A question about "minimum power" airspeed had me wondering, why not think of "maximum Lift/Drag ratio" as maximum Lift/Thrust ratio.

Gliders would be included, as they use gravity for propulsion.

It seems that the condition of maintaining linear flight using the least amount of fuel (altitude) per unit time is V min sink.

So what is Vbg? It seems to be the maximum Speed/Drag ratio, but it burns more fuel per unit time than V min sink.

Power = Thrust x Airspeed

Would it be more accurate to describe Vbg as maximum Speed/Drag or Power/Drag (Thrust = Drag at steady state) ratio?

Looking at the units for Lift and Drag uncovers something troubling:

Lift is a Force = ma = mv/s that is constant regardless of speed.

Drag = Thrust = mv/s varies with speed and Angle of Attack

"Glider thrust" = fuel burn = mg×delta h/s = mv$^2$/s.

You see, the units of lift and the units of fuel burn do not cancel to form a proper ratio (no units).

Lift × Speed = mv/s × v = mv$^2$/s. Now one can divide this by fuel burn to get a meaningful ratio.

So "Lift/Drag" appears to be a bit of a curiosity$^1$. Here, slower may be better.

$^1$ analysis of the Clark Y airfoil show similar L/D ratio between AOA of +3 and +7 degrees.

  • 1
    $\begingroup$ From a glider perspective, min sink is generally slower than best glide; in the former case a certain amount of forward speed is sacrificed to give a relatively small improvement in sink rate. With a glider, the energy that goes into maintaining flight comes from a loss of height (in still air) and so power is directly proportional to sink rate and so min sink is the point at which power is at its minimum, by definition. I think the essence of your question relates to the difference between energy per unit time vs energy per unit distance travelled $\endgroup$
    – Frog
    Commented Feb 6, 2022 at 20:27
  • 3
    $\begingroup$ Does this answer your question? Why is the L/D ratio numerically equal to the glide ratio? $\endgroup$
    – Sanchises
    Commented Feb 6, 2022 at 20:30
  • 2
    $\begingroup$ I think this answer covers the distinction quite well: Why would maximum fly-time (endurance) not coincide with max L/D operating point? $\endgroup$
    – ROIMaison
    Commented Feb 7, 2022 at 14:25
  • $\begingroup$ Not a duplicate- I'm pretty sure we've never before had a question about maximizing the quantity Speed/Drag. $\endgroup$ Commented Feb 7, 2022 at 16:42
  • $\begingroup$ "Best glide" in what sense? Time or distance? Assuming still air, or a tail/head wind. $\endgroup$
    – Max Power
    Commented Feb 7, 2022 at 21:18

2 Answers 2


This answer is for gliders, or airplanes with failed engines or engines producing zero thrust. To adapt for a powered airplane with operating engine, in a descending glide under partial power, simply substitute "Drag-Thrust" for "Drag", throughout the answer.

The speed for best glide ratio (in still air) is the speed for maximum L/D.

For reasonably flat glide angles it is easy to show that Lift is only slightly less than Weight, and nearly all the variation in the L/D ratio, as airspeed is varied, is due to changes in Drag. A graph of L/D versus airspeed will have a rather broad "plateau" where we can vary the airspeed above or below the optimum for best L/D, without making a huge change in L/D. By the logic above, we are also not making a huge change in D itself. Therefore it's very clear that if we are trying to maximize Airspeed/Drag, this will certainly come at a higher speed than the speed for maximum 1/D (i.e. minimum Drag), which is very nearly the same as the speed for maximum L/D.

Therefore the speed for best glide ratio (in still air), which is also the speed for maximum L/D, and is also very nearly the speed for maximum 1/D, cannot also be the speed for maximum Speed/Drag. The speed for best glide ratio (in still air) is always slower than the speed for maximum Speed/Drag.

(It's not obvious that the quantity "Speed/Drag" should have any particular practical significance to a glider pilot, or to a power pilot for that matter, but that need not prevent us from talking about it.)

PS as you read this answer, please don't confuse Lift and Drag with their respective coefficients. While it happens to be true that the ratio of L/D is also the ratio of the Lift coefficient to the Drag coefficient, in most applications we can't simply interchange Lift and Drag with their respective coefficients, and most of the statements about Drag in this answer would not also be true of the Drag coefficient.

  • $\begingroup$ If one grinds through some numbers maximum (ratio of) speed to a given amount of drag will be similar maximum L/D, too slow and induced (from AoA) is less efficient, too fast and form drag is too much. It somehow made more sense to equate drag to "Glider thrust". But you are right, speed is not the critical factor for distance, glide angle is. $\endgroup$ Commented Feb 7, 2022 at 19:26
  • $\begingroup$ Are yo using lift in the sense of normal to the surface of earth/gravity or normal to the fuselage? Most uses I see are normal to the fuselage. A force vector could be broken into into horizontal and vertical components(relative to earth), or parallel to flight path and perpendicular to the flight path. $\endgroup$
    – Max Power
    Commented Feb 7, 2022 at 21:28
  • $\begingroup$ @MaxPower - the accepted definition of Lift is well established. See related ASE questions about L/D ratio and glide ratio. Maybe I should add links to some of them, to this answer. PS in gliding flight Lift is not defined as normal to the earth, and also is not defined as normal to the fuselage. $\endgroup$ Commented Feb 7, 2022 at 23:10
  • $\begingroup$ @MaxPower -- see for example aviation.stackexchange.com/a/81831/34686 $\endgroup$ Commented Feb 8, 2022 at 14:05
  • $\begingroup$ This answer needs some more improvement-- it states that the speed for max L/D is not exactly the same as the speed for max (1/D) (i.e. the speed for min D). In fact you can see from the L-D-W vector triangle that these two speeds are identical. I've actually already stated that in a previous answer-- "The vector diagrams above demonstrate that as long as Weight is fixed, when we maximize L/D, we also minimize Drag."-- aviation.stackexchange.com/a/81831/34686. $\endgroup$ Commented Feb 8, 2022 at 14:10

One way to resolve the issue of units is to consider "glider thrust" to be the sin of the angle of descent × the weight vector. Now there are all force units as mv/s or simply pounds.

It then becomes clear that, allowing for minor differences in lift requirements for angles of descent, modeling follows reality in that:

Best speed for maximum distance glide, after satisfying the lift requirement, will occur at Vbg.

A steeper angle of descent is used at V min sink (the plane does not glide as far). This resolves as higher AoA and lower speed keep one in the air longer, but more drag is used for lift, and less on airspeed.

Lift is similar, gravitational "push" is greater, therefor L/D is less.

A steeper angle of descent at lower AoA produces the same result: a shorter glide path, will more drag than Vbg from greater speed.

sin angle of descent x weight translates as "Drag"

enter image description here

  • $\begingroup$ What is "best speed for distance"? Are you trying to maximize Airspeed / Sink rate? If so, that quantity is essentially the definition of the glide ratio or very nearly so, for reasonably flat glide angles. (Wouldn't work for a falling brick!) $\endgroup$ Commented Feb 7, 2022 at 16:46
  • $\begingroup$ Minimum sink speed corresponds to minimum power required, and thus, to best endurance speed, the ratio best range speed/best range speed is 1.317. More info here: aviation.stackexchange.com/questions/36440/… $\endgroup$
    – xxavier
    Commented Feb 7, 2022 at 17:07
  • 1
    $\begingroup$ Just trying to rationize it with consistant units. Essentially, the flattest glide angle makes most efficient use of gravity. $\endgroup$ Commented Feb 7, 2022 at 18:56

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .