OpenEuphoria

Re: Object movement..

• Posted by Joe Otto <jotto at NETZERO.NET> Jun 08, 1999
• 649 views

This sounds like a physics problem from way back when...

The only way the other objects can influence the primary object's velocity
is to exert a force on it.  Since you want the force to be inversely
proportional to distance, it should be k/d or even k/d^2 where k is a
constant associated with a specific object.  It may be worthwhile to
associate a mass with each object and use Newton's Law, F=GMm/d^2, to
calculate the force between the two objects.  The gravitational constant G
= 6.67 ? 10^-11 N m^2 kg^-2 would be the number to change to tweak how
strongly objects interact.

Remember though, the force between two objects is dependent on their mass
and distance - not on their velocity.

Once you've calculated the force on the primary object by all other
objects, you can add the force vectors to arrive at a net force.  Now,
since F=MA and you know F and M (the mass of the primary object), you can
rewrite it as A=F/M and figure out A, how much the force accelerates the
object.

Now you can use simple motion equations to calculate the new position and
velocity:

        Pnew=P+Vt+A(t^2)
        Vnew=V+At

If you arbitrarily make your time interval t=1, the multiplies fall right
out.  I'll leave the vectorization of these equations to you...  :)

Hope this helps...

Joe

-----Original Message-----
From:   Liquid-Nitrogen Software [SMTP:nitrogen_069 at HOTMAIL.COM]
Sent:   Monday, June 07, 1999 1:55 AM
To:     EUPHORIA at LISTSERV.MUOHIO.EDU
Subject:        Object movement..

Hi, anyone got any ideas on the following:

I have an object with a position {x,y} and velocity {x,y}

i also have other objects with position {x,y} these other objects attract
or
repell the first object, how do i caluclate a value to add to the first
objects velocity so that it will be pushed directly away from a repelling
object, or pulled toward an attracting object? The force should be less
further away from the object.

Thanks,
    -Mark.

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