I've been struggling looking for an understandable way to do this. I have four points, a StartPt, EndPoint, and Intersection points to represent the peak and valley in the bezier.
The BezierSegment in C# requires start, controlPoint 1, controlPoint 2, endpoint - however I don't have any control points I only have these two points that lie along the bezier curves (i'm calling them intersection points above)... how can I calculate the two control points?
Thanks in advance, this has been driving me crazy.
There's some kind of explanation here: http://www.tinaja.com/glib/nubz4pts1.pdf but it's written in postscript and that language makes no sense to me at all - it's over my head.
To find any point P along a line, use the formula: P = (1-t)P0 + (t)P1 , where t is the percentage along the line the point lies and P0 is the start point and P1 is the end point. Knowing this, we can now solve for the unknown control point.
To get C1 continuity, we need the two tangent vectors to be the same magnitude and opposite direction. That is, p3 - p2 = q1 - q0. 2.
Two obvious special cases: k=0: The Bezier curve starts at the first control point and stops at the last control point. (In general, it will not pass through any other control point.)
A matlab package for handling n-dimentional bezier curves. A bezier curve is parametrized by controlPts - which is [N x dim] for N control points of dimension dim. Note that we use matlab matrix ordering, so the first dimension will be treated as 'y' in the 2D case.
There are an infinite number of solutions to a curve passing through 4 points, but the best simple solution is to try to make the curve segment lengths proportional to the chord lengths. The code you link to is the a first order approximation that works well and is pretty fast.
Here's the C# translation of the PostScript code:
static class DrawingUtility
{
// linear equation solver utility for ai + bj = c and di + ej = f
static void solvexy(double a, double b, double c, double d, double e, double f, out double i, out double j)
{
j = (c - a / d * f) / (b - a * e / d);
i = (c - (b * j)) / a;
}
// basis functions
static double b0(double t) { return Math.Pow(1 - t, 3); }
static double b1(double t) { return t * (1 - t) * (1 - t) * 3; }
static double b2(double t) { return (1 - t) * t * t * 3; }
static double b3(double t) { return Math.Pow(t, 3); }
static void bez4pts1(double x0, double y0, double x4, double y4, double x5, double y5, double x3, double y3, out double x1, out double y1, out double x2, out double y2)
{
// find chord lengths
double c1 = Math.Sqrt((x4 - x0) * (x4 - x0) + (y4 - y0) * (y4 - y0));
double c2 = Math.Sqrt((x5 - x4) * (x5 - x4) + (y5 - y4) * (y5 - y4));
double c3 = Math.Sqrt((x3 - x5) * (x3 - x5) + (y3 - y5) * (y3 - y5));
// guess "best" t
double t1 = c1 / (c1 + c2 + c3);
double t2 = (c1 + c2) / (c1 + c2 + c3);
// transform x1 and x2
solvexy(b1(t1), b2(t1), x4 - (x0 * b0(t1)) - (x3 * b3(t1)), b1(t2), b2(t2), x5 - (x0 * b0(t2)) - (x3 * b3(t2)), out x1, out x2);
// transform y1 and y2
solvexy(b1(t1), b2(t1), y4 - (y0 * b0(t1)) - (y3 * b3(t1)), b1(t2), b2(t2), y5 - (y0 * b0(t2)) - (y3 * b3(t2)), out y1, out y2);
}
static public PathFigure BezierFromIntersection(Point startPt, Point int1, Point int2, Point endPt)
{
double x1, y1, x2, y2;
bez4pts1(startPt.X, startPt.Y, int1.X, int1.Y, int2.X, int2.Y, endPt.X, endPt.Y, out x1, out y1, out x2, out y2);
PathFigure p = new PathFigure { StartPoint = startPt };
p.Segments.Add(new BezierSegment { Point1 = new Point(x1, y1), Point2 = new Point(x2, y2), Point3 = endPt } );
return p;
}
}
I haven't tested it, but it compiles. Just call DrawingUtility.BezierFromIntersection
with the 4 points you have, and it will return a PathFigure
for drawing the curve.
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