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?


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 $D$.

enter image description here

| improve this answer | |
  • $\begingroup$ @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. $\endgroup$ – hmakholm left over Monica May 1 '14 at 14:33
  • $\begingroup$ @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? $\endgroup$ – DeltaLima May 1 '14 at 15:03
  • $\begingroup$ @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. $\endgroup$ – DeltaLima May 1 '14 at 15:12
  • $\begingroup$ Assuming a spherical aircraft in a vacuum... $\endgroup$ – Sean May 21 '18 at 17:22

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.

| improve this answer | |
  • $\begingroup$ This is a very simple, intuitive answer which most pilots should understand easily once they read it. +100 if I could! $\endgroup$ – Lnafziger May 1 '14 at 16:01
  • $\begingroup$ Is it correct for both propeller and jet aircraft? $\endgroup$ – Nikita Sep 18 '19 at 6:28
  • $\begingroup$ 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. $\endgroup$ – Nikita Sep 18 '19 at 6:37
  • $\begingroup$ @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. $\endgroup$ – Peter Kämpf Sep 18 '19 at 8:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy