1885.] 



Note on a previous Paper. 



325 



is a solid harmonic of the second degree, and therefore the present 

 solution will apply to this case also. 



If we extract the case i=2 from Tables I, II, III, and put i=2 in 

 (26), and substitute colatitude for latitude I, we have after some 

 simple redactions — 



19(P - W 2 ) = 16a 2 - (19 + 3 cos 2ey- 

 19(R- W 2 ) = -32a 2 + (29 + 3 cos 20)r 2 

 19(Q- W 2 ) = 16a 2 - (13 + 9 cos 20)r 2 

 19T = 3 sin 20r 2 



(a) 



[Note the introduction of W 2 in the P, Q, R, in accordance with 

 the first correction.] 



Let N 1? N 2 , N 3 , be the three principal stresses, each diminished by 

 TV 2 , so that — 



^^:} =i(p+R)± *^ (p - R)2+4T2 4 . . . (6). 



Ns+TTa =Q J 

 Then— 



^^j==-8a 2 +5r 2 +3y{64(a 2 -r 2 ) 2 +r 4 -16r 2 (a 2 -r 2 )cos2^}l ( 

 19N 2 = 16a 2 — 13r 2 -9r 2 cos2<9 J 



Now let us find the surfaces, if any, over which N 2 =Nj or N 3 . 

 They are obviously given by — 



24a 2 -18r 2 -9r 2 cos2^=±3v / {64(a 2 -r 2 ) 2 + . . - &c.}. 



This easily reduces to — 



r 2 (l-cos2^)[32a 2 -20r 2 -9r 2 (l+cos2^)]=0 . . (d). 



Thus the solutions are — 



r = 1 

 0=0 and tt } .... (e). 



and 32a 2 -20(aj 2 + 2/ 2 )-382 2 =0 J 



By trial it is easy to see that at the centre and all along the polar 

 axis N 2 =N 1 , and that inside of the ellipsoid 10(x 2 +'y%) +I9^ 2 =16a 2 , 

 N 2 is greater than N l5 and outside it is less. 



Hence inside of the ellipsoid N 2 — N 3 and outside of it N^— N 3 

 is the stress- difference. N 2 — N 3 nowhere vanishes so long as N 2 is 

 not equal to N l5 and N x — N 3 vanishes where r=f \/2 . a=*9428a and 

 0=0, which is inside of the region for which N x — N 3 is the stress- 

 difference. This is the only point in the whole sphere for which the 

 stress- difference vanishes. 



The ellipsoid of separation cuts the sphere in colatitude 35° 16'. 



