Draw a quadratic Bézier curve through three given

问题:

I have three points in 2D and I want to draw a quadratic Bézier curve passing through them. How do I calculate the middle control point (x1 and y1 as in quadTo)? I know linear algebra from college but need some simple help on this.

How can I calculate the middle control point so that the curve passes through it as well?

回答1:

Let P0, P1, P2 be the control points, and Pc be your fixed point you want the curve to pass through.

Then the Bezier curve is defined by

P(t) = P0*t^2 + P1*2*t*(1-t) + P2*(1-t)^2

...where t goes from zero to 1.

There are an infinite number of answers to your question, since it might pass through your point for any value of t... So just pick one, like t=0.5, and solve for P1:

Pc = P0*.25 + P1*2*.25 + P2*.25

P1 = (Pc - P0*.25 - P2*.25)/.5

   = 2*Pc - P0/2 - P2/2

There the "P" values are (x,y) pairs, so just apply the equation once for x and once for y:

x1 = 2*xc - x0/2 - x2/2
y1 = 2*yc - y0/2 - y2/2

...where (xc,yc) is the point you want it to pass through, (x0,y0) is the start point, and (x2,y2) is the end point. This will give you a Bezier that passes through (xc,yc) at t=0.5.

回答2:

I have used Nemos answer in my JavaFX apllication, but my goal was to draw the curve, so that the visual turning point of the curve always fits with the choosen fixed one (CP).

CP = ControlPoint
SP = StartPoint
EP = EndPoint
BP(t) = variable Point on BeziérCurve where t is between 0 and 1

To achieve this i made a variable t (not fix 0.5). If the choosen Point CP is no longer in the middle between SP and EP, you have to vary t up or down a little bit. As a first step you need to know if CP is closer to SP or EP: Let distanceSP be the distance between CP and SP and distanceEP the distance between CP and EP then i define ratio as:

ratio = (distanceSP - distanceEP) / (distanceSP + distanceEP);

Now we are going to use this to vary t up and down:

ratio = 0.5 - (1/3) * ratio;

note: This is still an approximation and 1/3 is choosen by try and error.

Here is my Java-Function: (Point2D is a class of JavaFX)

private Point2D adjustControlPoint(Point2D start, Point2D end, Point2D visualControlPoint) {
    // CP = ControlPoint, SP = StartPoint, EP = EndPoint, BP(t) = variable Point on BeziérCurve where t is between 0 and 1
    // BP(t) = SP*t^2 + CP*2*t*(1-t) + EP*(1-t)^2
    // CP = (BP(t) - SP*t^2 - EP*(1-t)^2) / ( 2*t*(1-t) )
    // but we are missing t the goal is to approximate t
    double distanceStart  = visualControlPoint.distance(start);
    double distanceEnd    = visualControlPoint.distance(end);
    double ratio          = (distanceStart - distanceEnd) / (distanceStart + distanceEnd);
    // now approximate ratio to be t
    ratio = 0.5 - (1.0 / 3) * ratio;

    double ratioInv = 1 - ratio;
    Point2D term2 = start.multiply( ratio * ratio );
    Point2D term3 = end.multiply( ratioInv * ratioInv );
    double  term4 = 2 * ratio * ratioInv;

    return visualControlPoint.subtract(term2).subtract(term3).multiply( 1 / term4);
}

I hope this helps.

回答3:

If you don't want the exact middle point, rather you want any value for t (0 to 1), the equation is:

controlX = pointToPassThroughX/t - startX*t - endX*t;
controlY = pointToPassThroughY/t - startY*t - endY*t;

Of course, this will also work for the mid point, just set t to be 0.5. Simple! :-)