对于一个具有四个常规点a、b、c和d的三次贝塞尔曲线,
给定一个值t,
如何最优雅地找到该点的切线?
曲线的切线就是它的导数。Michal使用的参数方程式:
P(t) = (1 - t)^3 * P0 + 3t(1-t)^2 * P1 + 3t^2 (1-t) * P2 + t^3 * P3
dP(t) / dt = -3(1-t)^2 * P0 + 3(1-t)^2 * P1 - 6t(1-t) * P1 - 3t^2 * P2 + 6t(1-t) * P2 + 3t^2 * P3
顺便提一下,在您之前的问题中,似乎是错误的。我认为您在那里使用的是二次贝塞尔曲线的斜率,而不是三次的。
从那里开始,实现一个执行此计算的 C 函数应该是微不足道的,就像 Michal 已经为曲线本身提供的那样。
以下是经过全面测试的可以复制和粘贴的代码:
它沿着曲线绘制近似等距离的点,并且绘制切线。
bezierInterpolation
找到这些点。
bezierTangent
找到切线。
提供了两个版本的bezierInterpolation
:
bezierInterpolation
完美地工作。
altBezierInterpolation
与bezierInterpolation
完全相同,但是它以扩展、清晰、说明性的方式编写。这使得算术更加易于理解。
使用这两个程序中的任何一个:结果都是相同的。
在两种情况下,使用bezierTangent
来查找切线。(注:Michal的精彩代码库在此处。)
还包括如何在drawRect:
中使用的完整示例。
// MBBezierView.m original BY MICHAL stackoverflow #4058979
#import "MBBezierView.h"
CGFloat bezierInterpolation(
CGFloat t, CGFloat a, CGFloat b, CGFloat c, CGFloat d) {
// see also below for another way to do this, that follows the 'coefficients'
// idea, and is a little clearer
CGFloat t2 = t * t;
CGFloat t3 = t2 * t;
return a + (-a * 3 + t * (3 * a - a * t)) * t
+ (3 * b + t * (-6 * b + b * 3 * t)) * t
+ (c * 3 - c * 3 * t) * t2
+ d * t3;
}
CGFloat altBezierInterpolation(
CGFloat t, CGFloat a, CGFloat b, CGFloat c, CGFloat d)
{
// here's an alternative to Michal's bezierInterpolation above.
// the result is absolutely identical.
// of course, you could calculate the four 'coefficients' only once for
// both this and the slope calculation, if desired.
CGFloat C1 = ( d - (3.0 * c) + (3.0 * b) - a );
CGFloat C2 = ( (3.0 * c) - (6.0 * b) + (3.0 * a) );
CGFloat C3 = ( (3.0 * b) - (3.0 * a) );
CGFloat C4 = ( a );
// it's now easy to calculate the point, using those coefficients:
return ( C1*t*t*t + C2*t*t + C3*t + C4 );
}
CGFloat bezierTangent(CGFloat t, CGFloat a, CGFloat b, CGFloat c, CGFloat d)
{
// note that abcd are aka x0 x1 x2 x3
/* the four coefficients ..
A = x3 - 3 * x2 + 3 * x1 - x0
B = 3 * x2 - 6 * x1 + 3 * x0
C = 3 * x1 - 3 * x0
D = x0
and then...
Vx = 3At2 + 2Bt + C */
// first calcuate what are usually know as the coeffients,
// they are trivial based on the four control points:
CGFloat C1 = ( d - (3.0 * c) + (3.0 * b) - a );
CGFloat C2 = ( (3.0 * c) - (6.0 * b) + (3.0 * a) );
CGFloat C3 = ( (3.0 * b) - (3.0 * a) );
CGFloat C4 = ( a ); // (not needed for this calculation)
// finally it is easy to calculate the slope element,
// using those coefficients:
return ( ( 3.0 * C1 * t* t ) + ( 2.0 * C2 * t ) + C3 );
// note that this routine works for both the x and y side;
// simply run this routine twice, once for x once for y
// note that there are sometimes said to be 8 (not 4) coefficients,
// these are simply the four for x and four for y,
// calculated as above in each case.
}
@implementation MBBezierView
- (void)drawRect:(CGRect)rect {
CGPoint p1, p2, p3, p4;
p1 = CGPointMake(30, rect.size.height * 0.33);
p2 = CGPointMake(CGRectGetMidX(rect), CGRectGetMinY(rect));
p3 = CGPointMake(CGRectGetMidX(rect), CGRectGetMaxY(rect));
p4 = CGPointMake(-30 + CGRectGetMaxX(rect), rect.size.height * 0.66);
[[UIColor blackColor] set];
[[UIBezierPath bezierPathWithRect:rect] fill];
[[UIColor redColor] setStroke];
UIBezierPath *bezierPath = [[[UIBezierPath alloc] init] autorelease];
[bezierPath moveToPoint:p1];
[bezierPath addCurveToPoint:p4 controlPoint1:p2 controlPoint2:p3];
[bezierPath stroke];
[[UIColor brownColor] setStroke];
// now mark in points along the bezier!
for (CGFloat t = 0.0; t <= 1.00001; t += 0.05) {
[[UIColor brownColor] setStroke];
CGPoint point = CGPointMake(
bezierInterpolation(t, p1.x, p2.x, p3.x, p4.x),
bezierInterpolation(t, p1.y, p2.y, p3.y, p4.y));
// there, use either bezierInterpolation or altBezierInterpolation,
// identical results for the position
// just draw that point to indicate it...
UIBezierPath *pointPath =
[UIBezierPath bezierPathWithArcCenter:point
radius:5 startAngle:0 endAngle:2*M_PI clockwise:YES];
[pointPath stroke];
// now find the tangent if someone on stackoverflow knows how
CGPoint vel = CGPointMake(
bezierTangent(t, p1.x, p2.x, p3.x, p4.x),
bezierTangent(t, p1.y, p2.y, p3.y, p4.y));
// the following code simply draws an indication of the tangent
CGPoint demo = CGPointMake( point.x + (vel.x*0.3),
point.y + (vel.y*0.33) );
// (the only reason for the .3 is to make the pointers shorter)
[[UIColor whiteColor] setStroke];
UIBezierPath *vp = [UIBezierPath bezierPath];
[vp moveToPoint:point];
[vp addLineToPoint:demo];
[vp stroke];
}
}
@end
to draw that class...
MBBezierView *mm = [[MBBezierView alloc]
initWithFrame:CGRectMake(400,20, 600,700)];
[mm setNeedsDisplay];
[self addSubview:mm];
CGFloat bezierPoint(CGFloat t, CGFloat a, CGFloat b, CGFloat c, CGFloat d)
{
CGFloat C1 = ( d - (3.0 * c) + (3.0 * b) - a );
CGFloat C2 = ( (3.0 * c) - (6.0 * b) + (3.0 * a) );
CGFloat C3 = ( (3.0 * b) - (3.0 * a) );
CGFloat C4 = ( a );
return ( C1*t*t*t + C2*t*t + C3*t + C4 );
}
CGFloat bezierTangent(CGFloat t, CGFloat a, CGFloat b, CGFloat c, CGFloat d)
{
CGFloat C1 = ( d - (3.0 * c) + (3.0 * b) - a );
CGFloat C2 = ( (3.0 * c) - (6.0 * b) + (3.0 * a) );
CGFloat C3 = ( (3.0 * b) - (3.0 * a) );
CGFloat C4 = ( a );
return ( ( 3.0 * C1 * t* t ) + ( 2.0 * C2 * t ) + C3 );
}
重要事实:
(1) 绝对事实是:非常遗憾,目前为止,苹果公司没有提供任何方法从UIBezierPath中提取点。截至2019年,这一点是真实的。
(2) 别忘了,沿着UIBezierPath轻松地实现动画效果。Google 许多例子。
(3) 许多人会问:“不能使用CGPathApply从UIBezierPath中提取点吗?” 不行,CGPathApply完全无关:它只是给出您“制作任何路径的指令”的列表(因此,“从这里开始”,“画一条直线到这个点”,等等)。 名称有些令人困惑,但CGPathApply与bezier路径完全无关。
对于游戏程序员而言 - 正如 @Engineer 所指出的那样,您可能需要切线的法线,幸运的是,苹果已经内置了向量数学:
https://developer.apple.com/documentation/accelerate/simd/working_with_vectors
https://developer.apple.com/documentation/simd/2896658-simd_normalize
以下是我精心优化过的Swift实现。
我尽力减少了所有冗余的数学操作,以提高速度。即尽量少调用数学运算,并且使用最少的乘法(比加法更昂贵)。
创建bezier曲线不需要任何乘法。 获取bezier曲线上一点需要3次乘法。 获取bezier曲线上一点的切线需要2次乘法。
struct CubicBezier {
private typealias Me = CubicBezier
typealias Vector = CGVector
typealias Point = CGPoint
typealias Num = CGFloat
typealias Coeficients = (C: Num, S: Num, M: Num, L: Num)
let xCoeficients: Coeficients
let yCoeficients: Coeficients
static func coeficientsOfCurve(from c0: Num, through c1: Num, andThrough c2: Num, to c3: Num) -> Coeficients
{
let _3c0 = c0 + c0 + c0
let _3c1 = c1 + c1 + c1
let _3c2 = c2 + c2 + c2
let _6c1 = _3c1 + _3c1
let C = c3 - _3c2 + _3c1 - c0
let S = _3c2 - _6c1 + _3c0
let M = _3c1 - _3c0
let L = c0
return (C, S, M, L)
}
static func xOrYofCurveWith(coeficients coefs: Coeficients, at t: Num) -> Num
{
let (C, S, M, L) = coefs
return ((C * t + S) * t + M) * t + L
}
static func xOrYofTangentToCurveWith(coeficients coefs: Coeficients, at t: Num) -> Num
{
let (C, S, M, _) = coefs
return ((C + C + C) * t + S + S) * t + M
}
init(from start: Point, through c1: Point, andThrough c2: Point, to end: Point)
{
xCoeficients = Me.coeficientsOfCurve(from: start.x, through: c1.x, andThrough: c2.x, to: end.x)
yCoeficients = Me.coeficientsOfCurve(from: start.y, through: c1.y, andThrough: c2.y, to: end.y)
}
func x(at t: Num) -> Num {
return Me.xOrYofCurveWith(coeficients: xCoeficients, at: t)
}
func y(at t: Num) -> Num {
return Me.xOrYofCurveWith(coeficients: yCoeficients, at: t)
}
func dx(at t: Num) -> Num {
return Me.xOrYofTangentToCurveWith(coeficients: xCoeficients, at: t)
}
func dy(at t: Num) -> Num {
return Me.xOrYofTangentToCurveWith(coeficients: yCoeficients, at: t)
}
func point(at t: Num) -> Point {
return .init(x: x(at: t), y: y(at: t))
}
func tangent(at t: Num) -> Vector {
return .init(dx: dx(at: t), dy: dy(at: t))
}
}
使用方法:
let bezier = CubicBezier.init(from: .zero, through: .zero, andThrough: .zero, to: .zero)
let point02 = bezier.point(at: 0.2)
let point07 = bezier.point(at: 0.7)
let tangent01 = bezier.tangent(at: 0.1)
let tangent05 = bezier.tangent(at: 0.5)
在我意识到参数方程中,(dy/dt)/(dx/dt) = dy/dx之前,我无法使任何一个例子起作用。
t
的增长而增长。这个可以帮助其他人做到这一点。 - Engineer