282 Trans. Acad. Sci. of St. Louis. 



or 



\iO TCOX = O, 



in which C is the constant of integration. This is the same 

 as equation (1). It shows that the analogy of considering 

 force as flowing is correct. 



The slope of a line of force at any point (x, y) is given by 

 equation (19), which written in another form is 



/ 1 \ \2 / 1 \ \2 1 X 2 / \2 



^-2™)+(* + 2*™)=4^( 1 + 5 ) < 20 >> 



or 



r = r Q (cos a) — S sin co) (20a). 



If /S is a constant this is the equation of a circle which passes 

 through O (x=0, y = 0) and / (x = 0, y = —). This 



\ 7T07 



shows that a circle passing through and /cuts lines of force 

 in points at which they (the lines of force) have the same 

 slope /S. The slope of a tangent to this circle at the point O 

 is jS. For S = equation (20) becomes 



V 2iro) 4 ttV' 



which is the equation of a circle whose diameter is 01 = r = 



This circle cuts lines of force in points at which they 



7r<r 



have no slope. Fig. 3 shows several of the circles. The 



perpendicular bisector of 01 is the locus of the centres of all 



the circles. 



Fig. 3 shows that all the lines of force which are outside the 



critical line have points of inflection. For the locus of such 



points of inflection 



dx 2 



= J^ _1 -T( x + 2/2ar1 )) 



2ttV 



~Y~ y (a 2 + y 2 ) — 2tt<7 (a 2 + 2/ 2 ) 



A* =° < 21 )- 



