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Drag Force Equation Useless in Finding Terminal Velocity!

  1. Jun 14, 2007 #1
    As most of you know, the drag force equation is Fd=1/2*p*Cd*V^2*Cd*Ac.
    At terminal velocity, Fd equals weight. If you find the value of all the variables except Cd and V, you still won't have anything. But the terminal velocity equation is Vt= sqrrt of 2weight/p*Cd*Ac. So if V is terminal velocity, it is reasonable to conclude that V^2 is equivalent to 2W/p*Cd*Ac. So if you replace V^2 with what I've stated above, you'd get W= 1/2p*Cd*2W/p*Cd*Ac.
    This equation is true, but only because it cancels out everything but W. You are left with W=W. This is useless for finding Cd. You could also do the opposite. Replace Cd with (1/V^2*2weight)
    pAcCd

    In this case everything cancels out too. In conclusion, this formula is useless in finding terminal velocity.
     
  2. jcsd
  3. Jun 14, 2007 #2

    russ_watters

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    Well, no. Clearly the forumla is useful for finding terminal velocity of an object with a known drag coefficient.
     
  4. Jun 14, 2007 #3

    FredGarvin

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    Since I am assuming that you are carrying this over from the previous thread about the quarter...

    I'm going to neglect the buoyant force...
    [tex]F_d = W[/tex]

    [tex]\frac{1}{2} \rho V^2 A C_d = m*g[/tex]

    For air you can use STP value of :
    [tex]\rho = 1.201 \frac{kg}{m^3}[/tex]

    [tex]\mu = 1.79 x 10^{-5} \frac{N*s}{m^2}[/tex]

    The quarter has values of:
    [tex]d = 24.15 mm[/tex]
    [tex]A = 4.522 x 10^{-4} m^2[/tex]
    [tex]m=5.67 x 10^{-3} kg[/tex]

    So now we have:
    [tex]\frac{1}{2}\left(1.201 \frac{kg}{m^3}\right)\left(V^2\right)\left(4.522 x 10^{-4} m^2\right)\left(C_d\right) = \left(5.67 x 10^{-3} kg\right)\left(9.81 \frac{m}{s^2}\right)[/tex]

    [tex]V^2 C_d = 206.22 \frac{m^2}{s^2}[/tex]

    Now, it just so happens, that if you look up the drag coefficient for a circular flat plate, perpendicular to flow,
    it is a relatively constant value of 1.1 which makes this a trivial solution.

    [tex]V = \sqrt{\frac{206.22 \frac{m^2}{s^2}}{1.1}}[/tex]

    [tex]V = 13.69 \frac{m}{s}[/tex]

    However, if you did not have a constant [tex]C_d[/tex] you would guess at the [tex]C_d[/tex] and calculate the resultant [tex]V[/tex].

    From that [tex]V[/tex] you would then calculate the Reynolds Number for that flow using

    [tex]Re = \frac{\rho*V*D}{\mu}[/tex]

    You would then use the calculated Reynolds Number and look up on a chart the corresponding [tex]C_d[/tex]. If that value you
    just looked up comes very close to your guess, you are done. If it isn't, use the new value of [tex]C_d[/tex] you just looked up
    and recalculate the velocity. Repeat the Reynolds Number calculation to get another [tex]C_d[/tex] value.

    Keep doing that until you converge on a solution. It shouldn't take that many iterations to solve it.

    The fact that you got W=W in your result means that you had circular reasoning in your analysis. You did some algebra on the original equation and then plugged it right back into the original equation. If you do that I should hope you get W=W.
     
    Last edited: Jun 14, 2007
  5. Jun 14, 2007 #4
    The fact is, if I already knew the Cd, I wouldn't have to use the Fd equation.
    I'd just use the terminal velocity equation of Vt= sqrt 2weight
    pAcCd
    And I did find the terminal velocity ofa quarter. It's 83.136 miles per hour (rounded to the nearest thousandth.) on a different not, How do I use the mathematical symbols? i can't find them.
     
  6. Jun 14, 2007 #5

    PhanthomJay

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    But the terminal velocity equation is derived from Fd = Weight. You're going in a circle again.
    You're way off, you slipped a digit or two somewhere. Per Fred's response, the terminal velocity of a quarter falling belly up is about 14m/s, or a mere 30 mph.
     
  7. Jun 14, 2007 #6
    I'm realize now that I placed the decimal point one space to the left while figuring out the cross-sectional area (oops). But for some reason I don't see where you got 30 m.p.h. If I plug in 1.17 for Cd, 11.34 g. for weight, .46353465 m^2 for Ac and 1.184 for air density, I get about 10.9914 miles per hour. My problem might be the units I'm using. If so, please correct me. thanks.
     
  8. Jun 15, 2007 #7

    FredGarvin

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    You need to use Newtons for the weight, not grams. Also, your area is off. Take a look at my post and you'll see that the area is [tex]A = 4.522 x 10^{-4} m^2[/tex]
     
  9. Jun 15, 2007 #8
    Thanks for the info. For the cross-sectional area, do I use centimeters or meters? And I still don't know where to find the mathematical symbols to put in my replies. Can you tell me where they are?
     
  10. Jun 15, 2007 #9

    FredGarvin

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  11. Jun 15, 2007 #10
    Thank you to all who helped me with this problem. And thanks for the link to the mathematical symbols.
     
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