在三次贝塞尔曲线上找到点的切线

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 ?

回答:

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 点, 绘制切线。

bezierInterpolation 找到要点

bezierTangent 找到切线

有 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 沿贝塞尔三次方 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 。(回想一下abcd通常称为四个 。)

当然,t从0到1,例如每0.05。

只需 调用这些例程 调用

希望它能对某人有所帮助!


重要事实:

(1) 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,

: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

以上是 在三次贝塞尔曲线上找到点的切线 的全部内容, 来源链接: utcz.com/qa/431990.html

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