多边形的重心

Question

我正在尝试编写一个PHP函数来计算多边形的重心.

我看过其他类似的问题,但我似乎无法找到解决方案.

我的问题是我需要能够计算规则和不规则多边形甚至自相交多边形的重心.

那可能吗？

我也读过:http://paulbourke.net/geometry/polyarea/ 但这仅限于非自相交的多边形.

我怎样才能做到这一点？你能指出我正确的方向吗？

Answer 1

重心(也称为"质心"或"质心")可以使用以下公式计算:

X = SUM[(Xi + Xi+1) * (Xi * Yi+1 - Xi+1 * Yi)] / 6 / A
Y = SUM[(Yi + Yi+1) * (Xi * Yi+1 - Xi+1 * Yi)] / 6 / A

从维基百科中提取:由n个顶点(x0,y0),(x1,y1),...,(xn-1,yn-1)定义的非自相交闭合多边形的质心是点(Cx, Cy),在哪里


其中A是多边形的有符号区域,

使用VBasic的示例:

' Find the polygon's centroid.
Public Sub FindCentroid(ByRef X As Single, ByRef Y As _
    Single)
Dim pt As Integer
Dim second_factor As Single
Dim polygon_area As Single

    ' Add the first point at the end of the array.
    ReDim Preserve m_Points(1 To m_NumPoints + 1)
    m_Points(m_NumPoints + 1) = m_Points(1)

    ' Find the centroid.
    X = 0
    Y = 0
    For pt = 1 To m_NumPoints
        second_factor = _
            m_Points(pt).X * m_Points(pt + 1).Y - _
            m_Points(pt + 1).X * m_Points(pt).Y
        X = X + (m_Points(pt).X + m_Points(pt + 1).X) * _
            second_factor
        Y = Y + (m_Points(pt).Y + m_Points(pt + 1).Y) * _
            second_factor
    Next pt

    ' Divide by 6 times the polygon's area.
    polygon_area = PolygonArea
    X = X / 6 / polygon_area
    Y = Y / 6 / polygon_area

    ' If the values are negative, the polygon is
    ' oriented counterclockwise. Reverse the signs.
    If X < 0 Then
        X = -X
        Y = -Y
    End If
End Sub

有关更多信息,请查看本网站或维基百科.

希望能帮助到你.

问候!

Answer 2

在冷c ++中,假设你有一个带有x和y属性的Vec2结构:

const Vec2 findCentroid(Vec2* pts, size_t nPts){
    Vec2 off = pts[0];
    float twicearea = 0;
    float x = 0;
    float y = 0;
    Vec2 p1, p2;
    float f;
    for (int i = 0, j = nPts - 1; i < nPts; j = i++) {
        p1 = pts[i];
        p2 = pts[j];
        f = (p1.x - off.x) * (p2.y - off.y) - (p2.x - off.x) * (p1.y - off.y);
        twicearea += f;
        x += (p1.x + p2.x - 2 * off.x) * f;
        y += (p1.y + p2.y - 2 * off.y) * f;
    }

    f = twicearea * 3;

    return Vec2(x / f + off.x, y / f + off.y);
}

并在JavaScript中:

function findCentroid(pts, nPts) {
    var off = pts[0];
    var twicearea = 0;
    var x = 0;
    var y = 0;
    var p1,p2;
    var f;
    for (var i = 0, j = nPts - 1; i < nPts; j = i++) {
        p1 = pts[i];
        p2 = pts[j];
        f = (p1.lat - off.lat) * (p2.lng - off.lng) - (p2.lat - off.lat) * (p1.lng - off.lng);
        twicearea += f;
        x += (p1.lat + p2.lat - 2 * off.lat) * f;
        y += (p1.lng + p2.lng - 2 * off.lng) * f;
    }
    f = twicearea * 3;
    return {
    X: x / f + off.lat,
    Y: y / f + off.lng
    };
}

或者在旧的c中,并假设你有一个带有x和y属性的Point结构:

const Point centroidForPoly(const int numVerts, const Point* verts)
{
    float sum = 0.0f;
    Point vsum = 0;

    for (int i = 0; i<numVerts; i++){
        Point v1 = verts[i];
        Point v2 = verts[(i + 1) % numVerts];
        float cross = v1.x*v2.y - v1.y*v2.x;
        sum += cross;
        vsum = Point(((v1.x + v2.x) * cross) + vsum.x, ((v1.y + v2.y) * cross) + vsum.y);
    }

    float z = 1.0f / (3.0f * sum);
    return Point(vsum.x * z, vsum.y * z);
}