Currently, I'm attempting to make multiple beziers have equidistant points. I'm currently using cubic interpolation to find the points, but because the way beziers work some areas are more dense than others and proving gross for texture mapping because of the variable distance. Is there a way to find points on a bezier by distance rather than by percentage? Furthermore, is it possible to extend this to multiple connected curves?
distance between P_0 and P_3 (in cubic form), yes, but I think you knew that, is straight forward.
Distance on a curve is just arc length:
fig 1 http://www.codecogs.com/eq.latex?%5Cint_%7Bt_0%7D%5E%7Bt_1%7D%20%7B%20|P'(t)|%20dt
where:
Probably, you'd have t_0 = 0, t_1 = 1.0, and dz(t) = 0 (2d plane).
This is called "arc length" parameterization. I wrote a paper about this several years ago:
http://www.saccade.com/writing/graphics/RE-PARAM.PDF
The idea is to pre-compute a "parameterization" curve, and evaluate the curve through that.
I know this is an old question but I recently ran into this problem and created a UIBezierPath extention to solve for an X coordinate given a Y coordinate and vise versa. Written in swift.
https://github.com/rkotzy/RKBezierMath
extension UIBezierPath {
func solveBezerAtY(start: CGPoint, point1: CGPoint, point2: CGPoint, end: CGPoint, y: CGFloat) -> [CGPoint] {
// bezier control points
let C0 = start.y - y
let C1 = point1.y - y
let C2 = point2.y - y
let C3 = end.y - y
// The cubic polynomial coefficients such that Bez(t) = A*t^3 + B*t^2 + C*t + D
let A = C3 - 3.0*C2 + 3.0*C1 - C0
let B = 3.0*C2 - 6.0*C1 + 3.0*C0
let C = 3.0*C1 - 3.0*C0
let D = C0
let roots = solveCubic(A, b: B, c: C, d: D)
var result = [CGPoint]()
for root in roots {
if (root >= 0 && root <= 1) {
result.append(bezierOutputAtT(start, point1: point1, point2: point2, end: end, t: root))
}
}
return result
}
func solveBezerAtX(start: CGPoint, point1: CGPoint, point2: CGPoint, end: CGPoint, x: CGFloat) -> [CGPoint] {
// bezier control points
let C0 = start.x - x
let C1 = point1.x - x
let C2 = point2.x - x
let C3 = end.x - x
// The cubic polynomial coefficients such that Bez(t) = A*t^3 + B*t^2 + C*t + D
let A = C3 - 3.0*C2 + 3.0*C1 - C0
let B = 3.0*C2 - 6.0*C1 + 3.0*C0
let C = 3.0*C1 - 3.0*C0
let D = C0
let roots = solveCubic(A, b: B, c: C, d: D)
var result = [CGPoint]()
for root in roots {
if (root >= 0 && root <= 1) {
result.append(bezierOutputAtT(start, point1: point1, point2: point2, end: end, t: root))
}
}
return result
}
func solveCubic(a: CGFloat?, var b: CGFloat, var c: CGFloat, var d: CGFloat) -> [CGFloat] {
if (a == nil) {
return solveQuadratic(b, b: c, c: d)
}
b /= a!
c /= a!
d /= a!
let p = (3 * c - b * b) / 3
let q = (2 * b * b * b - 9 * b * c + 27 * d) / 27
if (p == 0) {
return [pow(-q, 1 / 3)]
} else if (q == 0) {
return [sqrt(-p), -sqrt(-p)]
} else {
let discriminant = pow(q / 2, 2) + pow(p / 3, 3)
if (discriminant == 0) {
return [pow(q / 2, 1 / 3) - b / 3]
} else if (discriminant > 0) {
let x = crt(-(q / 2) + sqrt(discriminant))
let z = crt((q / 2) + sqrt(discriminant))
return [x - z - b / 3]
} else {
let r = sqrt(pow(-(p/3), 3))
let phi = acos(-(q / (2 * sqrt(pow(-(p / 3), 3)))))
let s = 2 * pow(r, 1/3)
return [
s * cos(phi / 3) - b / 3,
s * cos((phi + CGFloat(2) * CGFloat(M_PI)) / 3) - b / 3,
s * cos((phi + CGFloat(4) * CGFloat(M_PI)) / 3) - b / 3
]
}
}
}
func solveQuadratic(a: CGFloat, b: CGFloat, c: CGFloat) -> [CGFloat] {
let discriminant = b * b - 4 * a * c;
if (discriminant < 0) {
return []
} else {
return [
(-b + sqrt(discriminant)) / (2 * a),
(-b - sqrt(discriminant)) / (2 * a)
]
}
}
private func crt(v: CGFloat) -> CGFloat {
if (v<0) {
return -pow(-v, 1/3)
}
return pow(v, 1/3)
}
private func bezierOutputAtT(start: CGPoint, point1: CGPoint, point2: CGPoint, end: CGPoint, t: CGFloat) -> CGPoint {
// bezier control points
let C0 = start
let C1 = point1
let C2 = point2
let C3 = end
// The cubic polynomial coefficients such that Bez(t) = A*t^3 + B*t^2 + C*t + D
let A = CGPointMake(C3.x - 3.0*C2.x + 3.0*C1.x - C0.x, C3.y - 3.0*C2.y + 3.0*C1.y - C0.y)
let B = CGPointMake(3.0*C2.x - 6.0*C1.x + 3.0*C0.x, 3.0*C2.y - 6.0*C1.y + 3.0*C0.y)
let C = CGPointMake(3.0*C1.x - 3.0*C0.x, 3.0*C1.y - 3.0*C0.y)
let D = C0
return CGPointMake(((A.x*t+B.x)*t+C.x)*t+D.x, ((A.y*t+B.y)*t+C.y)*t+D.y)
}
// TODO: - future implementation
private func tangentAngleAtT(start: CGPoint, point1: CGPoint, point2: CGPoint, end: CGPoint, t: CGFloat) -> CGFloat {
// bezier control points
let C0 = start
let C1 = point1
let C2 = point2
let C3 = end
// The cubic polynomial coefficients such that Bez(t) = A*t^3 + B*t^2 + C*t + D
let A = CGPointMake(C3.x - 3.0*C2.x + 3.0*C1.x - C0.x, C3.y - 3.0*C2.y + 3.0*C1.y - C0.y)
let B = CGPointMake(3.0*C2.x - 6.0*C1.x + 3.0*C0.x, 3.0*C2.y - 6.0*C1.y + 3.0*C0.y)
let C = CGPointMake(3.0*C1.x - 3.0*C0.x, 3.0*C1.y - 3.0*C0.y)
return atan2(3.0*A.y*t*t + 2.0*B.y*t + C.y, 3.0*A.x*t*t + 2.0*B.x*t + C.x)
}
}
来源：https://stackoverflow.com/questions/10477/equidistant-points-across-bezier-curves