计算两条线之间的角度而不必计算斜率?(爪哇)

问题描述

我有两条线路:L1 和 L2.我想计算两条线之间的角度.L1 有点:{(x1, y1), (x2, y2)}，L2 有点:{(x3, y3), (x4, y4)}.

I have two Lines: L1 and L2. I want to calculate the angle between the two lines. L1 has points: {(x1, y1), (x2, y2)} and L2 has points: {(x3, y3), (x4, y4)}.

如何计算这两条线之间形成的角度，而不必计算斜率?我目前遇到的问题是有时我有水平线(沿 x 轴的线)并且以下公式失败(除以零异常):

How can I calculate the angle formed between these two lines, without having to calculate the slopes? The problem I am currently having is that sometimes I have horizontal lines (lines along the x-axis) and the following formula fails (divide by zero exception):

arctan((m1 - m2) / (1 - (m1 * m2)))

其中 m1 和 m2 分别是第 1 行和第 2 行的斜率.是否有公式/算法可以计算两条线之间的角度而不会出现除零异常?任何帮助将不胜感激.

where m1 and m2 are the slopes of line 1 and line 2 respectively. Is there a formula/algorithm that can calculate the angles between the two lines without ever getting divide-by-zero exceptions? Any help would be highly appreciated.

这是我的代码片段:

// Calculates the angle formed between two lines
public static double angleBetween2Lines(Line2D line1, Line2D line2)
{
    double slope1 = line1.getY1() - line1.getY2() / line1.getX1() - line1.getX2();
    double slope2 = line2.getY1() - line2.getY2() / line2.getX1() - line2.getX2();
    double angle = Math.atan((slope1 - slope2) / (1 - (slope1 * slope2)));
    return angle;
}

谢谢.

推荐答案

atan2 函数减轻了处理 atan 的痛苦.

The atan2 function eases the pain of dealing with atan.

声明为double atan2(double y, double x) 并将直角坐标(x,y) 转换为角度theta从极坐标 (r,theta)

It is declared as double atan2(double y, double x) and converts rectangular coordinates (x,y) to the angle theta from the polar coordinates (r,theta)

所以我将你的代码重写为

So I'd rewrite your code as

public static double angleBetween2Lines(Line2D line1, Line2D line2)
{
    double angle1 = Math.atan2(line1.getY1() - line1.getY2(),
                               line1.getX1() - line1.getX2());
    double angle2 = Math.atan2(line2.getY1() - line2.getY2(),
                               line2.getX1() - line2.getX2());
    return angle1-angle2;
}

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