Puncte echidistant între curbe Bezier
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?
distanta dintre P_0 si P_3 (in forma cubica), da, dar cred ca stiai asta, e drept.
Distanta pe o curba este doar lungimea arcului:
fig 1 http://www.codecogs.com/eq.latex?%5Cint_%7Bt_0%7D%5E%7Bt_1%7D%20%7B%20|P'(t)|%20dt
Unde:
Probabil, ați avea t_0 = 0, t_1 = 1.0 și dz (t) = 0 (planul 2d).
Aceasta se numește parametrizare "lungimea arcului". Am scris o lucrare despre asta acum câțiva ani:
http://www.saccade.com/writing/graphics/RE-PARAM.PDF
Ideea este de a calcula în prealabil o curbă de "parametrizare" și de a evalua curba prin aceasta.
Știu că aceasta este o întrebare veche, dar recent am intrat în această problemă și am creat o extensie UIBezierPath pentru a rezolva o coordonată X dată coordonatelor Y și viceversa. Scris rapid.
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)
}
}