Oval冲突检测不能正常工作

Question

因此，我正在尝试实现一个测试，其中一个椭圆可以连接一个圆，但它不起作用。

edist = (float) Math.sqrt(
    Math.pow((px + ((pwidth/2) )) - (bx + (bsize/2)), 2 ) + 
    Math.pow(-((py + ((pwidth/2)) ) - (bx + (bsize/2))), 2 )
);

下面是完整的代码(需要Slick2D)：

import org.newdawn.slick.AppGameContainer;
import org.newdawn.slick.BasicGame;
import org.newdawn.slick.Color;
import org.newdawn.slick.GameContainer;
import org.newdawn.slick.Graphics;
import org.newdawn.slick.Input;
import org.newdawn.slick.SlickException;

public class ColTest extends BasicGame{


    float px = 50;
    float py = 50;
    float pheight = 50;
    float pwidth = 50;



    float bx = 200;
    float by = 200;
    float bsize = 200;

    float edist;

    float pspeed = 3;
    Input input;

    public ColTest()
    {
        super("ColTest");
    }

    @Override
    public void init(GameContainer gc)
            throws SlickException {

    }

    @Override
    public void update(GameContainer gc, int delta)
            throws SlickException
    {
        input = gc.getInput();

        try{    
            if(input.isKeyDown(Input.KEY_UP))   
                py-=pspeed;

            if(input.isKeyDown(Input.KEY_DOWN)) 
                py+=pspeed;

            if(input.isKeyDown(Input.KEY_LEFT)) 
                px-=pspeed;

            if(input.isKeyDown(Input.KEY_RIGHT))    
                px+=pspeed;
        }

        catch(Exception e){}
    }

    public void render(GameContainer gc, Graphics g)
            throws SlickException
    {   
            g.setColor(new Color(255,255,255));
            g.drawString("col: " + col(), 10, 10);
            g.drawString("edist: " + edist + " dist: " + dist, 10, 100);

            g.fillRect(px, py, pwidth, pheight);
            g.setColor(new Color(255,0,255));
            g.fillOval(px, py, pwidth, pheight);
            g.setColor(new Color(255,255,255));
            g.fillOval(200, 200, 200, 200);

    }

    public boolean col(){

        edist = (float) Math.sqrt(Math.pow((px + ((pwidth/2) )) - (bx + (bsize/2)), 2) + Math.pow(-((py + ((pwidth/2)) ) - (bx + (bsize/2))), 2));

        if(edist <= (bsize/2) + (px + (pwidth/2)))
            return true;

        else
            return false;
    }

    public float rotate(float x, float y, float ox, float oy, float a, boolean b)
    {
         float dst = (float) Math.sqrt(Math.pow(x-ox,2.0)+ Math.pow(y-oy,2.0));

         float oa = (float) Math.atan2(y-oy,x-ox);

         if(b)
            return (float) Math.cos(oa + Math.toRadians(a))*dst+ox;

         else
            return (float) Math.sin(oa + Math.toRadians(a))*dst+oy;

    }

    public static void main(String[] args)
            throws SlickException
    {
         AppGameContainer app =
            new AppGameContainer( new ColTest() );

         app.setShowFPS(false);
         app.setAlwaysRender(true);
         app.setTargetFrameRate(60);
         app.setDisplayMode(800, 600, false);
         app.start();
    }
}

Answer 1

找到交叉口比你想象的要难。您的col()方法有点离谱，但该方法最多只能告诉您一个点是否在圆内。它将不能真正检测交叉点。

我用谷歌搜索了一些计算实际交叉点的代码。我发现one in JavaScript非常有趣而且非常复杂。看看the source吧。

如果你想要一些更简单(但不太精确)的东西，你可以检查椭圆周围的几个点，看看它们是否在圆内。

private boolean isInCircle(float x, float y) {
    float r = bsize / 2;
    float center_x = bx + r;
    float center_y = by + r;
    float dist = (float) Math.sqrt(Math.pow(x - center_x, 2) + Math.pow(y - center_y, 2));

    return dist < r;
}

public boolean col() {
    return 
        isInCircle(px + pwidth / 2, py              ) || // top
        isInCircle(px + pwidth    , py + pheight / 2) || // right
        isInCircle(px + pwidth / 2, py + pheight    ) || // bottom
        isInCircle(px             , py + pheight / 2);   // left
}

Answer 2

如果你计划实现更多的形状和/或需要形状之间的最小距离，你可以开始使用GJK：你只需要为每个新形状实现支持函数。如果计算时间也很关键，GJK绝对是你应该考虑的东西，但它肯定需要你这边更多的编程。

Answer 3

如果你能找到你的焦点，你可以用下面的伪代码检查冲突。警告:这只适用于两个椭圆碰撞(椭圆和圆碰撞也适用)。

r = length of semi major axis
a_x = x coordinate of foci 1 of the first ellipse
a_y = y coordinate of foci 1 of the first ellipse
b_x = x coordinate of foci 2 of the first ellipse
b_y = y coordinate of foci 2 of the first ellipse

c_x = x coordinate of foci 1 of the second ellipse
c_y = y coordinate of foci 1 of the second ellipse
d_x = x coordinate of foci 2 of the second ellipse
d_y = y coordinate of foci 2 of the second ellipse

p_x = (a_x+b_x+c_x+d_x)/4 // i.e. the average of the foci x values
p_y = (a_y+b_y+c_y+d_y)/4 // i.e. the average of the foci y values

if r >= ( sqrt( (p_x + a_x)^2+(p_y + a_y)^2 ) + sqrt( (p_x + a_x)^2+(p_y + a_y)^2 ) )
then collision

如果你真的想知道它的来源，请让我知道，我会提供的。但它使用的思想是，椭圆焦点和椭圆边缘上的任何点之间的距离之和是一组距离(半长轴)。并求解位于两个椭球体边缘上的点，如果存在一个点，则它们是碰撞。