Ball — Contributions In tlic Theory of Screws. 41 



where (12) is the right-liainlcd angle between the screws 1 and 2. This is 

 obvious from the fact that 



^^^ = 'Ailh + )h] cos (12) - (I,, sin (12) I . (30) 



By the help of the pitch-operator, we are able to obtain many formuUe ; 

 and we proceed to give an illustration. We shall first prove the following 

 theorem : — 



Let (1), (2), (3) be any three screws of a 2-system — i.e., a system of 

 vector-screws about which three twists can neutralize — it is required to show 

 that 



2h2h2h - Pi^'23 - Z'sT'^l - ^Wl2 + 2^2373311512 = 0. (31) 



If a, /3', 7' be the respective amplitudes of the three neutralizing twists, 

 and if i/ be any other screw whatever, then, as already shown, 



a'wi, + /S'toj,, + ^'^3^ = 0. 

 If we allow the screw 17 to come successively into coincidence with (1), 

 (2), (3), we have the three following equations : — 



api + /8'-CT,2 + 7'to,3 = 0, 

 a'CTji + /S'^Jj + 7'w,3 = 0, 

 a'ra,! + /S'ot,., + y']h = 0, 

 whence, on elimination of a, ;8', 7', we obtain the desired result. 



The equation (31) must remain true if operated upon by A ; and we thus 

 obtain 



Pijh + Ihlh + Ihlh - ra^3 - w-si - •^5^2 



- 22?i cos (23) ^23 - 2252 cos (13) ts^i - '2ih cos (12) 13,2 (32) 



4- 2 cos (23) TO3, rai2 + 2 cos (31) zs^ j^u 2 cos (12) zs^i 7^33 = 0. 

 But this must remain true if again operated upon by A, whence we get 

 Ih sin= (23) + ih sin' (31) + p^ sin' (12) -1- 2^23 (- cos (23) 

 + cos (31) cos (12)) + 2oti3 (- cos (13) + cos (23) cos (12)) 

 + 2!3,, (- cos (12) + cos (31) cos (23)) = 0. (33) 



Finally the third operation by A gives 



1 - cos2 (23) - cos' (31) - cos' (12) + 2 cos (23) cos (31) cos (12) = 0. (34) 



Of course (34) merely proves the well-known law that the screws (1), 



(2), (3) must be parallel to the same plane ; assuming this to be the case, 



we may make 



L (12) = 0,-d,; L (31) =0, «. ; z (23) = d,-ii,; 



and with this substitution (33) becomes after a little reduction 

 ^1 sin' (02 - hi) + ly. sin' (^3 - 0,) + ih sin' (0, - #3) 



+ 2ct,3 sin (03 - Wi) sin (0, - «,) + 2rjn sin (0, - i)^^ sin (0, - 0,) 

 + 2ro,2 sin (0., - 03) sin (0, - 0,) = 0. (35) 



L6*] 



