Form of Harmonic Analyser. 369 



where, to fix the sign, we have assumed D to lie initially to 

 the left of K and XX to lie above the disk, as in the figure. 

 Hence the area traced out by D is 



R x = \ydx — r . cos nir\ y d (cos n6) 



■i-r . cosmr ^sinnOdx — r 2 \ sinnd d (cos nO). 



The last two integrals vanish on taking them round a closed 

 curve. Nothing is added to either of the first two by con- 

 tinuing the integration from R back to P, as y is then zero. 

 Therefore, calling the area of the whole curve PQRa, we 

 have 



E 1 = a + cos nir . —r \ y sin n6 . dx. 



Similarly, if D had been initially on the vertical instead of on 

 the horizontal diameter, and below K, we should have had 



mir 



+i 



R 2 = a + cos nir . — — 1 y cos n6 . dx. 



1 J-i 



It will evidently be convenient to take r some multiple of 

 \jir units of length, say 10. We then have, rewriting the 

 last two equations, 



when r = 10jir 



~R Y = a-\- cos mr . lOrtB^, ..... (1) 



R 2 =<2 + cosn7r . K)nA n (2) 



Care must be taken with regard to the sign on the right-hand 

 side if any other initial position of disk and line be assumed 

 than that dealt with above. 



§ 3. These two equations contain the whole theory of my 

 instrument ; they show how to construct a curve the area of 

 which gives the required coefficients. The geometrical 

 mechanism seems to me to be somewhat interesting, and to 

 be possibly capable of generalization by the use of noncircular 

 disks. 



The area of the D-curve (in dealing with a material 

 instrument) might be obtained in two ways. We might put 

 a pencil through the disk at D, draw the curve, and integrate 

 it afterwards : or we might attach the pointer of an integrator 

 to D and let the integrating go on simultaneously with the 

 following of the curve. 



It is the latter alternative that I have adopted. The former 

 method would have some advantages, but would be slow and 

 would lead to mechanical difficulties. 



