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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.

Robert DiGiovanni
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    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 – Frog Feb 06 '22 at 20:27
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    Does this answer your question? Why is the L/D ratio numerically equal to the glide ratio? – Sanchises Feb 06 '22 at 20:30
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    I think this answer covers the distinction quite well: Why would maximum fly-time (endurance) not coincide with max L/D operating point? – ROIMaison Feb 07 '22 at 14:25
  • Not a duplicate- I'm pretty sure we've never before had a question about maximizing the quantity Speed/Drag. – quiet flyer Feb 07 '22 at 16:42
  • "Best glide" in what sense? Time or distance? Assuming still air, or a tail/head wind. – Max Power Feb 07 '22 at 21:18
  • @MaxPower distance, and, as your mentioned in comment, the horizontal component (earth referenced). – Robert DiGiovanni Feb 08 '22 at 04:51
  • @RobertDiGiovanni, I'm trying to understand what it is you're looking for exactly. For V minimum sink (i.e. staying in the air as long as possible, Maximum endurance), we don't care about the speed (we're not going anywhere) and hence we try to minimize drag for a given Lift equal to Weight. For the V best glide (i.e. getting as far as we can, max range), staying in the air alone is not enough, we need to minimize the power used, so we need to maximize drag times V for a given Lift equal to Weight. These goals are the same regardless of where the energy (initially) comes from. – ROIMaison Feb 08 '22 at 16:14
  • @ROIMaison sort of stumbled my way to an easy "pocket proof" the L/D max is at Vbg. Vmin sink had me confused, but using Force vectors (with gravity) w/out V (power) has provided a solution. Thanks. See answer w/ diagrams. – Robert DiGiovanni Feb 08 '22 at 17:07
  • @ROIMaison -- re "Maximum endurance), we don't care about the speed (we're not going anywhere) and hence we try to minimize drag for a given Lift equal to Weight." -- this is not actually true. See my answer, including my comments below (which I'll eventually integrate into answer)-- Drag is actually minimized at the speed for best L/D, not the speed for min sink. Min sink happens to be the speed where we minimize the quantity ((lift coefficient cubed)/(drag coefficient squared)). And best glide doesn't minimize Drag * V, it simply minimizes Drag. – quiet flyer Feb 08 '22 at 17:13

2 Answers2

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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.

quiet flyer
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  • 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. – Robert DiGiovanni Feb 07 '22 at 19:26
  • 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. – Max Power Feb 07 '22 at 21:28
  • @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. – quiet flyer Feb 07 '22 at 23:10
  • @MaxPower -- see for example https://aviation.stackexchange.com/a/81831/34686 – quiet flyer Feb 08 '22 at 14:05
  • 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."-- https://aviation.stackexchange.com/a/81831/34686. – quiet flyer Feb 08 '22 at 14:10
  • "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. "-- even if the graph had a very "peaked" shape, the speed that maximizes Speed/Drag will still always be higher than the speed that maximizes 1/Drag, which is the same as the speed that maximizes Lift/Drag. – quiet flyer Feb 08 '22 at 14:32
  • @quietflyer Exactly my point, it is defined dependent on context. I should have said relative to bulk airflow rather than fuselage, that was sloppy. If lift is defined as vertical (relative to earth) then wing lift of a glider will never be exactly equal to weight because there is also a vertical component of the drag vector. While in airfoil design lift is always normal to the bulk air flow; in this case total lift is the vector which gives a vertical component equal to weight minus the vertical component of drag, which could mean total lift is greater than weight. – Max Power Feb 09 '22 at 03:30
  • @MaxPower total lift is greater than weight it would not be steady state. The vertical lift vector triangle is closed by the component of the gravity vector pulling the plane forward. Drag simply is opposite the direction of flight. Reduction in Lift requirement is weight - (cosine angle of descent x weight). Factors involved in increase of drag at Vmin sink are: increased AoA, increased form drag, increased tail drag, greater tail downforce. At Vmin sink drag slows the plane down, even though it's angle of descent is steeper. – Robert DiGiovanni Feb 10 '22 at 00:28
  • By that logic flying in a circle cannot be steady state because the total lift is greater than the weight of the aircraft (Approximately 1.4*weight in a turn created by a 45degree bank to achieve a vertical component of lift equal to the weight.) Gravity pulls down and only down, not forward, it is the same with a sled sliding down a slope, gravity provides the energy and vertical force but it is the hill that applies the horizontal component of force. Likewise the wing's lift acts like the hill providing a horizontal forward force to counter drag. – Max Power Feb 10 '22 at 06:45
  • @MaxPower hear you there, one can consider drag against the angled bottom of the glider as the driving force. That is one way of seeing it. Yes, gravity only pulls "down", but from the plane reference it also provides a forward force component. I particularly like it because sin angle × W provides the "sum of all drags" in a glide. – Robert DiGiovanni Feb 10 '22 at 08:24
  • @RobertDiGiovanni Gravity is not providing any forward force in any reference. The forward force only comes from the lift vector being slightly forward. "Weight" is created by lift not by gravity. Gravity isn't really a force at all, the "force" is the ground or lift constantly accelerating you up to maintain a relative position. A weird effect of curved spacetime theoretically linked to the forward flow of time. When "falling" [in a vacuum] you are stationary in spacetime but spacetime its self is moving toward the earth... – Max Power Feb 15 '22 at 06:45
  • ... like being in a lifeboat, you really don't move relative to the water but you are going along with the general ocean current so you are moving relative to the stuff which is resisting the flow (and thus encountering a force). But with gravity the water is spacetime its self. This also helps explain time dilation in gravity wells and why GPS time is not equal to earth surface time, if the dilation was to accumulate it would cause gps nav error of 6miles per day. Clearly its not a perfect analogy, one is a 2D surface and the other is 4D. – Max Power Feb 15 '22 at 06:47
  • @MaxPower describing reality in mathematical terms and formulas can be local. In other words, "earth is flat" is good enough within a few km, but certainly not in orbit. In aviation, gravity is a force of constant magnitude and direction. When an aircraft has a vertical component towards earth, it has an acceleration = G × sin angle of descent × weight. – Robert DiGiovanni Feb 15 '22 at 08:53
  • @MaxPower. Draw the vectors, make it a 90 degree descent, gravity pulls forward, any lift curves the aircraft flight path. I was a big believer in the "forward lift vector" originally too. – Robert DiGiovanni Feb 15 '22 at 09:00
  • @MaxPower - re gravity not providing any forward force- depends on how define forward- food for thought : see https://aviation.stackexchange.com/a/56371/34686 and https://aviation.stackexchange.com/a/75480/34686 – quiet flyer Feb 15 '22 at 18:05
  • You are correct I should have said horizontal force not "forward", I was using forward in the broad sense of horizontal travel rather than local motion, the difference between a glider and a stone. In local relative motion there is no "forward" lift vector because lift by definition is a force perpendicular to flow direction. But there can be a horizontal lift vector if there is any non-zero vertical component of motion. Gravity still only has a vertical component and therefore cannot provide any horizontal force, but gravity can provide vertical motion and thus allow some horizontal lift. – Max Power Feb 19 '22 at 20:23
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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

Robert DiGiovanni
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  • 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!) – quiet flyer Feb 07 '22 at 16:46
  • 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: https://aviation.stackexchange.com/questions/36440/why-is-the-ratio-maximum-range-speed-maximum-endurance-speed-the-same-for-any-ai – xxavier Feb 07 '22 at 17:07
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    Just trying to rationize it with consistant units. Essentially, the flattest glide angle makes most efficient use of gravity. – Robert DiGiovanni Feb 07 '22 at 18:56