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Coefficient of drag

  1. Mar 25, 2014 #1

    462chevelle

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    Could someone explain to me the difference in calculating the coefficient of drag with thrust, and without. What causes the difference in application?
     
  2. jcsd
  3. Mar 26, 2014 #2

    BvU

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    Could you clarify a little ? Is there a difference? What difference ? What drag ? What difference in application ?
    And what is the further context ? What do you know already ? At what level, approximately ?
     
  4. Mar 28, 2014 #3

    462chevelle

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    Here is the problem, I finally figured out how to solve this. But it wasn't because my own independent research of coefficient of drag. I simply noticed someone else on the discussion section commenting about it. Whenever I discovered there was a problem that needed the coefficient of drag. I looked over the lecture again and didn't see anything about it other than a conceptual comment about it. So I started looking for formulas and explanations, and all of them I found stated 'without thrust'
    So naturally it made me think there would be something different about it conceptually. Or that the fact is that you can derive it in such a way to get rid of the Thrust variable. I couldn't figure it out just by looking at all of the formulas with it in it. So i decided to ask.
    Thanks.
    Also to add, I'm not great at physics, but with enough time i can understand most things. As far as my level of math, I can do simple calculus but nothing crazy. Good at algebra and trig. I'm not really taking this course expecting anything miraculous but I have noticed its really helping my problem solving skills. Most of the problems start off in the wrong units. Or you don't have defined formulas, meaning you have to derive and figure stuff out. I get a kick out of the challenge but its still hard to me sometimes.
     

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  5. Mar 29, 2014 #4

    AlephZero

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    I don't really understand your comment about formulas for drag "without thrust" (and presumably, other formulas for drag "with thrust"). If you can give a link or a picture that might help.

    The thrust and drag forces are independent of each other. Thrust comes from the engines. Drag is caused by the air flow over the plane and depends on the air speed and the angle of attack of the wings.

    In the question you attached, the plane is flying straight and level so thrust = drag. You can work out the thrust from the change in momentum of the air by the propeller. Them you can find the drag coefficient from the drag force, wing area, air speed, and air density.

    In real life, the propeller wash might have some effect on the drag coefficient. For a multi-engine propeller driven plane, the air flow over the wings would be slightly faster than the airspeed of the plane, especially at low airspeed and high engine power. (For a single engine plane the effect would probably be less). But I think that sort of effect would be too complicated to include in a "formula".
     
  6. Mar 29, 2014 #5

    BvU

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    Next time please start with the template. It really helps helping and saves time. For all.
    1. The problem statement, all variables and given/known data
    2. Relevant equations
    3. The attempt at a solution
    What was the argument for removing it altogether ? Ever seen the guidelines to counter those arguments?
     
  7. Mar 29, 2014 #6

    462chevelle

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    I didn't use it because I wasn't really looking for an answer to a question. Just conceptual help, so I didn't think it would be that big of a deal to start from scratch.
    That's how i solved the problem, making drag=thrust. Then using the equation, here is the link to the one that showed thrust/no-thrust as something to be considered.
    http://www.grc.nasa.gov/WWW/k-12/airplane/flteqs.html
    The equation for the horizontal is similar to the one i used and could probably be derived to get the equation i used. Off the top of my head i think it was
    Cd=(mdot)/(A*rho*V^2)
     
  8. Mar 29, 2014 #7

    462chevelle

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    I looked back at my notes, I used
    Cd=(T)/(.5*rho*V^2*A)
     
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