determined until after the return 

 of the expedition, and may thus be not altogether free from error, we will see what can be concluded 

 regarding height and strength of current, when we suppose that we know only the horizontal compo- 

 nents, and the direction, but not the strength, of the vertical components. 



Between i h 45 and 2 h o m , the horizontal components at Axeleen and Kaafjord are almost alike 

 in direction; and the outer field shows that the storm-centre during this time must be somewhere near 

 Spitsbergen. We will therefore take i h 52-5 m as the most favorable moment for determining the 

 strength and altitude of the current. At this point of time, the values are as follows: 



P," = i86y 

 P,' = 30 

 D = 896 km. 



(// = 70. 



We further introduce here a quantity x which is thus defined, 



_P," 



6.2 . 



If we divide the equation (4) by (3), we find 



and by employing equation (6) we obtain 



ft ^ 



D sin i// 1/ i x(5 2 



i -f- (J ' x i 



(8) 



