1. Assume first the flow caused by the sphere in unbounded fluid. Its potential is that of 

 a dipole at C or, as in Equation [127a], 



ar'lJ cos B 



where t denotes distance from C and the angle Q is measured from a line CQ drawn in the 

 direction of V. 



2. To satisfy the boundary condition on the wall, add an equal dipole at the mirror image C 

 of C in the wall, with axis CO' in the plane QCT but inclined at an angle rr - a to OT produced 

 backward; see Figure 215. The potential of this dipole is, similarly, 



a^V cos d' 

 T ^ 



where ^'is measured from C'Q'. 



3. At the sphere, since a/x is small, the partial flow due to 1,'jj'is practically uniform, 

 with components of velocity u in the direction CT and v m z. perpendicular direction CR lying 

 in the plane QCT where, from [127c,d], in which r = r'= 2a; and 6 - 6' = n -a , 



cosa a^U sin a 



2 



These two uniform components of flow, in interaction with the sphere, add a potential that may 



be obtained from [128a] by first replacing V by -u and Q by ^j, where Q is measured from CT, 



then replacing f/ by -t; and Q by Q^, where d^ is measured from Cff, and adding the results. 



The value of this potential will be needed only on the surface of the sphere, where r = a. 



There its value is 



ar'lJ Za I 1 \ 



4>2~ — ^^^ ^1 ^^^ a + — COS ^2 Sin a 



8^^ 2 V 2 } 



The potential 4>^ includes the variable part oi 4>{\ ^ constant part equal to the value of (jSj'at 

 the center C has been omitted, but this omission has no effect upon the result to be obtained. 



Now introduce also an angle cu, so that r, Q, co are polar coordinates with origin at C; 

 let (u be measured from the plane QCT; see Figure 216, where P is any point on the unit 

 sphere about C. By projection of CP on CT and CR it is seen that 



a24 



