Drag coefficient w/o terminal velocity

  • Thread starter picklefeet
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  • #1
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I know the equation for terminal velocity and drag coefficient. But I can't find one without the other. It's a real catch-22. I also don't know the Drag Force. HEEELP!!!
 

Answers and Replies

  • #2
58
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Do you have a specific question? Also, remember, when an object reaches terminal velocity, there is no net force acting on the object (as it isn't accelerating). Hence, the drag force is of equal magnitude to the object's weight.
 
  • #3
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An example would be finding the terminal velocity of a quarter. the weight is 5.67g. The cross-sectional area is 4.522 cm. (rounded to the nearest thousandth.) and the density is 1.184. I have the equation Fd= 1/2*p*V^2*Cd. I get 5.67=.592*Cd*V^2. (I can't remember if drag weight is measured in grams or kilograms.) You see my problem. I stil need the velocity of the object.
 
  • #4
FredGarvin
Science Advisor
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It will have to be an iterative process. Cd is a function of Reynolds Number. You would need to guess at an original Cd. Use that Cd to calculate the speed based on your equation of motion. Use that speed to calculate Reynolds number. Finally, use that Reynolds Number to check your original guess for the Cd. Eventually it will converge on a solution.
 
  • #5
russ_watters
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Cd is a function of Reynolds Number.
This is getting past what Aero I took, but isn't that only for situations where viscous drag changes a lot? Can't we just use this equation there: http://www.grc.nasa.gov/WWW/K-12/airplane/drageq.html

If your reference conditions aren't vastly different from the conditions you are studying (ie, if you find the Cd at 60mph and your terminal velocity is around 120), it should work out ok, shouldn't it?

Obviously, this doesn't help with the ultimate problem, of course: that drag coefficient is something that really needs to be found experimentally.
 
  • #6
FredGarvin
Science Advisor
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I was thinking in general terms. I think you're right though. The speed isn't going to be changing by that much so it should be pretty simple to get the answer. I guess we won't be covering a few orders of magnitude in Re for this.

I did look at a chart I had for a round plate perpendicular to the flow. It is a constant 1.1 over a very wide range so that makes this a very easy problem.
 
  • #7
russ_watters
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...easy if the flat plate (a quarter) falls at a stable position parallel to the ground...

I do believe that Mythbusters actually tested this, though....
 
  • #8
Wouldn't the quarter turn to the stable position in air therefore making the cross sectional area equivalent to the diameter times the width of the quarter?
 
  • #9
4,662
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There are two basic types of drag force; drag force at high Reynold's number (Re>1000), proportional to velocity squared, and Stokes drag force (Re<1000), linearly proportional to velocity. It is not obvious the the falling quarter will exceed Re=1000. The Reynold's number is proportional to velocity times sqrt(cross section), so a small size object would have to have a very high velocity.
See
http://en.wikipedia.org/wiki/Drag_(physics [Broken])
Also see
http://en.wikipedia.org/wiki/Reynolds_number
 
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