12 Scientific Proceedings, Royal Dublin Society. 



Now we must determine /3 so as to make P maximum or dP = 0, 



whicli gives 



dW 

 -^+(cot(i3-(/>)-tan(7-/3)^/3 = 0; 



but, denoting CD by r, we have, geometrically ^ 



dW=-gp'^d^: 



substituting tbis value for dW, we obtain gp — sin (j3 -^) cos (7 - j3) 



= W{coB {(5-(j)) cos (y-jS) - sin (jS -^) sin (7 -jS) = TTcos (7-^), (1) 



or 



5f|0 ^ sin (j3 - <^) cos { (a + ^') ~ {[5 - (p)} =W cos (a + 0'), 



or 



^io ^ sin (/3 - 0) {cos (/3 - 0) + sin (/3 - 0) tan (a + 0') } = W. 

 .0 



If, then, we suppose CD to be the line which solves the problem, 

 W will denote the weight of the cylinder of earth, of length 

 one foot, whose mean section is the area AZDC, so that W = 

 gp area AZDC ; and, if we draw DM perpendicular to CD, and 

 DE inclined to DM at the angle a + (p', the area 



CDE = i DM. CE = ^DM[CM+ ME) 



= - sin (j3 - 0) r cos (jS - 0) + r sin (j3 - 0) tan (a + tp') ; 



therefore area AZDC = area CD^, and therefore the line CD, 

 which solves the problem, is such that, drawing DE inclined to 

 CB at the complement of the angle a + 0', area AZDC = area 



ADE, while 



^ F-sin(3-(^) ^ 



cos (7-/3) ' 

 or, by equation (1), 



_ gpr^ sin^ (i3 - 0) 5'(or^ sin'^ (j3 - 0) 



2 cos [y - (p) 2 cos (a + (p') 



dp r.r..( . ..X ( r sin(/3-(^) )^ 

 -77- cos (a + d> ) { ; r- \ 



9P 



^ COS (a + 0') Z)^% 



