Torsion of a Helix on any Cylinder. 59 
Examples. 
1. In the common helix, p'=a=the radius of the circular 
section of the cylinder. 
.*. p = a cosec 2 a, 
a = a cosec a sec a. J 
These are well-known formula?. 
a 2 b 2 
2. In an elliptic cylinder p '= — r , where p is the per- 
pendicular from the axis onto the tangent plane at P, a and b 
being the semi-axes of the principal elliptic section of the 
cylinder. 
a 2 b 2 
p = — 3 cosec «, 
H P 
a 2 b 2 
<r = — r cosec a. sec a. 
The next step is to apply the above method to find the 
curvature and torsion of a helix on any surface of revolution. 
As before, I shall take a as the angle which the tangent at 
any point makes with the axis of z, and I shall further 
suppose that the axis of z is the axis of the surface. Let 
-=/"(>) be the equation of the generating curve, r being the 
distance of any point from the axis. 
Let P and Q be two neighbouring points on the helix, and 
let PQ=d5. 
Let the cylindrical coordinates of P and Q be (r, r, <£) and 
{z + dz, r + dr, (j> + d(f)) respectively. 
Then 
ds 2 = dz 2 + dr 2 + r 2 dcj> 2 ._ 
.', dz 2 sec 2 a = dz 2 + dr 2 + r 2 d(f> 2 , since dz = ds cos a. 
,\ dr 2 [tan 2 a . f (r) 2 — 1] = i^d<jr, since dz =/' (r)dr. 
dr 
'# \/tan 2 «./V) 2 -l 
the + or — sign being taken according as r increases or 
decreases with <f>. 
Draw the spherical diagram as before and let TZT' = d<o. 
It will not now be equal to d<f>. 
day = the angle between a vertical plane at P containing 
the tangent line and the corresponding plane at Q. 
Now the tangent plane at P makes an angle i/r with a 
