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Old Jun19-09, 11:20 AM                  #1
sullis3

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Radius of Curvature in 3 Dimensions

I have an equation for a curve that lies along the surface of a truncated cone. In polar coordinates:

theta(r) = K * [ U + arctan(1/U) - (Pi/2) ]

where:

U = SQRT[ ((r/r1)^2) - 1 ]
K = SQRT[ 1 + (H/(r2-r1))^2 ]
r = r1 + (r2-r1)(z/H)

r1 = minor radius of the truncated cone
r2 = major radius of the truncated cone
H = the height of the truncated cone

Physically, think of this as a string wrapped around the truncated cone from one end to the other, with its "angle of attack" varying along the height of the cone.

How can I go about calculating the radius of curvature of this 3D curve as a function of some parameter (preferably arc length)?
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Old Jun19-09, 11:28 AM                  #2
arildno

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Re: Radius of Curvature in 3 Dimensions

Well, you can check up on the Frenet-Serret formulas:
http://en.wikipedia.org/wiki/Frenet-Serret
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Old Jun19-09, 12:00 PM       Last edited by sullis3; Jun19-09 at 12:23 PM.. Reason: clarification            #3
sullis3

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Re: Radius of Curvature in 3 Dimensions

Thanks arildno - that is helpful. Here's my understanding of what I read:

Given a parameterized form of the equation, I can calculate the values T, N, and B. Furthermore I can calculate their derivatives with respect to s. This provides the inputs needed to solve for kappa and tau using the first two Frenet-Serret equations (with only two unknowns, only two of the equations are required, yes?).

Kappa is the curvature - the reciprocal of which would be the radius of curvature I'm looking for.

Does all that sound correct?

Assuming my understanding so far is on solid ground, this is very promising ... but I find myself still stuck at the beginning. How can I convert my original equation - theta as an unwieldy function of z - into a parameterized form that I can then work with? I'm unsure whether I should be converting this to Cartesian form and parameterizing it to r(x(t), y(t), z(t)) - and if so how to do that - or if there is a means to arrive at a parameterized form directly from the polar form.

Thanks ...

[edit] There are actually two equations I'd need to parameterize to fully define the curve: theta(z) as given above, but also r(z). These two equations taken together specify the curve in question.
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Old Jun19-09, 02:05 PM                  #4
arildno

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Re: Radius of Curvature in 3 Dimensions

You should check out the latter part of that essay,
starting at "4 Other expressions of the frame"

That will give you the expression for the radius of curvature with respect to an arbitrary paramater.
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