这是OpenLayers 2的实现函数。有一个顺时针多边形的条件是面积< 0，这是由这个参考确定的。

function IsClockwise(feature)
{
    if(feature.geometry == null)
        return -1;

    var vertices = feature.geometry.getVertices();
    var area = 0;

    for (var i = 0; i < (vertices.length); i++) {
        j = (i + 1) % vertices.length;

        area += vertices[i].x * vertices[j].y;
        area -= vertices[j].x * vertices[i].y;
        // console.log(area);
    }

    return (area < 0);
}

The cross product measures the degree of perpendicular-ness of two vectors. Imagine that each edge of your polygon is a vector in the x-y plane of a three-dimensional (3-D) xyz space. Then the cross product of two successive edges is a vector in the z-direction, (positive z-direction if the second segment is clockwise, minus z-direction if it's counter-clockwise). The magnitude of this vector is proportional to the sine of the angle between the two original edges, so it reaches a maximum when they are perpendicular, and tapers off to disappear when the edges are collinear (parallel).

因此，对于多边形的每个顶点(点)，计算两条相邻边的叉乘大小:

Using your data:
point[0] = (5, 0)
point[1] = (6, 4)
point[2] = (4, 5)
point[3] = (1, 5)
point[4] = (1, 0)

把边连续地标为 edgeA是从point0到point1的段 点1到点2之间的edgeB .．. edgeE在point4和point0之间。

那么顶点A (point0)在两者之间 edgeE[从点4到点0] 从point0到' point1'

这两条边本身就是向量，它们的x坐标和y坐标可以通过减去它们的起点和终点的坐标来确定:

edgeE = point0 - point4 = (1,0) - (5,0) = (- 4,0) and edgeA = point1 - point0 = (6,4) - (1,0) = (5,4) and

这两个相邻边的外积是用下面矩阵的行列式来计算的，这个矩阵是通过将两个向量的坐标放在表示三个坐标轴的符号(i, j， & k)下面来构造的。第三个(零)值坐标在那里，因为外积概念是一个三维结构，所以我们将这些2-D向量扩展到3-D，以便应用外积:

 i    j    k 
-4    0    0
 1    4    0    

假设所有的叉乘都产生一个垂直于两个向量相乘平面的向量，上面矩阵的行列式只有一个k(或z轴)分量。 计算k轴或z轴分量大小的公式为 A1 *b2 - a2*b1 = -4* 4 - 0* 1 = -16

这个值的大小(-16)是两个原始向量夹角的正弦值，乘以两个向量大小的乘积。 实际上，它值的另一个公式是 A X B(叉乘)= |A| * |B| * sin(AB)。

为了得到角度的大小你需要用这个值(-16)除以两个向量大小的乘积。

|A| * |B| = 4 *根号(17)= 16.4924…

所以sin(AB) = -16 / 16.4924 = -.97014…

这是一个度量顶点后的下一段是否向左或向右弯曲，以及弯曲的程度。不需要取arcsin函数。我们只关心它的大小，当然还有它的符号(正的还是负的)!

对闭合路径周围的其他4个点都这样做，并将每个顶点的计算值相加。

如果最终的和是正的，就顺时针，负的，逆时针。

为了它的价值，我使用这个mixin来计算谷歌Maps API v3应用程序的缠绕顺序。

该代码利用了多边形区域的副作用:顺时针旋转顺序的顶点产生一个正的区域，而逆时针旋转顺序的相同顶点产生一个负的区域。该代码还使用了谷歌Maps几何库中的一种私有API。我觉得使用它很舒服——使用风险自负。

示例用法:

var myPolygon = new google.maps.Polygon({/*options*/});
var isCW = myPolygon.isPathClockwise();

完整的单元测试示例@ http://jsfiddle.net/stevejansen/bq2ec/

/** Mixin to extend the behavior of the Google Maps JS API Polygon type
 *  to determine if a polygon path has clockwise of counter-clockwise winding order.
 *  
 *  Tested against v3.14 of the GMaps API.
 *
 *  @author  stevejansen_github@icloud.com
 *
 *  @license http://opensource.org/licenses/MIT
 *
 *  @version 1.0
 *
 *  @mixin
 *  
 *  @param {(number|Array|google.maps.MVCArray)} [path] - an optional polygon path; defaults to the first path of the polygon
 *  @returns {boolean} true if the path is clockwise; false if the path is counter-clockwise
 */
(function() {
  var category = 'google.maps.Polygon.isPathClockwise';
     // check that the GMaps API was already loaded
  if (null == google || null == google.maps || null == google.maps.Polygon) {
    console.error(category, 'Google Maps API not found');
    return;
  }
  if (typeof(google.maps.geometry.spherical.computeArea) !== 'function') {
    console.error(category, 'Google Maps geometry library not found');
    return;
  }

  if (typeof(google.maps.geometry.spherical.computeSignedArea) !== 'function') {
    console.error(category, 'Google Maps geometry library private function computeSignedArea() is missing; this may break this mixin');
  }

  function isPathClockwise(path) {
    var self = this,
        isCounterClockwise;

    if (null === path)
      throw new Error('Path is optional, but cannot be null');

    // default to the first path
    if (arguments.length === 0)
        path = self.getPath();

    // support for passing an index number to a path
    if (typeof(path) === 'number')
        path = self.getPaths().getAt(path);

    if (!path instanceof Array && !path instanceof google.maps.MVCArray)
      throw new Error('Path must be an Array or MVCArray');

    // negative polygon areas have counter-clockwise paths
    isCounterClockwise = (google.maps.geometry.spherical.computeSignedArea(path) < 0);

    return (!isCounterClockwise);
  }

  if (typeof(google.maps.Polygon.prototype.isPathClockwise) !== 'function') {
    google.maps.Polygon.prototype.isPathClockwise = isPathClockwise;
  }
})();

下面是基于这个答案的一个简单的Python 3实现(反过来，它是基于已接受答案中提出的解决方案)

def is_clockwise(points):
    # points is your list (or array) of 2d points.
    assert len(points) > 0
    s = 0.0
    for p1, p2 in zip(points, points[1:] + [points[0]]):
        s += (p2[0] - p1[0]) * (p2[1] + p1[1])
    return s > 0.0

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();



        } 
    }
}

我将提出另一个解决方案，因为它很简单，不需要大量的数学运算，它只是使用了基本的代数。计算多边形的带符号面积。如果是负的，点是顺时针的，如果是正的，点是逆时针的。(这与Beta的解决方案非常相似。)

计算带符号的面积: A = 1/2 * (x1*y2 - x2*y1 + x2*y3 - x3*y2 +…+ xn*y1 - x1*yn)

或者在伪代码中:

signedArea = 0
for each point in points:
    x1 = point[0]
    y1 = point[1]
    if point is last point
        x2 = firstPoint[0]
        y2 = firstPoint[1]
    else
        x2 = nextPoint[0]
        y2 = nextPoint[1]
    end if

    signedArea += (x1 * y2 - x2 * y1)
end for
return signedArea / 2

注意，如果你只是检查顺序，你不需要麻烦除以2。

来源:http://mathworld.wolfram.com/PolygonArea.html