368 



Mr. G. U. Yule on a Simple 



i f +* 



A = \ I y cos nd d6 1 



Ij-i 



iC +l 



B — - y sin 116 do. 

 n IJ—i' 



These are the integrals which any harmonic analyser has to 

 evaluate. 



Now suppose we have a circular disk, centre K (fig. 1), 



Ffr. 1. 



constrained to keep in contact with a straight line XX 

 parallel to PR, and capable of rolling along XX without 

 slip. Further, let X X be capable of motion in the plane in 

 a vertical but not a horizontal direction, so that every point 

 fixed in it describes a perpendicular to P R. Then we can 

 make the point K trace out any arbitrary curve by moving 

 X X and rolling the disk along it. 



Bring K over P, and then mark any point D on the 

 horizontal diameter at a distance r from the centre (not 

 necessarily inside the disk). Starting from P carry K right 

 round P Q R and back to P again by the motion just 

 described. Supposing the circumference of the disk to be an 

 aliquot part of P It, say 2l/n, let us find the area of the curve 

 traced out by D during this operation. As the disk turns 

 through an angle 2mr in rolling along a length 21 of X X, 

 it will turn through nirocjl in a distance x ; so if x, y be the 

 coordinates of K at some point on its journey, the corre- 

 sponding coordinates of D will be 



x — r. cos nir . cos n9, 

 y -\-r . cos nir . sin n$ } 



