A RIGID BODY IN ELLIPTIC SPACE. 
307 
Then 
Oa — Acq “I - 
Ob^=A8j —/r8 2 
Oc — Xc 1 — b gc^ 
Squaring and adding we get 
Of — A/) +/x/ 2 
Og=\y, +/xg :3 f 
Oh A h [ —h /x/q 
also 
O 2 = A 2 -bg 2 +2A/x cos 0 cos 8 ; 
fl 2 k=Xg sin 9 sin 8 
where 8 is the length of a common perpendicular to the two lines and 6 the angle 
between them. Further, from the iorm of the above equations, it follows that if any 
line of the rotation complex meet one axis of rotation, it will meet the other. Hence 
the two axes of rotation are conjugate polars with respect to the rotation complex. 
The axes of the screw are obtained by making the second line the polar of the first. 
This gives 
Oa=Arq-b/x/j 
Of=A f x +/xcq 
&c. 
and therefore 
A 2 +/x 2 =0 2 
\g — 0 2 & 
The axes of the screw are conjugate polars with respect to both complexes ; i.e ., 
they are the directrices of the congruance formed by lines common to the two 
complexes. 
37. We shall now find the surface corresponding to the Cylindroid. 
Let one twist be defined by the components Oa, Of, the others being zero; and a 
second twist by the components O'b', O'g', the rest being zero. Let the resultant 
motion be a rotation A about a line a, b, c,f, g, h, and a translation g along it; we have 
to find the surface generated by this line. As before 
Oa=Aa+/t/' 1 O b — Xb gg 
Of = A/ > +/raJ Vl'g'—\g-\-gb 
and 
c=0, h= 0. 
Hence the new axis always meets the lines (0, 0, 1,0, 0, 0), (0, 0, 0, 0, 0, 1), i.e., 
the new axis always meets the common perpendiculars to the axes of the given screws. 
From these equations we easily deduce 
2 e 2 
