EQUILIBRIUM OF A SPHERICAL ENVELOPE. 91 



diameter to that in which the forces are applied at the extremities of any 

 chord of the sphere, subtending at the centre an angle = 2a. The lines of tension 

 will be circles passing through the extremities of the chord. Let the angle 

 which one of these circles makes with the great circle through these points 

 be <j>. The angle (f> is the same as the corresponding angle in the inverse 

 surface. 



The lines of pressure, being orthogonal to these, will be circles whose 

 planes if produced pass through the polar of the chord. Let r,, r 2 be the 

 distances of any point on the sphere from the extremities of the chord, then 



7* 



is constant for each of these circles, and has the same value as it has in 

 ?\ 



the inverse surface. Hence, if we make 



=<j> (10), 



G and H will give the lines of principal stress, and the absolute value of the 

 stress at any point will be 



P sin a. (dG\* P sin a [dH\* 



p = - j~o ) = 7T~ iFcf ) (11)- 



Zrra \ctOj/ 2ira \ttOj/ 



If we draw tangent planes to the sphere at the extremities of the chord, 

 and if q v q t are the perpendiculars from any point of the spherical surface 

 on these planes, it is easy to shew that at that point 



p = Pa ^^.. .. (12) . 



2irq,q, 



If any number of forces, forming a system in equilibrium, be applied at 

 different points of a spherical envelope, we may proceed as follows. First 

 decompose the system of forces into a system of pairs of equal and opposite 

 forces acting along chords of the sphere. To do this, if there are n forces 

 applied at n points, draw a number of chords, which must be at least 3 (n 2), 

 so as to render all the points rigidly connected. Then determine the tension 

 along each chord due to the external forces. If too many chords have been 

 drawn, some of these tensions will involve unknown quantities. The n forces 

 will now be transformed into as many pairs of equal and opposite forces as 

 chords have been drawn. 



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