From my understanding, high lift to drag ratio could make an aircraft efficient during cruising conditions. This is when the aircraft is in equilibrium, lift is equal to weight, and thrust is equal to drag, and since there's less drag, less thrust is required. Is this correct and are there situations where having higher lift but higher drag as a consequence would be beneficial (e.g. higher camber has higher lift, but drag increases more leading to a lower lift to drag ratio)?

For example, if lift generated were to be around 10000N with 700N of drag (14.29 l/d) compared to less lift and drag, 8000N of lift with 600N of drag (13.33 l/d), would less drag at the expense of less lift be worth it in the context of cruising conditions (where thrust would need to equal drag to maintain constant velocity)?

Sidenote: I'm mainly thinking about this in terms of cruising conditions, not necessarily during climb or descent, but if you're happy to share any info on that, I would appreciate it too!

EDIT: Just to clarify (thanks for the comment quiet flyer), this is lift and drag generated by the wings.

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    $\begingroup$ This question could use some clarification. There are certainly cases where having a high lift coefficient, but a low L/D ratio, would allow the aircraft to come in on final approach at a low airspeed but a steep approach path. Clearly that could be good in some situations. In cruise-- not likely. Anyway, if you mean lift coefficient rather than just plain lift as a force in pounds or newtons, you should say so. $\endgroup$ – quiet flyer Oct 31 '20 at 23:54
  • $\begingroup$ Are you sure you want to specify absolute lift and not its coefficient? Realize that, once you take off, the only way to increase required lift is to pull g-forces (e.g. in a tight turn). So your question is really, "is there a situation where high g-forces are pulled where you want a lot of drag?". Is that your intended question? $\endgroup$ – Sanchises Nov 1 '20 at 8:30
  • $\begingroup$ Alright now I'm actually quite confused about the difference between lift and lift coefficient, but I'm not talking about g-forces, just generated lift by the wings. $\endgroup$ – SharkyPanda Nov 1 '20 at 10:49
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    $\begingroup$ I was waiting for the question to be clarified before answering, but it seems that the confusion between forces and their coefficients is not likely to be straightened out any time soon, plus there are now answers posted that would be invalidated by substantial changes, so here goes-- $\endgroup$ – quiet flyer Nov 2 '20 at 12:23
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    $\begingroup$ @SharkyPanda, if you are talking about cruise, do you realise that lower absolute lift implies lower weight of the aircraft? It's an absolutely valid thing to ask, but you need to clarify that the conditions are not the same. If, on the other hand, you have the same conditions (i.e. weight) in mind, then, as others have noted, you can't arbitrarily change [absolute] lift. Depending on your choice, one or the other of the already posted answers will be correct. $\endgroup$ – Zeus Nov 3 '20 at 0:10

Landing phase would benefit from high lift but low lift-to-drag ratio. At most phase of flight you need about the same amount of lift to keep the plane in the air. However during landing you need to slow down to landing speed. Hence you lower lift-to-drag ratio by keep the same amount of lift but increase amount of drag. This is usually accomplish by selection of large amount of flaps.


In cruise: No. Less drag means less thrust, which is always beneficial for the practical operation of an airplane.

There is only one condition except for approach and landing where high drag helps, an that is also not during cruise: In aerobatic airplanes during vertical maneuvers.

If, for example, the aerobatic display includes a vertical dive, high drag will reduce the dive speed the aircraft can reach, so the inertial loads at the end of the dive will be lower. Also, lower speeds means that maneuvers can be flown more tightly for the benefit of spectators.

Other than that: No, less drag is always better.

  • $\begingroup$ This answer is seemingly saying that having lower lift to have less drag is better, even if it means having a lower lift to drag ratio. Can you clarify this confusion please? $\endgroup$ – SharkyPanda Nov 2 '20 at 19:07
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    $\begingroup$ @SharkyPanda I never touched lift in that answer. It only deals with drag. Lift is non-negotiable: It must at least be equal to weight; in turns even higher. Less drag with equal lift is better. $\endgroup$ – Peter Kämpf Nov 2 '20 at 22:37
  • $\begingroup$ Well, by mentioning vertical maneuvers and specifically vertical dives, you kind of did touch lift-- would it be fair to say that the same characteristics that give high Drag in a vertical dive also tend to degrade the max possible L/D ratio that can be attained in cruising flight, where Lift is high? The point being that the tradeoff is acceptable in purpose-designed aerobatic biplanes etc-- $\endgroup$ – quiet flyer Nov 3 '20 at 12:48
  • $\begingroup$ @quietflyer: That is correct. Drag-producing designs tend to produce their drag with or without lift. In a pure vertical maneuver overall lift is zero, btw. $\endgroup$ – Peter Kämpf Nov 3 '20 at 17:09

High lift at the expense of even higher drag means that the plane will not be able to fly very fast, as drag rises sharply with speed. But the extra lift is still useful in several situations and is often provided by drag-creating high-lift devices. Some of these situations include:

  • STOL (short takeoff and landing) and low-speed flight performance, where high lift at low speed is the critical capability. High-lift devices such as large slats and/or large flaps are deployed, then often retracted for cruise.

  • Fast climb rate, where surplus lift is essential but forwards speed less so. Some planes will partially deploy their flaps to improve climb rate.

  • Extreme high-altitude planes. All planes are limited in altitude because the stalling speed increases with altitude and eventually reaches the plane's maximum speed. Reducing the stalling speed, typically by adding wing area, allows the plane to fly higher. At lower altitudes these huge wings will be draggy, even though the lift is the same at all altitudes.

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    $\begingroup$ Re "Fast climb rate, where surplus lift is essential.... Some planes will partially deploy their flaps to improve climb rate.-- well, we're back to the old coefficient versus force thing, which the asker of the question has already expressed some puzzlement about. A high lift coefficient may (?) help deliver a fast climb rate-- actually it is more important to a steep climb angle -- but lift is actually less than weight in a climb ( aviation.stackexchange.com/a/56476/34686). The suggested improvement being to consider replacing "lift" with "lift coefficient". $\endgroup$ – quiet flyer Nov 1 '20 at 14:32
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    $\begingroup$ @quiet flyer I beg to differ here. Aircraft that use wings for lift are far more efficient using them to climb than thrust. Any climb angle you pick, more lift is lost from the wing than gained from the prop. More thrust is used to maintain airspeed at higher wing AOA. We need to get this squared away to continue improving the state if our art. We can see raising AOA is what is actually done to climb. The lowest drag combination of thrust and AOA produces the best climb. One of your remembered quotes: at cruise, I just pull a little stick to climb. Why is this smart? $\endgroup$ – Robert DiGiovanni Nov 1 '20 at 15:58
  • $\begingroup$ BTW we do have "curved flight" from excess lift in a turn, but we don't in a side slip (because of the cross wind). What prevents upward "curving" from excess lift in a climb? Happy to continue working on this. We are getting better. One needs excess vertical Lift to climb. Upward acceleration is matched by vertical drag. This is a mirror image of gliding is it not? $\endgroup$ – Robert DiGiovanni Nov 1 '20 at 16:16
  • $\begingroup$ Let us continue this discussion in chat. $\endgroup$ – quiet flyer Nov 1 '20 at 17:03
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    $\begingroup$ @Guy Inchbald that's the whole fuzzy thing about coefficient of lift factor, it combines AOA and wing type/configuration. Generally, you see the bush planes with lots of horsepower to over come the drag of high lift configurations (gobs of camber, slots in front, slots in flaps), but it does get them off the ground at much lower V. Really the antithesis of the graceful sailplane. $\endgroup$ – Robert DiGiovanni Nov 2 '20 at 15:18

The question shows some confusion around the difference between forces and their coefficients.

Let's address forces first.

The key thing about forces is that in an unaccelerated state (which excludes turning flight) we have to be able to rearrange the force vectors into a closed triangle, square, or other closed figure.

As in the vector diagrams shown in these ASE answers:

Can we show through simple geometry rather than formulae or graphs that the best glide ratio occurs at the maximum ratio of Lift to Drag?

Can we show through simple geometry rather than formulae or graphs that the best glide ratio occurs at the maximum ratio of Lift to Drag?

Does lift equal weight in a climb?

Is excess lift or excess power needed for a climb?

You can see that in the simple case where the Thrust vector is parallel to the flight path (or is zero), Lift = Weight * (cosine glide or climb angle), so Lift is less than Weight if we're descending or climbing. Lift only equals Weight when we are flying level.

We'll continue to keep things simple by assuming that Thrust vector acts parallel to the flight path, and therefore directly opposite to the Drag vector, throughout the remainder of this answer. (See the last link above for more a more detailed treatment of the case where a significant upthrust or downthrust is in fact present, in relation to the direction of the flight path.)

For small climb or glide angles, the decrease in the Lift vector is small, but it is not zero.

Since a force is proportional to its coefficient * airspeed squared, there's no problem with the lift coefficient being higher in a slow climb than in fast cruise. Lift and drag coefficients are correlated to angle-of-attack. If we've increased the angle-of-attack, we know we've increased the lift coefficient. But not the Lift force. Excess Thrust, i.e. more Thrust than Drag, not excess Lift, is the key hallmark of a climb.

An excellent place to start learning more about lift and drag coefficients is section 4.5 of John Denker's "See How It Flies" on-line book.

See especially the green line on fig 4.14, lift coefficient versus alpha, fig 4.16, lift coefficient versus airspeed, and figure 4.17, Lift force versus airspeed. You can see how in level flight, Lift stays equal to Weight, even though the lift coefficient is changing as angle-of-attack and airspeed change.

After finishing reading this answer, the reader may have a good idea of how to modify these graphs for climbing flight at some given climb angle. The key changes are: the total Lift force is decreased by a factor equal to the cosine of the climb angle, and the airspeed for any given angle-of-attack is reduced by a factor equal to the square root of the cosine of the climb angle. All because some of the Weight is borne by the Thrust vector rather than the Lift vector.

Many people find the idea that Lift is less than Weight in a climb to be very counter-intuitive. What happens when we transition from high-speed cruise into a climb by pulling back on the control stick or yoke to slow to closer to Vx, with no change in Thrust?

The key hallmark of a climb is excess Thrust compared to Drag. If Thrust is constant, we have to reduce Drag if we want to climb1. How can we do that? By improving the L/D ratio. We increase the angle-of-attack, so that the lift coefficient increases. The drag coefficient increases too, but not as much, so the ratio of lift coefficient to drag coefficient improves. The L/D ratio is arithmetically equal to the ratio of lift coefficient to drag coefficient, so the L/D ratio improves too. When the airspeed is finished adjusting itself (decreasing) so that the Lift vector doesn't exceed the Weight vector, and more specifically becomes equal to Lift * cosine (climb angle), we find that Drag is now less than Thrust, and up we go.2

We can see from the vector diagrams in Does lift equal weight in a climb? that our climb angle will be equal to the arctangent of ((Thrust-Drag) / Lift)), which is also equal to the arcsine of ((Thrust-Drag) / Weight)). Note the latter expression-- it is important to understand that the climb angle can be expressed in a formula that makes no reference to the Lift vector at all. Optimizing the climb angle is all about maximizing the value of (Thrust-Drag).

A key point is that increasing the angle-of-attack doesn't actually improve the drag coefficient. So how can it reduce the Drag force? Because the increased lift coefficient causes a decrease in airspeed that causes a net reduction in the Drag force.

Of course, there's short interval immediately after we've moved the stick or yoke aft, but before the airspeed has had time to decrease substantially, where Lift is actually greater than Weight. During this time, the flight path is curving upward into the climb. This is an accelerated condition. The curve may be so gentle that the pilot can't even feel the extra G-loading, but some extra G-loading is in fact present during the transition. During this transition, we've also increased Drag. The increased Drag force, plus the fact that as the flight path starts to curve upward, the Weight vector starts to gain a component acting against the direction of the airspeed vector, are responsible for the loss of airspeed during this transition into the climb.

With that background under our belt, let's turn our attention to some specific aspects in the original question.

From my understanding, high lift to drag ratio could make an aircraft efficient during cruising conditions. This is when the aircraft is in equilibrium, lift is equal to weight, and thrust is equal to drag, and since there's less drag, less thrust is required. Is this correct


For example, if lift generated were to be around 10000N with 700N of drag (14.29 l/d) compared to less lift and drag, 8000N of lift with 600N of drag (13.33 l/d), would less drag at the expense of less lift be worth it in the context of cruising conditions (where thrust would need to equal drag to maintain constant velocity)?

Here's the tongue-in-cheek answer:

Absolutely! You've reduced Weight by 2000 Newtons, so you'll need less Thrust. Your Drag force is now only 600 Newtons, so that's how much Thrust you'll need. Reducing Weight is always helpful, if the goal is to minimize Thrust required in cruising flight.

But if the intention was that Weight actually remained constant, well-- hopefully by now the reader will understand that there's a flawed paradigm at play here. If Weight is constant, then no matter what we do to the L/D ratio, we aren't free to vary the size of the Lift vector in cruising flight. Lift is constrained to equal Weight. But we can optimize the ratio of lift coefficient to drag coefficient, i.e. the ratio of Lift to Drag, so that Drag is minimized and the Thrust requirement is therefore also minimized.

On the other hand, as other answers have noted, a low ratio of lift coefficient to drag coefficient, and therefore a low ratio of Lift to Drag, is beneficial to make the glide path steeper during approach. But we also want a low airspeed, and therefore a high lift coefficient, as we approach to landing. These goals are not inconsistent-- flaps will deliver them nicely. Flaps increase the lift coefficient, but they increase the drag coefficient even more.

In hindsight the idea of reducing Lift (without reducing Weight) in cruising flight may seem a bit foolish. But the truth is that pilots very often envision that Lift is greater than Weight in a climb, which is an equally flawed concept.3 So the confusion is understandable. What we actually often increase to enter a climb is not Lift, but lift coefficient. And the fundamental reason that we do this is not because we need more Lift, but rather because we want a better ratio of lift coefficient to drag coefficient and therefore a better ratio of Lift to Drag. All because we need to reduce Drag, to maximize our climb angle for a given amount of Thrust.

Likewise, pilots often speak of a "high-lift wing". It would be more clear, and might help to avoid some of the confusion contained in the original question, to speak of a wing with a high maximum lift coefficient.

A related question-- what is going on with a STOL bush plane with slats, lots of wing camber, etc-- if climbing is all about minimizing Drag rather than maximizing Lift, then how can all that Drag help the climb angle?

Answer-- the main purpose of those design features is to increase the lift coefficient and help the aircraft land and take off slowly. If thrust is constant, then the aircraft's maximum climb angle occurs at the maximum ratio of Cl/Cd and L/D, where Drag is minimized and (Thrust-Drag) is maximized. Many of those design features increase the drag coefficient so much that they decrease the maximum ratio of Cl/Cd and L/D. However, in the real world, for piston and turboprop engines, substantially more thrust is available at lower airspeed than at higher airspeed, so some of those features may offer a net improvement in the maximum climb angle after all.4 Consider also that there's a benefit in obstacle clearance if the maximum climb angle is achieved soon after takeoff, rather than after a long period of acceleration. But the fundamental reason that those features are there is to permit slow--i.e. short-- takeoffs and landings.


  1. Of course in reality, Thrust does not actually stay constant as we increase the angle-of-attack and slow to a lower airspeed, especially in a piston-engine aircraft, which tends to produces a roughly constant amount of power. More Thrust typically becomes available as we reduce the airspeed, which is another reason why the climb angle improves as we slow down to get closer to the Vx airspeed.

  2. See footnote 3 for a completely different-- and erroneous-- description of what happens when we move the stick or yoke aft to start climbing, taken from FAA flight training materials.

  3. For example, on page 3-16 of the FAA's "Airplane Flying Handbook" (2016), we read "When an airplane enters a climb, excess lift must be developed to overcome the weight or gravity. This requirement to develop more lift results in more induced drag, which either results in decreased airspeed and/or an increased power setting to maintain a minimum airspeed in the climb. An airplane can only sustain a climb when there is sufficient thrust to offset increased drag..." As discussed in this answer, this is simply not true. If we transition from level flight to a climb without doing something (like slowing to Vy or Vx) to reduce the Drag force to a lower value than we had a level flight, we certainly will have to increase Thrust, but that that excess Thrust is being used to help support the aircraft's Weight, not to counteract increased Drag due to increased Lift. FAA ground school training materials of this nature are often found to be poor sources of factual information about the actual forces present in various in-flight scenarios. See for example the poor depiction of the forces in gliding flight in the "Glider Flying Handbook" (2013) as discussed near the end of this related ASE answer, and the poor depiction of the forces in slipping or skidding flight in the "Pilot's Handbook of Aeronautical Knowledge" (2016) featured in this related ASE question.

  4. Sometimes so much power is available that the maximum available climb angle is simply not an issue of concern-- watch bush plane on steroids "Draco" on this You Tube video-- but still note the difference in configuration between approach and landing and takeoff and climbout.)

  • $\begingroup$ Much of this answer could be well adapted to a potential ASE question "What exactly happens to the magnitude of the lift vector as we transition from high-speed cruise into a moderate climb by moving the stick or yoke aft slightly, while holding Thrust exactly constant"? Staying faster than max L/D speed throughout. Maybe someone should ask that question. Maybe someone would like to attempt to offer an alternative theory of what happens in such a case. $\endgroup$ – quiet flyer Nov 2 '20 at 21:02
  • $\begingroup$ Such a question might arguably be a duplicate of aviation.stackexchange.com/q/40921/34686, except that question specifically says it is really asking about the sum of all vertical forces , which is not actually the same as Lift. $\endgroup$ – quiet flyer Nov 2 '20 at 21:04
  • $\begingroup$ If the OP insists on the absolute numbers (as opposed to coefficients), then the paragraph in bold is the answer, and there is nothing cheeky about it. I was about to post it until I saw yours, but I fear not everyone will reach that part... ;) $\endgroup$ – Zeus Nov 2 '20 at 23:58
  • $\begingroup$ @Zeus -- "but I fear not everyone will reach that part... ;" -- truer words have never been spoken-- $\endgroup$ – quiet flyer Nov 3 '20 at 3:27

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