java – 如何在2D中的2个点之间创建曲线并返回每隔d距离生成该曲线的点？

我的数学不好.

我在2D中有2个点,A(x1,y1)和B(x2,y2).

我需要创建一个从R点(半径)弯曲的A点到B点的虚拟路径,然后返回一个描述这个弯曲路径的点数组,而不是每个D(距离)彼此之间的距离.

在Java中我需要一个像这样的方法：

private ArrayList<PointF> generateCurve(PointF pFrom,PointF pTo,float pRadius,float pMinDistance){

    ArrayList<PointF> pOutPut = new ArrayList<PointF>();
    // ...generate result to pOutPut 

    return pOutPut;
}

这该怎么做 ？

解决方法

我创建了一个方法,您可以指定是否需要点之间最长的最长弧.

以下是对它的一些调用,产生的输出：

generateCurve(pFrom,pTo,100f,7f,false,false);

generateCurve(pFrom,true,true);

generateCurve(pFrom,true);

正如你所看到的,它就像一个魅力.这是代码：

package curve;

import java.awt.BasicStroke;
import java.awt.Color;
import java.awt.Graphics2D;
import java.awt.RenderingHints;
import java.awt.geom.Ellipse2D;
import java.awt.geom.Line2D;
import java.awt.geom.Rectangle2D;
import java.awt.image.BufferedImage;
import java.io.File;
import java.io.IOException;
import java.util.ArrayList;
import java.util.List;
import javax.imageio.ImageIO;

/**
 *
 * @author martijn
 */
public class Main
{

    /**
     * @param args the command line arguments
     */
    public static void main(String[] args) throws IOException
    {
        PointF pFrom = new PointF(-10f,30.0f);
        PointF pTo = new PointF(-100f,0.0f);
        List<PointF> points = generateCurve(pFrom,true);

        System.out.println(points);

        // Calculate the bounds of the curve
        Rectangle2D.Float bounds = new Rectangle2D.Float(points.get(0).x,points.get(0).y,0);
        for (int i = 1; i < points.size(); ++i) {
            bounds.add(points.get(i).x,points.get(i).y);
        }
        bounds.add(pFrom.x,pFrom.y);
        bounds.add(pTo.x,pTo.y);

        BufferedImage img = new BufferedImage((int) (bounds.width - bounds.x + 50),(int) (bounds.height - bounds.y + 50),BufferedImage.TYPE_4BYTE_ABGR_PRE);
        Graphics2D g = img.createGraphics();
        g.setRenderingHint(RenderingHints.KEY_ANTIALIASING,RenderingHints.VALUE_ANTIALIAS_ON);

        g.translate(25.0f - bounds.getX(),25.0f - bounds.getY());
        g.setStroke(new BasicStroke(1.0f));


        g.setColor(Color.DARK_GRAY);
        g.drawLine(-1000,1000,0);
        g.drawLine(0,-1000,1000);

        g.setColor(Color.RED);
        for (int i = 0; i < points.size(); ++i) {
            if (i > 0) {
                Line2D.Float f = new Line2D.Float(points.get(i - 1).x,points.get(i - 1).y,points.get(i).x,points.get(i).y);
                System.out.println("Dist : " + f.getP1().distance(f.getP2()));
//                g.draw(f);
            }

            g.fill(new Ellipse2D.Float(points.get(i).x - 0.8f,points.get(i).y - 0.8f,1.6f,1.6f));

        }
        g.setColor(Color.BLUE);
        g.fill(new Ellipse2D.Float(pFrom.x - 1,pFrom.y - 1,3,3));
        g.fill(new Ellipse2D.Float(pTo.x - 1,pTo.y - 1,3));

        g.dispose();

        ImageIO.write(img,"PNG",new File("result.png"));
    }

    static class PointF
    {

        public float x,y;

        public PointF(float x,float y)
        {
            this.x = x;
            this.y = y;
        }

        @Override
        public String toString()
        {
            return "(" + x + "," + y + ")";
        }
    }

    private static List<PointF> generateCurve(PointF pFrom,float pMinDistance,boolean shortest,boolean side)
    {

        List<PointF> pOutPut = new ArrayList<PointF>();

        // Calculate the middle of the two given points.
        PointF mPoint = new PointF(pFrom.x + pTo.x,pFrom.y + pTo.y);
        mPoint.x /= 2.0f;
        mPoint.y /= 2.0f;
        System.out.println("Middle Between From and To = " + mPoint);


        // Calculate the distance between the two points
        float xDiff = pTo.x - pFrom.x;
        float yDiff = pTo.y - pFrom.y;
        float distance = (float) Math.sqrt(xDiff * xDiff + yDiff * yDiff);
        System.out.println("Distance between From and To = " + distance);

        if (pRadius * 2.0f < distance) {
            throw new IllegalArgumentException("The radius is too small! The given points wont fall on the circle.");
        }

        // Calculate the middle of the expected curve.
        float factor = (float) Math.sqrt((pRadius * pRadius) / ((pTo.x - pFrom.x) * (pTo.x - pFrom.x) + (pTo.y - pFrom.y) * (pTo.y - pFrom.y)) - 0.25f);
        PointF circleMiddlePoint = new PointF(0,0);
        if (side) {
            circleMiddlePoint.x = 0.5f * (pFrom.x + pTo.x) + factor * (pTo.y - pFrom.y);
            circleMiddlePoint.y = 0.5f * (pFrom.y + pTo.y) + factor * (pFrom.x - pTo.x);
        } else {
            circleMiddlePoint.x = 0.5f * (pFrom.x + pTo.x) - factor * (pTo.y - pFrom.y);
            circleMiddlePoint.y = 0.5f * (pFrom.y + pTo.y) - factor * (pFrom.x - pTo.x);
        }
        System.out.println("Middle = " + circleMiddlePoint);

        // Calculate the two reference angles
        float angle1 = (float) Math.atan2(pFrom.y - circleMiddlePoint.y,pFrom.x - circleMiddlePoint.x);
        float angle2 = (float) Math.atan2(pTo.y - circleMiddlePoint.y,pTo.x - circleMiddlePoint.x);

        // Calculate the step.
        float step = pMinDistance / pRadius;
        System.out.println("Step = " + step);

        // Swap them if needed
        if (angle1 > angle2) {
            float temp = angle1;
            angle1 = angle2;
            angle2 = temp;

        }
        boolean flipped = false;
        if (!shortest) {
            if (angle2 - angle1 < Math.PI) {
                float temp = angle1;
                angle1 = angle2;
                angle2 = temp;
                angle2 += Math.PI * 2.0f;
                flipped = true;
            }
        }
        for (float f = angle1; f < angle2; f += step) {
            PointF p = new PointF((float) Math.cos(f) * pRadius + circleMiddlePoint.x,(float) Math.sin(f) * pRadius + circleMiddlePoint.y);
            pOutPut.add(p);
        }
        if (flipped ^ side) {
            pOutPut.add(pFrom);
        } else {
            pOutPut.add(pTo);
        }

        return pOutPut;
    }
}

请享用！
PS：我创建了两个关于数学的问题来解决你的问题：

> Analytic Geometry: Point coordinates,same distance from two points.
> Trigonometry: Solve (1cosα)2+sin2α=d2 for α

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