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If I have an aircraft, and I reduce the induced drag produced by it, does this mean I should operate it at a higher altitude and faster speed to take advantage of the drop in total drag, or slower at lower altitude to take advantage of the lower green dot drag?

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  • $\begingroup$ It depends on the numbers. "reduce the induced drag" is vague. And we know nothing about the engines (altitude matter). What about same altitude, same speed, less fuel consumption? Try with real numbers. $\endgroup$ Oct 9, 2023 at 9:30
  • $\begingroup$ @sophit What is an airliner's "green dot speed", and how does it vary with weight, and why? $\endgroup$
    – Bianfable
    Oct 9, 2023 at 16:59
  • $\begingroup$ @Bianfable: thanks, never heard about it 🫣🖖 $\endgroup$
    – sophit
    Oct 9, 2023 at 17:31
  • $\begingroup$ Would it be accurate to say that you specifically mean you reduced the coefficient of induced drag? $\endgroup$ Oct 10, 2023 at 7:26
  • $\begingroup$ Thanks people! This is all extremely helpful and really useful! $\endgroup$ Dec 22, 2023 at 16:28

3 Answers 3

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The easiest way to reduce induced drag in an aircraft is to make it lighter. However, I'll assume you're talking about design / configuration changes, not operational ones.

If the drag polar has the simple form:

$C_D=C_{D,0}+\frac{C_L}{\pi\,e\,AR}$

We can talk about the balance of induced and parasite drag.

If you change the balance such that induced drag coefficient goes down (say by improving $e$), you'll need to fly at higher lift coefficient to achieve best L/D.

$C_L=\frac{W}{q\S}$

$q=0.5\,\rho\,V^2$

So, if wing loading stays the same, to fly at higher $C_L$, you want to fly at lower $q$. This requires flying at lower speed or higher altitude. You probably want to maintain your cruise Mach number, so you'll fly at higher altitude.

This helps explain why the Global Hawk flies where it does.

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  • $\begingroup$ Thank you, very helpful! $\endgroup$ Dec 22, 2023 at 16:28
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We use the usual approximation that the drag coefficient for the entire airplane can be expressed via the well known:

$C_d=C_{d_0}+kC²_l$

The first term represents the "parasite drag" while the second term is the "induced drag".

Equating the power requested to win that drag $C_d$ with the power available from the engine(s), the results of the following table are obtained:

Speed best range Speed best loiter
Propeller $\sqrt{\frac{2W}{\rho S}\sqrt{\frac{K}{C_{d_0}}}}$ $\sqrt{\frac{2W}{\rho S}\sqrt{\frac{K}{3C_{d_0}}}}$
Jet $\sqrt{\frac{2W}{\rho S}\sqrt{\frac{3K}{C_{d_0}}}}$ $\sqrt{\frac{2W}{\rho S}\sqrt{\frac{K}{C_{d_0}}}}$

Being $K$ at the numerator, all the rest being the same a reduction in the induced drag always implies a reduction of the optimal speed (of whatever you'd like to optimise, either range or loiter).

If you want to compensate for this reduction and increase again the optimal speed, then you have either to fly at a lower density $\rho$ (i.e. higher altitude) and/or increase the wing loading $W/S$ and/or reduce the parasite drag $C_{d_0}$ as well.

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  • $\begingroup$ The other assumptions commonly left out for these 'best' conditions include: fuel consumption (BSFC or TSFC), propeller efficiency, and drag polar all constant with flight condition -- airspeed / Mach, altitude, throttle. In reality, none of these things are true. In practice, they are close enough as long as you're careful. The variation of max L/D with Mach number sets the cruise speed for a transport. The variation of TSFC with altitude TSFC=TSFC$_0\sqrt{\theta}$ drives transports to want to cruise at the Tropopause (11km, 36,089ft). $\endgroup$ Oct 10, 2023 at 17:01
  • $\begingroup$ @RobMcDonald: well, there are other simplification too other than what you've listed (a real polar has three terms at least; the weight changes during the flight; and so on). Anyway those simplified equations are enough to grasp the basic behaviour of an airplane in terms of speed optimisation. And... this is not Journal of Fluid Dynamics 😉 Thanks anyway for you comment, which applies equally well to your own answer 🖖 $\endgroup$
    – sophit
    Oct 10, 2023 at 17:58
  • $\begingroup$ That is why I left my answer at CL and best L/D. I was trying to stray away from the specificity that leads to needing the rest of the complexity. Weight change during flight is captured by your equations, you aren't ignoring anything there. The simplicity of the drag polar was clearly stated. The fact that you're solving for a best speed -- using information that in reality varies with Mach number is a hidden nuance. $\endgroup$ Oct 10, 2023 at 20:15
  • $\begingroup$ @RobMcDonald: if it's ok for you I'll expand my answer with you suggestions $\endgroup$
    – sophit
    Oct 10, 2023 at 20:28
  • $\begingroup$ By all means, go ahead. $\endgroup$ Oct 10, 2023 at 20:31
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You may notice airliners are designed to have their Vbg (best glide) very close to their desired cruising indicated airspeeds.

It is better to go high, because at higher flight altitudes True Airspeed is greater then Indicated Airspeed.

This means the plane goes farther for the same amount of "push".

All well and good until True Airspeed gets too close to Mach 1, the speed of sound. Then drag begins to rise.

So you want to fly "Green dot" indicated as high as you can until Mach number increases to the point where fuel economy is lost.

Subsonic airliners use swept, supercritical wings to get as close as they can to Mach 1 economically.

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  • $\begingroup$ Thank you, although most airliners are not operated at green dot but faster, since their total fuel consumption is lower over a given DISTANCE if you fly faster. $\endgroup$ Dec 22, 2023 at 16:29
  • $\begingroup$ @AlastairWyllie lowest fuel consumption per distance is at Vbg indicated. Airliners indeed fly faster True Air Speed at higher altitudes up to Mach limits, but for best fuel economy they want to be around Vbg indicated, just like gliders. $\endgroup$ Dec 22, 2023 at 19:54

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