正如这篇维基百科文章中所解释的曲线方向，给定平面上的3个点p, q和r(即x和y坐标)，您可以计算以下行列式的符号

如果行列式为负(即定向(p, q, r) < 0)，则多边形是顺时针方向(CW)。如果行列式为正(即定向(p, q, r) > 0)，则多边形是逆时针方向(CCW)。如果点p, q和r共线，行列式为零(即定向(p, q, r) == 0)。

在上面的公式中，由于我们使用的是齐次坐标，我们将1放在p, q和r的坐标前面。

一个计算上更简单的方法，如果你已经知道多边形内的一个点:

从原始多边形中选择任意线段，按此顺序选择点及其坐标。 加上一个已知的“内部”点，形成一个三角形。 根据以上三点计算CW或CCW。

我的c# / LINQ解决方案是基于下面@charlesbretana的交叉积建议的。你可以为线圈指定一个参考法线。只要曲线大部分在向上向量所定义的平面内，它就可以工作。

using System.Collections.Generic;
using System.Linq;
using System.Numerics;

namespace SolidworksAddinFramework.Geometry
{
    public static class PlanePolygon
    {
        /// <summary>
        /// Assumes that polygon is closed, ie first and last points are the same
        /// </summary>
       public static bool Orientation
           (this IEnumerable<Vector3> polygon, Vector3 up)
        {
            var sum = polygon
                .Buffer(2, 1) // from Interactive Extensions Nuget Pkg
                .Where(b => b.Count == 2)
                .Aggregate
                  ( Vector3.Zero
                  , (p, b) => p + Vector3.Cross(b[0], b[1])
                                  /b[0].Length()/b[1].Length());

            return Vector3.Dot(up, sum) > 0;

        } 

    }
}

使用单元测试

namespace SolidworksAddinFramework.Spec.Geometry
{
    public class PlanePolygonSpec
    {
        [Fact]
        public void OrientationShouldWork()
        {

            var points = Sequences.LinSpace(0, Math.PI*2, 100)
                .Select(t => new Vector3((float) Math.Cos(t), (float) Math.Sin(t), 0))
                .ToList();

            points.Orientation(Vector3.UnitZ).Should().BeTrue();
            points.Reverse();
            points.Orientation(Vector3.UnitZ).Should().BeFalse();



        } 
    }
}

Javascript实现的lhf的答案 (再次强调，这只适用于简单的多边形，即不适用于图8)

let polygon = [ {x:5,y:0}, {x:6,y:4}, {x:4,y:5}, {x:1,y:5}, {x:1,y:0} ] document.body.innerHTML += `Polygon ${polygon.map(p=>`(${p.x}, ${p.y})`).join(", ")} is clockwise? ${isPolygonClockwise(polygon)}` let reversePolygon = [] polygon.forEach(point=>reversePolygon.unshift(point)) document.body.innerHTML += `<br/>Polygon ${reversePolygon.map(p=>`(${p.x}, ${p.y})`).join(", ")} is clockwise? ${isPolygonClockwise(reversePolygon)}` function isPolygonClockwise (polygon) { // From http://www.faqs.org/faqs/graphics/algorithms-faq/ "How do I find the orientation of a simple polygon?" // THIS SOMETIMES FAILS if the polygon is a figure 8, or similar shape where it crosses over itself // Take the lowest point (break ties with the right-most). if (polygon.length < 3) { return true // A single point or two points can't be clockwise/counterclockwise } let previousPoint = polygon[0] let lowestPoint = polygon[1] let nextPoint = polygon[2] polygon.forEach((point, index)=>{ if (point.y > lowestPoint.y || (point.y === lowestPoint.y && point.x > lowestPoint.x)) { // larger y values are lower, in svgs // Break ties with furthest right previousPoint = polygon[(index-1) >= (0) ? (index-1) : (polygon.length-1)] lowestPoint = polygon[index] nextPoint = polygon[(index+1) <= (polygon.length-1) ? (index+1) : (0)] } }) // Check the angle between the previous point, that point, and the next point. // If the angle is less than PI radians, the polygon is clockwise let angle = findAngle(previousPoint, lowestPoint, nextPoint) return angle < Math.PI } function findAngle(A,B,C) { var AB = Math.atan2(B.y-A.y, B.x-A.x); var BC = Math.atan2(C.y-B.y, C.x-B.x); if (AB < 0) AB += Math.PI*2 if (BC < 0) BC += Math.PI*2 return BC-AB; }

求出这些点的质心。

假设有直线从这个点到你们的点。

求line0 line1的两条直线夹角

而不是直线1和直线2

...

...

如果这个角是单调递增的，而不是逆时针递增的，

如果是单调递减，则是顺时针递减

Else(它不是单调的)

你不能决定，所以这是不明智的

虽然这些答案是正确的，但它们在数学上的强度比必要的要大。假设地图坐标，其中最北的点是地图上的最高点。找到最北的点，如果两个点相等，它是最北的，然后是最东的(这是lhf在他的答案中使用的点)。在你的观点中，

点[0]= (5,0)

点[1]= (6,4)

点[2]= (4,5)

点[3]= (1,5)

点[4]= (1,0)

If we assume that P2 is the most north then east point either the previous or next point determine clockwise, CW, or CCW. Since the most north point is on the north face, if the P1 (previous) to P2 moves east the direction is CW. In this case, it moves west, so the direction is CCW as the accepted answer says. If the previous point has no horizontal movement, then the same system applies to the next point, P3. If P3 is west of P2, it is, then the movement is CCW. If the P2 to P3 movement is east, it's west in this case, the movement is CW. Assume that nte, P2 in your data, is the most north then east point and the prv is the previous point, P1 in your data, and nxt is the next point, P3 in your data, and [0] is horizontal or east/west where west is less than east, and [1] is vertical.

if (nte[0] >= prv[0] && nxt[0] >= nte[0]) return(CW);
if (nte[0] <= prv[0] && nxt[0] <= nte[0]) return(CCW);
// Okay, it's not easy-peasy, so now, do the math
if (nte[0] * nxt[1] - nte[1] * nxt[0] - prv[0] * (nxt[1] - crt[1]) + prv[1] * (nxt[0] - nte[0]) >= 0) return(CCW); // For quadrant 3 return(CW)
return(CW) // For quadrant 3 return (CCW)