在三次贝塞尔曲线上找到点的切线

Question

对于三次贝塞尔曲线，通常具有四个点a，b，c和d，

对于给定的值t

如何最优雅地找到 切线 ？

Answer 1

这是经过完全测试的代码，可以复制和粘贴：

它沿着曲线绘制 近似 点， 并 绘制切线。

bezierInterpolation 找到要点

bezierTangent 找到切线

有 两个版本 的bezierInterpolation下面提供：

bezierInterpolation 完美地工作。

altBezierInterpolation完全相同，但以扩展，清晰，解释性的方式编写。它使算法更容易理解。

使用这两个例程之一：结果相同。

在这两种情况下，bezierTangent都可用于查找切线。

drawRect:还包括如何使用with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];

这是两个用于计算 近似等距点以及 沿贝塞尔三次方 的切线的 例程。

为了清楚和可靠，这些例程以最简单，最具解释性的方式编写。

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 );
    }

四个预先计算的值C1 C2 C3 C4有时被称为贝塞尔 系数 。（回想一下abcd通常称为四个 控制点 。）

当然，t从0到1，例如每0.05。

只需 对X一次 调用这些例程 ，然后对Y分别 调用 一次。

希望它能对某人有所帮助！

重要事实：

（1） 绝对的事实 是：不幸的是，Apple确实没有提供任何方法来从UIBezierPath中提取点。截至2019年为准。

（2）不要忘记， 沿着 UIBezierPath
进行动画设置就像馅饼一样容易。Google的例子很多。

（3）许多人问： “不能使用CGPathApply从UIBezierPath中提取点吗？” 不， CGPathApply是完全不相关的
：它只是为您提供“进行任何路径的说明”的列表（因此，“从此处开始”，“到此处画一条直线”等）。名称令人困惑，但CGPathApply与贝塞尔曲线完全无关。

对于游戏程序员-正如@Engineer指出的那样，您可能想要切线的法线，幸运的是Apple内置了矢量数学：

https://developer.apple.com/documentation/accelerate/simd/working_with_vectors


https://developer.apple.com/documentation/simd/2896658-simd_normalize