不规则多边形的内角，其角度> 180 [英] interior angles of irregular polygon with angles > 180

问题描述

我正在尝试计算图中红色显示的值，即内角.

I'm trying to calculate the values shown in the picture in red i.e. the interior angles.

我得到了直线相交的点的数组，并尝试使用点积，但是它只返回最小的角度.我需要全范围的内角(0-359)，但似乎找不到满足此条件的东西.

I've got an array of the points where lines intersect and have tried using the dot-product but it only returns the smallest angles. I need the full range of internal angles (0-359) but can't seem to find much that meets this criteria.

推荐答案

假设您的角度为标准的逆时针格式，则应该可以进行以下操作:

Assuming your angles are in standard counterclockwise format, the following should work:

void angles(double points[][2], double angles[], int npoints){
    for(int i = 0; i < npoints; i++){
        int last = (i - 1 + npoints) % npoints;
        int next = (i + 1) % npoints;
        double x1 = points[i][0] - points[last][0];
        double y1 = points[i][1] - points[last][1];
        double x2 = points[next][0] - points[i][0];
        double y2 = points[next][1] - points[i][1];
        double theta1 = atan2(y1, x1)*180/3.1415926358979323;
        double theta2 = atan2(y2, x2)*180/3.1415926358979323;
        angles[i] = (180 + theta1 - theta2 + 360);
        while(angles[i]>360)angles[i]-=360;
    }
}

很明显，如果您要为点使用某种数据结构，则需要用对数据结构的引用替换double points[][2]及其引用.

Obviously, if you are using some sort of data structure for your points, you will want to replace double points[][2] and references to it with references to your data structure.

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