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1/2 factor in drag equation

  1. Dec 19, 2014 #1
    Why the 1/2 factor in the equation [itex]F_{R} = \frac{1}{2}\rho C_{d}A v^{2}[/itex]? Why not just divide every [itex]C_{d}[/itex] value found in tables by 2 and omit the 1/2 in the equation instead?
     
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  3. Dec 19, 2014 #2

    ShayanJ

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    The only reason I can think of is that the guy proposing this formula was thinking the force should be proportional to half of one of those things in it(I don't know why though!) and he wants to get sure people have a clue for that. I think the best candidate is the area.
     
  4. Dec 19, 2014 #3
    Energy of compressed spring is ##\frac{1}{2}kx^2##. So according to you I will do some magic and replace all k value of spring in every table to k/2.
    Now I have an equation ##F=kx##. This is the force exerted by a compressed /elongated spring. If I use the values given in the new table, I will get wrong answer. So I will have to multiply 2 every time I use it. Maybe ##C_d## has other uses and ##C_d## is more commonly used than ##C_d/2##.
     
  5. Dec 19, 2014 #4

    boneh3ad

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    In the sense of this equation, the drag scales with dynamic pressure, which is defined as ##\rho v^2/2##. The ##1/2## therefore comes from the fact that the equation contains the dynamic pressure. Without that factor, you could still just divide ##C_d## by two but the scaling with dynamic pressure would be wrong.
     
  6. Dec 19, 2014 #5
    Pretty much like boneh3ad said. Having the 1/2 makes the Aerodynamic coefficient wrap up in a nice manner ## C_L = \frac{L}{qS} ## with the dynamic pressure.

    It's really convenient this way especially when it comes to application in Flight mechanics, Stability and Control.
     
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