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About telcom_un

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  1. curvature of an arc

    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