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Need help with deriving air drag equation

  1. Nov 11, 2014 #1
    △Work = △KE + △PE
    Work = Force x Distance
    W=F△x Pressure = Force/Area
    W=PA△x Volume=Area x length
    W=PV Work = Pressure x Volume

    KE=(1/2)mv^2
    Density = mass/volume
    mass = density x volume m=DV
    KE=(1/2)DVv^2 (work done by air pressure?)

    PE=mgy
    PE=DVgy (work done by gravity and air pressure??)

    PV=(1/2)DVv^2 + DVgy
    P=(1/2)Dv^2 + Dgy
    F=PA
    F=(1/2)Dv^2A + DgyA (final equation?)   

    Final equation I was expecting was F= ½DCAv2
    I'm so confused right now.
    can you guys help me fix the equation?
     
  2. jcsd
  3. Nov 13, 2014 #2
    You cannot strictly "derive" the V^2 drag law from something like F = m*a. It is largely an empirical result. You can show that it works by dimensional analysis, but the drag coefficient is still an empirical value.
     
  4. Nov 13, 2014 #3

    A.T.

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    Not strictly, but the ~v2 relationship is based on momentum conservation.
     
  5. Nov 13, 2014 #4
    Really? That's news to me. I'd like to know how you think that is done.
     
  6. Nov 13, 2014 #5

    A.T.

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    - Number of collisions per second is proportional to speed
    - Mean velocity of collisions is proportional to speed
    - The force (momentum transfer) is proportional to the product of both above, so it's proportional to speed squared.
     
  7. Nov 13, 2014 #6
    That sounds good, but I'd like to examine these statements more closely.

    You say that "the number of collisions per second is proportional to speed" where we presume that this refers to the speed of the object subject to drag. So, does this mean that a moving car hits more gas molecules per unit of time than a stationary car does, remembering that the gas molecules are wandering around in random directions all the time? That seems questionable. Can you justify this, please?

    What is the "mean velocity of collisions"? In air, with a mix of molecules of various sorts, all whizzing around in random directions, this term does not seem to me to have much meaning. Can you define it more precisely, please?
     
  8. Nov 13, 2014 #7

    A.T.

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    If you look more closely, you will find that this simple drag formula doesn't really work. But if you want a simple formula you use a simplistic model.

    Since it's random in all directions, it doesn't produce any net force and can be ignored in a simplistic model.
     
  9. Nov 13, 2014 #8
    I'll bite: How closely do I have to look? This formula is certainly extremely widely used for all cases where the Reynold's Number is too high for a viscous model to apply. There are countless folks who think it works well enough to use it everyday, so a vast number of us need your wisdom and insight. Where do I look for a better formula?

    But, but, but .... wasn't this the basis for your first two statements at the beginning of our discussion?

    Nice dodge to avoid addressing both of my points!!
     
  10. Nov 13, 2014 #9

    A.T.

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    Nope. I was talking about a model which ignores Brownian motion, because it doesn't produce any net force.
     
  11. Nov 13, 2014 #10
    Looks to me like your original assertion, "Not strictly, but the ~v2 relationship is based on momentum conservation" has no basis in fact at all. You have steadfastly refused to answer any of my questions for clarification. Sounds like hot air to me!!
     
  12. Nov 13, 2014 #11

    A.T.

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    What is still unclear?
     
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