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I created for me a table from an aircraft for Mach=0.8 for alpha= 0° to 8°

Actually, I an alpha of alpha=4° should be sufficient, so that my lift coefficient is enough, to counterbalance the weight in steady horizontal flight. The corresponding lift/drag ratio is 8.

I could now choose also alpha=8° , with a lift/drag ratio of 12.

My question: regarding the range equation(breguet) it is better to have a higher lift-drag ratio for making more range.

But I am wondering: if I choose alpha=8° my lift coefficient is greater than the one I need, so in the end, my aircraft will climb ? This would not be a steady horizontal flight.

Am I right?

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  • $\begingroup$ You used the term "I am flying" to describe this airplane. I'm just curious, do you own or even fly an aircraft that cruises at mach 0.8? I'm just wondering because it is worded as if you are the pilot, but it is difficult for me to understand from a practical standpoint what you are asking if that is the case. $\endgroup$
    – Michael Hall
    Feb 26, 2021 at 18:54

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If your optimal lift coefficient for long range cruise flight is greater than what you need, you are flying too fast, or too low.

For cruise flight, your lift needs to equal the weight, otherwise the aircraft will accelerate upwards (in the body frame of reference).

If you know the lift coefficient that is optimal for cruise, you need to adjust the airspeed and density (altitude) to make sure that the resulting lift equals the weight.

$W = L = \frac{1}{2}\rho V^2 C_L S $

$\rho V^2 = \frac{2 W}{C_LS}$

Reduce speed and/or increase altitude until the lift and weight are equal at the optimal lift coefficient. That will give you the maximum range.

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    $\begingroup$ Or maybe you're flying too low? $\endgroup$
    – Abdullah is not an Amalekite
    Feb 26, 2021 at 15:10
  • $\begingroup$ @Abdullah or in too cold air... $\endgroup$
    – DeltaLima ♦
    Feb 26, 2021 at 15:12
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This related ASE answer may help -- "Does Lift equal Weight in a climb?". Note that a steady-state climb is associated with a smaller lift force than level flight, not a larger lift force. On the other hand, a loop is associated with a larger lift force than level flight.

If you increase the lift coefficient, you've increased the lift force -- if you keep airspeed constant. Since Mach is constant, and therefore airspeed is constant, it appears that your question is overly constrained. Not all your parameters can be met simultaneously if the flight path is to remain horizontal. In a particular aircraft in a particular configuration, only one value of the lift coefficient is consistent with horizontal flight at a particular indicated airspeed.

(Note-- actually we can decrease the indicated airspeed while holding the Mach constant, by choosing a higher cruising altitude. Since your question doesn't specify any particular altitude, this would appear to be a valid solution to your problem.)

On the other hand, if the altitude and Mach number are both constrained, constraining the indicated airspeed and giving "too much" lift at the optimal L/D ratio, and this is a design exercise, with the shape of the aircraft not yet "frozen", you could solve the "excess lift" problem by decreasing the wing area, allowing the aircraft to cruise at the optimal L/D ratio at the target airspeed and altitude.

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  • $\begingroup$ great answer quiet flyer! $\endgroup$
    – Lucas
    Mar 3, 2021 at 11:47

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