Is it possible to find the control points of a quadratic bezier curve, given the starting, ending and stationary point

bezier-curveconic sections

I know from this question on SO that it is possible to get the stationary point of a bezier curve given the control points, but I want to know wether the opposite is true:

If I have the start and end points of a Parabola, and I have the maximum point, is it possible to express this a quadratic bezier curve?

I have seen this answer explaining that you usually would need 6 points on a cubic to effectively describe a bezier curve, but looking at the question itself, it describes a few special cases where you can find the control points on a cubic, and I am curious on wether a quadratic bezier curve has such points, and wether a stationary point is one of them

The variable t of the stationary point is not avalible

I am coming from a computer science background, and although im not too weak in math, any answers/articles that don't use complicated maths and keep it simple would be appreciated

If this is not possible, would it help of the stationary point is always a maximum point and the x coordinate if the max point is the mean of the x coordinates of the start and end points?

By maximum point I mean the point with the maximum value for its y coordinate and by stationary I meant the point with the maximum or minimum value for y

Best Answer

Define the following:

The quadratic Bézier has control points $(x_0,y_0)$ (start), $(x,y)$, $(x_2,y_2)$ (end)
The third known point is at $(x_1,y_1)$ with parameter $t$ (in this part of the answer, this is the point with extremal $y$-coordinate)

The curve's equation is then $$x(t)=(1-t)^2x_0+2t(1-t)x+t^2x_2\tag1$$ $$y(t)=(1-t)^2y_0+2t(1-t)y+t^2y_2\tag2$$ and the constraints are $x(t)=x_1$, $y(t)=y_1$ and $y'(t)=0$. Eliminating $y$ in the last two equations gives $$\Delta y_2t^2-2\Delta y_1t+\Delta y_1=0\qquad(\Delta c_i=c_0-c_i)$$ $$t=a\pm\sqrt{a(a-1)},a=\frac{\Delta y_1}{\Delta y_2}\qquad(\Delta y_2\ne0)$$ $$\Delta y_2=0\implies t=\frac12$$ There is a solution in general iff $a\not\in(0,1)$ and only one choice of sign will give $t\in[0,1]$. Once $t$ has been found, it may be substituted back into $(1)$ and $(2)$ to find $x$ and $y$ respectively.

If $(x_1,y_1)$ is the parabola's vertex, the third equation may be obtained from this diagram:

The curve's tangent at the vertex is perpendicular to the line passing between the midpoint of the curve endpoints and the third control point, i.e. $$\begin{bmatrix}\frac{x_0+x_2}2-x\\\frac{y_0+y_2}2-y\end{bmatrix}\cdot\begin{bmatrix}x'(t)\\y'(t)\end{bmatrix}=0$$

A resultant computation followed by factoring out bogus solutions then shows that $t$ solves the cubic $$\begin{align} (\Delta x_2^2+\Delta y_2^2)&t^3\\ -3(\Delta x_1\Delta x_2+\Delta y_1\Delta y_2)&t^2\\ +(\Delta x_1(2\Delta x_1+\Delta x_2)+\Delta y_1(2\Delta y_1+\Delta y_2))&t\\ -(\Delta x_1^2+\Delta y_1^2)&=0 \end{align}$$

Related Solutions

[Math] Changing a bezier curve by dragging a point on the curve itself rather than a control point

I think the simplest thing that would work in your application is to show the user 4 special points on the parametric cubic curve, and allow the user to manipulate those 4 special points. (Allowing the user to pick any point on the curve, and move it, makes things more complicated).

I think this is the same as what Stephen H. Noskowicz calls "Cubic Four Point" representation, aka the quadratic Lagrange with t1 = 1/3 and t2 = 2/3.

While your user is moving those 4 special points U0, U1, U2, U3 around, periodically you find a cubic Bezier curve that goes through those 4 points using John Burkardt's approach:

P0 = U0
P1 = (1/6)*( -5*U0 + 18*U1 - 9*U2 + 2*U3 )
P2 = (1/6)*(  2*U0 -  9*U1 +18*U2 - 5*U3 )
P3 = U3.

That gives you the Bezier curve representation of the same cubic curve -- a series of 4 control points. You then feed those 4 control points (the endpoints P0 and P3, and the intemediate control points P1 and P2) into any Bezier curve plotter.

The resulting curve (usually) doesn't touch P1 or P2, but it will start at X0, go exactly through X1 and X2, and end at X3.

(This uses the special points at t=0, 1/3, 2/3, and 1. It's possible to, instead, use the special points at t=1, 1/4, 3/4, and 1, as shown at How do I find a Bezier curve that goes through a series of points? . Or, I suppose, any 4 distinct t values. But I suspect the 0, 1/3, 2/3, 1 values are used most often, and I don't see any advantage to using any other fixed values).

Bezier Curve – Control Points of Offset Bezier Curve

The simplest heuristic approximation method is the one proposed by Tiller and Hanson. They just offset the legs of the control polygon in perpendicular directions: bezier-offset

The blue curve is the original one, and the three blue lines are the legs of its control polygon. We offset these three lines, and intersect/trim them to get the points A and B. The red curve is the offset. It's only an approximation of the true offset, of course, but it's often adequate.

If the approximation is not good enough for your purposes, you split the curve into two, and approximate the two halves individually. Keep splitting until you're happy. You will certainly have to split the original curves at inflexion points, if any, for example.

Here is an example where the approximation is not very good, and splitting would probably be needed: bad offset

There is a long discussion of the Tiller-Hanson algorithm plus possible improvements on this web page.

The Tiller-Hanson approach is compared with several others (most of which are more complex) in this paper.

Another good reference, with more up-to-date materials, is this section from the Patrikalakis-Maekawa-Cho book.

For even more references, you can search for "offset" in this bibliography.

Best Answer

Related Solutions

[Math] Changing a bezier curve by dragging a point on the curve itself rather than a control point

Bezier Curve – Control Points of Offset Bezier Curve

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