I'm stuck: my new I/O Particle-System uses adaptive iteration to calculate
the path for a particle, along with object interaction. Now, when Particles
would normally roll on a surface, they oscillate heavily in my System:
Gravity pulls them down, object interaction puts them back on surface, only
the speed perpendicular to the surface-normal is retained. Thus, a particle
won't slow down when rolling along plane, but it will "hop" on the surface
hundreds of times (two of my particles reached more than 2500 iterations
that way...).

To avoid this, I want a particle to switch to "Rolling" when enough
iterations have been done, and thus the particle will just get moved along
the surface without "real" interaction. Now, if I'd just be dealing with a
plane, that'd be no problem, but since particles may also roll uphill, I
want to calculate how much velocity they'd lose when doing that. I know the
angle of the surface, the initial velocity and the distance the particle
will travel, now what's the formulae to calculate how much velocity is lost?

For simplicities sake, use a ball with 10km/h that starts rolling a 45
degree upward slope for 10 meters. Gravity is 10m/s. What'd be the velocity
after 5 meters? What's the formula for that?

Links, direct answers, anything will do. I've tried googling, but it seems
I'm using the wrong keywords...

Regards,
Tim

-- 
"Tim Nikias v2.0"
Homepage: <http://www.nolights.de>
Email: tim.nikias (@) nolights.de

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> Ignoring friction and rotational energy, you'd simply be checking that
> your potential energy matches your kinetic energy, wouldn't you? The
> formulae for both are pretty straightforward.

Ehm, yeah, sure... Point is: I can't find those formulae. That aside, I just
want the velocity approach, not energies etc, cause I don't know anything
else of that particle beside it's velocity. Might be that I'm talking BS
because I don't know which formulae you're talking about, but, for example,
I've got a few formulae like "Height after object is thrown with that slope
and that velocity with earth's gravity". What I want is something along the
lines of "Velocity after object is rolled up/down a slope with that angle,
velocity and gravity".

I'm in a rush now, but I'll check on your energy-clues when I'm back, might
be that they'll lead me to exactly what I was looking for.

Regards,
Tim

-- 
"Tim Nikias v2.0"
Homepage: <http://www.nolights.de>
Email: tim.nikias (@) nolights.de

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<"Tim Nikias v2.0" <tim.nikias (@) nolights.de>> wrote:

> For simplicities sake, use a ball with 10km/h that starts rolling a 45
> degree upward slope for 10 meters. Gravity is 10m/s. What'd be the velocity
> after 5 meters? What's the formula for that?

1) Does the ball have an initial rotation?
2) Should it be completely rolling (or is partial sliding allowed)?
3) if 1=true and 2=false, is a direction of rotation that is not
   parallel to vcross(velocity,middleOfBall-pointWhereBallTouchesSurface)
   allowed?

If 3 is true, the solution can get quite complicated, I think.

Lutz-Peter

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Among other things, Tim Nikias v2.0 wrote:

>> Ignoring friction and rotational energy, you'd simply be checking that
>> your potential energy matches your kinetic energy, wouldn't you? The
>> formulae for both are pretty straightforward.
> 
> Ehm, yeah, sure... Point is: I can't find those formulae. That aside, I
> just want the velocity approach, not energies etc, cause I don't know
> anything else of that particle beside it's velocity. Might be that I'm
> talking BS because I don't know which formulae you're talking about, but,
> for example, I've got a few formulae like "Height after object is thrown
> with that slope and that velocity with earth's gravity". What I want is
> something along the lines of "Velocity after object is rolled up/down a
> slope with that angle, velocity and gravity".

Kinetic energy: 1/2 * m * v^2
Potential energy: g * m * h

Total energy must be constant.

If the particle has velocity v1 at height h1, then at height h2 it should 
have:

1/2*m*(v1)^2 + g*m*(h1) = 1/2*m*(v2)^2 + g*m*(h2)
1/2*m*(v2)^2 = 1/2*m*(v1)^2 - g*m*(h2-h1)
(v2)^2 = (v1)^2 - 2*g*(h2-h1)
v2 = sqrt[ (v1)^2 - 2*g*(h2-h1) ]

-- 
light_source{9+9*x,1}camera{orthographic look_at(1-y)/4angle 30location
9/4-z*4}light_source{-9*z,1}union{box{.9-z.1+x clipped_by{plane{2+y-4*x
0}}}box{z-y-.1.1+z}box{-.1.1+x}box{.1z-.1}pigment{rgb<.8.2,1>}}//Jellby

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Jellby <jel### [at] M-yahoocom> wrote:

> Kinetic energy: 1/2 * m * v^2
> Potential energy: g * m * h

This is the correct solution for a sliding (or slipping? 
what is the correct term?) particle, but not for a rolling ball.

Lutz-Peter

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You could use the law of conservation of energy if you want:

Ek = m*pow(v,2)
Ep= F*Height , F = m*a

E = Ek + Ep == constant

where m is a mass, v is the velocity, a is the gravity acceleration and at
any time E must be constant in a frictionless system. This way, you can
alway tell what should eb the speed at any point without simulating.

In practice, you may want to apply mechanics equation like this

v(n) = v(n-1) - at
pos(n) = pos(n-1) + vt + 0.5a*pow(t,2)

where t is the time step, a is the length of the horizontal component of the
gravity acceleration (cos A if A = 0 for horizontal plane) given the normal
of the surface.

If you dig in the vectorial aspect of Newtonian mechanics (High school level
I think), you should be able to solve your rolling problem by assigning a
matching vector component to the normal of the surface the particule is
rolling on in the first place.

Hope this helps, as well as the other posts sent while I was writing
this(...)

bongo

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Tim Nikias v2.0 wrote:
> I'm stuck: my new I/O Particle-System uses adaptive iteration to calcul
ate
> the path for a particle, along with object interaction. Now, when Parti
cles
> would normally roll on a surface, they oscillate heavily in my System:
> Gravity pulls them down, object interaction puts them back on surface, 
only
> the speed perpendicular to the surface-normal is retained. Thus, a part
icle
> won't slow down when rolling along plane, but it will "hop" on the surf
ace
> hundreds of times (two of my particles reached more than 2500 iteration
s
> that way...).
> 
> To avoid this, I want a particle to switch to "Rolling" when enough
> iterations have been done, and thus the particle will just get moved al
ong
> the surface without "real" interaction. Now, if I'd just be dealing wit
h a
> plane, that'd be no problem, but since particles may also roll uphill, 
I
> want to calculate how much velocity they'd lose when doing that. I know
 the
> angle of the surface, the initial velocity and the distance the particl
e
> will travel, now what's the formulae to calculate how much velocity is 
lost?

In a particle system you usually describe the particles as point masses 
so they won't roll (rolling means rotation and a point can't rotate). 
The effect you get (hopping) is pretty normal but it will diminish if 
you add damping (just imagine a glass marble falling on a carpet floor 
instead of a massive steel block).  Damping can be modelled in different 

ways - you can either continuously diminish the kinetic energy or you 
slow down the particles during every collision.

> For simplicities sake, use a ball with 10km/h that starts rolling a 45
> degree upward slope for 10 meters. Gravity is 10m/s. What'd be the velo
city
> after 5 meters? What's the formula for that?


y 
depends on the friction between the ball and the surface (if you really 
want it rolling you also have to take the moment of inertia into account)
.

Christoph

-- 
POV-Ray tutorials, include files, Sim-POV,
HCR-Edit and more: http://www.tu-bs.de/~y0013390/
Last updated 11 Jan. 2004 _____./\/^>_*_<^\/\.______

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I'm not sure I understand...

If, every time the particle is pulled down by gravity, it is brough back
up immediately (before the positions are represented on the screen),
then why is the hopping visible in the animation?

And if it isn't visible, then what is the reason that you want to avoid
it?

Rune
--
3D images and anims, include files, tutorials and more:
rune|vision:  http://runevision.com **updated Jan 29**
POV-Ray Ring: http://webring.povray.co.uk

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Among other things, Lutz-Peter Hooge wrote:

>> Kinetic energy: 1/2 * m * v^2
>> Potential energy: g * m * h
> 
> This is the correct solution for a sliding (or slipping?
> what is the correct term?) particle, but not for a rolling ball.

It's a quick "ad-hoc" solution I wrote assuming no friction and no 
rotational energy :)

-- 
light_source{9+9*x,1}camera{orthographic look_at(1-y)/4angle 30location
9/4-z*4}light_source{-9*z,1}union{box{.9-z.1+x clipped_by{plane{2+y-4*x
0}}}box{z-y-.1.1+z}box{-.1.1+x}box{.1z-.1}pigment{rgb<.8.2,1>}}//Jellby

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> It's a quick "ad-hoc" solution I wrote assuming no friction and no
> rotational energy :)

Which is what I'm after, thank you very much. It's too late now, gotta go to
bed, but I'll sort through that complex-looking formula and see what I can
do.

Regards,
Tim

-- 
"Tim Nikias v2.0"
Homepage: <http://www.nolights.de>
Email: tim.nikias (@) nolights.de

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