EQUILIBRIUM OF A SPHERICAL ENVELOPE. 93 



= r u angle APB = XAT=<$>. Bisect AB in C, and draw PD perpendicular 

 to AB. Then the Hue of tension at P is a circle through A and B, for which 

 ( is constant, and the line of pressure is an orthogonal circle for which the 

 ratio of r t to r 3 is constant, and the angle <f> and the logarithms of the ratio 

 of r, to r t differ from the corresponding quantities in the sphere only by con- 

 stants. We may therefore put 



_ Psina (dG\ _ Psina tdH\ _ 

 Pu ~ ~~ '' (dSJ '' ~ P *' 



where the values of G and H are the same as in equations (8) and (10). 

 Transforming these principal stresses into their components, we get 



dx 



dG*\ 

 dy 



.(16), 



Psina dGdG , } 



P *~ 27ra Z ~dxdy (17)> 



Psma/dGl" d<Gft\ 



Puu = I T- T- 



2ira \dy\ dx\J 



From the relations between G and H we have 



dG dH . dG dH 



-j- = j- and -7 = -j- (19), 



dx dy dy dx 



#G#G . d*H d*H 

 whence ^ + ^ = ^ d d* + ^ = ( 2 )' 



The values of the component stresses, being expressed as functions of the 

 second degree in G, cannot be compounded by adding together the corresponding 

 values of G, we must, therefore, in order to express them as linear functions, 

 find a value of F, such that 



Psina (fdG\* 



dy 1 ' 



We shall then have, by differentiation with respect to x, and integration 

 with respect to y, 



(22), 



> (23). 



dxdy 2ira J dx dy 

 d'F Psina 



