232 Mr. HOPKINS, ON THE TRANSPORT OF ERRATIC BLOCKS, 



or if 



6-og- 



18. To estimate numerically the dimensions and weights of the blocks which may be moved in 



each of the preceding cases by a current of given velocity, let us take — = 4, which gives a 



velocity of about 16 feet per second, or about 10^ miles an hour. 



The triangular prism will never roll ; when it is just on the point of sliding we shall have 

 (the section being equilateral) 



This involves the unknown quantity fx. Suppose the friction such that the body would just 

 slide down the surface on which it rests, if that surface were inclined at an angle of 45" to the 

 horizon ; then fx = I, and 



a = 1,68 feet. 

 If M = ,5 



a = 2,84 feet. 

 In the parallelopiped of which the section is a square, wt have, when it is on the point of rolling, 



a = 2| feet. 



If the length = a the body will be a cube containing nearly 19 cubic feet, and weighing 

 nearly 1^ ton. 



For the pentagonal prism on the point of rolling, 



a = 2,268 feet. 



If the length of the prism be 2o, the volume will be about 40 cubic feet, and the weight nearly 

 3 tons. 



In the hexagonal prism on the point of rolling, 



a = 2,28 feet. 



If the length of the prism = 2a, and therefore do not differ much from its height, its volume 

 will be upwai'ds of 60 cubic feet, and its weight between 4 and 5 tons. 



When the body is approximately spherical, let ra = 4 ; then 



o = |- feet. 

 If we estimate the volume as equal to that of a sphere whose radius a' is a mean between 

 a and b, we find 



a' = |- feet, 

 and the volume about 52 cubic feet. The weight will be about 4 tons. 



If w = 6, 

 o = 4 feet, 



«' = ii «' 

 and the volume may be taken at nearly 200 cubic feet. Its weight will be 14 or 15 tons. 



