两条线之间的内角 [英] Inner angle between two lines

问题描述

我有两行：Line1和Line2。每一行由两个点（P1L1（x1，y1），P2L1（x2，y2）和 P1L1（x1，y1），P2L3 x2，y3））。我想知道这两条线定义的内角。

I have two lines: Line1 and Line2. Each line is defined by two points (P1L1(x1, y1), P2L1(x2, y2) and P1L1(x1, y1), P2L3(x2, y3)). I want to know the inner angle defined by these two lines.

为此，我用横坐标计算每条线的角度：

For do it I calculate the angle of each line with the abscissa:

double theta1 = atan(m1) * (180.0 / PI);
double theta2 = atan(m2) * (180.0 / PI);

了解角度后，我计算如下：

After to know the angle I calculate the following:

double angle = abs(theta2 - theta1);

我有问题或怀疑是：有时候我得到正确的角度，角度（对于我外面）。如何知道何时减去180º知道内角？有什么算法更好吗？因为我尝试了一些方法：dot product，
下面的公式：

The problem or doubt that I have is: sometimes I get the correct angle but sometimes I get the complementary angle (for me outer). How can I know when subtract 180º to know the inner angle? There is any algorithm better to do that? Because I tried some methods: dot product, following formula:

result = (m1 - m2) / (1.0 + (m1 * m2));

但是我总是有同样的问题;我从来不知道当我有外角或内角！

But always I have the same problem; I never known when I have the outer angle or the inner angle!

推荐答案

我认为你要找的是< a href =http://en.wikipedia.org/wiki/Inner_product>内部产品（您可能还需要查看点积条目）。在你的情况下，它由：

I think what you're looking for is the inner product (you may also want to look over the dot product entry) of the two angles. In your case, that's given by:

float dx21 = x2-x1;
float dx31 = x3-x1;
float dy21 = y2-y1;
float dy31 = y3-y1;
float m12 = sqrt( dx21*dx21 + dy21*dy21 );
float m13 = sqrt( dx31*dx31 + dy31*dy31 );
float theta = acos( (dx21*dx31 + dy21*dy31) / (m12 * m13) );

回答以弧度表示。

编辑：这是一个完整的实现。替换p1，p2和p3中的有问题的值，让我知道你得到什么。点p1是两条线相交的顶点，根据您对两条线的定义。

Here's a complete implementation. Substitute the problematic values in p1, p2, and p3 and let me know what you get. The point p1 is the vertex where the two lines intersect, in accordance with your definition of the two lines.

#include <math.h>
#include <iostream>

template <typename T> class Vector2D
{
private:
    T x;
    T y;

public:
    explicit Vector2D(const T& x=0, const T& y=0) : x(x), y(y) {}
    Vector2D(const Vector2D<T>& src) : x(src.x), y(src.y) {}
    virtual ~Vector2D() {}

    // Accessors
    inline T X() const { return x; }
    inline T Y() const { return y; }
    inline T X(const T& x) { this->x = x; }
    inline T Y(const T& y) { this->y = y; }

    // Vector arithmetic
    inline Vector2D<T> operator-() const
        { return Vector2D<T>(-x, -y); }

    inline Vector2D<T> operator+() const
        { return Vector2D<T>(+x, +y); }

    inline Vector2D<T> operator+(const Vector2D<T>& v) const
        { return Vector2D<T>(x+v.x, y+v.y); }

    inline Vector2D<T> operator-(const Vector2D<T>& v) const
        { return Vector2D<T>(x-v.x, y-v.y); }

    inline Vector2D<T> operator*(const T& s) const
        { return Vector2D<T>(x*s, y*s); }

    // Dot product
    inline T operator*(const Vector2D<T>& v) const
        { return x*v.x + y*v.y; }

    // l-2 norm
    inline T norm() const { return sqrt(x*x + y*y); }

    // inner angle (radians)
    static T angle(const Vector2D<T>& v1, const Vector2D<T>& v2)
    {
        return acos( (v1 * v2) / (v1.norm() * v2.norm()) );
    }
};

int main()
{
    Vector2D<double> p1(215, 294);
    Vector2D<double> p2(174, 228);
    Vector2D<double> p3(303, 294);

    double rad = Vector2D<double>::angle(p2-p1, p3-p1);
    double deg = rad * 180.0 / M_PI;

    std::cout << "rad = " << rad << "\tdeg = " << deg << std::endl;

    p1 = Vector2D<double>(153, 457);
    p2 = Vector2D<double>(19, 457);
    p3 = Vector2D<double>(15, 470);

    rad = Vector2D<double>::angle(p2-p1, p3-p1);
    deg = rad * 180.0 / M_PI;

    std::cout << "rad = " << rad << "\tdeg = " << deg << std::endl;

    return 0;
}

上述代码产生：

rad = 2.12667   deg = 121.849
rad = 0.0939257 deg = 5.38155

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