ACTED ON BY NO EXTERNAL FOKCES. 769 



Similarly, if ZIZ, be the angular displacement of the plane ZI measured in the same 

 direction as before, 



dt 



(cos /3)2 



ZIZ, = Li; -tr--, 1?™' 



(cos a)' + (cos p)'' 



and the rotation about Z causes no displacement of the plane in question. Hence if 

 the horary angles, as they may be called, which measure the angular deviations of the 

 planes XI, YI, ZI from a fixed meridian plane through I be called |, ^, ^, we have 



(cos|3)* (cos 7)* (cos 7)* (cos «)' (cos «)* (cos /3)* 



dt~ (cos ^)^+ {cos yy^' dt~^ {cosy)- + {cos u)'^' rf/~(cosa)«+ (cos/S)* 



If, now, preserving L constant we replace —, ^,1 ^i M. by 



A. o \j 



j-X; ^--K; 1-X; M-XL', 

 the equations (1), (2), (3) remain unaltered, and the right-hand sides of equations (4) 



* By combining this with the system of equations previously found, both i) and 5 may readily be obtained 

 under the form of elliptic functions of the third kind in terms of cos y, but ij — J or the angle I in the quadrantal 

 Bpherical triangle XIY of fig. 3 will also be expressible as a function of a, /3, and therefore of y. The compa- 

 rison of the forms of ij— J given by the two methods respectively, leads therefore to a theorem in elliptic 

 functions ; Professor Catley has worked this out, and finds that it is the well-known theorem which expresses 

 the dependence between two elliptic functions of the third order, the product of whose parameters is equal to the 

 square of the modulus. I subjoin an extract from his letter, in which I have only introduced some slight 

 changes in the lettering : — 



« Writing 



Ap=+B2'+Cr'=M, 



Ay+BV+CV=P, 



your theorem is 



■where 



Jp-Ay Jp-BY 



d«: 



tan-'(. . , .), 

 Cdr 



Whence expressing everything in terms of r, this is 



write for shortness. 





AM-L==0, BM-L-=6, 

 B-C=a, C-a=/3, A-B=y. 



Then we have 



or if x^= — 



By^=a+CBr'; -A.yp'=h-Car ; 



