56 



Proceedings of the Royal Irish Academy. 



>. 



(90) 



Hence, X must be parallel to the vector part of the product of these two 

 vectors X and Y. iat which we have the expression 



F;i,^ (234) (345) 1 



r,fc^ (341) (452) - (345) (412) 



F^,j,. (412) (523) - (123)(452) 



ravi.(234)(123) 



Fm^, (234) (234) 



F;i,^,(234)(425) 



F;^a..(234)(421) 



F^^, (234)(314) 



V^,^ (314) (523) - (123) (354) 



F;i^,(234)(532) j 



where for b^e^-ity (234) is written instead of SkO<^. 



In its present form this expression is not symmetrical, though it is obvious 

 that the vector parallel to the screw reciprocal to five screws should be 

 symmetrical with regard to those five screws. The desired form can be 

 obtained by the following formuke, easily verified by well-known rules in 

 vector manipulation: — 



5A,A.A,.5X^J(. - .SAjX^k, . 5A J^.A, = 5AAA4.5AsA,A.. 

 SX^,A,..SA^:Aj - .SA.A.X,.5A.A»A, = ^A.XA. - 5A,X.A„ 

 -SXA.X, . 6A»A,A, - 5A»A,A« . S\,\X = ^XJ^A. . 6'X,XiXj- 



Introducing tbeae values into (90), and rejecting the factor ^,X>X4, because 

 we assume that the problem has not doners ted by ha\-ing three of the five 

 screws parallel to a plane, we obtain the symmetrical expression 



X = ♦• I'^i/ij. SA)X,Aj 

 + F|ujuj. .9A,AjX, 

 ^ F/i^i . SX^iX] 

 + VutfLi. •S'A,A»Xi 

 + F|Uj^, . SA^jA, 



- Vft^fi^. SXjXjA, 



- Fju^, . SX,X«X, 



- y^Litli ■ SA,XiX, 



- Fjut^. SA,A^s 



- yfuni . SAjXiA] 



The order of the euttices in the several terms in A will be seen as follows : — 



Writing the digits 1, 2, 3, 4, 5 round the circumference of a circle con- 

 secutively, we notice that in each of the five positive terms the five digits 



>. 



(91) 



