我需要一种算法来绘制任意多边形的回声路径。如果多边形是凸的,则很容易解决。要了解我的意思,请查看波纹管,其中黑色是原始多边形,红色是从原始多边形生成的回声多边形。
d
是给定的回声路径之间的距离
β
角度很容易计算已知顶点的坐标
如您所见,我们可以为每个顶点进行计算L
,从而为下一个回声路径提供新的顶点。
问题是,当我们在某个点具有凹面多边形时,会得到一张自相交多边形的丑陋图片。请看这张照片。
我想做的是生成没有自相交部分的回波多边形,即没有带有虚线的部分。算法或java
代码将非常有帮助。谢谢。
编辑
只需添加一段代码即可生成注释中要求的凸多边形的回波路径。
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;
}
好的,找到了一个可以满足我需要的库。它叫做快船队
如果有人感兴趣,这里也有Java实现。
使用Java库,几行代码可以解决问题
Path originalPath = new Path();
for (PointF areaPoint:pointsOriginal){
originalPath.add(new LongPoint((long)areaPoint.x, (long)areaPoint.y));
}
final ClipperOffset clo = new ClipperOffset();
Paths clips = new Paths();
Paths solution = new Paths();
clips.add(originalPath);
clo.addPaths( clips, Clipper.JoinType.SQUARE, Clipper.EndType.CLOSED_LINE );
float encoderEchoDistance = (float) UnitUtils.convertInchOrMmUnitsToEncoderUnits(this, inchOrMm, appliqueEchoDistance);
clo.execute( solution, encoderEchoDistance );
// Now solution.get(0) will contain path that has offset from original path
// and what is most important it will not have self intersections.
它是开源的,因此我将深入了解实施细节。感谢所有尝试提供帮助的人。
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