So far in my particle system I have used a simple collision model where
the outcoming angle is equal to the incoming angle and the outcoming
velocity is smaller than the incoming velocity due to the energy loss at
the collision. ( See my illustration I in povray.binaries.images in the
message "Collision with energy loss" ) But I wonder if that is a very
good model to use...

The energy loss is often quite big, like 80%, because many things when
they bounce only reach a small fraction of the height they were dropped
from.

When the incoming angle is close to 90 degrees the model can look quite
fine. But imagine that the incoming angle is very small. Then the energy
loss of 80% will seem very unnatural, like the particle slows down to
20% speed for no reason.

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.

What are your thoughts on this?

Collisions in the real world are probably way more complicated, but I'm
looking for a very simplistic model which is still as realistic as
possible...

Rune
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Hi Rune,

Shouldn't the energy loss be based upon the angle of collision. ie if the
angle is 90 degrees, then maximum energy loss occurs. If the angle is
shallow i.e the particle is moving almost parallel to a surface and touches
that surface, the resulting slowdown will be small (things like the
surface's friction will come into play here - the particle may slide along
the surface)

After typing that, I re-read your post and realised that the above was your
third option! Third option gets my vote then ;-)

All the best,

Andy Cocker

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"Rune" <run### [at] mobilixnetdk> wrote in message
news:3cf200ad@news.povray.org...
> So far in my particle system I have used a simple collision model where
> the outcoming angle is equal to the incoming angle and the outcoming
> velocity is smaller than the incoming velocity due to the energy loss at
> the collision. ( See my illustration I in povray.binaries.images in the
> message "Collision with energy loss" ) But I wonder if that is a very
> good model to use...

When an object collides at angle, only the portion of energy perpendicular
to the collision vector has an effect, while the portion parallel has no
effect (discounting friction, but that can easily be applied if so desired)

The best way to do this is to split your equations into x, y and z vectors,
do your conservation and then do your conservation of energy equation and
then use the resulting vectors.

I don't have the equations and such off hand right now but I can get them
later if you would like.

-tgq

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Ok, here's why I'm not satisfied with the third option I proposed.

I want to simulate not just bouncing but also sliding/flowing of
particles. I define flowing as when the particle moves perpendicular
with the surface or almost perpendicular.

When a particle hits a surface with a big incoming angle, it should
bounce, but after some bounces it should eventually begin to slide (or
flow if you want). But if the outcoming angle is always perpendicular to
the incoming angle, then the sliding will never occur, and the particle
will just continue making smaller and smaller bounces until the bounces
get insignificantly small. That is not realistic.

Again see the thread in povray.binaries.images called "Collision with
energy loss", where I have posted a second image.

So somehow the outcoming angle should become a little smaller for each
bounce, but I don't know according to which rules. My second proposal
was a speculation in this direction, but it doesn't seem plausible,
because it would make the outcoming angle be completely different from
the incoming angle, which is not realistic either...

Any ideas how the outcoming (reflected) movement vector should be
calculated?

Rune
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TinCanMan wrote:
> When an object collides at angle, only the portion
> of energy perpendicular to the collision vector has
> an effect, while the portion parallel has no effect
> (discounting friction, but that can easily be
> applied if so desired)

That sounds similar to my second idea. It just seems to me that this
would cause the outcoming angle to be very different from the incoming
angle, but maybe that's indeed the case...

> The best way to do this is to split your equations
> into x, y and z vectors, do your conservation and
> then do your conservation of energy equation and
> then use the resulting vectors.
>
> I don't have the equations and such off hand right
> now but I can get them later if you would like.

I would like that very much. :)
I just hope I'll be able to understand them...

Rune
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> That sounds similar to my second idea. It just seems to me that this
> would cause the outcoming angle to be very different from the incoming
> angle, but maybe that's indeed the case...

Actually the *magic* of math and physics resolves everything very easily.
My code is in an old version of visual basic which I don't have here at work
so I'll try to clean it up and post it when I get home.

-tgq

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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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On Mon, 27 May 2002 14:39:07 +0200, "Rune"
<run### [at] mobilixnetdk> wrote:

<snip>
>
>When a particle hits a surface with a big incoming angle, it should
>bounce, but after some bounces it should eventually begin to slide (or
>flow if you want). But if the outcoming angle is always perpendicular to
>the incoming angle, then the sliding will never occur, and the particle
>will just continue making smaller and smaller bounces until the bounces
>get insignificantly small. That is not realistic.
>
<snip>

I reckon the smaller and smaller bounces idea is correct - if you've
ever seen a small ball bearing bounce on a horizontal piece of glass
it seems to bounce many many times getting smaller and smaller until
it stops bouncing. Also, think about those Newton's Cradle executive
toys ... you know, the ones with 5 or 6 ball bearing suspended in a
line - they bounce for ages.

Scott

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Scott Moore wrote:
> I reckon the smaller and smaller bounces
> idea is correct

But this will never lead to sliding or flowing I think...

> - if you've ever seen a small ball bearing
> bounce on a horizontal piece of glass it
> seems to bounce many many times getting
> smaller and smaller until it stops bouncing.

But that's not the point here. If I have understood you correctly, ball
bearings don't have freedom of movement so it doesn't make sense to talk
about the incoming and outcoming angle... :/

You're right though that the bouncing often continues many many times,
but after the last tiny bounce the ball or particle should still roll or
flow or slide or whatever you want to call it. Did you see that image in
povray.binaries.images I mentioned?

Rune
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In article <3cf223fa$1@news.povray.org>, Tin### [at] hotmailcom 
says...

> When an object collides at angle, only the portion of energy perpendicular
> to the collision vector has an effect,

Energy can't be perpendicular to anything, because it isn't a vector.
What you mean is velocity.

The reason why a ball loses energy is because of friction inside it, and 
the amount of friction is dependent on the deformation, wich is dependent 
on the acceleration/change of velocity.

Probably this friction mainly (maybe only) slows down the velocity 
perpendicular to the surface, but if the ball has a velocity-component 
parallel to the surface it will slide (or roll) on it while in contact 
with it, resulting in friction between the surface and the ball wich 
slows it down in this direction, too.

How about letting the user specify the ratio between horizontal and 
vertical friction?

Lutz-Peter

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Rune wrote:
> 
> So far in my particle system I have used a simple collision model where
> the outcoming angle is equal to the incoming angle and the outcoming
> velocity is smaller than the incoming velocity due to the energy loss at
> the collision. ( See my illustration I in povray.binaries.images in the
> message "Collision with energy loss" ) But I wonder if that is a very
> good model to use...
> 
> [...]

As you probably know my own physics simulations use an elasticity based
method to calculate collisions.  Therefore math is probably quite
different although there are similarities of course.  

As Scott said, the collision damping is strongly influenced by the
material properties of the colliding objects, but not necessarily only by
the elasticity.  A well known example for very few energy loss is a glass
ball on a massive metal surface, but a rubber ball (quite high elasticity)
looses much less enery than a wood ball for example.

The source of collision damping is the dissipation of energy during
deformation of objects.  Therefore you are right that shallow collisions
result in less energy loss, but you will have to consider friction and
rolling (although rolling is something not considered in a particle
system by definition since the particles have no size and therefore no
moment of inertia).

I'm not totally sure about the direction of the damping force, the method
i use in my system is having a viscous damper (i.e. force proportional to
the speed) parallel to the collision elasticity which is orthogonal to the
surface.  This is a common method, but i have to say i'm not always
content with the results of my system.  In general i would say daming
forces should always be against the direction of current speed which would
result in both an orthogonal and tangential damping component.

Christoph

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There's more to bouncing than just a simple decrease in vertical/horizontal
speed. In the case of a ball, for example, bouncing also gives the ball a
rotation (if the collision angle isn't 90 degrees). When the ball stops
bouncing, the ball's rotational momentum keeps it rolling.

Anders

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> Energy can't be perpendicular to anything, because it isn't a vector.
> What you mean is velocity.
>
> The reason why a ball loses energy is because of friction inside it, and
> the amount of friction is dependent on the deformation, wich is dependent
> on the acceleration/change of velocity.
>
> Probably this friction mainly (maybe only) slows down the velocity
> perpendicular to the surface, but if the ball has a velocity-component
> parallel to the surface it will slide (or roll) on it while in contact
> with it, resulting in friction between the surface and the ball wich
> slows it down in this direction, too.
>
> How about letting the user specify the ratio between horizontal and
> vertical friction?
>

Yes, this is all true, I was just trying to simplify the matter without
getting too technical.  By energy i really meant (and should have said)
momentum.  The amount of energy lost is dependent on the plasticity of both
objects involved in the collision, for simplistic purposes you can express
this as a percentage.  I also like to assign energy loss in two parts,
elastic and plastic, (maybe not the most correct terms) with elastic energy
loss being a percent and plastic being an absolute value (this keeps the
ball from bouncing forever as it would if you used a pure percentage). If
you want to use friction, I would apply it separate (not a ratio) as it
would depend on the material of the two impacting objects, and it would
apply to the parallel vectors.

Also, I have noticed I misstepped a bit in my initial explanation.  When
resolving the momentum vectors into x y z components, these are according to
a local frame of reference aligned to the plane of impact, not the global x
y z, so I suppose I should call them u v w instead, with u being the
perpedicular vector and v and w being (perpendicular) vectors lying on the
plane of impact.

Let me know if further clarification is needed.  I will explain a bit more
later when I rehash my code.

-tgq

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On Mon, 27 May 2002 15:54:13 +0200, "Rune"
<run### [at] mobilixnetdk> wrote:

>But that's not the point here. If I have understood you correctly, ball
>bearings don't have freedom of movement so it doesn't make sense to talk
>about the incoming and outcoming angle... :/
>
>You're right though that the bouncing often continues many many times,
>but after the last tiny bounce the ball or particle should still roll or
>flow or slide or whatever you want to call it. Did you see that image in
>povray.binaries.images I mentioned?
>
>Rune

Actually, I hadn't looked at the images - have now though 8o)

If you're trying to keep things simple then maybe you could treat the
vertical and horizontal energy loss coefficients separately (maybe
coefficients isn't the right word but I'll use it anyway!) By vertical
I mean perpendicular to the tangent of the collision point and by
horizontal I mean the tangent of the collision point. The horizontal
energy loss is handled by the material's friction coefficient and the
vertical energy loss is handled by the elasticity coefficient. When
either component gets below a certain threshold you can consider that
component to have zero energy - this could stop the (near) infinite
smaller bounces problem?

Basically, with this you will be able to get the sliding to come to a
half after the bouncing has stopped.

Don't take any of this too seriously - I am probably way of base here
... just a bunch of ideas ...

Bye,

Scott

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Scott Moore wrote:
> If you're trying to keep things simple then maybe you could treat the
> vertical and horizontal energy loss coefficients separately (maybe
> coefficients isn't the right word but I'll use it anyway!) By vertical
> I mean perpendicular to the tangent of the collision point and by
> horizontal I mean the tangent of the collision point. The horizontal
> energy loss is handled by the material's friction coefficient and the
> vertical energy loss is handled by the elasticity coefficient.

Yes, this is what some others have suggested too, and I think I'll use
this approach. :)

> When either component gets below a certain threshold you can
> consider that component to have zero energy - this could stop
> the (near) infinite smaller bounces problem?

Well, the infinite bouncing problem shouldn't be a problem at all when
the perpendicular loss of momentum is greater than the tangential,
because the bounces will become more and more flat, thus being the same
as sliding in effect. A threshold is what I want to avoid, because it
means one setting less for the user to worry about. :)

Thanks for the feedback. :)

Rune
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Rune wrote:

> Scott Moore wrote:
> > If you're trying to keep things simple then maybe you could treat the
> > vertical and horizontal energy loss coefficients separately (maybe
> > coefficients isn't the right word but I'll use it anyway!) By vertical
> > I mean perpendicular to the tangent of the collision point and by
> > horizontal I mean the tangent of the collision point. The horizontal
> > energy loss is handled by the material's friction coefficient and the
> > vertical energy loss is handled by the elasticity coefficient.
>
> Yes, this is what some others have suggested too, and I think I'll use
> this approach. :)

This is the approach I took for my system and it works ok. Good luck with it.

MJL

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Lutz-Peter wrote:
> How about letting the user specify the ratio
> between horizontal and vertical friction?

That seems to be what people generally suggest. :)
Yes, I've already recoded my system to use this method, and the first
tests look promising.

TinCanMan wrote:
> I also like to assign energy loss in two parts,

> plastic being an absolute value (this keeps the
> ball from bouncing forever as it would if you
> used a pure percentage).

I don't think it matters that the particles keep bouncing forever, when
the bounces get flatter and flatter, so that it is in effect the same as
sliding.

> Also, I have noticed I misstepped a bit in my
> initial explanation. When resolving the momentum
> vectors into x y z components, these are according
> to a local frame of reference aligned to the plane
> of impact, not the global x y z, so I suppose I
> should call them u v w instead, with u being the
> perpedicular vector and v and w being
> (perpendicular) vectors lying on the plane of impact.

That's what I assumed, and I think that only two vectors are needed for
this, the second being the tangential vector.

Thanks for the feedback everyone! :)

Rune
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...
> 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?

This is IMHO the best option, and the most physically based.
I've implemented in the past a particle system, and my bounces followed this
law.
A damping factor could be applied to the tangential component of the
movement factor to take into a first account friction between surfaces. A
more complex model would modulate this tangential damping factor according
to the angle of incidence as in your third option.

I found a lot of inspiration in the particle system library by David
McAllister found at www.cs.unc.edu/~davemc/Particle/, I suggest you to dig
into the source of that API which is available for free.

Alex

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

>What are your thoughts on this?

Rune,

sorry I don't have the time to read the whole thread right now so I
apologize in advance if what I write is nothing new.

There are two laws which you should observe. One is the law of
conservation of energy and the other is the law of conservation of
momentum.

Now, I know you know these laws, but I'll write them down to make my
points more clear.

The law of conservation of energy states that:

E_b,t = sum(E_b,k) + sum(E_b,p) = E_a,t = sum(E_a,k) + sum(E_a,p) + Q

where the index "b" indicates the time before collision and the index
"a" indicates the time after the collision. Also, "k" is for kinetic,
"p" is for potential, "t" stands for "total" and "Q" is heat produced
in the collision (you may also add m*c^2 to the right hand side of the
equation if there is some mass converted into energy, as in the
collision of several close-to-critic-mass pieces or plutonium... but
that's probably overkill). The sums indicate summing of the respective
energy components of all bodies.

Kinetic energy E_k comes from two components - linear and angular.
They are equal to m*[v]^2/2 and J*omega^2/2, respectively, where m is
mass, v is linear velocity, J is moment of inertia about the rotation
axis, and omega is angular velocity. Brackets indicate vector
quantities.

The law of conservation of momentum states that:

[P_b,t] = sum([P_b,r]) + sum([P_b,l]) = [P_a,t] = sum([P_a,r]) +
sum([P_a,l])

where r stands for "rotational" (should be angular, but 'a' is used
already) and "l" stands for linear. Moreover,

P_l = m*[v]
P_a = J*omega

The last thing which should be mentioned is that reaction forces are
always normal to the common tangent plane of the two objects at the
point of collision.

The most simple collision model is the completely elastic collision
without rotation. The only things you have to observe are linear
motion kinetic energy and momentum. In partially inelastic collisions
without rotation, you have to define the loss either as a loss of
momentum or as a loss of energy (they are linked together through
linear velocity).

If you throw in rotation, you'll have to account for the conservation
of angular momentum and the kinetic energy of rotational motion.
Things get more dicey in this case, but allow for very cool
simulations such as a rubber ball bouncing around with a top- or side
spin, or a pool table simulation. Again, if you want to have inelastic
collisions, you have to define the loss as either loss of momentum or
of energy.

Potential energy doesn't make much sense in collision models except in
very special cases such as prolonged (i.e. not instantaneous)
collision with material deformation, for example a ball falling on a
stretched rubber sheet. In this case the kinetic energy of the ball is
converted into potential energy of deformation of the material, and
after reaching zero, the reverse process begins. The main losses here
are due to internal friction loss in the deformed material, usually
represented using a viscuous friction damping model in either a
mass-spring (iterative) or a finite element (matrix) model. Things are
*really* dicey here and even the real pros (David Baraff, Andrew
Witkin, James O'Brien etc.) don't have all the answers.

Losses can also occur due to friction. The losses then are best
expressed in terms of a friction coefficient, because you're summing
up forces anyway. The friction force is parallel to the common tangent
and is proportional to the friction coefficient and the normal
component of the forces of motion,

[F_fr] = -k_fr * |[F_n]| * F_t / |[F_t]|

where the bars indicate length, "fr" is "friction", "t" is
"tangential" and "n" is "normal."

How friction affects angular motion, my high-school physics does not
tell. I only know Euler's friction between two bodies with a common
axis and contacting along a circular arc,

F_fr = -e^(k * phi)

where "k" is the coefficient of friction and "phi" is the angle of the
arc along which the two bodies contact. IIRC the formula indicates
both static and dynamic friction and is most commonly used when
dealing with ropes or belts contacting with drums, as well as with
hinges.

Enough of high-school physics :) About the only thing I want to
mention, which should be obvious since you already have a working
system, is that there is a shortcut in calculations in the case of a
finite-mass body colliding with an infinite-mass body.

Hope this helps. Sorry again if this is no news or has been mentioned.


Peter Popov ICQ : 15002700
Personal e-mail : pet### [at] vipbg
TAG      e-mail : pet### [at] tagpovrayorg

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Christoph Hormann wrote:

> 
> rolling (although rolling is something not considered in a particle
> system by definition since the particles have no size and therefore no
> moment of inertia).

What if you want to program a 10-dimensional p-brane system?  <g>

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The collision only robs the ball of vertical momentum.  Since the normal 
force is (by definition) vertical, it cannot change the ball's 
horizontal momentum.  Any horizontal change comes from friction or 
conversion to rotational energy, which doesn't exist in a particle 
system anyway.

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