Radius of curvature of a helix.

Leon.Smith

unread,

Sep 1, 2002, 1:32:17 PM9/1/02

to

I have two books and a paper, that all simply state at the beginning of a
problem,

"It can be shown that: r = R/(CosQ)^2"

Where r is radius of curvature of a helix traced on a circular cylinder of
radius R. Q being the angle that the helix descends at. Then that result is
used to solve other problems. However, they don't say where that initial
result comes from. I thought it must be a fairly well known result, since
one of the books was published way back in 1935, the most recent was 1995,
but I've since exhausted my college library trying to find how it was
derived.

Anyone know how to derive that, or have a source for it that I can follow
up?

Leon

Lee Rudolph

unread,

Sep 1, 2002, 1:49:17 PM9/1/02

to

"Leon.Smith" <Leon....@blueyonder.co.uk> writes:

>I have two books and a paper, that all simply state at the beginning of a
>problem,
>
>"It can be shown that: r = R/(CosQ)^2"
>
>Where r is radius of curvature of a helix traced on a circular cylinder of
>radius R. Q being the angle that the helix descends at. Then that result is
>used to solve other problems. However, they don't say where that initial
>result comes from.

It comes, via trivial calculation, from the definitions of the terms
involved. Surely the two books (and perhaps even the paper?), since
they include "a problem", do so because they have previously given
an exposition of the terms *in* the problem, the reader's mastery of
which they are (hopelessly) trying to verify?

>I thought it must be a fairly well known result, since
>one of the books was published way back in 1935, the most recent was 1995,
>but I've since exhausted my college library trying to find how it was
>derived.
>
>Anyone know how to derive that,

Find the definition (or post-definition calculation) of "curvature" (or
of "radius of curvature") of a parametrized space curve in one or more
of those books, then apply the definition to your helix, parametrized
any way you see fit (the obvious parametrization will make things
easiest).

>or have a source for it that I can follow
>up?

I would no more expect to find a "source" for that result than
I would expect to find a "source" for the result that 132+97=229;
the authors' phrase "It can be shown that" is a bit of a dead
giveaway, don't you think?

Lee Rudolph

The Last Danish Pastry

unread,

Sep 1, 2002, 2:20:19 PM9/1/02

to

"Leon.Smith" <Leon....@blueyonder.co.uk> wrote in message
news:iasc9.3428$rd2...@news-binary.blueyonder.co.uk...

Let p=R*(cos(theta), sin(theta), theta*tan(Q)) be the position vector of a
point on the helix.

t=dp/ds is a unit vector in the direction of the tangent to the curve at p
(s being distance measured along the helix).

So, t=R*(-sin(theta), cos(theta), tan(Q)) * d theta/ds

Since t is a unit vector we have
t=cos(Q)*(-sin(theta), cos(theta), tan(Q))
so
R * d theta/ds = cos(Q)
so
d theta/ds = cos(Q)/R

For any space curve dt/ds = kappa n, where kappa is the curvature and n is
the unit normal vector.

Now, dt/ds=cos(Q)*(-cos(theta), -sin(theta), 0) * d theta/ds
=(-cos(theta), -sin(theta), 0)*cos(Q)^2/R

So, kappa = |dt/ds| = cos(Q)^2/R

So, the radius of curvature, r = 1/kappa = R/cos(Q)^2

Clive Tooth
http://www.clivetooth.dk

Leon.Smith

unread,

Sep 2, 2002, 7:39:16 PM9/2/02

to

"The Last Danish Pastry" wrote:
>
> So, the radius of curvature, r = 1/kappa = R/cos(Q)^2
>

Thanks, much appreciated!

Leon

RM Mentock

unread,

Sep 6, 2002, 9:50:59 PM9/6/02

to

Lee Rudolph wrote:
>
> a bit of a dead giveaway, don't you think?

absolutely, thanks

--
RM Mentock

C. K. Monet, c'est moi