以给定角度在矩形上查找点

Question

我正在尝试在矩形对象中绘制渐变,具有给定角度(Theta),其中渐变的末端接触矩形的周长.

我认为使用切线会起作用,但是我很难解决问题.有一个简单的算法,我只是缺少？

最终结果

因此,这将是(angle,RectX1,RectX2,RectY1,RectY2)的函数.我希望它以[x1,x2,y1,y2]的形式返回,以便渐变将在正方形上绘制.在我的问题中,如果原点是0,那么x2 = -x1和y2 = -y1.但它并不总是在起源上.

Answer 1

让我们调用a和b你的矩形边,和(x0,y0)你的矩形中心的坐标.

您需要考虑四个区域:

    Region    from               to                 Where
    ====================================================================
       1      -arctan(b/a)       +arctan(b/a)       Right green triangle
       2      +arctan(b/a)        π-arctan(b/a)     Upper yellow triangle
       3       π-arctan(b/a)      π+arctan(b/a)     Left green triangle
       4       π+arctan(b/a)     -arctan(b/a)       Lower yellow triangle

通过一点trigonometry-fu,我们可以得到每个区域中所需交点的坐标.

因此 Z0是区域1和3 的交叉点的表达式
而Z1是区域2和4的交叉点的表达式.

期望的线根据区域从(X0,Y0)传递到Z0或Z1.所以记住Tan(φ)= Sin(φ)/ Cos(φ)

    Lines in regions      Start                   End
    ======================================================================
       1 and 3           (X0,Y0)      (X0 + a/2 , (a/2 * Tan(φ))+ Y0
       2 and 4           (X0,Y0)      (X0 + b/(2* Tan(φ)) , b/2 + Y0)

只需注意每个象限中Tan(φ)的符号,并始终从THE POSITIVE x轴ANTICLOCKWISE测量角度.

HTH!

Answer 2

好的,哇!,我终于得到了这个.

注意:我的基础是belisarius的真棒答案.如果你喜欢这个,请也喜欢他.我所做的就是将他所说的内容转化为代码.

这是Objective-C中的样子.它应该很简单,可以转换为您喜欢的语言.

+ (CGPoint) edgeOfView: (UIView*) view atAngle: (float) theta
{
    // Move theta to range -M_PI .. M_PI
    const double twoPI = M_PI * 2.;
    while (theta < -M_PI)
    {
        theta += twoPI;
    }

    while (theta > M_PI)
    {
        theta -= twoPI;
    }

    // find edge ofview
    // Ref: http://stackoverflow.com/questions/4061576/finding-points-on-a-rectangle-at-a-given-angle
    float aa = view.bounds.size.width;                                          // "a" in the diagram
    float bb = view.bounds.size.height;                                         // "b"

    // Find our region (diagram)
    float rectAtan = atan2f(bb, aa);
    float tanTheta = tan(theta);

    int region;
    if ((theta > -rectAtan)
    &&  (theta <= rectAtan) )
    {
        region = 1;
    }
    else if ((theta >  rectAtan)
    &&       (theta <= (M_PI - rectAtan)) )
    {
        region = 2;
    }
    else if ((theta >   (M_PI - rectAtan))
    ||       (theta <= -(M_PI - rectAtan)) )
    {
        region = 3;
    }
    else
    {
        region = 4;
    }

    CGPoint edgePoint = view.center;
    float xFactor = 1;
    float yFactor = 1;

    switch (region)
    {
        case 1: yFactor = -1;       break;
        case 2: yFactor = -1;       break;
        case 3: xFactor = -1;       break;
        case 4: xFactor = -1;       break;
    }

    if ((region == 1)
    ||  (region == 3) )
    {
        edgePoint.x += xFactor * (aa / 2.);                                     // "Z0"
        edgePoint.y += yFactor * (aa / 2.) * tanTheta;
    }
    else                                                                        // region 2 or 4
    {
        edgePoint.x += xFactor * (bb / (2. * tanTheta));                        // "Z1"
        edgePoint.y += yFactor * (bb /  2.);
    }

    return edgePoint;
}

另外,这是我创建的一个小测试视图,用于验证它是否有效.创建此视图并将其放置在某个位置,它将在边缘周围进行另一个小视图.

@interface DebugEdgeView()
{
    int degrees;
    UIView *dotView;
    NSTimer *timer;
}

@end

@implementation DebugEdgeView

- (void) dealloc
{
    [timer invalidate];
}


- (id) initWithFrame: (CGRect) frame
{
    self = [super initWithFrame: frame];
    if (self)
    {
        self.backgroundColor = [[UIColor magentaColor] colorWithAlphaComponent: 0.25];
        degrees = 0;
        self.clipsToBounds = NO;

        // create subview dot
        CGRect dotRect = CGRectMake(frame.size.width / 2., frame.size.height / 2., 20, 20);
        dotView = [[DotView alloc] initWithFrame: dotRect];
        dotView.backgroundColor = [UIColor magentaColor];
        [self addSubview: dotView];

        // move it around our edges
        timer = [NSTimer scheduledTimerWithTimeInterval: (5. / 360.)
                                                 target: self
                                               selector: @selector(timerFired:)
                                               userInfo: nil
                                                repeats: YES];
    }

    return self;
}


- (void) timerFired: (NSTimer*) timer
{
    float radians = ++degrees * M_PI / 180.;
    if (degrees > 360)
    {
        degrees -= 360;
    }

    dispatch_async(dispatch_get_main_queue(), ^{
        CGPoint edgePoint = [MFUtils edgeOfView: self atAngle: radians];
        edgePoint.x += (self.bounds.size.width  / 2.) - self.center.x;
        edgePoint.y += (self.bounds.size.height / 2.) - self.center.y;
        dotView.center = edgePoint;
    });
}

@end

Answer 3

Javascript版本:

function edgeOfView(rect, deg) {
  var twoPI = Math.PI*2;
  var theta = deg * Math.PI / 180;
  
  while (theta < -Math.PI) {
    theta += twoPI;
  }
  
  while (theta > Math.PI) {
    theta -= twoPI;
  }
  
  var rectAtan = Math.atan2(rect.height, rect.width);
  var tanTheta = Math.tan(theta);
  var region;
  
  if ((theta > -rectAtan) && (theta <= rectAtan)) {
      region = 1;
  } else if ((theta > rectAtan) && (theta <= (Math.PI - rectAtan))) {
      region = 2;
  } else if ((theta > (Math.PI - rectAtan)) || (theta <= -(Math.PI - rectAtan))) {
      region = 3;
  } else {
      region = 4;
  }
  
  var edgePoint = {x: rect.width/2, y: rect.height/2};
  var xFactor = 1;
  var yFactor = 1;
  
  switch (region) {
    case 1: yFactor = -1; break;
    case 2: yFactor = -1; break;
    case 3: xFactor = -1; break;
    case 4: xFactor = -1; break;
  }
  
  if ((region === 1) || (region === 3)) {
    edgePoint.x += xFactor * (rect.width / 2.);                                     // "Z0"
    edgePoint.y += yFactor * (rect.width / 2.) * tanTheta;
  } else {
    edgePoint.x += xFactor * (rect.height / (2. * tanTheta));                        // "Z1"
    edgePoint.y += yFactor * (rect.height /  2.);
  }
  
  return edgePoint;
};