如何为凹多边形生成回波路径

Question

d是给出的回波路径之间的距离

？角很容易计算，知道我们所拥有的顶点坐标

因此，正如您所看到的，对于每个顶点，我们可以计算L，从而为下一个回波路径提供新的顶点。

我想做的是生成没有自交叉部分的回波多边形，即没有带有虚线的部分。一个算法或Java代码将非常有帮助。多谢了。

编辑

只是添加了一段代码，生成凸多边形的回波路径，就像注释中要求的那样。

public List<MyPath> createEchoCoCentral( List<Point> pointsOriginal, float encoderEchoDistance, int appliqueEchoCount){

    List<Point> contourPoints = pointsOriginal;
    List<MyPath> echoPaths = new ArrayList<>();
    for (int round = 0; round < appliqueEchoCount; round++) {

        List<Point> echoedPoints = new ArrayList<>();
        int size = contourPoints.size()+1;//+1 because we connect end to start

        Point previousPoint = contourPoints.get(contourPoints.size() - 1);
        for (int i = 0; i < size; i++) {
            Point currentPoint;
            if (i == contourPoints.size()) {
                currentPoint = new Point(contourPoints.get(0));
            } else {
                currentPoint = contourPoints.get(i);
            }
            final Point nextPoint;
            if (i + 1 == contourPoints.size()) {
                nextPoint = contourPoints.get(0);
            } else if (i == contourPoints.size()) {
                nextPoint = contourPoints.get(1);
            } else {
                nextPoint = contourPoints.get(i + 1);
            }
            if (currentPoint.x == previousPoint.x && currentPoint.y == previousPoint.y) continue;
            if (currentPoint.x == nextPoint.x && currentPoint.y == nextPoint.y) continue;

            // signs needed o determine to which side of polygon new point will go
            float currentSlope = (float) (Math.atan((previousPoint.y - currentPoint.y) / (previousPoint.x - currentPoint.x)));
            float signX = Math.signum((previousPoint.x - currentPoint.x));
            float signY = Math.signum((previousPoint.y - currentPoint.y));
            signX = signX == 0 ? 1 : signX;
            signY = signY == 0 ? 1 : signY;

            float nextSignX = Math.signum((currentPoint.x - nextPoint.x));
            float nextSignY = Math.signum((currentPoint.y - nextPoint.y));
            nextSignX = nextSignX == 0 ? 1 : nextSignX;
            nextSignY = nextSignY == 0 ? 1 : nextSignY;

            float nextSlope = (float) (Math.atan((currentPoint.y - nextPoint.y) / (currentPoint.x - nextPoint.x)));
            float nextSlopeD = (float) Math.toDegrees(nextSlope);

            //calculateMidAngle - is a bit tricky function that calculates angle between two adjacent edges
            double S = calculateMidAngle(currentSlope, nextSlope, signX, signY, nextSignX, nextSignY);
            Point p2 = new Point();

            double ew = encoderEchoDistance / Math.cos(S - (Math.PI / 2));
            p2.x = (int) (currentPoint.x + (Math.cos(currentSlope - S)) * ew * signX);
            p2.y = (int) (currentPoint.y + (Math.sin(currentSlope - S)) * ew * signX);

            echoedPoints.add(p2);
            previousPoint = currentPoint;


        }

        //createPathFromPoints just creates MyPath objects from given Poins set
        echoPaths.add(createPathFromPoints(echoedPoints));
        //remove last point since it was just to connect end to first point
        echoedPoints.remove(echoedPoints.size() - 1);
        contourPoints = echoedPoints;
    }
    return echoPaths;
}

Answer 1

这个问题叫做计算多边形偏移。解决这个问题有两种常见的方法：

1）最有效的方法是通过计算缠绕数来计算偏置多边形（据我了解，这个算法是Clipper库使用的）

2）计算直线骨架图，帮助您建立偏移多边形

关于这个主题的有趣文章：

Chen，通过计算缠绕数实现多边形偏移

计算直线骨架Felkel算法

Answer 2

你在找笔直的骷髅：

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