124 SCIENCE PROGRESS 



(2) To prove thsit a = K''^o = K^^-W. For 



K^o = cos a.o + sin a . Vr^ — o^ = r sin a = a; 



K'^~\ = cos {a— i) .r + sin (a — i) . Vr^ — r^ = y sin a= a. 



If a be a right chord the angle of rotation from y to a is a negative angle, 

 namely i — a reversed. If a be a left chord, denote it by a' and its chord 

 angle by a . Then a' — i is a positive angle. 



We can obtain either of the equations in the enunciation from the other. 



For if a = i^"~V, then, since r = Kp, a = K')-^K\o = K^p ; and if a = i^^o 



it also equals K^K~^r. 



(4) To prove that if a, b, c, ... be any chords of the same or of 

 equal circles and a, jS, y, . . . be their respective chord-angles, then 

 K- "a = K- ^b = K- ^c, etc. For 



a = K^o, b = Kfo. c = K-^o. etc. 



therefore o = R- "a = R- ^b = R- ^c = etc., 



and r = R\-''a = R\-^b = R\-'<c^etc. 



(5) To prove that, if a and b be any two chords of the same or of equal 

 circles and 6 be the angle between them, then a = R^^b. 



For R- "a = R- ^b, and a^R'^-^b; 



and (9 = a - /3. 



(6) To prove certain relations between counter-chords. Let a and a' 

 be counter-chords and a and a their respective angles. Then by 4*0, 

 a + a =2, and 



a' = R^'-^a = K'-'"a = - iiC- V 



a = R^-^'a' = Rl-'-^'a' = - R-'^'^'a'; 



a' = K\-''r, r = R\-''a: 



Ry = R^'a: 

 a' = R^-^o, a = R^o. 



From this last equation we have it that the length of a' is r sin (2 — a), and 

 of o is y sin a. These lengths are equal, but this does not mean that the chords 

 themselves are equal (see 4-0). 



4'2. Remarks. (i) There seems to be a discrepancy here ; for if 

 a = y sin a and a' = r sin (2 — a), then a = a'. We might have used Hamilton's 

 tensor-symbol and have written 



Ta = TR'^o = r sin a ; 



but really we should not employ the trigonometrical ratios at all until we 

 finally come to put our operative analysis into scalars ; and I have employed 

 them here merely for the purpose of giving a simple exposition. We can 

 save unnecessary symbols by remembering that sin Q = sin (2 — B) and 

 cos 6 = cos (— 6) only for scalar values. 



(2) In place of o and — o for the zero-counter-chords we can write t and 

 t' or even dt and dt'. I prefer the first. 



(3) It seems curious that a reversed chord is not a negative chord ; but 

 the supposition that it is so leads to immediate discrepancies. The reversed 

 chord is really the counter-chord. 



(4) The effects of repeated operations with Rr will be easily gathered 

 from Figure i. If a be a right chord, R/i will be a left chord drawn from the 



