Particle on the Surface of a Smooth Rotating Globe. 469 



The smallest positive root is -^=*881 and, for this value 

 of i/r, tanh -^ = 1/V'2 = *707. Accordingly, the half-width 

 is 



R0 = RX o [tanh^-ii|r] =1324 x *266 km. 

 = 352 km. 

 The position of the double point is found by solving 

 tanh \jr=^. 



The positive root of this equation is -yfr = 1*915, which 

 corresponds with A, = *2884\ and R\ = 382 km. 



The tangents at the double point are inclined to the 

 meridian at an angle 



cos -1 (2 sinh -vjr sech 2 -v|r), where ^=1-915. 



This angle is equal to 5 6°' 5. 



The time taken in describing the loop is 2-yjrj\ (o, or 

 2-9 days. 



Case III. — If the particle is projected eastwards at 

 10 m./s. at a point whose distance from the equator exceeds 

 1324 km. the track lies entirely in one hemisphere and 

 consists of loops. 



A simple example is found by choosing //, in the notation 

 of § 9 as W2. It follows that 



/4V 



A= 1/^/2, X = y / s -=-294, ^\o=-208, 



\ R=1870km. and ^X R = 1320km. 



The last two of these data are the distances from the equator 

 of the points where the motion is eastwards and westwards 

 respectively. 



The period for the velocity is 2K/A &> or 48 hr. This is 

 the time from the north of one loop to the north of the next. 



The change in longitude during this period is 



X [(1 + ^ 2 )K-2E] towards the west. 



On substitution of the values of E and K, viz. E = 1*3506 

 and K = 1*8541, we find that the distance between the most 

 northerly points of successive loops is 150 km. 



The positions of the double points, which have the 



