在一个给定的角度上寻找矩形上的点

Question

我试图在一个矩形物体上画一个梯度，有一个给定的角度（θ），梯度的末端接触矩形的周长。

我原以为使用tangent是可行的，但我很难把扭结弄出来。有没有一个简单的算法我只是错过了？

最终结果

所以，这是一个函数（角度，矩形1，矩形2，矩形1，矩形2）。我想要它以[x1，x2，y1，y2]的形式返回，这样梯度就可以画出正方形。 在我的问题中，如果原点是0，那么x2=-x1和y2=-y1。但它并不总是在原点上。

Answer 1

好的，呼！我终于找到了这个。

注意：我是根据贝利萨里奥斯令人敬畏的回答得出这个结论的。如果你喜欢这个，也请喜欢他的。我所做的就是把他说的变成代码。

这是Objective-C的外观，它应该足够简单，可以转换成您最喜欢的语言。

+ (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，这里有一个我创建的小测试视图来验证它是否有效。创建此视图并将其放置在某个位置，它将使另一个小视图围绕边缘快速移动。

@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 2

让我们调用矩形边和矩形中心的坐标。

有四个地区需要考虑：

    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

通过一点三角函数，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的交点的表达式
并且Z1是区域2和区域4的交点的表达式

所需的线从（X0，Y0）传递到Z0或Z1，具体取决于区域。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)

只要注意每个象限中Tan（φ）的符号，并且角度总是从正x轴逆时针方向测量。

啊！

Answer 3

Javascript版本：

&13；

&13；

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;
};

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