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Need the 2D equation for calculateing the collision with stright lines at differant angles

Posted on 2004-10-06
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Hello Experts,

I Need the 2D equation for calculateing the collision with stright lines at differant angles.(makeing a game in shockwave)
I will explain in a diagram.....

X
                         


                                      Y

Ok do you see X and Y, imagine there is a line joining them,
I need an equation or equations to calculate if a point hits the line.
Idealy if this equation would only work on the line width height and roataion it wolud be cool , however I would be very thankful just for the equation.

Also if you could throw in somemore colision detection for more roatated shapes like roatated rectangles, that would be good.
k thanks

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Question by:FatNewbie
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jimbobmcgee earned 500 total points
ID: 12286988
You have a line, on a grid (the screen co-ords) between point A and point B.  You want to know when point P will collide with this line.  

Bear in mind that the collision between point P and line AB will only occur in two spots along that line.  The x coord of one collision will be the same as the x coord of P (Px).  The y coord of the second collision will be the same as the y coord of P (Py).  As both the point and the line moves, these coords will chage, but the relationship remains the same: The collisions will occur at (Px, Uy) and (Ux, Py), where U is an unknown.  U can be thought of, now, as a coord, U(x,y), which must be solved.

You first need the gradient of that line:

    m = difference in y / difference in x
    m = (By - Ay) / (Bx - Ax)

Then, to get Ux, you would run through all the points between Ay and Py, adding the gradient, m, to an initial value of Ax (example in BASIC):

    Ux = Ax
    For i = Ay to Py
        Ux = Ux + m
    Next i

As you should see, this is the same as multiplying the difference between Ay and Py by m and adding the result to Ax.  This gives the following formula:

    Ux = Ax + (Py - Ay)m

or, as a coded equation:

    Ux = Ax + ((Py - Ay) * m)

You can now adapt this function to find Uy:

                             Px = Ax + (Uy - Ay)m
                      Px - Ax = (Uy - Ay)m  
              (Px - Ax) / m = Uy - Ay
    ((Px - Ax) / m) + Ay = Uy

Therefore, in longhand form:

   Collision #1 = [Px, (Ax + ((Py - Ay) * ((By - Ay)/(Bx - Ax)))]
   Collision #2 = [((Px - Ax) / ((By - Ay) / (Bx - Ax)) + Ay), Py]

Now all you need to do is clean up the code a little:

   If Px is not within the boundaries of AB, Collision #1 will never occur
   If Py is not within the boundaries of AB, Collision #2 will never occur
   If By = Ay then the gradient calculation will cause a divide by zero error; get around this
        by making Ay = Ay + 0.001 (it will cause slight errors in your results but it should not
        be noticable on screen).

Hope this helps you get started.  

J.

(NB:  Depending on the screen coord layout, you may find you have to swap/invert the method for finding the gradient,  e.g., instead of m = (By - Ay) / (Bx - Ax):

    m =  (Bx - Ax) / (By - Ay)     or
    m = -(By - Ay) / (Bx - Ax)     or
    m = -(Bx - Ax) / (By - Ay)     etc

Use some trial and error to get this to work to your liking)
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LVL 16

Expert Comment

by:jimbobmcgee
ID: 12287075
I wrote the following in QBASIC (for speed, you understand!!) to backup the theory I posted above.  I hope you can follow it

      SCREEN 13

      TYPE coord
              x       AS DOUBLE
              y       AS DOUBLE
      END TYPE

      DIM A AS coord                  'LinePointer
      DIM B AS coord                  'LinePointer
      DIM P AS coord                  'Point
      DIM C AS coord                  'Collision
      DIM M AS DOUBLE                 'Gradient

      A.x = 300:      A.y = 10
      B.x = 100:      B.y = 70
      P.x = 230:      P.y = 30

      LINE (A.x, A.y)-(B.x, B.y), 14
      CIRCLE (P.x, P.y), 1, 4

      M = (B.x - A.x) / (B.y - A.y)

      C.x = A.x + ((P.y - A.y) * M)
      C.y = A.y + ((P.x - A.x) / M)

      LINE (P.x, P.y)-(P.x, C.y), 12
      LINE (P.x, P.y)-(C.x, P.y), 12

If you can run it, see what it does and if it helps you...

J
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LVL 16

Expert Comment

by:jimbobmcgee
ID: 12287275
If you _do_ manage to run the above code, try changing the values of A.x, A.y, B.x, B.y, P.x and P.y...

J.
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