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On the Airbus A320, the two engines produce about 220 kN in cruise, whose weight is about 700 kN (about 70 tons). The lift / drag ratio in cruise is about 15 (likely higher), meaning in cruise lift = 700 kN and drag = 47 kN (or less).

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What is the accurate explanation of the figures above? How does 220 kN of thrust create the 700 kN required to maintain altitude and the 47 kN required to maintain speed?


marked as duplicate by Federico, Community Sep 20 '17 at 21:13

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  • $\begingroup$ Please note that drag increases with the square of velocity. And that airliners have operational Mach number limits. $\endgroup$ – Gürkan Çetin Sep 20 '17 at 20:04

I can't produce any equations, but the forward motion is what creates the lift. If you had a completely drag-free glider and no wind it could stay in straight flight forever with all forces in balance. Gravity counters lift. And no drag, so no thrust. The reason it must eventually come down is because the drag force slows it down. Less speed equals less lift. In a powered plane the thrust simply needs to balance against the drag. Then all forces are in balance again, so the only force the engines need to balance is the drag force.

  • $\begingroup$ I know this is the usual explanation and maybe you are just right to restate what I've read a thousand of times, but I can't make sense of that easily. The implication for the drag-free glider, is indeed surprising, but very logical. $\endgroup$ – mins Sep 20 '17 at 21:31
  • $\begingroup$ To be honest, I have trouble with this too. Sometimes I think I've got it, then I start to doubt. Maybe later on I can reword this so it makes more sense $\endgroup$ – TomMcW Sep 20 '17 at 22:07
  • $\begingroup$ Key is to rememeber that work is force scalar speed. It is all clear then. $\endgroup$ – Caterpillaraoz Sep 21 '17 at 7:13
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    $\begingroup$ To be fair, the drag free glider thought experiment is probably the most counterintuitive thing for me since lift creates drag, besides any further parasite drag from the glider body. $\endgroup$ – AEhere supports Monica Sep 21 '17 at 7:57

The engines only need to overcome the induced drag to remain in the air, which is a result of the lift but usually much smaller on the order of 1/20 for a modern jet. This ratio is specified as lift to drag or L/D. It is a vital performance statistic of an aircraft. Why? Wings are weird and complicated, there is not a simple 100% correct answer.

Only a rocket needs thrust equaling gravitational force since it has no wings! (L/D = 1/1)


the two engines produce about 220 kN in cruise

A single A320 engine (V2527-A5) produces 120 kN at SLC (sea level conditions). Both will be 240 kN. In cruise it will be much less thrust. 25% of that as a ballpark figure.

A solid example: The CF6-80C2B1F of a 747 produces 57,160 lbf at SLC, and 12,820 in cruise (22% of SLC). The thrust is much lower in cruise (less air, less thrust).

In any straight-and-level flight regardless of altitude, thrust = drag. Except if the flight is too slow with a high nose up then some of the thrust counters the weight, but for all intents and purposes, assume they are the same and you won't be wrong.

L/D is not a fixed value too. It changes with speed.

At low altitudes the drag is higher (denser air). In cruise the true airspeed is higher. Lift depends on the true, and not the indicated, airspeed. That's how less thrust can do more lift up there where the air is thin and the plane can reach fast true airspeeds.

The form of velocity applicable to the lift equation is the true airspeed. True airspeed is defined as the actual speed of the aircraft through the air and includes corrections for density, compresibility, and instrumentation error.

The main issue: The pulley system (now removed from question) is a one-body problem, an airplane is a two-body problem (airplane and air). The air does the lifting.

  • $\begingroup$ Also forward velocity is a factor that decreases the thrust of an engine. (Static thrust is larger than thrust at speed). $\endgroup$ – Gürkan Çetin Sep 20 '17 at 20:18
  • $\begingroup$ Then you have a supersonic airliner :) $\endgroup$ – Gürkan Çetin Sep 20 '17 at 20:20
  • $\begingroup$ "the air acts back on the wing but in a different direction (the two-body system)" that's not really what is difficult for me to understand. The difficult part is why 40 kN of thrust are sufficient to make this air act "in a different direction" in order to balance 700 kN without external energy. While I know what you would answer (this is L/D ratio), it's difficult to make sense of it. $\endgroup$ – mins Sep 20 '17 at 21:39
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    $\begingroup$ @mins Your question is a rephrasing of, "How do wings work anyways?", the answer to which is often explained simply but also subtly incorrectly. The physics textbook would be fluid dynamics. Before computers things just worked because empirical data said they did. $\endgroup$ – user9394 Sep 20 '17 at 22:55

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