如何创建2D 2点之间的曲线,并取回点,使该曲线每条D距离是多少? [英] How to create a curve between 2 points in 2D and get back Points that makes that curve every d distance?
问题描述
我不是数学不错。
我有2点, A(X1,Y1)
和 B(x2,y2)的2D
。
我需要创建点的虚拟路径 A
到 B
弯曲为R(半径),然后返回的是描述这种曲线路径指向数组,不是所有可能每D(距离)彼此。
I need to create a virtual path from point A
to B
curved at R(radius), and then return an array of points which are describing this curved path, not all maybe every D(distance) from each other.
在Java中,我需要一种方法是这样的:
In Java I need a method like this:
private ArrayList<PointF> generateCurve(PointF pFrom,PointF pTo,float pRadius,float pMinDistance){
ArrayList<PointF> pOutPut = new ArrayList<PointF>();
// ...generate result to pOutPut
return pOutPut;
}
如何做到这一点?
How to do this ?
推荐答案
我没有放弃,我一直在做这个工作,为几个小时。这里是结果:
I didn't gave up and I've been working on it for a few more hours. And here is the result:
我创建了一个方法,你可以指定,如果你想在最短的两点之间最长的电弧。
I created a method where you can specify if you want the shortest of the longest arc between the points.
下面是一些调用它,与生成的输出:
Here are some calls to it, with the produced output:
generateCurve(pFrom, pTo, 100f, 7f, false, false);
generateCurve(pFrom, pTo, 100f, 7f, true, false);
generateCurve(pFrom, pTo, 100f, 7f, false, true);
generateCurve(pFrom, pTo, 100f, 7f, true, true);
正如你所看到的,这是工作就像一个魅力。这里是code:
As you can see, it is working like a charm. Here is the code:
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, pTo, 100f, 7f, true, 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, 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, 0, 1000, 0);
g.drawLine(0, -1000, 0, 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, 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, PointF pTo, float pRadius, 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;
}
}
享受!
PS:我在上数学两个问题要解决你的问题:
Enjoy!
PS: I created two questions on Mathematics to solve your question:
- <一个href="http://math.stackexchange.com/questions/50661/analytic-geometry-point-coordinates-same-distance-from-two-points">Analytic几何图形:点坐标,同样的距离两点
- <一个href="http://math.stackexchange.com/questions/50678/trigonometry-solve-1-cos-alpha2-sin2-alpha-d2-for-alpha">Trigonometry:解决(1-cosα)2 +sin2α= D2为α
- Analytic Geometry: Point coordinates, same distance from two points.
- Trigonometry: Solve (1−cosα)2+sin2α=d2 for α
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