Path: chuka.playstation.co.uk!scea!peter_alau@playstation.sony.com
From: shade@dragonshadow.com (Scott Cartier)
Newsgroups: scee.yaroze.programming.2d_graphics
Subject: Re: What do'you mean I have to brush up on me maths!
Date: Tue, 06 Jul 1999 16:53:52 GMT
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>Hi Rad,
>
>First find the distance between the two objects:
>
>    dx = x1 - x2
>    dy = y1 - y2
>
>Next find the length of the line between the two objects:
>
>    dxy = square_root( dx*dx + dy*dy )
>
>And now the vector:
>
>    vx = (dx * speed) / dxy
>    vy = (dy * speed) / dxy
>
>
>I think that'll do the trick.  Due to the integer maths, there will be
>accuracy problems if the speed is low, or if the two objects are very close,
>but this can be fixed by using fixed point if necessary.
>
>Herbs


You can avoid the squareroot by doing an inverse tangent instead.
Whether this is faster or not I'm not sure.  First, find dx & dy like
above.  Then find the angle of this line:

  Angle = inverse_tan(dy / dx)

Once you have the angle you can get the vx & vy components using sin
and cos:

  vx = cos(angle)
  vy = sin(angle)

I hope I didn't get those switched.

Use a look-up table for the tan, sin, and cos to speed things up.  The
inverse tangent takes a bit longer despite using a look-up table.
This may or may not actually be faster than doing the sqrt() as Alex
suggests.  That would be an interesting experiment (and one I probably
should have performed before blindly going with the inverse tan
method).

Scott