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Rune wrote:
>
> How can I simulate air resistance to the movement of an object? If the
> object is moving with the velocity V, what will happen to the velocity (or
> speed) over time? In which way will it decrease? Linearly, exponentially, or
> something different?
Well, a few years ago, some english chap name Newton came up with the
silly idea that a force apllied to an object resulted in a acceleration
proportional to its mass, of f=ma for short.
the force caused by air resistence is computed by the following formula:
F = (Cd)A(V^2) in the direction opposing the movement.
Where:
- Cd is the drag coefficient ( 1 for a box, 0.5 for a sphere and around
0.30 for most modern cars)
- A is the cross-sectional area of the object (area "facing" the air
flow)
- V is the speed of the object in that direction.
Provided that the shape and mass of your object dont change, you can see
that the force, hence the decelleration, will decrease at a rate which
is proportional to the square of the current speed.
>
> Just to check if my general understanding of the terms is correct: I'd say
> that gravity increases the speed of an object linearly (assuming the object
> is falling downwards). Is that correct?
For all intents and purposes, let's say that the force of gravitation is
constant (the mass and size of your object are negligeable compared to
the mass and size of the Earth) and that the mass of your object is also
constant, hence the acceleration will be constant. On Earth, it is 9.81
m/s^2 (or 32.2 ft/s^2 for the metric-impaired)
So yes, the downward velocity of your falling object will increase
linearly.
--
Francois Labreque | //\\ Wear an ASCII ribbon!
flabreque | || ||
@ | \\// Support the campain
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Francois Labreque wrote:
>
> Well, a few years ago, some english chap name Newton came up with the
> silly idea that a force apllied to an object resulted in a acceleration
> proportional to its mass, of f=ma for short.
>
> the force caused by air resistence is computed by the following formula:
>
> F = (Cd)A(V^2) in the direction opposing the movement.
>
> Where:
> - Cd is the drag coefficient ( 1 for a box, 0.5 for a sphere and around
> 0.30 for most modern cars)
> - A is the cross-sectional area of the object (area "facing" the air
> flow)
> - V is the speed of the object in that direction.
>
> Provided that the shape and mass of your object dont change, you can see
> that the force, hence the decelleration, will decrease at a rate which
> is proportional to the square of the current speed.
>
You should mention, that Cd is in fact not a constant but is a highly
complicated function of the reynolds number and thereby also depends on
the speed. Values given for cars etc. are usually only valid for a
limited range of speed.
BTW with some shapes (like the wings of a plane) it is even more
difficult, because the force no more only goes in direction opposite to
the air flow.
Christoph
--
Christoph Hormann <chr### [at] gmx de>
IsoWood include, radiosity tutorial, TransSkin and other
things on: http://www.schunter.etc.tu-bs.de/~chris/
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You have to take into account that air resistance can cause uneven forces
in different parts of the object (for example if the density of the object is
not constant along its volumen), which can cause the object to rotate.
--
char*i="b[7FK@`3NB6>B:b3O6>:B:b3O6><`3:;8:6f733:>::b?7B>:>^B>C73;S1";
main(_,c,m){for(m=32;c=*i++-49;c&m?puts(""):m)for(_=(
c/4)&7;putchar(m),_--?m:(_=(1<<(c&3))-1,(m^=3)&3););} /*- Warp -*/
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"Greg M. Johnson" wrote:
> In my systems,
> #declare velocity=velocity-0.1*y;
> is how I usually do gravity.
IMHO I think my advice is the best here for frame-by-frame pov coding of gravity.
> #declare velocity=velocity-dragconstant*velocity^2;
> where dragconstant=0.005; //for starters
Oops! an error above.
// for a frame-by-frame system, experiment with small numbers for drag; toss
textbook values in dustbin.
#declare dragconstant=0.005;
// Wind is some varying vector.
#declare wind=foo;
// Wind blowing at craft means double effect; blowing with craft negates
effect.
#declare netdirection=wind-velocity;
// I forgot that velocity was a vector in last version I shared.
#declare
velocity=velocity+vnormalize(netdirection)*dragconstant*vlength(netdirection)^2;
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Warp wrote:
>
> You have to take into account that air resistance can cause uneven forces
> in different parts of the object (for example if the density of the object is
> not constant along its volumen), which can cause the object to rotate.
>
It's got nothing to do with wind resistence per se, but with the uneven
density of the object. the Force IS constant over the area, but since
the density isn't, the decelleration won't be, so you have to divide you
surface into smaller subsurfaces and divide the force applied to each
individual subsurface to take into account the change of density.
Of course, if your shape starts to rotate, its cross-sectional area will
also change, hence the Cd will also chance and the wind resistence will
also change.
Welcome to the fun world of finite element analysis.
> --
> char*i="b[7FK@`3NB6>B:b3O6>:B:b3O6><`3:;8:6f733:>::b?7B>:>^B>C73;S1";
> main(_,c,m){for(m=32;c=*i++-49;c&m?puts(""):m)for(_=(
> c/4)&7;putchar(m),_--?m:(_=(1<<(c&3))-1,(m^=3)&3););} /*- Warp -*/
--
Francois Labreque | //\\ Wear an ASCII ribbon!
flabreque | || ||
@ | \\// Support the campain
videotron.ca \\ against HTML e-mail
//\\ and news!
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Thank you all! Your replies were all helpful.
I needed just a simple approximation, nothing complicated. And what I need
to simulate it on is not really objects, but different kinds of particles
such as smoke, water drops, sparks and other things a particle system could
be used for.
I suppose that doesn't change anything; the basic drag formula should still
work the same, right? It's just a matter of finding the right drag
coefficient, drag constant or whatever you want to call it.
Rune
--
\ Include files, tutorials, 3D images, raytracing jokes,
/ The POV Desktop Theme, and The POV-Ray Logo Contest can
\ all be found at http://rsj.mobilixnet.dk (updated January 28)
/ Also visit http://www.povrayusers.org
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Francois Labreque <fla### [at] videotronca> wrote:
: It's got nothing to do with wind resistence per se, but with the uneven
: density of the object.
The density of the object can be constant thorough its volume, but the
shape of the object can affect the air resistance so that it makes the
object rotate.
A windmill can have a constant density. It rotates. :)
--
char*i="b[7FK@`3NB6>B:b3O6>:B:b3O6><`3:;8:6f733:>::b?7B>:>^B>C73;S1";
main(_,c,m){for(m=32;c=*i++-49;c&m?puts(""):m)for(_=(
c/4)&7;putchar(m),_--?m:(_=(1<<(c&3))-1,(m^=3)&3););} /*- Warp -*/
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Warp wrote:
>
> Francois Labreque <fla### [at] videotronca> wrote:
> : It's got nothing to do with wind resistence per se, but with the uneven
> : density of the object.
>
> The density of the object can be constant thorough its volume, but the
> shape of the object can affect the air resistance so that it makes the
> object rotate.
> A windmill can have a constant density. It rotates. :)
Yes, but that's not what you said. you said "(for example if the
density of the object is
not constant along its volume)".
--
Francois Labreque | //\\ Wear an ASCII ribbon!
flabreque | || ||
@ | \\// Support the campain
videotron.ca \\ against HTML e-mail
//\\ and news!
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Francois Labreque <fla### [at] videotronca> wrote:
: Yes, but that's not what you said. you said "(for example if the
: density of the object is
: not constant along its volume)".
It was one example where the force of the air resistance is not constant.
There are other examples as well.
--
char*i="b[7FK@`3NB6>B:b3O6>:B:b3O6><`3:;8:6f733:>::b?7B>:>^B>C73;S1";
main(_,c,m){for(m=32;c=*i++-49;c&m?puts(""):m)for(_=(
c/4)&7;putchar(m),_--?m:(_=(1<<(c&3))-1,(m^=3)&3););} /*- Warp -*/
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I believe the more precise concept is "specific surface area," defined in units
like m^2/g.
CASE ONE: dumbells with one sphere made of W (19 g/cc), the other of Al (3?
g/cc).
Of course the wind is going to push the less dense sphere more than the other.
Either description (uneven density or uneven specific surface area) are equally
good at predicting this effect.
CASE TWO: dumbells made of tungsten, one end is a sphere, the other end is a
mace with long spikes; each end's mass is equal.
Of course the wind is going to push the higher surface area end more than the
other.
Here, the superiority of the term "specific surface area" over "density" wins
out.
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