数学计算两点之间的夹角找回？ [英] Math Calculation to retrieve angle between two points?

问题描述

可能重复：结果
的如何计算相对于水平轴的两个点之间的角度θ

我一直在寻找这样的年龄，它只是真的很烦我，所以我决定只问...

I've been looking for this for ages and it's just really annoying me so I've decided to just ask...

所提供的，我有两个点（即X1，Y1，和X2，Y2），我想以计算这两个点之间的角度，假设，当Y1 == y 2和X1> X2角度为180度...

Provided I have two points (namely x1, y1, and x2, y2), I would like to calculate the angle between these two points, presuming that when y1 == y2 and x1 > x2 the angle is 180 degrees...

我有下面的代码，我一直在与（使用从高中的知识），我只是似乎无法。产生期望的结果。

I have the below code that I have been working with (using knowledge from high school) and I just can't seem to produce the desired result.

float xDiff = x1 - x2;
float yDiff = y1 - y2;
return (float)Math.Atan2(yDiff, xDiff) * (float)(180 / Math.PI);

谢谢，我得到如此沮丧...

Thanks in advance, I'm getting so frustrated...

推荐答案

从我收集了，你想以下举行：

From what I've gathered, you want the following to hold:

• 水平分割线： P1 P2 -------- => 0°


• 水平分割线： P2 -------- P1 => 180°

• Horizontal line: P1 -------- P2 => 0°
• Horizontal line: P2 -------- P1 => 180°

您说，您想要的角度，​​顺时针方向增加。

You said, you want the angle to increase in clockwise direction.

旋转此行 P1 P2 -------- ，使得 P1 超过 P2 中，角度必须从而在90°

Rotating this line P1 -------- P2 such that P1 is above P2, the angle must thus be 90°.

如果，然而，我们在相反的方向转动时， P1 将低于 P2 和角度是90°或270°。

If, however, we rotated in the opposite direction, P1 would be below P2 and the angle is -90° or 270°.

依据：考虑到 P1 为原点和测量 P2 相对的角度原点，那么 P1 P2 -------- 将正确产生 0

Basis: Considering P1 to be the origin and measuring the angle of P2 relative to the origin, then P1 -------- P2 will correctly yield 0.

float xDiff = x2 - x1;
float yDiff = y2 - y1;
return Math.Atan2(yDiff, xDiff) * 180.0 / Math.PI;

然而， ATAN2 让我们的角度增加CCW方向。
逆时针方向绕原点旋转，是经过以下值：

However, atan2 let's the angle increase in CCW direction. Rotating in CCW direction around the origin, y goes through the following values:

• Y = 0


• Y> 0


• Y = 0


• Y'LT; 0


• Y = 0

• y = 0
• y > 0
• y = 0
• y < 0
• y = 0

这意味着，我们可以简单地反转的迹象是翻转方向。但由于C＃的坐标从从上到下依次增加，符号已经计算时， yDiff 逆转。

This means, that we can simply invert the sign of y to flip the direction. But because C#'s coordinates increase from top to bottom, the sign is already reversed when computing yDiff.

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