On Mon, 27 May 2002 11:43:41 +0200, "Rune"
<run### [at] mobilixnetdk> wrote:

<snip>

>So how should energy loss be applied? One Idea I had was that the energy
>loss should only apply to that part of the movement vector that is
>perpendicular to the surface the particle collides against. ( See
>illustration II in the before-mentioned image. ) But this would mean
>that the outcoming angle would not be equal to the incoming angle. In
>reality is the incoming angle always equal to the outcoming angle, also
>when energy loss is taken into consideration?
>
>A third option would be to still use the same model as in illustration
>I, but vary the energy loss so that it is greatest when the
>incoming=outcoming angle is close to 90 degrees and so that there's
>almost no energy loss when the incoming=outcoming angle is close to 0
>degrees.
>

<snip>


Hi Rune,

I think the first paragraph sounds about right to me (although school
physics lessons were a long time ago 8o) i.e. outcoming=incoming angle
although I guess that would only hold for non-spinning particles?
(Ever thrown a backwards-spinning rubber ball away from you only to
find it bounces right back?)

Hmm, that reminds me ... won't the energy loss be related to the
material's ... ummm ... elasticity? (not sure if that's the right
word) e.g. a particle system of rubber balls will probably lose less
energy per bounce than a system of ball bearings?

I assume you're taking into account some kind of drag coefficient?
(Crikey, that almost sounds like I know what I'm talking about!) So
that even if a particle has an incoming angle close to 0 degrees it
will still be losing energy by the drag of the 'atmosphere' it's
travelling through? (Hey wouldn't it be cool if your particle system
were tied into POV's media!! ... Ooh! Ooh! How about an explosion
under water!! ... Ooh! Ooh! Maybe you could get your particle system
into POV4 ...  <calm down Scott ... one step at a time>

Great explosion animation on your webpage btw!

Cheers,

Scott

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