Understanding Mathematics behind Java Code: 2D Binary Space Partitioning

Question

I was trying to understand the Binary Space Partitioning method implementation that is given here (this code is also available as an applet here). I have understood most of the code, but can not understand the method:

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);
    }

I am not being able to grasp the co-ordinate geometry behind implementation of this method. Can you please explain it to me?

Answer 1

The first half computes the angle of the line with respect to where the virtual eye should be. It also computes whether it is visible at all using the tangent. The second part resizes the line to be proportional to the screen. It seems to me that they set the origin of the coordinate system in the middle of the screen.

I only skimmed the code, so some of these details may be off.

Take a look at this. It may start you in the right direction http://www.cs.uic.edu/~jbell/CourseNotes/ComputerGraphics/2DTransforms.html