DUB TO THE WEIGHT OE CONTINENTS. 
217 
Then co becomes nearly equal to and k is small compared with v and u>. 
becomes nearly equal to 2 i(i— l)co. Hence in this case 
Also K 
T 
1 
(7/2 
~2{i- 
■Iff 
^ M 
1 
2 
2(4 — 
■ Vy ~ 
’ dxdz 
(49) 
Now if we suppose It 7 to be a zonal harmonic of degree i, and only consider the 
state of stress at the equator, immediately underneath the centre of the elevation, 
then T is zero and by (15) 
And 
R= 
2(*-l) 
(a 2 — r~)v l 
:-2 
A=P-R= 
2(^-1) 
If i be large A =i(a ?— 
In § 7 we have found this identical result, under the like conditions, when the solid 
is incompressible. 
Now take the case of the 2nd harmonic, so that 
W = r 2 (| — cos 2 6 )=^(x~-\-y 2 —2 z 2 ) 
And we have 
P= -J(a 2 —r 2 ) 
* Q= Ua 2 —r 2 ) 
11= —-|(a 2 —r 2 ) 
T= 0 
Thus throughout the spheroid, the principal stress-axes are parallel and perpen¬ 
dicular to the polar axis ; also 
A = (a 2 —r 2 ) 
Hence the central stress-difference is a 2 . If the solid be incompressible it was found 
to be 8a 2 . Hence infinite compressibility largely relieves the stress-difference due to 
ellipticity of figure. Next take the case when the harmonic is of the 4th order. 
Then at the equator we have by (50) 
A = —-t 1 (a 2 — v 2 )= 1 3 iL r 2 (a 2 —r 2 ) 
The maximum is reached, when r 2 =|a 3 or r=707a, and is equal to |= IT67. 
Comparing this with the Table Y. (a) § 7, we see that infinite compressibility makes 
very little difference, for 707 differs little from 714 and 1*167 from 1*118. 
MDCCCLXXXII. 2 F 
