Particle on the Surface of a Smooth Rotating Globe. 467 



Case I. e. — Further increase of Ao leads to looped tracks 

 in which the motion is, on the whole, from east to west. 

 As an example, we take the case in which the equator is 

 crossed at 150°. 



^=cot 75° = '268, x o = ^/|^sin 75° = '201, 



X R = 1280 km., k = sin 75°= -966. 



The tables give K= 2-7681, E = 1'0764. 



Hence by (7*6) and (7-32) the period = 203 hr.= 8£ days and 

 the wave-length =1630 km. 



To determine the size of the loops we require the positions 

 of the double points. These are found by solving the 

 equation 



zn 



obtained by writing <f> = in (7*33). 



The root is found to be approximately -^ = 2*03. The 

 latitude of the double points is given by 



\=X cn^ = , 208 \ . 



Hence \R=266 km. 



The width of a loop can be determined by investigating 

 the points where u, the easterly component of the velocity, 

 vanishes. The condition (derived from (7*27)) may be 

 expressed in the form 



cn^r=[i(l-^)] 



1/2. 



This equation is satisfied in this particular case by tJt=JK, 

 and on substitution in (7*33) we find that the half-width of 

 the loop is 368 km. 



Case II. — As the interval between the loops is increased, 

 we approach the limiting case in which the equator is 

 asymptotic to the track. 



/2V 



In this case \ = \ / p— ='208 (by writing a — it in (7*5)), 



and therefore X R = 1324 km. 



To determine the size of the loop, its width and also the 

 position of the double point must be found. To find the 



