Torsion of a Helix on any Cylinder, 
57 
principal normal, and the osculating plane contains the tangent 
and the principal normal. Hence it follows that the osculating 
plane at any point of a helix on any surface makes an angle 
~ — a, with the plane of xy. 
The simplest helix is that described on a circular cylinder. 
This is known as the common helix or spiral staircase. Its 
curvature and torsion can easily be found by several methods. 
I shall not consider this curve at all. 
The first tj'pe of helix I shall discuss is that described on 
any cylinder, the axis of the cylinder being taken as axis of z. 
In this case the tangent line makes an 
angle a with the axis of z, and therefore 
with any generator. Hence the curve cuts 
the generators at the constant angle a. 
Then angle between the tangent-line and 
the axis =a, and the angle between the 
7T 
binormal and the axis =-— a. 
<C^ ]^ 
Q 
p N 
' : d_ 
^? 
Let P and Q be two neighbouring points 
on the curve, and let VQ = ds. Draw the 
generators through P and Q, and let the 
generator through Q cut the normal section 
of the cylinder through P in N. 
Let p ! be the radius of curvature of this 
normal section at P, and let defy be the angle 
between the tangent planes to the cylinder at P and Q, 
t. e. at P and N. 
Then ¥m=p'd$ = dssm a, from the figure. 
z Draw a unit sphere with any 
point as centre. Draw OZ 
parallel to the axis of the 
cylinder. Draw OB parallel 
to the binormal at P and OT 
parallel to the tangent line. 
Let B', T' be the points cor- 
responding to the binormal and 
tangent line at Q. 
T' 
Then 
7T 
ZB = ZB'=~-a, 
ZT=ZT'=«, BT=B'T': 
7T 
Now TT' = (i0=the angle between the tangents to the 
curve at P and Q. 
Also 
and Q. 
BB / = <;^ = the angle between the binomials at P 
