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I'm given a periodic parametric curve $P = ( x(t),y(t) )$, where $t \in [0, 2\pi)$ and $P(2\pi)=P(0)$. I have a point $F = (x,y)$ that is not on that surface. Could someone tell me how to find the point $B = (x_p,y_p)$ on that surface that is normal to F? Is there a closed form expression I could use?

Note: I have the same problem in 3D, with a surface $P = ( x(u,v),y(u,v),z(u,v) )$, where $u \in [0, 2\pi)$ and $v \in [0, \pi)$.

Thank you for your time!

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up vote 1 down vote accepted

This link, may point you in the right direction. It uses an iterative approach, so I'm not sure if it is ideal for your situation. Initially, I assumed there would be a quick solution, analogous to projection onto a line, however I couldn't find such a solution.

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That would definitely work in my case. To provide some context, the parametric curve I'm working on was originally defined as a set of piecewise quadratics- in that case, one can simplify this down to a cubic equation. However, as a consequence of my research, the curve now has a more general description using a lot more than three coefficients, so I couldn't simplify it down to such a neat closed form expression. Thank you for the quick answer! – vergere6 Apr 15 '12 at 1:47

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