How can I find t in a Bezier Curve when I know the Y coordinate?

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To find the parameter t in a Bezier curve when the Y coordinate is known, one must solve the equation Y(t) - Ygiven = 0, where Y(t) is the polynomial representing the curve. This polynomial's degree corresponds to the number of control points, and multiple t values may exist within the interval [0,1]. For lower degree curves, established methods can be applied, while for higher degrees, iterative methods like Newton's can be effective. Additionally, if the polynomial can be factored easily, the roots can be directly obtained from the factored form. Understanding these approaches is crucial for accurately determining t based on a known Y value.
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Hello

I have a little problem with bezier curve. There is a bezier curve, with n degree, with some control points. And the problem is, that i want to know the t, and i know the y of that point. So there's a curve, in this curve there is a point, i know that place, and i want to know, what is t in this bezier.

With this algorithm(which i attached) i know the point, if i know the t
8ce08ca45bf146d5ec689f731367b326.png


but reverse?

i know, that the t can be more numbers. I want just one of them...
 
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The Y(t) coordinate of the 2-dimensional B(t) equation you posted is a polynomial of degree n; what you want are the solutions of the equation Y(t) - Ygiven = 0, that is, the roots of the polynomial Y(t) - Ygiven, where Ygiven is your known Y value. The expression for Y(t) is, of course, the same as in the image you posted, substituting each point P with the Y value of that point. Among the multiple values for t, you want the one(s) that are in the interval [0,1].

If n is small (like 2) there are known methods. Otherwise, a possibility is to use an iterative method like Newton's, since the derivative of the polynomial should be easy to write down. See, for example, http://en.wikipedia.org/wiki/Newton%27s_method" . Or maybe the polynomial turns out to be easy to factor, in which case the roots can be taken immediately from the factored expression.
 
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