Commenter Timo wants to know how to get a nice shape for a reverse engineered Bézier curve.

The question started from calculating the control points of a cubic Bézier curve if you’re given 4 points that lie on the curve, assuming the first and last given points were also the first and last control points. I wrote a similar article for a quadratic Bézier curve too.

To sum up, you have 1 degree of freedom when working on a quadratic Bézier curve. In the case of a cubic Bézier curve, you have 2 degrees of freedom, meaning 2 variables in the calculation that *you* have to decide on. Now this posed a problem, because there are an infinite number of solutions. How do you decide on a numeric value?

Well, for a quadratic Bézier curve, the simplest and obvious option is to choose the second given point (since the first and last control points are determinable) to be at the halfway mark. For a cubic Bézier curve, the second and third given points are chosen to be at the 1/3 and 2/3 mark along the curve respectively. Now this may or may not be suitable, but it *does* give you something to start with.

I want to state right now, that I had not been doing formal mathematics for a while. There is a limit to what I know, and I’m not an expert. I just know enough to figure out how to solve problems. Sometimes, it’s not enough. Keep that in mind.

### “Nice curve” is subjective

Now Timo’s problem is getting a better shaped cubic Bézier curve from those calculations. Since the 4 given points are fixed, and the first and last control points are also fixed, the only thing you can manipulate are the second and third control points. Which in turn means deciding on values of u and v to get a “nice” cubic Bézier curve in the end.

This, “niceness”, is a subjective criteria. **How do you determine if a cubic Bézier curve looks nice?** Remember that we don’t have the control points yet. So we don’t know how the curve looks like. So we don’t even know if manipulating the second and third control point to *not* be at 1/3 and 2/3 will result in a nice curve. It’s a chicken and egg problem.

### Apportioned chord length

During the correspondence with Timo, some solutions were discussed. The next simplest set of values to try for u and v are calculated based on the given points.

Let d1 be the distance between the first and second given point.

Let d2 be the distance between the second and third given point.

Let d3 be the distance between the third and last (fourth) given point.

Then let u = d1/(d1 + d2 + d3) and v = (d1 + d2)/(d1 + d2 + d3). This should result in a curve that’s “better shaped” than the (u,v) pair (1/3, 2/3). When I wrote that article for the cubic curve version, this was the next default set of values for u and v, but I didn’t want to add too much more. Well, nobody asked, so I left it as it was.

It turns out that Don Lancaster already wrote about it. He called it “apportioned chords” method.

http://www.tinaja.com/glib/nubz4pts1.pdf

### Inflection points

Then Timo had another problem. He wanted the second and third given points to also lie at the “loop tips”. What are loop tips? After some clarification, I believe Timo is referring to the inflection points on the curve. An inflection point

is a point on a curve at which the curvature (second derivative) changes signs. The curve changes from being concave upwards (positive curvature) to concave downwards (negative curvature), or vice versa.

A cubic Bézier curve can have 0, 1 or 2 inflection points. If it’s a straight line, it has 0. If it’s U-shaped, it has 1. And if it zigzags, it has 2. And Adrian Colomitchi proved that there are at most 2 inflection points on a cubic Bézier curve.

**UPDATE**: The following diagram is wrong. Please refer to this article for the correct version. I was thinking of points where the second derivative was zero, not when it changed signs (as defined above).

As you can see, inflection points don’t necessarily have to coincide with the given points.

By the way, I used Paint.NET for the illustration. I took a screenshot of me drawing the curve, still with the given points visible (noted by the small squares). Paint.NET appears to have succeeded in doing the very thing Timo wants, to render a cubic Bézier curve using 4 given points. Of course, I’m assuming the image editor is using cubic Bézier curves…

### The math paper

I found another reference with a more explicit mathematical formulation to help Timo.

http://www.cis.usouthal.edu/~hain/general/Publications/Bezier/bezier%20cccg04%20paper.pdf

As of this writing, page 3 of that paper explicitly shows the calculation needed to find the inflection points of a cubic Bézier curve, **if they exist**. Let me emphasise that again. The curve might have 2 inflection points, only 1 inflection point, or **none at all** (a straight line, the trivial version of a Bezier curve, has no inflection points).

### Maximum curvature

Then Timo found another forum posting for the fast calculation of maximum curvature. An inflection point would have the “maximum curvature”. The problem with that solution is that it assumes we have the control points.

So my suggestion was to do iteration. Use my method to get a set of control points. Perhaps use the apportioned chords to start off with a good set of control points. Then apply the maximum curvature solution to find the inflection points and the associated values of t. With those values of t, pump them back into my method to find a new set of control points. Pump *those* control points into the maximum curvature solution to find inflection points and their t values. Iterate till the t values between iterations are within acceptable margins of error.

**Caveat**: I don’t know if this combination of 2 algorithms in an iterative manner will converge. I have not tested this. Use at your own risk.

### Scaling up to 3 dimensions

Anyway, Timo found another solution himself (he didn’t say what though). He still needed to handle that cusp point. What cusp point, you ask? It’s in page 3 of that paper I mentioned above. That paper is for 2 dimensional cubic Bézier curves. The degree of the curve is independent of the degree of the dimensions. Timo wants to know how the 3 dimensional case will look like.

Now, the method of finding inflection points is to do a cross product of the first derivative and second derivative of the cubic Bézier curve equation. The Bézier curve is parametrised into

x(t) = ax * t^3 + bx * t^2 + cx * t + dx

y(t) = ay * t^3 + by * t^2 + cy * t + dy

and using the Bézier basis matrix, the coefficients are

ax = -x0 + 3*x1 – 3*x2 + x3

bx = 3*x0 – 6*x1 + 3*x2

cx = -3*x0 + 3*x1

dx = x0

ay = -y0 + 3*y1 – 3*y2 + y3

by = 3*y0 – 6*y1 + 3*y2

cy = -3*y0 + 3*y1

dy = y0

Set the cross product to 0. The inflection points are found at values of t when you solve that equation:

x’ * y” – x” * y’ = 0

where x’ and x” are the first and second derivatives of x(t). Similarly for y’ and y”. The solution is in that paper I mentioned before.

This is if the curve is in 2D. The cross product of 2D vectors is a scalar. And we set the scalar to 0 to solve for t. The cross product of 3D vectors is a *vector*, and so we’re solving with a zero vector.

So with

z(t) = az * t^3 + bz * t^2 + cz * t + dz

for the third dimension, we have

x’ = 3 * ax * t^2 + 2 * bx * t + cx

y’ = 3 * ay * t^2 + 2 * by * t + cy

z’ = 3 * az * t^2 + 2 * bz * t + cz

x” = 6 * ax * t + 2 * bx

y” = 6 * ay * t + 2 * by

z” = 6 * az * t + 2 * bz

The cross product is the determinant of the following

where **i, j, k** are the unit vectors. I’ll leave it to you to find out the formula for the determinant of a 3 by 3 matrix.

So we’re going to solve this:

**0** = (y’ * z’’ – z’ * y’’) * **i** – (x’ * z’’ – z’ * x’’) * **j** + (x’ * y’’ – y’ * x’’) * **k**

where **0** is the zero vector.

This implies that

y’ * z’’ – z’ * y’’ = 0

x’ * z’’ – z’ * x’’ = 0

x’ * y’’ – y’ * x’’ = 0

This time the zeroes are scalars. We now have 3 times the number of equations to solve when compared to the 2D case. This means there are potentially 6 values of t for the inflection points to check. Hopefully, there will be repeated values of t. Hopefully, the number of unique values of t is 2 or less (remember Adrian’s proof?).

If a t value is repeated, it’s probably an inflection point. What if we get 6 unique values? Is a 6-unique-value case even possible? I don’t know. You’ll have to interpret the values in the best way you can, based on some assumptions and arguments.

### What do you think?

So after reading through that entire article, what do you think? Comments on the methods I described? Do you have a new method? Your thoughts on whether this problem is even solvable? That I’m a complete idiot?

Let me know in a comment. Or better, write a blog post and tell me about it. Because if it took me that long to explain the solutions, your solution is probably just as long. It’s a Bézier curve. A picture might be appropriate.

Thanks for summing it up, and helping out! Currently I’m on vacation with somewhat limited connectivity, as my macbook has decided it’s ok to completely turn off the display backlight as soon as you have logged in XP.. Later I can post some screenshots and my implementation in C#, although I haven’t yet finished all parts

What I initially wanted to achieve is done – now user can grab any point from the curve and move that. So I could totally hide the control points if I wanted.

You’re welcome Timo. Have fun on your vacation, and I’d love to see your program.

I put a minimalistic page up with some source, binary and a screenshot: http://neure.ath.cx/curve.html

Unfortunately it needs Windows and OpenGL 3.2 or later.

You can click on any point on the curve, that will show up edit handle near the curve point you clicked. You can also click on the control points. When control point or edit handle is selected, you can move them by dragging the manipulator cones.

The manipulator is slightly buggy, if the view is nearly aligned with the axis. I will fix this eventually.