with the Flow of Electricity in a Plane. 43 



Hence 



and 

 Q=^P 1 P 2 = _^L F 0(3O°)e i (6O°) (mod. angle 75°), 



writing ®i(V) for —. — H(V). 



From the British- Association Tables 



log— L= = 1-4091498986, 



log © (30°) = 1-9223413942, 



log0 1 (6O°)= -0491498189; 

 so 



log 10 Q =1-7528522478, 

 or 



Q= rA64 about ' 



and the resistance of the triangle is 



E =i lo g(f' Q ) ^ 



Manner in which the image-method fails for an angle of 120°. 



§ 26. We found in § 5 that the only isosceles triangles which 

 could be treated by the method of images were those with the 

 equal angles either 0°, 45°, 60°, 90°, or 180° ; and of these all 

 except the first have been already done in the course of this 

 paper (§§ 23, 25, 19, 10), and the first shall be considered in 

 the next section. But suppose we attempted the triangle with 

 equal angles 30°, which is apparently a very simple case ; the 

 images of the source A (on one of the equal angles) would be 

 readily placed on the corners of regular hexagons ; but they 

 would be symmetrically situated with respect to the middle 

 line of the triangle, and hence when the images of the sink B 

 came to be placed they would coincide with those of A and 

 would blot them all out of the sheet. Or if we overlook this 

 and apply the expression (/3) to the images of A, then, because 

 of their symmetry with respect to the triangle, the product Q 

 (§§ 8 & 22) will equal 1 ; which can be proved to be wrong 

 (§ 27). This failure, caused by the presence of the angle 120°, 

 is one that cannot, even apparently, be got over by silvering both 



