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.

Answer 1

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

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.