Warp wrote:
> triple_r <nomail@nomail> wrote:
>> In all fairness, there are better methods.  Think of dividing all the particles
>> up into an octree.  A whole group of particles at a distance can be lumped into
>> one mass.
> 
>   But you have to calculate the center of mass of the whole group, and
> since each element of the group can change independently of the others,
> that would mean you have to re-calculate this center of mass each time
> anything changes.
> 
>   Also another problem is that groups don't stay the same. If a particle
> not belonging to the group enters the "inside" of the group, it cannot
> be calculated anymore against that group as if it was just one single
> mass located at the center of mass of the group.

The point that accuracy requires the O(n^2) solution was rather blithely 
ignored, too.

If the forces involved follow the 1/(d^2) rule (such strange forces as, 
oh, gravity), then the non-linearity causes the averaging to yield a 
different total force than the sum of the forces exerted by the 
different masses.  For instance, if particle A is at distance x_a 
(relative to the affected point), and particle B is at distance x_b 
(likewise relative), then using an average yields a force of:

   2/(x_a/2 + x_b/2)^2, whereas the actual force is

   1/(x_a^2) + 1/(x_b^2) = (x_b*x_b)+(x_a*x_a))/(x_a^2 * x_b^2)

If x_a = 1 and x_b = 1.01, then the forces are:

   2/( 1.005^2) = 2/1.010025 = 1.9801490062127175069924011781887

whereas the other sum is

   2.0201/1.0201 =             1.9802960494069208901088128614842

and this is an application where an error of this magnitude most 
certainly does make a difference.

Unless you just want to make pretty pictures.

Regards,
John

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