378 Mr. T, H. Blakesley on Logarithmic 



Some geometrical matters more immediately arising from 

 this view of the curve may be introduced. 



The problem of finding the pole when two consecutive 

 chords are given is solved thus : — Let AB, BC, be the 

 consecutive chords given. Complete the parallelogram, and 

 let BD be the diagonal through B. Make the angle BCP 



Fur. 1. 



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equal to the angle DBC, and make the angle ABP equal to 

 the same angle. Then P is the pole of the spiral in which 

 AB, BC are consecutive chords. 



As an alternative to the setting off o£ one of the angles 

 BCP or ABP, the angle BAP may be made equal to DBA. 



Or, as it may be shown that the product of PB . DB is 

 equal to that of AB . BC : 



BP may be easily calculated from the data, viz., the values 

 of AB and BC, and the angle between them. 



If one of the two chords (say BC) is maintained in position, 

 but the other BA is made to turn round B, so as to vary the 

 angle between the chords, the pole P will describe a circle 

 whose centre is in the line CB, produced if required 



If 6 is the angle between the two consecutive chords 

 externally, the characteristic angle of the spiral (a) will be 

 given by 



tan a == 







BC 



l °Sab 



In any cases therefore in which — - has the same 



value, the spirals are similar. °& AB 



It follows that, in any mechanical construction of linkages 



BC 

 &c, if we can keep -r-~ constant, but can at the same time 



cause 6 to vary, we have the power of changing a, that is to 

 say the one thing which settles the character of the equi- 

 angular spiral. 



