Checking if a player passed two gates

Question

I have two 3D Gates (points) in a 3D world, and I would like to check if the player has passed between the two gates or passed the right gate or left gate.

So I constructed a vector V (distance) between the two points, and constructed a plane out of the that vector but (Vx, Vz).

I can check now the current distance of the player and know if the player intersects with the plane, but how would I know if he passed the right gate or the left one ?

Answer 1

If I get it right (not 100% sure from your drawing) then it is like this:

1. do not test single point

   • use actual and last player's position instead
   • compute the intersection point G between axis given by P0,P1
   • and line P2,P3 which is also player movement vector
   • if no intersection then no passing
   • if none or infinite intersections then player moving along gate not passing (this edge case need to be handled)

2. one intersection

   • this is non edge case of player passing so
   • if |P0-G|<=d then player is pasing through gate point P0
   • if |P1-G|<=d then player is pasing through gate point P1
   • d is some accuracy distance
   • else compute q=|G-P0|/|P1-P0|
   • if 'q<0' then player is passing on left side
   • if 'q>|P1-P0|' then player is passing on right side
   • if (q>=0) && (q<=|P1-P0|) then player is passing in the middle

some C++ code:

    double x0=?,x1=?,y0=?,y1=?; // input points (axis)
    double x2=?,x3=?,y2=?,y3=?; // input points (line)

    double xx0,yy0,xx1,yy1;
    double kx0,ky0,dx0,dy0,t0;
    double kx1,ky1,dx1,dy1,t1;
    kx0=x0; ky0=y0; dx0=x1-x0; dy0=y1-y0;
    kx1=x2; ky1=y2; dx1=x3-x2; dy1=y3-y2;
    t1=divide(dx0*(ky0-ky1)+dy0*(kx1-kx0),(dx0*dy1)-(dx1*dy0));
    xx1=kx1+(dx1*t1);
    yy1=ky1+(dy1*t1);
    if (fabs(dx0)>=fabs(dy0)) t0=divide(kx1-kx0+(dx1*t1),dx0);
    else                        t0=divide(ky1-ky0+(dy1*t1),dy0);
    xx0=kx0+(dx0*t0);
    yy0=ky0+(dy0*t0);
    // check if intersection exists
    if (fabs(xx1-xx0)<=1e-6) // intersection in both lines are the same
     if (fabs(yy1-yy0)<=1e-6)               
      if ((t1<0.0)||(t1>1.0)) // t1 is in <0,1> ... line bounds
       {
       if (t0<0.0) return "left pass";
       if (t0>1.0) return "right pass";
       return "middle pass";
       }
    return "no pass";

divide:

double divide(double x,double y) { if ((y>=-1e-30)&&(y<=+1e-30)) return 0.0; return x/y; }

[notes]

• you do not need normal (I first thought that it would be necessary but it isn't)
• intersection code is extracted from Finding holes in 2d point sets
• so if I forget to copy something there you can find it
• if your player does not go through the axis (instead goes below or above it)
• then you have to modify the intersections and/or project the players points to P0,P1,normal plane before computations

Answer 2

Add two more planes. One to the right and one to the left. Check if the player passes through the plane.

Answer 3

You could extend the plane so that the player will collide with it whichever side of the gates they go. Then check the X position of the player and compare it with the X position of the gates when the player collides with the plane.

Answer 4

I would put two trigger box colliders at both gates. Then I would keep track of which gate was passed through with a simple class that keeps track of this information.

public class Gate
{
   void OnTriggerEnter(Collider other) 
   {
      GateKeeper.setPassedGateCount(this);
   }
}

Then this GateKeeper class.

public static class GateKeeper
{
    private int count = 0;

    public static void setPassedGateCount(GateKeeper instance)
    {
        count++;
    }
}

The reason I pass the instance of the Gate is so that if you want to, you can check by name, for example, which gate was passed.