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Physics in schools teaches two contradictory and mutually exclusive things:

  1. That the upward lift force on an airplane in flight equal its weight (Lift = Weight = Mass x Gravity). This is based on applying Newtons 2nd law of motion (F = ma) to the airplane in flight.

  2. However, modern commercial airplanes like the Boeing 747–400 can fly with thrust-to-weight ratio as low as 0.3. Here the engine thrust is only 0.3x the weight of the airplane, but this thrust is sufficient to push the airplane forward and generate enough lift to fly.

Therefore the upward force required for lift and flight must be a lot less than 0.3x the weight of the Boeing 747-400 (Lift < Thrust < Weight). Both statements cannot be true. Which is correct?


marked as duplicate by Manu H, Juan Jimenez, bogl, DeltaLima, ymb1 Jun 19 at 12:40

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    $\begingroup$ You may read how it flies?. You are mixing 2 notions: thrust and lift. Keep in mind that thrust is not necessary to stay aloft (you can glide). Moreover you may rephrase the title of your question so that it is easier to browse through all questions speaking of physics of flight. $\endgroup$ – Manu H Jun 18 at 7:48
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    $\begingroup$ There are some problems with your question. It may help you to read the related answer aviation.stackexchange.com/questions/40921/… . In a steep climb the 747 is surely producing much more thrust than in level flight, but in neither case does the thrust need to be anywhere near equal to weight. Lift is LESS than weight in steady-state powered climb btw. $\endgroup$ – quiet flyer Jun 18 at 11:12
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    $\begingroup$ "With enough thrust, even a brick can fly." The only time that the thrust-to-weight is at a 1:1 ratio with lift is when you're talking about bricks, rockets, pianos, and other things that don't have wings. $\endgroup$ – Ghedipunk Jun 18 at 18:17
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    $\begingroup$ @Ghedipunk even a brick generates some lift if you throw it at the right angle. $\endgroup$ – alephzero Jun 18 at 22:48
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    $\begingroup$ @alephzero, very true (and have a comment uptick for pointing that out), as evidenced by the Space Shuttle. It makes its point better than talking about spherical cows that graze on frictionless planes, though. $\endgroup$ – Ghedipunk Jun 18 at 22:52

Your misunderstanding lies in your thought that lift is smaller than thrust, while in fact, lift is much larger than thrust.

The lift is provided by the wings. Their purpose is exactly to create a lift force (upwards force) while requiring relatively little thrust (forwards force). How well they do this is expressed by their lift-to-drag ratio (L/D ratio). A modern airplane can have an L/D of 20 or more. This means that for every newton of thrust, you can lift 20 newtons of weight!

You can think of wings the same way you think of an inclined plane: you need a lot less force to push a car up a gentle hill than up a steep hill. Indeed, if the hill is gentle, most people can push the car up the hill, while most people cannot single-handedly lift a car up. Wings work by pushing air down, and by pushing the air down at a relatively shallow angle (by going forwards real fast), you need less force to go forwards than to lift the plane.

If you want to know more than this 'intuitive' explanation, perhaps I can recommend you this answer of mine. I will present a terse derivation below, with the aim to calculate the L/D (neglection parasitic drag) purely from first principles.

Lift is due to a certain mass flow $\dot{m}$ being given a certain velocity $v$ downwards: $$L=\dot{m}v.$$ This mass flow is due to the wings encountering a certain amount of air; as an approximation, you can think that the airplane only affects a circular 'tube' of air with the diameter equal to its wingspan. The energy per unit of time (power) necessary to impart this downwards momentum equals $$P=\frac{1}{2}\dot{m}v^2.$$

This power is provided by the combination of thrust $T$ and forwards velocity, $$P=Tu.$$ This gives us an expression for the thrust: $$T=\frac{P}{u} = \frac{1}{2}\dot{m}\frac{v^2}{u}.$$ Note that the thrust $T$ equals drag force $D$, so we can now calculate the L/D:

$$\frac{L}{D}=\frac{L}{T} = \frac{\dot{m}v}{\frac{1}{2}\dot{m}\frac{v^2}{u}} = 2\frac{u}{v}. $$

In other words, like I said before, the L/D ratio is exactly how 'shallow' the relative velocity vector of the affected airflow is. It follows that the forwards speed $u$ should be as large as possible to maximize the L/D. However, I only took into account the so-called "lift induced drag" which is purely the horizontal force it takes to create a vertical lift force. In reality, there is also parasitic drag, which scales with $u^2$ and will at some point dominate over the reduction in L/D.

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    $\begingroup$ This is an outstanding answer in every regard. I enjoy that it starts out with a simple analogy and then gets more serious, and finally shows how the serious analysis supports the analogy. Also points out where the analysis breaks down. Very nicely done. $\endgroup$ – Wayne Conrad Jun 18 at 20:58

Because a wing produces way more lift than drag.

Thrust must equal drag, not lift.

...thrust-to-weight ratio as low as 0.3.

That's a drag-to-lift ratio, actually. And that is why you find fixed wing aircraft everywhere, otherwise all aircraft would be VTOLs

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    $\begingroup$ Drag-to-lift ratio is around 0.05. 0.3 is about right for a thrust-to-weight at full power, which allows the aircraft to climb quite fast. $\endgroup$ – Jan Hudec Jun 18 at 21:49

For any airplane to fly level at a constant altitude, the amount of lift generated by its wings must be equal to its weight. If lift exceeds weight, the plane starts to climb. If weight exceeds lift, the plane starts to descend. In a constant rate climb/descent, lift equals weight; a simplification that ignores the propulsive force vector.

The engine & prop in a light plane like a 4-seat Cessna develops no more than a couple hundred pounds of thrust at takeoff which is much less than the weight of the plane. But that amount of thrust is enough to move the plane's wings through the air at a speed sufficient to make enough lift to get the plane off the ground.

Note that the propulsive thrust does not have to equal the plane's weight in order to make it fly; this is only true if the plane had no wings and you were trying to hoist it off the ground by pointing the plane straight up and "hanging" it on its propeller.

  • $\begingroup$ Starting from level flight, if lift exceeds weight, this doesn't cause a climb but rather causes an upward curvature of the flight path-- like an entry to a loop. Likewise weight exceeding lift doesn't cause a descent but rather causes a downward curvature of the flight path. Lift is LESS than weight in a steady-state climb. See related answer aviation.stackexchange.com/questions/40921/… $\endgroup$ – quiet flyer Jun 18 at 11:06
  • $\begingroup$ "upward curvature of the flight path" aka a climb? Unless a climb at increasing rate is not climby enough in your Book of Climbs. $\endgroup$ – AEhere Jun 18 at 13:07
  • $\begingroup$ @AEhere -- sure it's plenty climby but it cannot be sustained for more than a few seconds -- you will either run out of steam as you approach vertical or you will go around past inverted in which case you are no longer climbing! $\endgroup$ – quiet flyer Jun 18 at 14:06
  • $\begingroup$ "If lift exceeds weight, the plane starts to climb. If weight exceeds lift, the plane starts to descend." This is not accurate. Lift (or, well, total upward force, part of which could come directly from thrust) = weight is true whenever the aircraft is not accelerating vertically. A climb at constant vertical speed will have upward force = weight, as will a descent at constant vertical speed. Slowing a decent rate will have upward force > weight, even though the aircraft is still descending. Slowing a climb rate will have upward force < weight, even though the aircraft is still climbing. $\endgroup$ – reirab Jun 18 at 16:23

When you push your car on a flat area you need very little force to move it, much less than its weight

With the same force you are pushing the car, you notice that at the beginning it will get an accelerated motion and afterwards due to the drag mainly, the speed becomes steady even though keeping the same pushing effort

Therefore in order to make an aircraft accelerate you need just to overcome the drag

Acceleration means increasing the speed, thus the drag, but all you need is to overcome the drag not the weight.

To lift off you just need to get the necessary speed for the wings to provide, as a function of the angle of attack, that lift that is just above the aircraft weight

  • $\begingroup$ I agree with this answer, except for the statement of not having to overcome weight. A car can sit on a flat surface, but the aircraft must stay at an incline to remain afloat. This incline may be small (like for a glider) but it means the plane must have some amount of energy expended at all times to fly. The amount of energy used is strongly linked to weight, in fact it's a squared component in the lift/drag equation. So I think that could be reworded, otherwise great answer! $\endgroup$ – YAHsaves Jun 18 at 10:15
  • $\begingroup$ Not a disagreement but rather a clarification that you may wish to include in your answer. Yes lift may slightly exceed lift at takeoff but only because the flight path is CURVING upwards, i.e. bending up into a climb. A curvature is a from of acceleration and requires force to accomplish, i.e. an excess of lift over weight. Once the climb is established and stabilized, lift is actually a bit less than weight. See related answer aviation.stackexchange.com/questions/40921/… $\endgroup$ – quiet flyer Jun 18 at 11:17
  • $\begingroup$ @YAHsaves, thank you, my answer if for beginners as you may imagine from the question content, so I had to explain the matter as intuitively as possible. $\endgroup$ – user40476 Jun 18 at 11:47
  • $\begingroup$ @quiet flyer, thank you, my answer if for beginners as you may imagine from the question content, so I had to explain the matter as intuitively as possible $\endgroup$ – user40476 Jun 18 at 11:48

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