我正在编写一种计算直线和圆的交集的方法，作为编写某种光线投射演示的第一步。如果计算了交点，它将获得到将成为碰撞点的两个交点的最短距离，然后重复新线从碰撞点开始的过程。
我被这段激光击中不同圆圈的视频所激励。
该方法从 JavaFX 接收直线的角度、它的来源点、窗口的大小、圆的半径、圆的中心数组和 GraphicsContext 对象。
该方法有几个布尔值来确定是否发生了冲突，还有一个 ArrayList 来存储稍后将在 JavaFX 画布上绘制的冲突。
在 while 循环中，直线的方程定义为 y = m*x + b。然后检查哪个圆的中心和线之间的距离小于线的半径，这是用这里解释的方法计算的：math.stackexchange.com。
如果到中心的距离小于半径，则与该圆发生碰撞。据我所知，要找到直线和圆之间的交点，您需要求解方程组：y = m*x + b，(x-x1)^2 + (y-y1)^2 = r^2，我通过替换求解。这得到一个二阶极性方程，该方程有一个实解，如果：p1*p1>= 4*p0*p2。
到原点距离最短的解决方案是线首先到达的解决方案，并且是我们问题的解决方案。使用圆心、碰撞点和原点计算新角度。这样，将定义一条新行并重复循环，直到没有计算与圆的碰撞，从而计算与窗口边界的碰撞的情况。
最后，for 循环绘制所有定义为 collisionList 内点对的线。
这是代码，我试图尽可能地注释它：
private void extendPoint(double angle, Point origin, double x, double y, double radius, ArrayList<Point> pointList) {
double newAngle = angle; //Angle that defines the direction of the line
//This is used if the line does not hit a circle
double angle11 = Math.atan2(origin.getY(), origin.getX());
double angle_11 = Math.atan2(origin.getY(), -origin.getX());
double angle_1_1 = angle11 + Math.PI;
double angle1_1 = angle_11 + Math.PI;
boolean noCollision = true; //Will be true if the line does not hit a circle
boolean repeat = true; //If no collision has been made the while loop stops with this
Point currentPoint = Point.copy(origin); // (x0, y0)
Point collision = new Point(-1,-1); //Stores the collision point
Point newDirection = new Point(-1,-1); //Stores the new direction after a collision, returns(magnitud, angle) of a vector
ArrayList <Point> collisionList = new ArrayList<>(); //ArrayList of collision points that will be drawn later
collisionList.add(origin); //The origin point is added as a collision for representation purposes
while(repeat == true) {
//Line equation that passes through a point with an angle
//y = a*x - a*x0 + y0; -> y = m*x + b;
double m = Math.tan(-newAngle);
double a = m;
double b = -m*currentPoint.getX() + (currentPoint.getY());
for(int i = 0; i < pointList.size(); i++) {
Point gridPoint = pointList.get(i); //(x1, y1)
//From: https://math.stackexchange.com/questions/2552687/distance-between-line-and-point
//Given a line defined as A*x + B*y + C = 0
//x*(y1-y0)+y*(x1-x0)+(-y0*(x1-x0)-x0*(y1-y0)
double A = gridPoint.getY()-currentPoint.getY();
double B = gridPoint.getX()-currentPoint.getX();
double C = -currentPoint.getY()*B + currentPoint.getX()*A;
// double d_cp_gp = Math.abs(m*gridPoint.getX()-b*(gridPoint.getY()))/(Math.sqrt(m*m + 1));
double d_cp_gp = Math.abs(A + B + C)/Math.sqrt(A*A + B*B);
if(d_cp_gp < radius) {
System.out.println("radio " + d_cp_gp);
//The intersection between a line and a circunference:
//Circunference: (x-x1)^2 + (y-y1)^2 = r^2
//Line: y = tan(alpha)*(x-x0)+y0 -> y = a*x + b; a = tan(alfa), b = -tan(alfa)*x0 + y0
//Substituting the line equation in the circunference equation:
//x^2*(1+a^2) + x*(-2x1 + 2*a*b) + 2*a*b + x1^2+b^2-r^2 = 0
double p2 = 1 + a*a;
double p1 = -2*gridPoint.getX() + 2*a*b;
double p0 = gridPoint.getX()*gridPoint.getX() + b*b - radius*radius;
double p0_ = 4*p0*p2;
System.out.println(p1*p1 + " " + p0_);
//Check if the second order equation has solutions
if(p1*p1 >= p0_) {
System.out.println("IT HAS SOLUTION");
//Solution
double root = Math.sqrt(p1*p1 - p0_);
double sol1x = (-p1 + root)/(2*p2);
double sol2x = (-p1 - root)/(2*p2);
double sol1y = a*sol1x - a*currentPoint.getX() + currentPoint.getY();
double sol2y = a*sol1x - a*currentPoint.getX() + currentPoint.getY();
//The line will intersect twice with the circle, we want the solution
//with the shortest distance to currentPoint (x0,y0)
double distSol1 = Math.sqrt(Math.pow(currentPoint.getX()- sol1x, 2) +
Math.pow(currentPoint.getY() - sol1y, 2));
double distSol2 = Math.sqrt(Math.pow(currentPoint.getX()- sol2x, 2) +
Math.pow(currentPoint.getY() - sol2y, 2));
//The collision point is the point that the line hits first
if(distSol1 < distSol2) {
collision.setXY(sol1x, sol1y);
}
else {
collision.setXY(sol2x, sol2y);
}
//newAngle returns a vector with the form (magnitude, angle)
newDirection = newAngle(currentPoint, gridPoint, collision, radius);
currentPoint = collision;
//The new line after the collision is defined here
m = Math.tan(-newDirection.getY());
a = m;
b = -m*collision.getX() + (collision.getY());
collisionList.add(collision);
System.out.println("A collision has been calculated successfully: " + collision.toString());
//If a collision
noCollision= false;
}
}
//If no collisions have been detected at the end of the for loop exit the while loop
if(i == pointList.size() - 1 && noCollision == true) {
repeat = false;
}
}
//If no collision has been calculated with the circles this
//calculates the collision with the limits of the window
if(noCollision == true && repeat == false) {
if(angle<angle11 || angle > angle1_1) {
collision.setXY(x, m*x + b);
}
else if(angle > angle11 && angle < angle_11){
collision.setXY((0 - b)/m, 0);
}
else if(angle > angle_11 && angle < angle_1_1) {
collision.setXY(0, m*0 + b);
}
else if(angle> angle_1_1 && angle < angle1_1) {
collision.setXY((y - b)/m, y);
}
collisionList.add(collision);
}
}
System.out.println("Number of collisions: " + (int)(collisionList.size() - 1));
}
我的主要问题是到圆的最短距离似乎没有正确计算，如果其余代码正常工作，这直接困难。
我尝试了不同的方法来找到最短的距离，这是我最喜欢的方法，因为我发现它很容易理解，但是实现无法正常工作。我认为这可能是因为 JavaFX 坐标系(x 向右增加，y 向底部增加)，但我不确定，在这一点上我有点迷茫。
谢谢你的时间。
编辑： 正如建议的那样，我添加了一些额外的代码以促进可重现性。
点类和向量类定义如下：
public class Point {
private double x;
private double y;
public Point(double x, double y) {
this.x = x;
this.y = y;}
public double getX() {
return x;}
public double getY() {
return y;}
public void setX(double x) {
this.x = x;}
public void setY(double y) {
this.y = y;}
public void setXY(double x, double y) {
this.x = x;
this.y = y;}
@Override
public String toString() {
return("(" + this.x + "," + this.y + ")");
}
public static Point copy(Point a) {
return new Point(a.getX(), a.getY());
}
}
public class Vector {
private double vx;
private double vy;
private double ptoApX;
private double ptoApY;
private double angle;
private double modulo;
public Vector(double vx, double vy) {
this.vx = vx;
this.vy = vy;
this.ptoApX = 0;
this.ptoApY = 0;
this.angle = angle(vx,vy);
this.modulo = modulo(vx,vy);
}
//Getters
public double getVx() {
return this.vx;
}
public double getVy() {
return this.vy;
}
public double getPtoApX() {
return this.ptoApX;
}
public double getPtoApY() {
return this.ptoApY;
}
public double getAngle() {
return this.angle;
}
public double getModulo() {
return this.modulo;
}
//Setters
public void setVx(double vx) {
this.vx = vx;
}
public void setVy(double vy) {
this.vy = vy;
}
public void setPtoApX(double ptoApX) {
this.ptoApX = ptoApX;
}
public void setPtoApY(double ptoApY) {
this.ptoApY = ptoApY;
}
public void setAngle(double angle) {
this.angle = angle;
}
public void setModulo(double modulo) {
this.modulo = modulo;
}
//To String
@Override
public String toString() {
return "("+this.getVx()+","+this.getVy()+")";
}
public static double dotProduct(Vector a, Vector b) {
return a.getVx()*b.getVx() + a.getVy()*b.getVy();
}
public static Vector escalarProduct(Vector v, double n) {
return new Vector(n*v.getVx(), n*v.getVy());
}
public static Vector vectorWith2Points(Point a, Point b) {
Point p = Point.resta(a,b);
return new Vector(p.getX(),p.getY());
}
public static Vector vectorPointAngle(Point a, double angle, double modulo) {
double angleRadians = Math.toRadians(angle);
Point b = new Point(Math.cos(angleRadians)*modulo, Math.sin(angleRadians)*modulo);
return vectorWith2Points(a,b);
}
public static double modulo(double vx, double vy) {
return Math.sqrt(vx*vx + vy*vy);
}
public static double angle(double vx, double vy) {
return Math.atan2(vy, vx);
}
public static Vector normalize(Vector v) {
return new Vector(v.getVx()/v.getModulo(),v.getVy()/v.getModulo());
}
public static double angle2vectors(Vector u, Vector v) {
double argument = dotProduct(u,v)/(u.getModulo()*v.getModulo());
return Math.acos(argument);
}
public static Point polar2cart(double r, double angle) {
return new Point(r*Math.cos(angle), r*Math.sin(angle));
}
public static Point cart2polar(Point p) {
return new Point(modulo(p.getX(), p.getY()), angle(p.getX(), p.getY()));
}
}
以及碰撞后获得新角度的方法：
private Point newAngle(Point origin, Point center, Point c, double radius) {
//Normal vector
Vector n = Vector.vectorWith2Points(c, center);
Vector nNorm = Vector.normalize(n);
//Incident vector
Vector d = Vector.vectorWith2Points(c, origin);
//Tangent vector
Vector tg = new Vector(-nNorm.getVy(), nNorm.getVx());
//Reflected vector
double product = Vector.dotProduct(d,tg);
Vector r = new Vector(d.getVx()-2*product*tg.getVx(),
d.getVy() - 2*product*tg.getVy());
return new Point(r.getModulo(), r.getAngle());
}
应检测碰撞的不同角度的代码示例：
double x = 600;
double y = 400;
double radius = 10;
Point origin = new Point(x/2, y/2);
ArrayList<Point> pointList = new ArrayList<>();
pointList.add(new Point(40,40));
pointList.add(new Point(500,100));
pointList.add(new Point(40,330));
pointList.add(new Point(450,300));
//This should return a solution
extendPoint(0.4363323129985824, origin, x, y, radius, pointList);
extendPoint(2.6179938779914944, origin, x, y, radius, pointList);
//this returns a solution when it should not
extendPoint(1.5707963267948966, origin, x, y, radius, pointList);
extendPoint(-1.5707963267948966, origin, x, y, radius, pointList);
我写了一个类，其中包含运行代码所需的一切：https://pastebin.com/wMjUh9pZ
我认为你应该创建一个表示射线交叉点的类。
class Intersection{
double distance;
Point loc;
double normal;
}
这样，距离是沿着射线的，法线是相交物体的法线。
然后我会有一种找到圆和一个点的交集的方法。
List<Intersection> lineAndCircle( Point org, double angle, Point center, double radius){...}
你似乎有一个类似的方法，但你在其中做了更多的工作。
然后，您还需要检查屏幕的边缘。
Intersection lineAndBoundary( Point org, double angle){ ... }
你有一个非常相似的方法，但你似乎在方法上做了更多的工作。这样，您就可以测试不同的方法。然后你的算法工作。
1 穿过圆并找到交叉点.
2 获取与边界的交点.
3 找到最近的交点(最小距离大于 0 )
这样做会使它更具可扩展性。首先，我们的射线被大量重复使用。让我们做一个类。
class Ray{
Point origin;
double angle;
}
然后我们将一条射线与多个物体碰撞。
interface Interceptable{
List<Intersection> intercepts(Ray r);
}
然后我们可以使用不同的类。
class Circle implements Interceptable{
Point pos;
double radius;
@Override
List<Intersection> collides(Ray r){
...
}
}
现在您可以正确碰撞并可测试。
Circle a = new Circle( new Point( 40, 40 ), 5 );
List<Intersection> yes = a.collides( new Ray( new Point(0, 0), 3.14/4 ) );
List<Intersection> no = a.collides( new Ray( new Point(0, 0), 0) ) );
然后，您可以将示例缩小到。 "如何编写碰撞方法？"或"为什么我的碰撞方法不适用于此射线/圆对？我希望它能击中两点，但它错过了"等等。
下面是一个完整的可运行示例，用于创建摆动窗口。我有点喜欢制作这样的玩具程序。
请注意，我使用了可相交的接口。所以现在是圆圈，但它可以是任何返回Intersection列表的东西
import javax.swing.*;
import java.awt.Graphics;
import java.awt.Dimension;
import java.awt.Color;
import java.awt.event.*;
import java.util.*;
public class RayAndCircle{
public static void main(String[] args){
List<Intersectable> circles = new ArrayList<>();
for(int i = 0; i<250; i++){
double r = Math.random()*50 + 50;
double x = 2048*Math.random();
double y = 2048*Math.random();
circles.add( new Circle( r, new double[]{x,y}));
}
List<LineSegment> segments = new ArrayList<>();
JFrame frame = new JFrame("Ray caster");
JPanel panel = new JPanel(){
@Override
public Dimension getPreferredSize(){
return new Dimension(2048, 2048);
}
@Override
public void paintComponent( Graphics g){
g.setColor(Color.RED);
for( Intersectable c: circles ){
c.draw(g);
}
g.setColor(Color.BLACK);
for( LineSegment segment: segments){
g.drawLine( (int) segment.a[0], (int) segment.a[1],(int)segment.b[0], (int)segment.b[1]);
}
}
};
panel.addMouseListener( new MouseAdapter(){
@Override
public void mouseClicked( MouseEvent evt ){
double x = evt.getPoint().getX();
double y = evt.getPoint().getY();
double theta = Math.random() * Math.PI * 2;
double dx = Math.cos( theta );
double dy = Math.sin( theta );
Ray ray = new Ray( new double[] {x, y}, new double[]{ dx, dy } );
int count = 500;
Intersectable last = null;
while( ray != null && count > 0 ){
Intersection hit = null;
Intersectable next = null;
for(Intersectable c: circles){
if(c == last){
continue;
}
List<Intersection> intersections = c.intersects(ray);
for(Intersection i : intersections){
if( hit == null ){
hit = i;
next = c;
} else{
if( hit.s > i.s ){
hit = i;
next = c;
}
}
}
}
if(hit != null){
last = next;
segments.add( new LineSegment( ray.origin, new double[]{ hit.pos[0], hit.pos[1] } ) );
count--;
//reflected portion of ray.
double dot = hit.normal[0]*ray.direction[0] + hit.normal[1]*ray.direction[1];
double rx = ray.direction[0] - 2 * hit.normal[0]*dot;
double ry = ray.direction[1] - 2 * hit.normal[1]*dot;
double z = Math.sqrt(rx*rx + ry*ry);
ray = new Ray(hit.pos, new double[] { rx/z, ry/z});
} else{
ray = null;
}
}
panel.repaint();
}
});
frame.setContentPane(panel);
frame.pack();
frame.setVisible(true);
frame.setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE);
}
}
class Ray{
double[] origin; double[] direction;
public Ray( double[] origin, double[] direction){
this.origin = new double[]{origin[0], origin[1]};
this.direction = new double[]{direction[0], direction[1]};
}
}
class Intersection{
double s;
double[] pos;
double[] normal;
Circle b;
public Intersection(double s, double[] pos, double[] normal){
this.s = s;
this.pos = pos;
setNormal(normal);
}
public void setNormal(double[] normal){
double m = Math.sqrt(normal[0]*normal[0] + normal[1]*normal[1]);
if( Double.isNaN(m) || m == 0) throw new RuntimeException("Invalid normal! Magnitude of" + m);
this.normal = new double[] { normal[0]/m , normal[1]/m };
}
}
interface Intersectable{
List<Intersection> intersects(Ray ray);
void draw(Graphics g);
}
class Circle implements Intersectable{
double[] origin;
double radius;
public Circle( double radius, double[] origin){
this.radius = radius;
this.origin = new double[]{origin[0], origin[1]};
}
Intersection intersectionAt(Ray ray, double s){
//intersection.
double locx = ray.origin[0] + s*ray.direction[0];
double locy = ray.origin[1] + s*ray.direction[1];
double nx = (locx - origin[0])/radius;
double ny = (locy - origin[1])/radius;
return new Intersection( s, new double[]{ locx, locy }, new double[]{nx, ny} );
}
public List<Intersection> intersects(Ray ray){
double rx = origin[0] - ray.origin[0];
double ry = origin[1] - ray.origin[1];
double m2 = rx*rx + ry*ry;
double m = Math.sqrt(m2);
//position along ray that is closest to circles origin.
double s = rx*ray.direction[0] + ry*ray.direction[1];
//closest distance to circle.
double approach = Math.sqrt(m2 - s*s);
List<Intersection> result = new ArrayList<>();
if( approach < radius ){
//two intersections at points on circle.
//radius is hypotenuse and approach is one of the lengths.
double l = Math.sqrt( radius*radius - approach*approach);
double s1 = s - l;
if(s1 > 0){
result.add( intersectionAt(ray, s1) );
}
double s2 = s + l;
if(s2 > 0){
//intersection!
result.add(intersectionAt(ray, s2) );
}
} else if(approach == radius){
//one intersection tangent.
if( s > 0 ){
result.add( intersectionAt(ray, s) );
}
} else{
//miss.
}
return result;
}
public void draw(Graphics g){
g.fillOval(
(int)(origin[0] - radius),
(int)(origin[1] - radius),
(int)radius*2,
(int)radius*2
);
}
}
class LineSegment{
double[] a, b;
public LineSegment( double[] a, double[] b){
this.a = new double[]{a[0], a[1]};
this.b = new double[]{b[0], b[1]};
}
}
您可能最感兴趣的是 Circle 类的相交方法，以及隐藏在计算反射光线的mouseClicked方法中的一小段代码。
如果您只想知道给定圆的线是否相交，请创建一条第二条线，该线始于给定圆的中心，方向是初始线旋转 90 度的方向。然后计算两条线的交点。如果交点和圆心之间的距离较小，则半径较小，两者相交。
不久前我写了一个小的几何库，我删除了与您相关的部分，这是我的代码：
线类
public class Line {
final Vector2D positionVector;
final Vector2D directionVector;
public Line(final Vector2D positionVector, final Vector2D directionVector) {
this.positionVector = positionVector;
this.directionVector = directionVector;
}
public OptionalDouble computeIntersection(final Line line) {
final double numerator = line.getPositionVector().subtract(this.positionVector).cross(this.directionVector);
final double denominator = this.directionVector.cross(line.directionVector);
if (Math.abs(numerator) < 1e-10 && Math.abs(denominator) < 1e-10) {
// collinear
return OptionalDouble.of(Double.POSITIVE_INFINITY);
} else if (Math.abs(denominator) < 1e-10) {
// parallel
return OptionalDouble.empty(); // Lines are parallel.
}
final double t = line.getPositionVector().subtract(this.positionVector).cross(line.directionVector) / denominator;
return OptionalDouble.of(t);
}
public Vector2D getPositionVector() {
return positionVector;
}
public Vector2D getDirectionVector() {
return directionVector;
}
public Point2D getClosestPointOnLine(final Point2D point) {
final Line line = new Line(new Vector2D(point.getX(), point.getY()), this.directionVector.turn90DegreeClockwise());
final OptionalDouble intersection = this.computeIntersection(line);
final Vector2D result = this.positionVector.add(this.directionVector.lerp(intersection.getAsDouble()));
return new Point2D(result.getX(), result.getY());
}
}
交集功能
public static PointResult intersection(final Line l1, final Circle c1) {
final Point2D intersection = l1.getClosestPointOnLine(c1.getCenter());
final double dist = intersection.distance(c1.getCenter());
if (Math.abs(dist - c1.getRadius()) < 1e-10) {
final List<Point2D> result = new LinkedList<>();
result.add(intersection);
return new PointResult(Collections.unmodifiableList(result));
} else if (dist < c1.getRadius()) {
// we have two points
final double adjacentLeg = Math.sqrt(c1.getRadius() * c1.getRadius() - dist * dist);
final Point2D pt1 = intersection.pointAt(l1.getDirectionVector().angle(), adjacentLeg);
final Point2D pt2 = intersection.pointAt(l1.getDirectionVector().angle() + Math.PI, adjacentLeg);
final List<Point2D> result = new LinkedList<>();
result.add(pt1);
result.add(pt2);
return new PointResult(Collections.unmodifiableList(result));
}
return new PointResult();
}
测试用例
@Test
void testIntersectionLineCircleTwoPoints() {
final Point2D ptCircleCenter = new Point2D(2.0, 5.0);
final Point2D ptLineCircleIntersection = new Point2D(5.0, 2.0);
final Point2D pt1 = new Point2D(3.0, 0.0);
final Point2D pt2 = new Point2D(7.0, 4.0);
final double a = Math.sqrt((2.0 * 2.0) + (2.0 * 2.0));
final double b = ptCircleCenter.diff(ptLineCircleIntersection).norm();
final double radius = Math.sqrt((a * a) + (b * b));
final Line l1 = new Line(pt1, pt2);
final Circle circle = new Circle(ptCircleCenter, radius);
PointResult intersection = GeometryOperation.intersection(l1, circle);
assertTrue(intersection.getPoints().isPresent());
assertEquals(2, intersection.getPoints().get().size());
assertEquals(7.0, intersection.getPoints().get().get(0).getX(), 1e-10);
assertEquals(4.0, intersection.getPoints().get().get(0).getY(), 1e-10);
assertEquals(3.0, intersection.getPoints().get().get(1).getX(), 1e-10);
assertEquals(0.0, intersection.getPoints().get().get(1).getY(), 1e-10);
}
我没有添加Circle、Vector2D和Point2D类，因为它们是微不足道的。类PointResult只是一个列表。