以给定角度在矩形上查找点 [英] Finding points on a rectangle at a given angle

问题描述

我试图在一个矩形对象中绘制一个具有给定角度（Theta）的渐变，渐变的末端触及矩形的周长。

我认为使用正切会工作，但我无法解决问题。有没有简单的算法，我只是失踪？

最终结果

所以，这将是（角度，RectX1，RectX2，RectY1，RectY2）的函数。我希望它以[x1，x2，y1，y2]的形式返回，以便渐变将在广场上绘制。
在我的问题中，如果原点为0，那么x2 = -x1和y2 = -y1。但它并不总是在原点上。 b 矩形边，（x0，y0）矩形中心的坐标。

您有四个地区需要考虑：

 
 Region from至Where 
 ===== ================================================== ============= 
 1 -arctan（b / a）+ arctan（b / a）右绿色三角形
 2 + arctan（b / a）π - arctan（b / a）上面的黄色三角形
 3-pictan（b / a）π+ arctan（b / a）左边的绿色三角形
 4 pi + arctan（b / a） - arctan（b / a）较低的黄色三角形
 

使用一点三角函数fu，我们可以得到每个所需交点的坐标地区。

所以 Z0 是区域1和3的交点表达式

和 Z1 是区域2和4的交点

所需线根据区域从（X0，Y0）传递到Z0或Z1。所以记住Tan（&）= Sin（φ）/ Cos（φ）

 
 
结束
 ============================================= ========================= 
 1和3（X0，Y0）（X0 + a / 2，（a / 2 * Tan （&））+ Y0 
 2和4（X0，Y0）（X0 + b /（2 * Tan（ph）），b / 2 + Y0）
 
 
 
 
 
只要知道每个象限中的符号（&），并且角度总是从THE POSITIVE x轴ANTICLOCKWISE测量。
 
 
 
 HTH！
 
I'm trying to draw a gradient in a rectangle object, with a given angle (Theta), where the ends of the gradient are touching the perimeter of the rectangle.



I thought that using tangent would work, but I'm having trouble getting the kinks out. Is there an easy algorithm that I am just missing?

End Result

So, this is going to be a function of (angle, RectX1, RectX2, RectY1, RectY2). I want it returned in the form of [x1, x2, y1, y2], so that the gradient will draw across the square. 
In my problem, if the origin is 0, then x2 = -x1 and y2 = -y1. But it's not always going to be on the origin.
解决方案
Let's call a and b your rectangle sides, and (x0,y0) the coordinates of your rectangle center.   

You have four regions to consider:

    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
With a little of trigonometry-fu, we can get the coordinates for your desired intersection in each region.  

  

So  Z0 is the expression for the intersection point for regions 1 and 3

And Z1 is the expression for the intersection point for regions 2 and 4 

The desired lines pass from (X0,Y0) to Z0 or Z1 depending the region. So remembering that 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)

Just be aware of the signs of Tan(φ) in each quadrant, and that the angle is always measured from THE POSITIVE x axis ANTICLOCKWISE.

HTH!

                        
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