javascript - 在多边形中找到凸 Angular 和凹 Angular

Question

我正在尝试检测任意多边形中的 Angular 是凹的还是凸的。我创建了下面的函数来计算所有边对之间的 Angular 。然而，人们永远不知道它返回的是内 Angular 还是外 Angular 。我不知道该怎么做。任何帮助表示赞赏!!!!

function findConvexCorner (pt){
var isCornerConvex = [];
for (var i =0; i < pt.length ;i++)
{
    var lastPt = pt.length -1;
    if (i==0){
        var vec1 = vec3.createXYZ( pt[lastPt].x - pt[i].x , pt[lastPt].y - pt[i].y ,0.0);
        var vec2 = vec3.createXYZ( pt[i].x - pt[i+1].x , pt[i].y - pt[i+1].y ,0.0);
        vec3.normalize(vec1);vec3.normalize(vec2);
            isCornerConvex.push(Math.acos(vec3.dot(vec1,vec2))*180/Math.PI);}
    else if(i == lastPt){
        var vec2 = vec3.createXYZ( pt[i-1].x - pt[i].x , pt[i-1].y - pt[i].y ,0.0);
        var vec1 = vec3.createXYZ( pt[0].x - pt[i].x , pt[0].y - pt[i].y ,0.0);
        vec3.normalize(vec1);vec3.normalize(vec2);
            isCornerConvex.push(Math.acos(vec3.dot(vec1,vec2))*180/Math.PI);}
    else{
        var vec1 = vec3.createXYZ( pt[i-1].x - pt[i].x , pt[i-1].y - pt[i].y ,0.0);
        var vec2 = vec3.createXYZ( pt[i+1].x - pt[i].x , pt[i+1].y - pt[i].y ,0.0);
        vec3.normalize(vec1);vec3.normalize(vec2);
            isCornerConvex.push(Math.acos(vec3.dot(vec1,vec2))*180/Math.PI);}
}
console.log("Angle: "+ isCornerConvex);
}

Answer 1

下面是一些计算凹 Angular 和凸 Angular 的代码:

// this assumes nextEdge and previousEdge are vectors pointing out of a vertex and to the next one
var angle = ((Math.atan2(nextEdge.x, nextEdge.y) - Math.atan2(previousEdge.x, previousEdge.y) + Math.PI * 2) % (Math.PI * 2)) - Math.PI;

if (angle > 0) {
   corner.type = 'convex';
} else if (angle < 0) {
    corner.type = 'concave';
} else {
    corner.type = 'straight';
}