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Drag Equation Derivation:

  1. Feb 17, 2005 #1
    Hi Guys,
    I am doing an extended essay on Terminal Velocity and I need the derivation for the drag force equation: 1/2*C*A*P*v^2
    Can you please post the derivation for me?
    Thanks a lot.
    Tsering :confused:
     
  2. jcsd
  3. Feb 17, 2005 #2

    Clausius2

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    There is no derivation. It is a DEFINITION.

    The coefficient of Drag is defined as:

    [tex] C_D=\frac{P}{1/2 \rho U^2} [/tex]

    Of course, the left term is COUPLED with the flow around the body considered. I mean, to obtain Cd, you'll have to solve the field of Pressures P around the body. And believe me, that's not a simple task. :wink:
     
  4. Feb 17, 2005 #3

    Andrew Mason

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    The drag force equation is a constructive theory based on the experimental evidence that drag force is proportional to the square of the speed, the air density and the effective drag surface area. The drag coefficient is simply the proportionality constant that relates drag force to these factors for a given object, so there is no way to 'derive' it other than by experiment.

    AM
     
  5. Feb 19, 2005 #4
    Thanks guys,
    But there is also the theory of the linear dependency of drag force on the velocity of the falling object.
    Can you explain that bit as well?
    Tsering
     
  6. Feb 19, 2005 #5

    Clausius2

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    Linear dependence??

    There is neither linear dependence on velocity, nor cuadratic dependence on velocity. In fact, the drag coefficient is not constant at all. The dependence of drag force on velocity can be much more complex. In usual industrial problems the drag coefficient depends on non dimensional parameters which are a function of the geometry, velocity, flow physical properties and time.

    However, most of physics books do not tell nothing about this complexity, but if you advance in the study of aerodynamics you will hear again about what I have just told you.
     
  7. Feb 19, 2005 #6

    Andrew Mason

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    For small velocities, the drag force is roughly proportional to velocity. Look up Stokes law. But, as Clausius says, it is just a rough approximation.

    If dF/dv = constant, k, then [itex]dF/dt = dF/dv * dv/dt = ka = Fk/m[/itex]. So F must be an exponential function of time ([itex]F \propto v \propto c + e^{kt/m}[/itex])

    Once the drag force reaches the weight of the object, there can be no further acceleration and the object falls at constant speed - terminal velocity, which is the asymptote of the velocity-time graph.

    AM
     
  8. Feb 21, 2005 #7
    Actually, there is a derivation for the drag formula. You have to write down the momentum contained in a fluid element and differentiate with respect to time, after which you integrate over dV. This will give you the formula without the drag coefficient, which is plugged into the formula afterwards. The Cd makes it possible that only a portion of the momentum is transferred to the body. It works just like the friction coefficient together with the normal force and is strongly simplified.
     
  9. Feb 21, 2005 #8

    Clausius2

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    Could you write here the derivation?. I don't believe that. I know there is a way to relate Drag force and free stream velocity in the case of two dimensional incompressible and steady flow, being the proportional scalar a function of the velocity profile downstream.

    In industrial problems sure it doesn't work. There is only a few simplified cases where Cd can be derived as I posted above.
     
  10. Feb 22, 2005 #9
    Okay, I'll do the derivation in terms of energy because it's easier than with momentum and doesn't even involve calculus.

    Let's start with the kinetic energy of the fluid in front of the body:
    [tex] E=\frac{1}{2}m v^2 [/tex]

    The mass is given by the area times an arbitrary length times the density:
    [tex] m=A s \rho [/tex]
    [tex] E=\frac{1}{2}A s \rho v^2 [/tex]

    Since work or energy is also:
    [tex] E=F s \rightarrow F=\frac{E}{s} [/tex]

    It follows that:
    [tex] F=\frac{1}{2}A \rho v^2 [/tex]

    It's even easier if you take the dynamic pressure,
    you can simply multiply by the frontal area to get the force.
    [tex] p_{dyn}=\frac{1}{2}\rho v^2 [/tex]

    You plug in a coefficient because you don't want all of the force to act on the body.
    [tex] F_{drag}=p_{dyn} A C_D = \frac{1}{2}\rho A C_D v^2 [/tex]

    As stated before, this is only a very rough approximation. It is for turbulent flows past a free body with high Reynolds numbers. It uses the fact that the fluid must be displaced.
     
  11. Feb 22, 2005 #10

    Clausius2

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    :surprised :surprised And you call this a DERIVATION??????

    If some PF mentor read this, please for the sake of the forum delete that post.
     
  12. Feb 22, 2005 #11

    FredGarvin

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    Perhaps this may help. I hadn't read this in a long time so I had to look this one up:

    [tex] D = \int dF_{x} = \int p cos \theta dA + \int \tau_{w} sin \theta dA [/tex]

    Where:
    [tex] \theta [/tex] is a function of the geometry of the object
    [tex] \tau_{w}[/tex] is the wall shear stress distribution
    [tex]p[/tex] is the normal stresses due to pressure distribution

    It essentially boils down to two P/A components added together. The tough part is in describing the distributions.
     
  13. Feb 22, 2005 #12

    dextercioby

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    Clausius,don't feel offended,but i think the proof was not mean to be rigurous.Assume a 15yr old boy asks you:if i stay put and the wind blows in my face,what force does my skin feels...??He doesn't know calculus or fluid mechanics,just elementary classical particle mechanics...What formulas /what derivation would u use...?:bugeye:

    Daniel.
     
  14. Feb 22, 2005 #13

    arildno

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    It is, of course true, that a friction coefficient can be defined as the ratio between the drag force and the product between some typical area and a measure of the dynamic pressure.
    We can even make a graph out of this! :smile:
    However, as Clausius2 has pointed out, the functional dependency of the coefficient (in, say, the velocity) is, in general impossible to gain from analytical considerations.
     
  15. Feb 22, 2005 #14

    Clausius2

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    I'm not a professor. But if you don't know nothing about geometry and ask me what is the shape of the earth, I'd never say "plane". I'd say: "you only should know the shape is a bit more complex", without giving a mathematical definition of an sphere.

    Once Einstein said that a good physicist must know how to explain his achievements to every people no matter what previous knowledge about the stuff they have. Sometimes I have to admit I am unable to do that, because I do not agree with Einstein (this is the unique thing I do not agree with him by the way :rolleyes: ) and because there are some things which cannot be explained on an easy way. If you try to reduce complex subjects to easy ones, surely you will find yourself lying.
     
  16. Feb 22, 2005 #15

    dextercioby

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    Yes,Arildno,it is true,however,the force in the example i have described is proportional to the dynamic pressure...And no matter how unrigurous it may seem,the derivation shown by Sergey proves it...

    Daniel.

    P.S.Einstein knew what he was talking about...:wink:
     
  17. Feb 22, 2005 #16
    All I wanted to show is that there is a way to come up with the drag formula, nothing more. Perhaps someone knows the true origin, if I didn't capture it. This formula works amazingly well for what it's intended, and it's used quite often for simple models on a level where students don't know fluid mechanics. I don't claim that this formula is useful for any complex flow problem.
     
  18. Feb 22, 2005 #17

    arildno

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    As long as everybody agrees that the friction coefficient is not necessarily a constant independent of the velocity, then I think most which should be said has been said.
    As examples:
    In the Stokes flow regime (Re<<1) for a sphere, the friction coefficient goes as the inverse velocity (giving the familiar linear force/velocity relationship), whereas the drag crisis in turbulent flow represents an upper limit of the regime where the friction coefficient is roughly constant.
     
  19. Feb 22, 2005 #18

    dextercioby

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    Of course I agree.However,the things you mentioned are "advanced" they can be explained by using the laws of (viscous) fluid dynamics.Sergey mentioned an elementary proof that there is a connection between the dynamical pressure & the drag force.

    Daniel.
     
  20. Feb 22, 2005 #19

    arildno

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    I wouldn't call it an elementary proof of anything (we are always free to define whatever ratios we like, I don't see what proof has to do with this); rather, I regard it as the result of a very intelligent modellling procedure.
     
  21. Feb 23, 2005 #20

    Clausius2

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    Some elementary proofs are false, due to its "elementarity".
     
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