RE: [Algorithms] Where is the enemy?

[Aldo . <al...@ho...>]

It works fine, now!
Thanks a lot, Steve.
And thanks to the other messages, too, guys!

Aldo



>From: Steve Wood <Ste...@im...>
>Reply-To: gda...@li...
>To: "'gda...@li...'"  
><gda...@li...>
>Subject: RE: [Algorithms] Where is the enemy?
>Date: Mon, 24 Jul 2000 10:48:44 -0700
>
>Come to think of it...the vector can be eliminated since it's in both
>equations...just this test need be done:
>
>if Px + Py + Pz > P1x + P1y + P1z then it's to the right
>if Px + Py + Pz < P1x + P1y + P1z then it's to the left
>if Px + Py + Pz = P1x + P1y + P1z then it's straight ahead
>
>R&R
>
> > From: Aldo . [mailto:al...@ho...]
> > How can I decide wich side must I turn?
> >
> > I've tried cross product, but it doesn't worked fine.
> >
> > Any sugestions?
> >
>
>Hmm, how about using the equation of a plane which has at it's normal a
>vector V pointing at a right angle to the up and forward vectors and
>positive to the right:
>
>VxP1x + VyP1y + VzP1z = C1
>
>Where C1 would be an index to compare with C in
>
>VxPx + VyPy + VzPz = C
>
>that tells you if point P is to the right (C>C1) or left (C<C1).
>
>R&R
>
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