Joining bezier curves

Question

I have a problem. Suppose we have a single cubic bezier curve defined by four control points. Now suppose, the curve is cut from a point and each segment is again represented using cubic bezier curves. So, now if we are given two such beziers B1 and B2, is there a way to know if they can be joined to form another bezier curve B? This is to simplify the geometry by joining two curves and reduce the number of control points.

Using the fact that a Bézier curve always passes through the first and last control points (that is you need it to go through the end points of B1 and B2) you should be able to calculate the candidate curve B (this assumes you're only not actually looking for the curve that best matches B1 + B2). — user786653
How do you "each segment is again represented using cubic bezier curves". If you can do that your question is already answered! — Dr. belisarius
To be clear, you're asking not just if the tangents (first derivative) are aligned across the overlapping points, but also if the curvature (second derivative) is the same, right? — Phrogz
Yes, to be more clear, I have a set of bezier curves. and due to noise in the source, It might happen that a long bezier curve might be traced into segments. So, as an optimization and smoothing requirement, I want to look in the neighborhood of every curve and join them if the final curve B can approximate B1 + B2. Hope I am clear now. — Aarkan
@Aarkan: Perhaps if you could edit your post with a small pseudo-code function to show how you're representing the data and included an example it would be a lot easier to give concrete advice. Something like bool match(const ControlPoint B1[4], const ControlPoint B2[4], ControlPoint B[4]) { Approx(B1,B2,B); return (Compare(B1,B2,B) } with values for B1,B2 and B for an example where you want this to return true and explaining in words what Approx and Compare should do. — user786653

MBo MBo · Accepted Answer · 2011-11-10T04:46:40

Some thoughts about this problem. I suggest there was initial Bezier curve P0-P3 with control points P1 and P2

enter image description here

Let's make two subdivisions at parameters ta and tb. We have now two subcurves (in yellow) - P0-PA3 and PB0-P3. Blue interval is lost. PA1 and PB2 - known control points. We have to find unknown P1 and P2.

Some equations

Initial curve:

C = P0*(1-t)^3+3*P1(1-t)^2*t+3*P2*(1-t)*t^2+P3*t^3

Endpoints:

PA3 = P0*(1-ta)^3+3*P1*(1-ta)^2*ta+3*P2*(1-ta)*ta^2+P3*ta^3

PB0 = P0*(1-tb)^3+3*P1*(1-tb)^2*tb+3*P2*(1-tb)*tb^2+P3*tb^3

Control points of small curves

PA1 = P0*(1-ta)+P1*ta => P1*ta = PA1 – P0*(1-ta)

PB2 = P2*(1-tb)+P3*tb => P2(1-tb) = PB2 – P3*tb

Now substitute unknown points in PA3 equation:

**PA3***(1-tb) = **P0***(1-ta)^3*(1-tb)+3*(1-ta)^2*(1-tb)*(**PA1** – **P0***(1-ta))+3*(1-ta)*ta^2*( **PB2** – **P3***tb)+**P3***ta^3*(1-tb)

(some multiplication signs have been lost due to SO formatting)

This is vector equation, it contains two scalar equations for two unknowns ta and tb

PA3X*(1-tb) = P0X*(1-ta)^3*(1-tb)+3*(1-ta)^2*(1-tb)*(PA1X – P0X*(1-ta))+3*(1-ta)*ta^2*( PB2X – P3X*tb)+P3X*ta^3*(1-tb)

PA3Y*(1-tb) = P0Y*(1-ta)^3*(1-tb)+3*(1-ta)^2*(1-tb)*(PA1Y – P0Y*(1-ta))+3*(1-ta)*ta^2*( PB2Y – P3Y*tb)+P3Y*ta^3*(1-tb)

This system might be solved both numerically and analytically (indeed Maple solves it with very-very big cubic formula :( )

If we have points with some error, that makes sense to build overdetermined equation system for some points (PA3, PB0, PA2, PB1) and solve it numerically to minimize deviations.

Joining bezier curves

2 Answers