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Figure 10-5 of the FAA's Pilot's Handbook of Aeronautical Knowledge shows:

I didn't think $L/D_{MAX}$ coincided with $D_{MIN}$. Is this Figure accurate?

  • Closely related, but couched in terms of gliding flight-- could be adapted to address your question-- the answer is "yes", whether we are flying horizontally or gliding -- https://aviation.stackexchange.com/a/81790/34686 – quiet flyer Oct 20 '20 at 15:04
  • Also related -- https://aviation.stackexchange.com/a/87933/34686 . The answer is "yes"-- and note that we need not assume that Lift is constant, i.e. exactly equal to Weight. – quiet flyer Jun 25 '21 at 16:47

2 Answers2

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Well, for all L/D curves and D curves, the assumption is that the weight of the aircraft is constant and that there is no acceleration. Therefore the lift equals the weight (neglecting the small vertical component of thrust). So lift is a constant in these curves.

The rest is simple mathematics; the maximum of $\frac{1}{f(x)}$ occurs at the minimum of $f(x)$ (when $f(x) > 0$), so the maximum of $\frac{L}{D}$ coincides with the minimum of $D$.

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DeltaLima
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  • @user2168: As long as your trajectory through the air is reasonably horizontal, the component of gravity perpendicular to it (which will have to be canceled out by lift) does not change much. On the other hand, the component of gravity in the direction of your motion (which is what must outbalance drag) can change by much more as you climb or descend. – hmakholm left over Monica May 01 '14 at 14:33
  • @user2168 As Henning Makholm says, during glide lift $\approx$ weight as long as the glide is fairly horizontal and the vertical acceleration is zero. The chart that you pulled from the book is in the section 'Straight and level flight'. It mentions 'steady flight condition' and 'condition of equilibrium' and 'a lift equal to the aircraft weight' when the chart is discussed. How much more explicit would you like it to be? – DeltaLima May 01 '14 at 15:03
  • @user2168 From your comments on this answer and the other answer, it seems to me that you think that during glide the lift is unequal to the weight. Make sure you have a good understanding of the balance of all four forces (lift, drag, weight, thrust) acting on an aircraft in steady symmetric flight. The equilibrium of these forces holds also during steady glide and steady climb. I'll be happy to assist if needed. – DeltaLima May 01 '14 at 15:12
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    Assuming a spherical aircraft in a vacuum... – Vikki May 21 '18 at 17:22
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Well, your lift equals weight, or the airplane drops out of the sky or climbs into orbit. Therefore, lift is constant. Then the point with minimum drag must be the one where L/D reaches it's maximum.

Peter Kämpf
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    This is a very simple, intuitive answer which most pilots should understand easily once they read it. +100 if I could! – Lnafziger May 01 '14 at 16:01
  • Is it correct for both propeller and jet aircraft? – Nikita Sep 18 '19 at 06:28
  • I heard that in a propeller-driven aircraft speed for (L/D)max is a little bit greater than speed for minimum drag, because of the thrust being not directly produced by the engine. – Nikita Sep 18 '19 at 06:37
  • @Nikita: What you probably mean is that maximum endurance speed for propeller aircraft is lower than the speed of optimum L/D. Since thrust is inverse to speed, flying at higher drag but lower speed reduces power for sustained flight to its minimum at a point where drag is not minimal. At maximum L/D range has its maximum for propeller aircraft and endurance for pure jets. Turbofans are in between. – Peter Kämpf Sep 18 '19 at 08:09
  • Lift equal weight, or there will be a vertical acceleration, i.e., your vertical velocity will be *changing! Many assume, (incorrectly), that weight greater than lift causes a descent, and weight less than lift causes climb. NO! these conditions do not cause vertical velocity, they cause vertical acceleration*! – Charles Bretana Oct 31 '20 at 02:15
  • @CharlesBretana: Correct. Now let that acceleration last some time. What do you get? – Peter Kämpf Oct 31 '20 at 05:25
  • Not sure where you're going with this. Are you just focusing on secondary effects rather than primary effects? But, the answer depends on orientation of the acceleration vector and how it is changing. Many possibilities exist. Is the direction of vector oriented vertically and constant? Then you get a loop or immellman. Is it constant with horizontal component? Then a climbing or descending turn. Is it rotating? Then you get a spiral. But I upvoted your answer. It is spot on correct. I just wanted to include the primary effect. It should at least be mentioned. – Charles Bretana Oct 31 '20 at 12:48
  • @CharlesBretana Acceleration integrated over time is speed. No pitching involved, only an accelerated vertical motion. Wait a while and you impact the ground or rise up into space (disregarding the fact that lift without atmosphere is zero). That is all I wanted to express. – Peter Kämpf Oct 31 '20 at 18:31
  • So what? Your statement is only true when you are integrating a constant acceleration. Acceleration is always a *function* of time, {A = f(t)}, not a constant . And so your second sentence ("Wait a while and you impact the ground or rise up into space"). is manifestly not generally true. What if acceleration function is Ax = A cos (t), and Ay = A sin (t)? - or any one of a thousand other variations? Granted, wait long enough and eventually you will impact the ground or rise into space, but that incontrovertible fact explains nothing. – Charles Bretana Nov 01 '20 at 00:33
  • @CharlesBretana: Come on - gravitational acceleration is constant for all practical purposes. A cyclic acceleration (like the sin or cos function you mention) is zero when averaged over time. Clearly, then lift is equal to weight and the premise of the answer is not fulfilled. Go somewhere else to bicker. Discussion closed. – Peter Kämpf Nov 01 '20 at 03:42
  • The point of my original comment is that Forces (and lift and weight *are* forces) do not cause velocity, they cause *changes* in velocity. That is a fundamental concept that should not be glossed over, no matter what might happen when you "average over time" or "integrate", or differentiate, or perform any of the many other secondary functions you can apply... Focusing on secondary, long-term effects of a condition/trigger without even mentioning the direct, first order consequences is misleading. And you are obfuscating and deliberately pretending to miss this point. – Charles Bretana Nov 02 '20 at 01:41