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'Particle' behaviour

  1. Dec 17, 2004 #1
    Hi, i am building a basic particle engine and i need to know what affects a 'particle' (a bit of dust for instance), i got these questions:

    What affects the rate at which the air slows down the particle when moving? It's volume and force (velocity*mass)?

    What affects gravity? It's volume?

    Are there any other important aspects that i should take into account?

    Thanks in advance.
  2. jcsd
  3. Dec 17, 2004 #2


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    I do not understand your question, please clarify.
    For starters, what is a "particle engine"?
  4. Dec 17, 2004 #3
    A particle engine is a computer program that can simulate the behaviour of particles (as used in special effects for water, smoke, dust and fire).

    To build such a program i need to know what affects the behaviour of these particles. I know all the intersection math i will need, i just need to know what physics i should take into account.

    The most obvious affection would be gravity, what aspect(s) of the particle (volume, mass, anything else?) affects the effect of gravity on it?

    I'd like to know what affects the effect of air too.

    I hope i am clear now.

    I think that's all i need to know for now, if i'm wrong please say so.
  5. Dec 17, 2004 #4


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    Gravity's effect comes only through the particle's mass.Shape or density is irrelevant.So you should take into account for each particle the force of gravity:[itex]F=mg [/itex].Of course,this in the simple case that the particles are not very heavy and very close together and that the gravitationa field of the earth is assumed constant.If not,then certain corrections need to be made.

    As for the friction with air,assume air is a Newtonian viscous fluid and that the particles are spherical with not a big radius and do not move very fast.Then u can use Stokes formula which gives the kinetic friction force:[itex] \vec{F}_{fr}=6\pi\eta r\vec{v} [/itex],where [itex] \eta [/itex] is one of the two viscosity coefficients (i don't know which,kinetic/dynamical),[itex] r [/itex] is the radius of the particle assumed sphere,and [itex] \vec{v} [/itex] is the particle's velocity vector.

  6. Dec 17, 2004 #5
    Ah ok, thanks.

    I don't know the term viscosity (at least, i don't know what it is in dutch), i found a translation and it's translated as stickyness, i think i will search for some viscosity tables to get a better idea of what it exactly means.

  7. Dec 17, 2004 #6
    I don't really understand the formula:

    You say it's:

    [itex] \vec{F}_{fr}=6\pi\eta r\vec{v} [/itex]

    So for aparticle with a radius of one cm in air ([tex]\eta[/tex]=0.15 cm[tex]^2 s^-1[/tex]) that would be:
    [tex] \vec{F}_{fr}=6\pi *0.15*1.0*\vec{v} [/tex]

    That doesn't seem right, did i do anything wrong?
    Last edited: Dec 17, 2004
  8. Dec 17, 2004 #7


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    Yep,but you have to put the velocity with CGS units,i.e.in cm/s.


    EDIT:The Stokes formula is with density (of the particle/sphere) instead of radius.U could have seen the units didn't match.Sorry. :blushing: I have't used it i decades.
    I believe now it's okay.
  9. Dec 17, 2004 #8
    Thanks, np.

    So that means particles of the same density, no matter what size, slow down at the same rate?
  10. Dec 17, 2004 #9


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    No,it's radius after all.I'm sorry,i have't worked with fluid dynamics in a while and i have forgotten that they use other convention for forces.I believe Clausius can tell you more about Stokes force.

    Anyway,here's some help.

  11. Dec 17, 2004 #10
    Thanks, that helped, i still got a question though, the final equation is:

    6*PI*viscosity*radius* 'free stream velocity'

    How can it be that the particle mass isn't used, that would mean any particle (paper or rock) will go through just as easy. I don't know what the 'free stream velocity' really is so that could be my problem, what does it mean?
  12. Dec 17, 2004 #11


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    I guess it's the velocity of the particle in the air assumed motionless.There were two formulas on the page.One for a motionless particle and moving fluid,and one for a moving particle in static (viscous) fluid.I guess you can make the symplifying assumption that the fluid is motionless.In the end,it's tha particles behavior that you're interested in,right??
    Apparently for a RIGID sphere only size matters in computing the force.It's independent of mass or density,but rather of volume through the sphere's radius.

  13. Dec 18, 2004 #12
    Ok thanks, i got these questions left:

    As a reference, what is the terminal velocity of a particle of sand in the air?

    And, in <a href="http://scienceworld.wolfram.com/physics/TerminalVelocity.html">this [Broken] article</a> there is/are the costant(s) C and D or CD, what are they?

    Last edited by a moderator: May 1, 2017
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