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我试图了解此处给出的二进制空间分区方法实现（此代码也可作为小程序在此处获得）。大部分代码我都看懂了，但是看不懂方法：

public void renderLine(int[] l){
        double x1=l[2];
        double y1=l[3];
        double x2=l[0];
        double y2=l[1];
        double pCos = Math.cos(eye_angle);
        double pSin = Math.sin(eye_angle);
        int[] x = new int[4];
        int[] y = new int[4];
        double pD=-pSin*eye_x+pCos*eye_y; //What is this line doing?
        double pDp=pCos*eye_x+pSin*eye_y; //And this?
        double rz1,rz2,rx1,rx2;
        int Screen_x1=0,Screen_x2=0;
        double Screen_y1,Screen_y2,Screen_y3,Screen_y4;
        rz1=pCos*x1+pSin*y1-pDp;     //perpendicular line to the players
        rz2=pCos*x2+pSin*y2-pDp;     //view point
        if((rz1<1) && (rz2<1))
            return;
        rx1=pCos*y1-pSin*x1-pD;
        rx2=pCos*y2-pSin*x2-pD;
        double pTan = 0;
        if((x2-x1) == 0)
            pTan = Double.MAX_VALUE;
        else
            pTan = (y2-y1)/(x2-x1);
        pTan = (pTan-Math.tan(eye_angle))/(1+
            (pTan*Math.tan(eye_angle)));
        if(rz1 < 1){
            rx1+=(1-rz1)*pTan;
            rz1=1;
        }if(rz2 < 1){
            rx2+=(1-rz2)*pTan;
            rz2=1;
        }
        double z1 = m_width/2/rz1;
        double z2 = m_width/2/rz2;
        Screen_x1=(int)(m_width/2-rx1*z1);
        Screen_x2=(int)(m_width/2-rx2*z2);
        if(Screen_x1 > m_width)
            return;
        if(Screen_x2<0)
            return;
        int wt=88;
        int wb=-40;
        Screen_y1=(double)m_height/2-(double)wt*z1;
        Screen_y4=(double)m_height/2-(double)wb*z1;
        Screen_y2=(double)m_height/2-(double)wt*z2;
        Screen_y3=(double)m_height/2-(double)wb*z2;
        if(Screen_x1 < 0){
            Screen_y1+=(0-Screen_x1)*(Screen_y2-Screen_y1)
                /(Screen_x2-Screen_x1);
            Screen_y4+=(0-Screen_x1)*(Screen_y3-Screen_y4)
                /(Screen_x2-Screen_x1);
            Screen_x1=0;
        }if(Screen_x2 > (m_width)){
            Screen_y2-=(Screen_x2-m_width)*(Screen_y2-Screen_y1)
                /(Screen_x2-Screen_x1);
            Screen_y3-=(Screen_x2-m_width)*(Screen_y3-Screen_y4)
                /(Screen_x2-Screen_x1);
            Screen_x2=m_width;
        }if((Screen_x2-Screen_x1) == 0)
            return;
        x[0] = (int)Screen_x1;
        y[0] = (int)(Screen_y1);
        x[1] = (int)Screen_x2;
        y[1] = (int)(Screen_y2);
        x[2] = (int)Screen_x2;
        y[2] = (int)(Screen_y3);
        x[3] = (int)Screen_x1;
        y[3] = (int)(Screen_y4);
        double_graphics.setColor(new Color(l[4]));
        double_graphics.fillPolygon(x,y,4);
    }

我无法掌握实施此方法背后的坐标几何。你能给我解释一下吗？

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1 回答 1

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前半部分计算线相对于虚拟眼睛应该在的位置的角度。它还使用切线计算它是否完全可见。第二部分调整线条的大小以与屏幕成比例。在我看来，他们将坐标系的原点设置在屏幕中间。

我只是浏览了代码，所以其中一些细节可能会被关闭。

看看这个。它可能会让你朝着正确的方向前进 http://www.cs.uic.edu/~jbell/CourseNotes/ComputerGraphics/2DTransforms.html

于 2012-06-01T02:10:20.403 回答