从UIBezierPath获取积分 [英] Get points from a UIBezierPath
问题描述
我通过执行以下操作绘制了上面的BezierPath: //位置是用户触摸屏幕的位置. //位置将是图表的最大值 CGPoint原点= CGPointMake(xStart，620.0); CGPoint endpt = CGPointMake(xEnd，620.0); CGPoint midpt1 = midPointForPoints(来源，位置)； CGPoint midpt2 = midPointForPoints(location，endpt);
I drew the above BezierPath by doing: // location is where the user touches screen. // location will be the maximum of the graph CGPoint origin = CGPointMake(xStart, 620.0); CGPoint endpt = CGPointMake(xEnd, 620.0); CGPoint midpt1 = midPointForPoints(origin, location); CGPoint midpt2 = midPointForPoints(location, endpt);
UIBezierPath *path = [UIBezierPath bezierPath];
[path moveToPoint:origin];
[path addQuadCurveToPoint:location controlPoint:CGPointMake(midpt1.x, midpt1.y+50)];
[path addQuadCurveToPoint:endpt controlPoint:CGPointMake(midpt2.x, midpt2.y+50)];
[shapeLayer setPath:path.CGPath];
现在，我想检索路径上某些x坐标的y坐标.例如，给定x = 0.0，我想得到y = 0.0，或者给定x = 300.0，y = 50.0.
Now, I want to retrieve y-coordinates for certain x-coordinates that lie on the path. For example, given x = 0.0, I want to get y = 0.0, or given x = 300.0, y = 50.0.
我查看了一些类似这样的参考文献问题和
I looked at some references like this question and sample code I am still not sure. Update: basically, I want to do something like this.
更新: 遵循@Fang的建议:
Update: Following @Fang's advice:
给出方程式
X = (1-t)^2*X0 + 2*t*(1-t)*X1 + t^2 *X2
我解决t
t = ((2.0 * x0 - x1) + sqrt(((-2.0 * x0 + x1) ** 2.0)
- ((4 * (x0 - 2.0 * x1 + x2)) * (x0 - x)))) / (2.0 * (x0 - 2.0 * x1 + x2))
或
t = ((2.0 * x0 - x1) - sqrt(((-2.0 * x0 + x1) ** 2.0)
- ((4 * (x0 - 2.0 * x1 + x2)) * (x0 - x)))) / (2.0 * (x0 - 2.0 * x1 + x2))
使用此值，找到与X对应的Y(我们使用X来找到上述t值)
Using this value, find Y that corresponds to X (we used X to find the above t value)
Y = (1-t)^2*Y0 + 2*t*(1-t)*Y1 + t^2 *Y2
按照上面的公式，我应该得到贝塞尔曲线上的点的y值，但我得到的点离正确的点很远.任何进一步的帮助将不胜感激..
Following the above equation, I am supposed to get the y-value of the point that lies on the Bezier curve but I get a point that's far from the right one. Any further help will be very much appreciated..
关注点:我认为一个可能的问题是，我用两个控制点两次调用addQuadCurveToPoint()，而不是用两个控制点一次.这是否意味着我绘制了两条贝塞尔曲线路径并将其合并?我也在查看
Concern: I think one possible problem is that I am calling addQuadCurveToPoint() twice with two control points instead of once with two control points. Does it mean I draw two Bezier paths and combine them? I am also looking at this to see what's wrong with my computation and the only difference seems to be that he uses two control points when calling addQuadCurveToPoint().
经过方的密切咨询之后的更新:
- (float)getYFromBezierPath:(float)x location:(CGPoint)location ctrlpt1:(CGPoint)ctrlpt1 ctrlpt2:(CGPoint)ctrlpt2 endpt:(CGPoint)endpt {
float yVal;
float tVal;
if (x <= location.x) {
tVal = [self getTvalFromBezierPath:x x0Val:0.0 x1Val:ctrlpt1.x x2Val:location.x];
yVal = [self getCoordFromBezierPath:tVal origin:0.0 p1Val:ctrlpt1.y p2Val:location.y];
} else {
// THIS PART IS THE PROBLEM //
tVal = [self getTvalFromBezierPath:x x0Val:location.x x1Val:ctrlpt2.x x2Val:endpt.x];
yVal = [self getCoordFromBezierPath:tVal origin:location.y p1Val:ctrlpt2.y p2Val:endpt.y];
}
return yVal;
}
- (float)getTvalFromBezierPath:(float)x x0Val:(float)x0 x1Val:(float)x1 x2Val:(float)x2 {
float tVal = (x-x0)/(2*(x1-x0));
return tVal;
}
- (float)getCoordFromBezierPath:(float)t origin: (float)origin p1Val: (float)p1 p2Val: (float)p2 {
// tVal = (sqrt((-2.0 * x * x1) + (x * x0) + (x * x2) + pow(x1, 2) - (x0 * x2)) + x0 - x1) / (x0 - (2.0 * x1) + x2);
return (pow((1-t),2) * origin) + (2 * t * (1-t) * p1) + (pow(t,2) * p2);
}
最后一个问题: 对于第二个Bezeir路径，y值应随t值的增加而减小.目前，y值一直在增加.我该如何解决?经过大量调试后，我仍未找到发生这种情况的原因，因为所有内容都符合该文档.
Last question: for the second Bezeir path, y-value should decrease as t-value increases. Right now, y-value keeps increasing. How should I fix this? After intensive debugging I haven't found why this is happening because everything conforms to the document.
推荐答案
应该可以沿着Bezier路径获取点，因为addQuadCurveToPoint会将二次Bezier线段添加到路径中.因此，您的第一个二次贝塞尔曲线的三个控制点是(请参阅原始文章中的代码段)
It should be possible to get points along a Bezier path as addQuadCurveToPoint is to add a quadratic Bezier segment into the path. So, the three control points of your first quadratic Bezier curve are (refer to the code piece in original post)
P(0)=起源
P(1)=(midpt1.x，midpt1.y + 50)
P(2)=位置
P(0) = origin
P(1) = (midpt1.x, midpt1.y+50)
P(2) = location
您可以通过将参数t从0更改为1任意小的增量值，来根据需要在此二次Bezier曲线上计算任意数量的点
You can compute as many points on this quadratic Bezier curve as you want by varying the parameter t from 0 to 1 by any small increment value as
C(t)=(1-t)^ 2 * P(0)+ 2 * t *(1-t)* P(1)+ t ^ 2 * P(2)
C(t) = (1-t)^2*P(0) + 2*t*(1-t)*P(1) + t^2 *P(2)
要从给定的X值中获取Y值，您将必须从t的二次多项式中的给定X值中求解t值:
To get the Y value from a given X value, you will have to solve for the t value from the given X value from this quadratic polynomial of t:
X =(1-t)^ 2 * X0 + 2 * t *(1-t)* X1 + t ^ 2 * X2
X = (1-t)^2*X0 + 2*t*(1-t)*X1 + t^2 *X2
其中X0，X1和X2是P(0)，P(1)和P(2)的X坐标，这意味着X0 = origin.x，X1 = midpt1.x和X2 = location.x.
where X0, X1 and X2 are the X coordinates of P(0), P(1) and P(2), which means X0=origin.x, X1=midpt1.x and X2=location.x.
由此，我们可以得到一个二次方程式
From this, we can obtain a quadratic equation
(X0-2 * X1 + X2) t ^ 2 + 2 (X1-X0)* t +(X0-X)= 0
(X0-2*X1+X2)t^2 + 2(X1-X0)*t + (X0-X) = 0
您可以使用二次公式求解t.如果您的X0，X1和X2值恰好使t ^ 2项的系数为零，则可以直接求解t，因为t =(X-X0)/(2 *(X1-X0)).
You can solve for t using the quadratic formula. If your X0, X1 and X2 values happens to make the coefficient of t^2 term zero, you can solve for t directly as t = (X-X0)/(2*(X1-X0)).
一旦有了t值，就可以轻松评估相应的Y值.
Once you have the t value, you can easily evaluate the corresponding Y value.
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