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Maximum range/endurance of an aeroplane

  1. May 3, 2005 #1
    An airplane in flight is subject to an air resistance force proportional to the square of its speed [itex]v[/itex]. But there is an additional resistive force because the airplane has wings. Air flowing over the wings is pushed down and slightly forward, so from Newton's third law the air exerts a force on the wings and airplane that is up and slightly backward. The upward force is the lift force that keeps the airplane aloft, and the backward force is called induced drag. At flying speeds, induced drag is inversely proportional to [itex]v^2[/itex], so that the total air resistance force can be expressed by [itex]F_{air}=\alpha v^2 + \beta/v^2[/itex], where [itex]\alpha[/itex] and [itex]\beta[/itex] are positive constants that depend on the shape and size of the airplane and the density of the air. To simulate a Cessna 150, a small single-engine airplane, use [itex]\alpha[/itex] and [itex]\beta[/itex]. In steady flight, the engine must provide a forward force that exactly balances the air resistance force.

    Calculate the speed (in km/h) for which the airplane will have the maximum endurance (that is, will remain in the air the longest time).

    Since the plane is in steady flight, the thrust of the engines equals the drag force from the air. The power of the engines is thus

    [itex]P = F_{air} v[/itex].

    I assume that the power is directly proportional to the fuel flow rate, which is inversely proportional to the endurance. Letting the derivative wrt v equal zero gives

    [itex]\frac{\mathrm{d}P}{\mathrm{d}v} = 0[/itex]
    [itex]v = \sqrt[4]{\frac{\beta}{3\alpha}}[/itex].

    Which is not correct. What am I doing wrong?


  2. jcsd
  3. May 3, 2005 #2


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    Try assuming the fuel flow rate is proportional to the thrust instead of to the power.
  4. May 3, 2005 #3
    I tried to minimise the thrust instead of the power to give


    which is still not correct. Any ideas?
  5. May 3, 2005 #4


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    Power is the the rate of doing work or converting energy. Fuel is stored energy. Assuming constant efficiency, over the duration of the flight the rate of fuel energy consumption is proportional to the power. If the flight lasts a time T, then if E represents useful energy

    [itex]E = P(v)T [/itex]

    [itex]T = \frac{E}{P(v)}[/itex]

    Maximize the time with respect to velocity, given that E is a constant. That seems like it should give the correct result. But setting

    [itex]\frac{\mathrm{d}T}{\mathrm{d}v} = 0[/itex]

    will lead to the same result as

    [itex]\frac{\mathrm{d}P}{\mathrm{d}v} = 0[/itex]

    I thought it was odd that your first result did not give you the correct answer because it seems you are making the correct assumption relating power to fuel use.

    The only other thing I can think of is that what they really mean by endurance is the distance of the flight rather than the time. In that case, you would want to maximize distance instead of T

    [itex]d = Tv = \frac{Ev}{P(v)}[/itex]

    [itex]d = \frac{E}{F(v)}[/itex]

    which is where the force idea came from.

    I do find it odd that there is a force term that is inversely proportinal to velocity squared. As v goes to zero, that force becomes infinite. But the problem does say "at flying speeds" so maybe it is OK.
    Last edited: May 4, 2005
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