312 The Rev. 0. Fisher on Mr. Mallet's 



and upon a square mile of this section , 



P 2 



~ke. 

 fi 



Mr. Mallet, at p. 8 of his paper in this Magazine for July, 

 states that friction may be as great as f of the pressure. Let 

 us, then, suppose first that the nucleus is solid, and that the 

 force requisite to move a column of the crust horizontally over 

 the nucleus is f of the weight of the column, and P is the weight 

 of a column 2000 miles long. Hence 



3 p k 



4 2000 ' 

 whence, if ^ = 400 miles, 



3P 

 ^=4 5' 



Hence the work upon a square mile of section cannot be so 

 great as 



which 



=fPex (2000 miles). 



Now in order to calculate the heat by means of Joule's equi- 

 valent, we require to have the force expressed in pounds and 

 the space in feet. Let /be the number of feet in a mile. 



The weight of a cubic foot of granite, as given by Mr. Mallet, 

 is 178*3392 pounds (say 200 pounds). Hence the weight of a 

 cubic mile of such rock will be/ 3 x 200 lbs. ; and of a column 

 2000 miles high, or P, / 3 200 x 2000 lbs. • 



.'. P=/ 3 x4xl0 5 lbs. 



The coefficient of compression for one year will be the same as 

 that of the radial contraction, which, on the data assumed by 

 Mr. Mallet (Addition &c, Table II.), will be, for 400 miles of 



crust, jqc; ^q q ; 



__9 [}_ 



" e ~"4 10 12 ' 



We have therefore, expressing the space in feet, 



Work in foot-pounds per square mile of vertical section at A 



= |/ 3 x 4 x 10 5 x | x -L x 2000 x /. 



And in foot-pounds per square foot, dividing by/ 2 , 



