252 Mr. L. F. Richardson on a Freehand Graphic way 
Example of the type of symmetry when V 25 independent of 
the radius in spherical coordinates. — " On a uniform spherical 
shell there are equal sources at the north and south poles 
and equal sinks at the extremities of a diameter lying in the 
equatorial plane. The sources and sinks send out and receive 
uniformly in all directions. The flux has no divergence except 
at the sources and sinks and no curl anywhere. Find the dis- 
tribution of potential on the surface.'"' To do this we might draw 
orthogonal lines on the surface of a globe so as to make the 
chequers ratio constant. Or because, in Mercator's projection, 
any small part on the globe transforms into a small part of 
the same shape on the map ? we may transform the boundary 
conditions and obtain the required solution by drawing 
chequers of constant chequer ratio on the map. Blank 
Mercator projections suitable for this work may be obtained 
from George Philip & Son, Fleet Street. In the present 
example the lines of flow radiating from the pole become lines 
straight, parallel and equidistant at infinity. And as the 
graph progressed it w^as found that by their symmetry 
with the sinks on the equator, the foregoing condition must 
be very nearly satisfied at 10° from the poles, a region which 
is within the confines of the map. Again, in this case it is 
only necessary to determine V in one octant of the sphere, 
and symmetry helps us in other ways. The accompanying 
graph (fig. 4) is the best of four or five separate attempts. 
The time taken to make these was collectively four hours. 
Special attention was given to the equipotential curve which 
passes mid-way between the two equatorial sinks, and as the 
result of the aforesaid trials it was found to pass through a 
point 44° due north of the sink on the equator. This suggested 
that the true value should be 45 °, and on looking at a sphere 
this is seen to follow from symmetry although it was not 
obvious on the map. Thus again we have a confirmation of 
the passable accuracy of the graphic method — the error here 
is 1 degree in 90 or 1*1 per cent, of the range. 
So far we have only treated the problem as relating to a 
spherical shell. But we may next suppose the sphere solid 
and V to be independent of the radius. We will then have 
a solution of Laplace's equation in space. Since the chequer 
ratio is constant, the magnitude of the flux is inversely pro- 
portional to the linear dimensions of the chequer (on the 
sphere not on the map) and is consequently proportional to 
- along any guiding line, r being the radius. But if we draw 
any small cone enclosing the polar axis — which is now a line 
source — We see that the outflow between two spheres r and 
