Center of gravity of a polygon

问题:

I am trying to write a PHP function that will calculate the center of gravity of a polygon.

I've looked at the other similar questions but I can't seem to find a solution to this.

My problem is that I need to be able to calculate the center of gravity for both regular and irregular polygons and even self intersecting polygons.

Is that possible?

I've also read that: http://paulbourke.net/geometry/polyarea/ But this is restricted to non self intersecting polygons.

How can I do this? Can you point me to the right direction?

回答1:

The center of gravity (also known as "center of mass" or "centroid" can be calculated with the following formula:

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

Extracted from Wikipedia: The centroid of a non-self-intersecting closed polygon defined by n vertices (x0,y0), (x1,y1), ..., (xn−1,yn−1) is the point (Cx, Cy), where


and where A is the polygon's signed area,

Example using VBasic:

' 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

For more info check this website or Wikipedia.

Hope it helps.

Regards!

回答2:

in cold c++ and while assuming that you have a Vec2 struct with x and y properties :

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);
}

and in javascript :

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
    };
}

or in good old c and while assuming that you have a Point struct with x and y properties :

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);
}

回答3:

Swift 4, based on the c answer given above

/// Given an array of points, find the "center of gravity" of the points
/// - Parameters:
///     - points: Array of points
/// - Returns:
///     - Point or nil if input points count < 3
static func centerOfPoints(points: [CGPoint]) -> CGPoint? {
    if points.count < 3 {
        return nil
    }

    var sum: CGFloat = 0
    var pSum: CGPoint = .zero

    for i in 0..<points.count {
        let p1 = points[i]
        let p2 = points[(i+1) % points.count]
        let cross = p1.x * p2.y - p1.y * p2.x
        sum += cross
        pSum = CGPoint(x:((p1.x + p2.x) * cross) + pSum.x,
                       y:((p1.y + p2.y) * cross) + pSum.y)
    }

    let z = 1 / (3 * sum)
    return CGPoint(x:pSum.x * z,
                   y:pSum.y * z)
}

回答4:

This was my implementation in Java of the accepted solution, I added an extra conditional check because some of my polygons were flat and had no area, and rather than giving me the midpoint, it was returning (0,0). Thus in this case, I reference a different method which simply averages the vertices. The rounding at the end is because I wanted to keep my output object as integers even though it is imprecise, but I welcome you to remove that bit. Also, since all of my points were positive integers, the check made sense for me, but for you, adding an area check == 0 would also make sense.

private Vertex getCentroid() {

        double xsum = 0, ysum = 0, A = 0;
        for (int i = 0; i < corners.size() ; i++) {

            int iPlusOne = (i==corners.size()-1)?0:i+1;

            xsum += (corners.get(i).getX() + corners.get(iPlusOne).getX()) * (corners.get(i).getX() * corners.get(iPlusOne).getY() - corners.get(iPlusOne).getX() * corners.get(i).getY());
            ysum += (corners.get(i).getY() + corners.get(iPlusOne).getY()) * (corners.get(i).getX() * corners.get(iPlusOne).getY() - corners.get(iPlusOne).getX() * corners.get(i).getY());
            A += (corners.get(i).getX() * corners.get(iPlusOne).getY() - corners.get(iPlusOne).getX() * corners.get(i).getY());
        }
        A = A / 2;
        if(xsum==0 &&ysum==0)
        {
            area = averageHeight/2;
            return getMidpointCenter();
        }
        double x = xsum / (6 * A);
        double y = ysum / (6 * A);
        area = A;


        return new Vertex((int) Math.round(x), (int) Math.round(y));
    }

回答5:

In php:

// Find the polygon's centroid.
function getCenter($polygon)
{ 
    $NumPoints = count($polygon);

    if($polygon[$NumPoints-1] == $polygon[0]){
        $NumPoints--;
    }else{
        //Add the first point at the end of the array.
        $polygon[$NumPoints] = $polygon[0];
    }

    // Find the centroid.
    $X = 0;
    $Y = 0;
    For ($pt = 0 ;$pt<= $NumPoints-1;$pt++){
        $factor = $polygon[$pt][0] * $polygon[$pt + 1][1] - $polygon[$pt + 1][0] * $polygon[$pt][1];
        $X += ($polygon[$pt][0] + $polygon[$pt + 1][0]) * $factor;
        $Y += ($polygon[$pt][1] + $polygon[$pt + 1][1]) * $factor;
    }

    // Divide by 6 times the polygon's area.
    $polygon_area = ComputeArea($polygon);
    $X = $X / 6 / $polygon_area;
    $Y = $Y / 6 / $polygon_area;

    return array($X, $Y);
}


function ComputeArea($polygon)
{ 
    $NumPoints = count($polygon);

    if($polygon[$NumPoints-1] == $polygon[0]){
        $NumPoints--;
    }else{
        //Add the first point at the end of the array.
        $polygon[$NumPoints] = $polygon[0];
    }

    $area = 0;

    for ($i = 0; $i < $NumPoints; $i++) {
      $i1 = ($i + 1) % $NumPoints;
      $area += ($polygon[$i][1] + $polygon[$i1][1]) * ($polygon[$i1][0] - $polygon[$i][0]);
    }

    $area /= 2;
    return $area;
}

Read more at:

PHP: How to determine the center of a Polygon

回答6:

Since we are all having so much fun implementing this algo in different languages, here is my version I knocked up for Python:

def polygon_centre_area(vertices: Sequence[Sequence[float]]) -> Tuple[Sequence[float], float]:
    x_cent = y_cent = area = 0
    v_local = vertices + [vertices[0]]

    for i in range(len(v_local) - 1):
        factor = v_local[i][0] * v_local[i+1][1] - v_local[i+1][0] * v_local[i][1]
        area += factor
        x_cent += (v_local[i][0] + v_local[i+1][0]) * factor
        y_cent += (v_local[i][1] + v_local[i+1][1]) * factor

    area /= 2.0
    x_cent /= (6 * area)
    y_cent /= (6 * area)

    area = math.fabs(area)

    return ([x_cent, y_cent], area)