156 
Transactions of the Royal Society of South Africa. 
For the determination of 4'i we therefore have 
= — Sc sm' e. 
On integrating we get 
= c (3 cos e — cos-^ ^) 
2z^ + 3^r^ 
- (1^) 
Section 7. 
Doublet. 
In considering the motion due to a doublet we must take the doublet 
on the z-axis, and the axis of the doublet must coincide with the 2;-axis 
in order that the resulting motion may be symmetrical about the z;-axis. 
P 
Let there be a sink at o and an equal source at A, the strength being c. 
Let P be any point (p, 9). 
Now, since the differential equation for the velocity-potential, equation 
11, is linear, we find the velocity-potential at P due to the source and sink 
by superposing the potentials due to each of them. 
Hence, if is the velocity-potential of the doublet, then 
A 
-> 0 
