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)

一些建议的方法在非凸多边形(如新月形)的情况下会失败。这里有一个简单的方法，它可以用于非凸多边形(它甚至可以用于自相交的多边形，如数字8，告诉你它是否主要是顺时针)。

对边求和，(x2−x1)(y2 + y1)如果结果是正的，曲线是顺时针的，如果结果是负的，曲线是逆时针的。(结果是封闭面积的两倍，采用+/-惯例。)

point[0] = (5,0)   edge[0]: (6-5)(4+0) =   4
point[1] = (6,4)   edge[1]: (4-6)(5+4) = -18
point[2] = (4,5)   edge[2]: (1-4)(5+5) = -30
point[3] = (1,5)   edge[3]: (1-1)(0+5) =   0
point[4] = (1,0)   edge[4]: (5-1)(0+0) =   0
                                         ---
                                         -44  counter-clockwise

c#代码实现lhf的答案:

// https://en.wikipedia.org/wiki/Curve_orientation#Orientation_of_a_simple_polygon
public static WindingOrder DetermineWindingOrder(IList<Vector2> vertices)
{
    int nVerts = vertices.Count;
    // If vertices duplicates first as last to represent closed polygon,
    // skip last.
    Vector2 lastV = vertices[nVerts - 1];
    if (lastV.Equals(vertices[0]))
        nVerts -= 1;
    int iMinVertex = FindCornerVertex(vertices);
    // Orientation matrix:
    //     [ 1  xa  ya ]
    // O = | 1  xb  yb |
    //     [ 1  xc  yc ]
    Vector2 a = vertices[WrapAt(iMinVertex - 1, nVerts)];
    Vector2 b = vertices[iMinVertex];
    Vector2 c = vertices[WrapAt(iMinVertex + 1, nVerts)];
    // determinant(O) = (xb*yc + xa*yb + ya*xc) - (ya*xb + yb*xc + xa*yc)
    double detOrient = (b.X * c.Y + a.X * b.Y + a.Y * c.X) - (a.Y * b.X + b.Y * c.X + a.X * c.Y);

    // TBD: check for "==0", in which case is not defined?
    // Can that happen?  Do we need to check other vertices / eliminate duplicate vertices?
    WindingOrder result = detOrient > 0
            ? WindingOrder.Clockwise
            : WindingOrder.CounterClockwise;
    return result;
}

public enum WindingOrder
{
    Clockwise,
    CounterClockwise
}

// Find vertex along one edge of bounding box.
// In this case, we find smallest y; in case of tie also smallest x.
private static int FindCornerVertex(IList<Vector2> vertices)
{
    int iMinVertex = -1;
    float minY = float.MaxValue;
    float minXAtMinY = float.MaxValue;
    for (int i = 0; i < vertices.Count; i++)
    {
        Vector2 vert = vertices[i];
        float y = vert.Y;
        if (y > minY)
            continue;
        if (y == minY)
            if (vert.X >= minXAtMinY)
                continue;

        // Minimum so far.
        iMinVertex = i;
        minY = y;
        minXAtMinY = vert.X;
    }

    return iMinVertex;
}

// Return value in (0..n-1).
// Works for i in (-n..+infinity).
// If need to allow more negative values, need more complex formula.
private static int WrapAt(int i, int n)
{
    // "+n": Moves (-n..) up to (0..).
    return (i + n) % n;
}

我认为为了使某些点顺时针方向，所有的边都必须是正的而不仅仅是边的和。如果一条边是负的，则逆时针方向给出至少3个点。

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。

解决方案R确定方向和反向如果顺时针(发现这是必要的owin对象):

coords <- cbind(x = c(5,6,4,1,1),y = c(0,4,5,5,0))
a <- numeric()
for (i in 1:dim(coords)[1]){
  #print(i)
  q <- i + 1
  if (i == (dim(coords)[1])) q <- 1
  out <- ((coords[q,1]) - (coords[i,1])) * ((coords[q,2]) + (coords[i,2]))
  a[q] <- out
  rm(q,out)
} #end i loop

rm(i)

a <- sum(a) #-ve is anti-clockwise

b <- cbind(x = rev(coords[,1]), y = rev(coords[,2]))

if (a>0) coords <- b #reverses coords if polygon not traced in anti-clockwise direction