Finding the curvature of a 2nd degree equation

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Ricardo Fuertes
Ricardo Fuertes 2018-6-18
回答： Guillaume 2018-6-18
I have a 2nd degree equation that represents the trayectory of a moving object (in x and y). What i want to find is the curvature ( the reciprocal of the radius of curvature) of said trayectory. I was thinking about finding 3 points in the curve and finding the osculating circle, but I wasnt sure if there was a faster way to do it in mathlab.

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Guillaume
Guillaume 2018-6-18
Not sure what a second degree equation is. Is it a quadratic curve?
Anyway, the wikipedia article on curvature gives you the equations.
If your curve is indeed a quadratic curve, then you can even calculate the curvature analytically. Otherwise, assuming your function is of the form y = f(x),
K = gradient(gradient(y, x), x) ./ (1 + gradient(y, x).^2).^(3/2)

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