the Problem of Columbus. 473 



The necessary condition £01* this is that the numerator of 

 (17) should be positive when /a = 0. 



The only positive term in this is A 2 K, but this contains a 

 term with in it ; we have assumed nothing about O , so we 

 can make K as large as we please. 



This condition is also sufficient ; for having /n (} at our dis- 

 posal, we can take it between and the next root. 



Also we see that fj, need not be small at-this point, so that 

 6 need not be ; F 2 will thus not become small : so that we can 



choose our initial conditions so that 1 — _ — L > any number 



R 



we please. It follows from this that with any material in 

 actual nature, under certain initial conditions, the egg must 

 slip at some time, and accordingly this solution will then fail. 

 13. To illustrate the motion I took a numerical example. 

 The egg was taken as a hemisphere with half a prolate 

 spheroid with major axis double of its minor, and its minor 

 axis equal to the diameter of the hemisphere, all supposed 

 homogeneous. In this case we have 



When we form equation (15) we see that a divides out of 

 every term except those containing g, and there we have — : 

 for convenience of calculation I took « = 4*90 ... cms., so that 

 ^ =200, taking one second as the unit of time; I also took 



/u, = /I = 0. 



As the top is not a complete sphere, fi must be positive 

 for the solution to apply. 



Referring to (18), we see that the condition tor this is 



a/ -+A + ., 



