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curvature of an arc


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#1 telcom_un   Members   -  Reputation: 101

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Posted 19 September 2012 - 02:34 AM

Hi everybody,

I got a problem and I don't know how to figure it out.

In short I have an arc r(t)=[x(t), y(t), z(t)]
i'm gonna calculate the radius of this arc. As we know the curvature is the inverse of radius. If we consider the following formula:
K(t)=||r'(t) x r''(t)|/||r'(t)||^3
when ' means the first derivative to time and '' means the second derivative to the time and the the X is cross product.

So by having this equation and considering one simple example: A circle at the origin of a 2D space:

x(t)= R (1-t^2)/(1+t^2)
y(t)=R 2t/(1+t^2)
z(t)=0
when tE[0,1] and R is the radius. (EQ1)

By the formula of curvature K(t) is expected to be 1/R but when I compute it strangely for t=0 I get 4*R and for the t=1 I get R and for the time between two margins the K(t) is a non-linear distribution from 4R to R.

I have no idea why it happens. It's very simple example and I am wondering whether I do a silly mistake in my calculations or it's just a wrong assumption in my calculation;.

I created one Maple file to do so. And the result is flawless in sense of numerical and parametric calculation I guess.

In addition, by selecting another form of a circle equation by sinusoidal ones. I get better consistency but it's not even the answer. So if I select

x(t)=R.cos(t)
y(t)=R.sin(t)
z(t)=0

t [0,pi/2]
(EQ2)

The comuted curvature for the whole range of t is R!!! it's wrong because we expect 1/R not R.

So I looked for any possible example. Finally I found one in a literature. They change the parameters and the new circle equation is:

x(t)=R.cos(t/R)
y(t)=R.sin(t/R)
z(t)=0
t [0,pi/2]
(EQ3)

And the result is correct we get the curvature by this equation 1/R!!

So I'm really confused why three different answers I get by the same formula when the resulting geometrical shape is the same. Please help me out thank you in advance.

all the best

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#2 RulerOfNothing   Members   -  Reputation: 1164

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Posted 19 September 2012 - 03:09 AM

With the second equation, I get r'(t)=[-Rsin(t), Rcos(t), 0] and r''(t)=[-Rcost(t),-Rsin(t),0], taking the cross product gives [0,0,R^2.sin^2(t)+R^2.cos^2(t)] which is equal to [0,0,R^2] which has a magnitude of R^2. r'(t) has a magnitude of R, so when I divide ||r'(t) X r''(t)|| by the cube of that magnitude I get 1/R. How are you computing the cross product?




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