问题
I have an app that draws a bezier curve in a UIView and I need to find the X intersect when I set a value for Y. First, as I understand, there isn’t a way to find a point directly of a UIBezierPath but you can locate a point of a CGPath.
First, if I “stroke” my UIBezierPath (as I have done in my code) is this actually creating a CGPath or do I need to take further steps to actually convert this to a CGPath?
Second, I want to find the curves intersect at X by providing the value for Y.
My intention is to automatically calculate X for the given value of Y as the user moves the slider (which moves the curve left or right respectively).
My starting display.
What happens when I currently adjust slider.
What I want my display too look like.
GraphView.h
#import <UIKit/UIKit.h>
@interface GraphView : UIView
{
float adjust;
int x;
int y;
}
- (IBAction)sliderChanged:(id)sender;
- (IBAction)yChanged:(id)sender;
@property (weak, nonatomic) IBOutlet UISlider *sliderValue;
@property (weak, nonatomic) IBOutlet UITextField *xValue;
@property (weak, nonatomic) IBOutlet UITextField *yValue;
@end
GraphView.m
#import "GraphView.h"
@interface GraphView ()
@end
@implementation GraphView
@synthesize sliderValue, xValue, yValue;
- (id)initWithCoder:(NSCoder *)graphView
{
self = [super initWithCoder:graphView];
if (self) {
adjust = 194;
y = 100;
}
return self;
}
- (IBAction)sliderChanged:(id)sender
{
adjust = sliderValue.value;
// Calcualtion of the X Value and setting of xValue.text textField goes here
[self setNeedsDisplay];
}
- (IBAction)yChanged:(id)sender
{
y = yValue.text.intValue;
[self setNeedsDisplay];
[self resignFirstResponder];
}
- (void)touchesBegan:(NSSet *)touches withEvent:(UIEvent *)event {
UITouch * touch = [touches anyObject];
if(touch.phase == UITouchPhaseBegan) {
y = yValue.text.intValue;
[self setNeedsDisplay];
[yValue resignFirstResponder];
}
}
- (void)drawRect:(CGRect)rect
{
UIBezierPath *lines = [[UIBezierPath alloc] init];
[lines moveToPoint:CGPointMake(0, y)];
[lines addLineToPoint:CGPointMake(200, y)];
[lines addLineToPoint:CGPointMake(200, 280)];
[lines setLineWidth:1];
[[UIColor redColor] setStroke];
float dashPattern[] = {2, 2};
[lines setLineDash:dashPattern count:2 phase:0.0];
[lines stroke];
UIBezierPath *curve = [[UIBezierPath alloc] init];
[curve moveToPoint:CGPointMake(0, 280)];
[curve addCurveToPoint:CGPointMake(280, 0) controlPoint1:CGPointMake(adjust, 280) controlPoint2:CGPointMake(adjust, 0)];
[curve setLineWidth:2];
[[UIColor blueColor] setStroke];
[curve stroke];
}
@end
回答1:
A cubic Bézier curve is defined by 4 points
P0 = (x0, y0) = start point,
P1 = (x1, y1) = first control point,
P2 = (x2, y2) = second control point,
P3 = (x3, y3) = end point,
and consists of all points
x(t) = (1-t)^3 * x0 + 3*t*(1-t)^2 * x1 + 3*t^2*(1-t) * x2 + t^3 * x3
y(t) = (1-t)^3 * y0 + 3*t*(1-t)^2 * y1 + 3*t^2*(1-t) * y2 + t^3 * y3
where t runs from 0 to 1.
Therefore, to calculate X for a given value of Y, you first have to calculate a
parameter value T such that 0 <= T <= 1 and
Y = (1-T)^3 * y0 + 3*T*(1-T)^2 * y1 + 3*T^2*(1-T) * y2 + T^3 * y3 (1)
and then compute the X coordinate with
X = (1-T)^3 * x0 + 3*T*(1-T)^2 * x1 + 3*T^2*(1-T) * x2 + T^3 * x3 (2)
So you have to solve the cubic equation (1) for T and substitute the value into (2).
Cubic equations can be solved explicitly (see e.g. http://en.wikipedia.org/wiki/Cubic_function) or iteratively (for example using the http://en.wikipedia.org/wiki/Bisection_method).
In general, a cubic equation can have up to three different solutions. In your concrete case we have
P0 = (0, 280), P1 = (adjust, 280), P3 = (adjust, 0), P4 = (280, 0)
so that the equation (1) becomes
Y = (1-T)^3 * 280 + 3*T*(1-T)^2 * 280
which simplifies to
Y/280 = 1 - 3*T^2 + 2*T^3 (3)
The right hand side of (3) is a strictly decreasing function of T in the interval [0, 1], so it is not difficult to see that (3) has exactly one solution if 0 <= Y <= 280.
Substituting this solution into (2) gives the desired X value.
回答2:
It took me a while but the code below is how I solved finding a point on a bezier curve. The math only finds one of the potential 3 values so I suspect if there is more than one it will fail, but in my circumstance my bezier should only ever have one solution since my curve should never cross the same X or Y plane more than once. I wanted to share what I have and I welcome any questions, comments, or suggestions.
#import "Calculation.h"
@implementation Calculation
@synthesize a, b, c, d, xy;
- (float) calc
{
float squareRootCalc =
sqrt(
6*pow(xy,2)*b*d
+4*a*pow(c,3)
-3*pow(b,2)*pow(c,2)
+9*pow(xy,2)*pow(c,2)
-6*a*c*b*d
+6*a*xy*c*b
-18*pow(xy,2)*b*c
+6*a*pow(xy,2)*c
-12*a*xy*pow(c,2)
-2*pow(a,2)*xy*d
+pow(a,2)*pow(d,2)
+4*pow(b,3)*d
+pow(xy,2)*pow(d,2)
-4*pow(b,3)*xy
-4*pow(c,3)*xy
+pow(a,2)*pow(xy,2)
+6*c*b*d*xy
+6*a*c*d*xy
+6*a*b*d*xy
-12*pow(b,2)*d*xy
+6*xy*c*pow(b,2)
+6*xy*b*pow(c,2)
-2*a*pow(xy,2)*d
-2*a*xy*pow(d,2)
-6*c*d*pow(xy,2)
+9*pow(xy,2)*pow(b,2)
-6*a*pow(xy,2)*b)
;
float aCalc = 24*c*d*xy + 24*a*pow(c,2) - 36*xy*pow(c,2) + 4 * squareRootCalc * a;
float bCalc = -12 * squareRootCalc * b;
float cCalc = 12 * squareRootCalc * c;
float dCalc = -4 * squareRootCalc * d;
float xyCalc =
24*xy*a*b
-24*xy*b*d
-12*b*a*d
-12*c*a*d
-12*c*b*d
+8*xy*a*d
+8*pow(b,3)
+8*pow(c,3)
+4*pow(a,2)*d
+24*pow(b,2)*d
-4*xy*pow(a,2)
-4*xy*pow(d,2)
+4*a*pow(d,2)
-12*c*pow(b,2)
-12*b*pow(c,2)
-12*a*b*c
-24*xy*a*c
+72*xy*c*b
-36*xy*pow(b,2)
;
float cubeRootCalc = cbrt(aCalc + bCalc + cCalc + dCalc + xyCalc);
float denomCalc = (a-3*b+3*c-d);
float secOneCalc = 0.5 * cubeRootCalc / denomCalc;
float secTwoCalc = -2 * ((a*c - a*d - pow(b,2) + c*b + b*d - pow(c,2)) / (denomCalc * cubeRootCalc));
float secThreeCalc = (a - 2*b + c) / denomCalc;
return secOneCalc + secTwoCalc + secThreeCalc;
}
- (Calculation *) initWithA:(float)p0 andB:(float)p1 andC:(float)p2 andD:(float)p3 andXy:(float)xyValue
{
self = [super init];
if (self) {
[self setA:p0];
[self setB:p1];
[self setC:p2];
[self setD:p3];
[self setXy:xyValue];
}
return self;
}
- (void) setA:(float)p0 andB:(float)p1 andC:(float)p2 andD:(float)p3 andXy:(float)xyValue
{
[self setA:p0];
[self setB:p1];
[self setC:p2];
[self setD:p3];
[self setXy:xyValue];
}
@end
来源：https://stackoverflow.com/questions/16755823/finding-a-point-on-a-path