5) 
is the angle between the tangent to the path and the axis 
Ox, we have on integrating (2) 
y = hb + (4b? + $a? + 4a? cos 6)? (3) 
We choose the axis Oy so that x is zero when y is a maximum, 
and measure the arc s from this point. Then 
2 268 
L= (1 — zs + Spam(=*); y =%b + Gane); (4) 
where the modulus of the elliptic functions is =(1+6?/4a?)-?. 
In terms of elliptic integrals which are usually tabulated, 
namely F(i,¢) and E(4,) and the corresponding complete 
integrals K and E, we find the following results: the letters 
refer to the symmetrical] curve in fig. 2. 
(i) At the point B. y=3b+a/k; p= fka. 
(ii) At C, the widest part of the loop. y= bo +2V (1 —k?); 
2 Oy or 
eeikka \ (1 — SFU 45% + GE 45°) } 
p= Ha/V 0 — 3). 
D) 9 
(iii) Atanend pointA. y=b; 7= biea\(1 — 7K +25} ; 
p= a?/206. 
These data are generally sufficient for drawing the curve 
with considerable accuracy. From the periodicity of the 
elliptic functions we can also’ write down the total length of 
the path ABCD; it is equal to kKa. As a numerical 
example, one finds that the total distance covered by a 
particle initially at a distance a from the axis, as the cylinder 
_moves from an infinite distance behind to an infinite distance 
in front of the particle, is approximately 2a; this is the 
curve denoted by 1:0 in Fig. 2. It need hardly be pointed 
out that although the limiting length of path is finite, the 
time involved becomes infinite. 
In Fig. 2, some curves are drawn for various values of the 
ratio of 6 toa; except in one case, only half of each complete 
path is shown. For 6 zero, the path is infinite in length and 
Is given by 
x=atanh(2s/a)—s; y =asec h(2s/a). 
84 
