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Aerodynamic Drag Problem

  1. Feb 3, 2005 #1
    I need some help. Our company is holding a paper airplane constest for distance and I've been trying to work out the math to find the best mass, wing-span, wing-chord, launch angle, etc. combination. I've worked through a bunch of the math but I've just been stumped by what appears to be a pretty simple problem. In essense, the current speed equals the previous speed minus the drag:

    [itex]V_{t+\Delta t} [/itex] is the velocity at the current time,
    [itex]V_{t} [/itex] was the velocity at the previous time,
    [itex]\mu [/itex] is a mindless coefficient (includes mass, surface area, etc.),
    [itex]\Delta t[/itex] is the difference in time between [itex]V_{i} [/itex] and [itex]V_{i+1} [/itex].

    [itex]V_{t+\Delta t} = V_{t} - \mu (V_{t}^2) \Delta t[/itex]

    So, given the initial speed, at [itex]t = 0 [/itex] will be, let's say, 20m/s, I want a continuous function that will tell me the velocity as a function of time. I know there is a slope-integration-trapezoidish thing that will solve this in a heart beat, but I just can't seem to think of it.
     
  2. jcsd
  3. Feb 3, 2005 #2

    Tide

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    If [itex]\mu[/itex] is a constant then you can turn your difference equation into a differential equation in the limit of small [itex]\Delta t[/itex] and you should have no difficulty demonstrating that

    [tex]V = \frac {V_0}{1 + \mu V_0 t}[/tex]

    is the required solution.
     
  4. Feb 11, 2005 #3
    Alright. I've had quite the difficulty figuring out how you got from my difference equation to the solution. I've hunted through my math books and through numerous sources on the internet. I really want to know how to do this, so a couple of intermediate steps would be nice to see. I've found my equation to be incorrect as well. The equation is:

    [itex]
    V_{t+ \Delta t} = V_{t} - \mu(V_{t}^2) \Delta t - \beta \Delta t
    [/itex]

    where mu and beta are constants.
     
  5. Feb 11, 2005 #4

    dextercioby

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    Wht do you know about calculus,then...?U know that you can approximate
    [tex] V_{t+\Delta t}-V_{t} [/tex]

    with the differential of V(t) and
    [tex] \Delta t [/itex]

    with the differential of 't'.Then u'd have to separate variables & integrate...

    Daniel.
     
  6. Feb 11, 2005 #5
    [tex]V = \frac {V_0}{1 + \mu V_0 t}[/tex]

    is the solution to the previous problem [Thanks Tide!] and I want a to know how to do it. I can't get figure out how to get this solution by separating and integrating. When I do that, I get:

    [itex]
    V = \frac {tan( t \sqrt \mu \sqrt \beta) * \sqrt \mu} {\sqrt \beta} + C
    [/itex]

    Which is WAY different than the above solution. If you can do it, dextercioby, I'd love to see it so I can understand how to do these problems. But if you can't or aren't willing to walk through some of the steps to actually help me understand, I'd appreciate it if you just let someone else answer the question. Thanks.
     
  7. Feb 11, 2005 #6

    dextercioby

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    It is,because it's ANOTHER EQUATION.As u may have noticed (i DID),adding a constant term changes the solution dramatically.
    So yes,your solution containing "tangent" of time is correct,but only for the modified (second posted) equation...

    Daniel.
     
  8. Feb 11, 2005 #7
    Problem is, the tan() equation is wrong. It does not produce realistic results. With an initial speed of 20m/s, and t = 2 the equation produces a velocity of 19.84m/s. Which is not possible. The actual velocity should be closer to 5.7m/s as predicted by the previous solution. There is a problem somewhere, and I'm betting on the approach.
     
  9. Feb 11, 2005 #8

    dextercioby

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    Then it must be another equation to begin with.I don't know why you've changed the first,as it had yielded good results...

    Daniel.
     
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