A few requests for the algorithm behind the Bezier y-at-x and x-at-y methods have been received. The code is self-contained in the com.degrafa.geometry.AdvancedQuadraticBezier and AdvancedCubicBezier classes. Both classes accept control points in their constructors and use classes from com.degrafa.utilities.math . The latter classes are independent from any internal Degrafa architecture. So, people who maintain their own Bezier class libraries have a path to include the y-at-x and x-at-y capability.
Finding the y-coordinates at a given x or the x-coordinates at a given y is an exercise in root-finding. Suppose the Bezier curve is B(t) which is quadratic or cubic in t. For the cubic Bezier, the vector equation for the curve produces two scalar equations, one for the x-coordinates and one for the y-coordinates, i.e.
Bx(t) = c0x + c1xt + c2xt2 + c3xt3
By(t) = c0y + c1yt + c2yt2 + c3yt3
For the y-at-x case, the required y-coordinates are found at the t-parameters corresponding to the intersection of B(t) with the vertical line x = a, or
c0x + c1xt + c2xt2 + c3xt3 – a = 0
The problem reduces to one of finding roots of a cubic polynomial. This starts by finding one root. Synthetic division is used to factor out the single root leaving a quadratic polynomial. The two possible roots of that polynomial are found with the venerable quadratic formula. Only roots in [0,1] are valid solutions to the y-at-x problem. The roots (values of t) are used to compute the y-coordinates using the equation for By(t). With a quadratic Bezier, roots are found directly with the quadratic formula.
The x-at-y problem is similar in that the required x-coordinates are the intersection of B(t) with the horizontal line y = a. There is one degenerate case where the Bezier curve is a segment of a horizontal or vertical line and the single parameter describing the line is the input value. For example, a cubic Bezier curve is the line x = a and the y-coordinates for x = a are desired. Theoretically, there are an infinite number. The Degrafa code does not currently handle this case. Some argue that no values should be returned while otheres argue that perhaps t = 0, t = 1/2 and t = 1 should be returned.
Given the rarity of this case actually occurring, the argument is postponed in lieu of more important developments. Tomorrow, I’ll post some code from the demos.
3 thoughts on “Bezier Y at X Algorithm”
How do you compute the initial root?
First, use bisection to either show that no root exists in [0,1] or get a better bound on one root. Then, use any root finder (I use Jack Crenshaw’s TWBRF) with the new bounds.
Thank you so much for this. I’ve been trying to understand this for a long time. I’m working on the simpler quadratic beziers in flash and I think this finally helped me to get it.
I still don’t understand the cubics and TWBRF, but baby steps…
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