It's been a while since I tried to read a little about flight mechanics and aerodynamics, but I remember one question I had back then:

Is the thrust of a common passenger aircraft (A320, B737) greater than its lift-off weight and thus could it take off vertically?


  • Max take-off weight 73.5 (78) tonnes
  • Thrust range 98 (120) kN

73,5 tonnes = 73500 kg * 9.81 m/s^2 = 721035 N = 721 kN

If my calculations are correct, it could clearly not (a result like what one would expect intuitively)!

But how can the dynamic lift be larger than the weight using comparatively little thrust? Rephrased: How can the dynamic lift be larger than the thrust?

Is there an invisible equivalent to gears at work here?

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    $\begingroup$ A complex math at work here but in simple form, thrust is used to generate forward movement. Movement of air above wings then create lift. Lift from wing then overcome weight for airplane to takeoff. Design of wings and angle of attack play more role than thrust. You can have tons of thrust but no lift with bad wing design and vice versa $\endgroup$ – vasin1987 Sep 22 '14 at 15:54
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    $\begingroup$ Since trust serves to overcome drag, the basic question boils down to 'how can lift be higher than drag?'. $\endgroup$ – florisla Feb 19 '15 at 15:56
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    $\begingroup$ I know this question is old, but it just popped up on the recent activity list. So, one thought: I had a somewhat related question at Physics.SE: "What does a wing do that an engine can't?" $\endgroup$ – yshavit May 24 '16 at 4:41
  • $\begingroup$ GLIDERS have lift with ZERO thrust. $\endgroup$ – Charles Bretana Aug 8 '19 at 1:43
  • $\begingroup$ @CharlesBretana The glider pushes itself with a gravitational component. $\endgroup$ – enbin Jan 12 '20 at 1:46

Thrust is needed to overcome drag, and a good airplane design can create a lot of lift for little drag. In the case of an A-320, the lift-to-drag ratio is 18 in cruise (a little less during take-off), so to lift those 78 tons needs only 4.33 tons of thrust. All it needs is the right amount of forward speed, and the engines only need to maintain this speed.

Gliders are even more efficient, their L/D can approach 60. A glider of maybe 450 kg mass (= 4412 N weight) will create a drag force of only 74 N, admittedly at a much slower speed than an A-320.

To be more precise, in a climb the thrust will actually compensate a little of the weight, in addition to the drag, depending on the climb angle. The thrust needs to grow by weight $\cdot$ sin($\gamma$), where $\gamma$ is the flight path angle. If the A-320 had enough thrust to lift all its weight, it could take off vertically, flying in a nose-up, vertical attitude. Modern fighter aircraft do really have that much thrust installed.

Forces on a climbing aircraft

This drawing shows a climbing aircraft. There are four forces at work:

  1. Lift L (blue arrow)
  2. Drag D (red arrow)
  3. Thrust T (green arrow), and
  4. Weight m$\cdot$g (black arrow)

For a trimmed state, the concatenated arrows should form a closed trapezoid (lightly shaded arrows). Note that both thrust and drag are drawn longer here in relation to weight and lift than they are in reality.

To answer the why part needs a few more lines. Lift is created by deflecting the airstream over the wings downwards. This change in momentum is lift, and creating it would be free of drag if the wing's span would be infinite and the air inviscid. In reality, flying causes these drag components:

  1. Lift creation incurs a small backwards-pointing component which is biggest at slow speed and decreases with speed. To be precise: This drag component is proportional to the aircraft's span loading (weight divided by wingspan) and the inverse square of airspeed. It is called induced drag because it could be successfully calculated first with the same equations that describe electric induction.
  2. Flying through air causes friction. This friction drag grows almost with the square of airspeed and is proportional to air density and surface area.
  3. In inviscid flow, the sum of pressures around a body would just cancel out. In reality, the pressure on the forward-facing sections is normally higher than that on the aft-facing sections. This drag component is called pressure drag and is also proportional to the square of airspeed at the same angle of attack.

At the speed of best L/D the sum of 2 and 3 is equal to 1, and 2 and 3 are of the same magnitude for a well-designed airplane. High subsonic and supersonic flight adds a fourth component, called wave drag, and it's onset is the reason why the A-320 prefers to fly not much faster than Mach 0.78.

  • $\begingroup$ mathsisfun.com/definitions/trapezoid.html -- maybe "trapezium" for speakers of British English? $\endgroup$ – quiet flyer Aug 7 '19 at 15:15
  • $\begingroup$ @quietflyer A bit attention to the colors would also add a lot of understanding. Note that the light arrows are shifted copies of the darker arrows, which of course all have labels. $\endgroup$ – Peter Kämpf Aug 7 '19 at 17:58
  • $\begingroup$ @quietflyer I am shocked - shocked! - that you failed to spot the obvious mistake with the dash in "A-320". This completely invalidates the whole answer - don't you agree? $\endgroup$ – Peter Kämpf Aug 7 '19 at 18:05
  • $\begingroup$ "Note that the light arrows are shifted copies of the darker arrows"- The way that part of the red drag arrow overlies the curve of the wing airfoil tends to give the illusion that it is shorter than it really is and creates the appearance of a big difference in length between it and the light red copy of it at the bottom of the trapezoid. This was really the root of my confusion; my mistake. In fact in my comment I mis-identified what the light red copy was supposed to represent. I still think care should be taken to make sure the Lift vector (and its copy) is shorter than the Weight vector. $\endgroup$ – quiet flyer Aug 8 '19 at 12:45

Peter's answer is very good, but to explain a little more of the 'why,' consider the forces that are at play. In order for a mass to not accelerate in a given direction, the sum of the forces acting on it in that direction must be zero (or the mass infinite, which brings about a whole new set of problems...) In the case of an aircraft that is cruising, there are 3 important categories of forces to consider: gravity, thrust, and the force of the aircraft hitting the air in which it's flying. The latter category is, of course, the most complicated one.

The sum of the forces applied to the aircraft by the air that it's running into can be modeled as 3 component forces which are all perpendicular to each other: the net sum of forces pushing the plane backwards (drag,) the net sum of the forces pushing the plane upwards (vertical lift,) and the net sum of the forces pushing the plane to the left or right (I'll call this horizontal lift.) When the plane is not turning, the horizontal lift is zero.

In order to maintain a given altitude, the vertical lift must be of equal magnitude to the weight of the airplane (the force applied to it by gravity). In order to maintain a given forward speed, the thrust must be of equal magnitude to the drag (assuming, for the sake of simplicity, that the thrust is pushing the aircraft straight forward.) So, the important thing to notice is that the thrust needs only to counter the backward component of the forces applied by air hitting the aircraft, not the total force applied by air hitting the aircraft. In a well-designed aircraft, the upward component of the force of air hitting the aircraft will be much larger than the backward component.


When the wing is moving horizontally, the air produces a force $F$ on the wing, and $F$ has two components, $L$ and $D$. Because the angle of attack of the wing is generally much less than $45$ degrees, $D$ is also much smaller than $L$. Since the thrust of the aircraft is only balanced with $D$. Therefore, with a small thrust, a large force $L $ can be generated, $L$ can be used to balance the weight of the aircraft.

enter image description here


Lift is greater than thrust because for a plane able to fly it can be demonstrated experimentally (in a wind tunnel) that Lift_constant > Drag_constant.

The formulas for lift and drag are:

1) Lift = Weight = Lift_constant x Density x Reference_surface x V^2

2) Drag = Thrust = Drag_constant x Density x Reference_surface x V^2

From 1 and 2 it follows that:

Lift = (Lift_constant / Drag_constant) x Thrust

If the plane is built in such a way that Lift_constant / Drag_constant > 1 then Lift > Thrust, otherwise it is not.

  • $\begingroup$ This answer could be improved by replacing "constant" with "coefficient". Also by deleting the first sentence. Also by specifying that the last two sentences apply only to linear, horizontal, constant-speed, steady-state flight. $\endgroup$ – quiet flyer Mar 25 '20 at 10:50
  • $\begingroup$ This answer could also be improved by noting that an aircraft with a L/D ratio of less than 1 would be quite a monstrosity and definitely an oddball/ outlier case. $\endgroup$ – quiet flyer Mar 25 '20 at 12:22

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