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;
}

另一个解决方案是;

const isClockwise = (vertices=[]) => {
    const len = vertices.length;
    const sum = vertices.map(({x, y}, index) => {
        let nextIndex = index + 1;
        if (nextIndex === len) nextIndex = 0;

        return {
            x1: x,
            x2: vertices[nextIndex].x,
            y1: x,
            y2: vertices[nextIndex].x
        }
    }).map(({ x1, x2, y1, y2}) => ((x2 - x1) * (y1 + y2))).reduce((a, b) => a + b);

    if (sum > -1) return true;
    if (sum < 0) return false;
}

把所有的顶点作为一个数组;

const vertices = [{x: 5, y: 0}, {x: 6, y: 4}, {x: 4, y: 5}, {x: 1, y: 5}, {x: 1, y: 0}];
isClockwise(vertices);

从其中一个顶点开始，计算每条边对应的角度。

第一个和最后一个将是零(所以跳过它们);对于其余部分，角度的正弦值将由归一化与(点[n]-点[0])和(点[n-1]-点[0])的单位长度的叉乘给出。

如果这些值的和是正的，那么你的多边形是逆时针方向绘制的。

找出y最小的顶点(如果有平手，则x最大)。假设顶点是A，列表中的前一个顶点是B，列表中的下一个顶点是c。现在计算AB和AC的叉乘的符号。

引用:

如何确定一个简单多边形的方向?在 常见问题:计算机。图形。算法。 维基百科的曲线定位。

Sean的答案的JavaScript实现:

function calcArea(poly) { if(!poly || poly.length < 3) return null; let end = poly.length - 1; let sum = poly[end][0]*poly[0][1] - poly[0][0]*poly[end][1]; for(let i=0; i<end; ++i) { const n=i+1; sum += poly[i][0]*poly[n][1] - poly[n][0]*poly[i][1]; } return sum; } function isClockwise(poly) { return calcArea(poly) > 0; } let poly = [[352,168],[305,208],[312,256],[366,287],[434,248],[416,186]]; console.log(isClockwise(poly)); let poly2 = [[618,186],[650,170],[701,179],[716,207],[708,247],[666,259],[637,246],[615,219]]; console.log(isClockwise(poly2));

我很确定这是对的。这似乎是有效的:-)

这些多边形看起来是这样的，如果你想知道的话: