36 Mr. T. H. Blakesley on a Geometrical Determination 



AB equal to e, and cut off from it BM equal to /. Describe 

 the F circle upon AB, and set off BAD as before, an angle 



having its tangent equal to 7p-« 



D is on the F circle. Draw MN perpendicular to AD, 

 and with centre D at distance DN describe a circle. This 

 is the E circle, for its centre is at D, and its radius 

 = DN=BMcosDAB. 



FE 

 the ratio -7-^ a maximum 

 AJb 



Through D draw the radius DH parallel to AB, H being 

 on the same side of D that M is of A. Join AH by a straight 

 line cutting the E circle in E and the F circle in F. Then 

 these are the particular positions on the circles which make 



FE 



i. e. -tt^ as here found, is the 

 AF' 



maximum efficiency. And if DE be joined, the angle ADE 

 will give the phase-difference between the electromotive forces 

 to give this efficiency. 



FF 

 For, in the first place, it is clear that the ratio -r-^ has some 



maximum (not a minimum) value between the positions where 

 E and F coincide, viz. when the two circles intersect. And 

 because DH is parallel to AB, and AH cuts them, therefore 

 the angle DHE = the angle BAF. 



But DH, BA pass through the centres of the two circles 

 respectively; therefore AH, making the same angle with BA 

 as it does with DH, must cut the circumferences of the E and 

 F circles at the same angle. Thus at E and F, the points 

 where AH cuts the E and F circles respectively, the arcs of 

 those circles are parallel. Therefore an elemental displace- 



