Does lift equal weight in level flight? Or does it equal weight + down pressure on the tail?

Added) Let's say the total weight of the aircraft on the ground was 100,000 lbs, and when it took off and leveled off, the down force from the tail was 10,000 lbs. Then what's the total lift required to support the level flight?

  • $\begingroup$ Lift should be considered as a whole, as the lift of the entire aircraft, wings, tail and body... $\endgroup$ – xxavier Jul 11 '17 at 10:23
  • $\begingroup$ @xxavier Thank you for your reply but I'm not sure your comment is an answer to my question $\endgroup$ – lemonincider Jul 11 '17 at 10:36
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    $\begingroup$ That's why I wrote it as a comment, and not as an answer... $\endgroup$ – xxavier Jul 11 '17 at 10:47

When you consider the aircraft as a whole, lift equals weight in level flight. This must be true following the laws of physics.

However, if you break down the forces, the lift produced by the wings does not necessary equal to the weight of the aircraft. Lift is also generated by the body (albeit very inefficient) and by the tail. Most horizontal stabilizers generate negative lift, but some (e.g. the A380) generate positive lift.

If we ignore the fact that fuselage produces lift, and assume a configuration where the tail produces downward force, then you are correct. But the fuselage does produces lift. Modern aircraft designs use software to take that into account.

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    $\begingroup$ Can you clarify "Modern aircraft designs use software to take that [fuselage lift] into account." Do you mean the lift predictions in the flight control software use fuselage lift, or that modern aircraft shape and structural design includes calculations of fuselage lift? $\endgroup$ – Cody P Jul 11 '17 at 16:12
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    $\begingroup$ "Most horizontal stabilizers generate negative lift" Why is that? $\endgroup$ – asawyer Jul 11 '17 at 18:03
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    $\begingroup$ @asawyer so that if power is lost the airplane is in nose down attitude, which is safer, rather than nose up, which can lead to stall aka unsafe. $\endgroup$ – vasin1987 Jul 11 '17 at 18:09
  • $\begingroup$ @vasin1987 Ah that makes sense. Thank you! $\endgroup$ – asawyer Jul 11 '17 at 18:14
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    $\begingroup$ @David yes it would cause upward pitching moment. However when airplane losing speed, say engine out, the upward pitching moment would be lesser make airplane to pitch down and gain airspeed. $\endgroup$ – vasin1987 Jul 12 '17 at 2:20

What exactly you mean by "lift" is important to answer this question. In unaccelerated flight (steady speed, constant rate of climb or descent, including level flight), all forces on the airplane are balanced - forward equals backward, up equals down, and left equals right. Weight is generally a downward force, lift upward, drag backward, thrust forward, and sideways force is generally minimal, but comes from things like rudder trim or the various forces involved in P-Factor (this also changes based on how you define "up", as in a turn lift will become a partially sideways force if "up" is defined by something outside the airplane, e.g. earth).

If by lift you mean the sum of all up/down aerodynamic forces (which is how I tend to think of it, in general, though this is not technically correct - see footnote or linked article) and by "up" you mean "directly away from earth's center of gravity", then yes, lift equals the aircraft's weight in unaccelerated flight. There may be some negative lift (as you say in your example, the 10,000 pounds of tail-down force), but this will be balanced by additional upward lift over the airplane's weight - that is, if the sum of all downward aerodynamic forces is 10,000 pounds and the aircraft weighs 100,000 pounds, then the sum of all upward aerodynamic forces must be 110,000 pounds. However, when speaking of the sum of forces, we must remember to treat them like the vectors they are - so 110,000 pounds of upward lift plus 10,000 pounds of downward lift equals 100,000 pounds of net upward lift - the airplane's weight.

If your definition of lift does not include that created by the engines (i.e., "vertical component of thrust"), or really, if you use any other definition without restricting it to special cases, then this may or may not be true - if the vertical component is zero (it's usually small, but probably not quite zero), then total lift will equal aircraft weight for unaccelerated flight. If it is nonzero, then lift plus the vertical component of thrust will equal weight, but lift alone will not equal weight in unaccelerated flight.

Footnote on the definition of "lift": the technical definition of the lift force is the component of force exerted on a body by fluid flowing past it which is perpendicular to the flow of the fluid - so my definition would include the drag of a falling body as a "lift" force, which it is not.

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    $\begingroup$ Lift is clearly defined. Why do you need to make up your own definition (which is neither logical nor intuitive) for it? Would you say the statement "lift needed to perform a 60° bank turn is equal to the lift needed for level flight" is true? $\endgroup$ – Gypaets Jul 12 '17 at 7:39
  • $\begingroup$ I'm not so sure lift is clearly defined. Sure, in the easy case of an infinite fluid flow without turbulence, the direction of fluid flow is well-defined. Reality is somewhat more complex, with wind and convection. And I'm not sure lift is even defined at all for that "falling body" example. There are two directions perpendicular to the fluid flow, after all. $\endgroup$ – MSalters Jul 12 '17 at 12:43
  • $\begingroup$ Lift as a scientific term is clearly defined, but in practical terms for myself as a pilot, answering questions like this, that definition often leads to more confusion than clarity. Yes, the true lift forces on an airplane rarely equal its weight; a propeller's thrust, and much of a jet engine's as well, comes from the lift force, but it opposes drag instead of weight, and no one ever refers to thrust as "forward lift." It is much easier to speak of lift as "the force that opposes gravity" in many cases. $\endgroup$ – Stankinator Jul 12 '17 at 15:59
  • $\begingroup$ @Stankinator I have no problem using simplifications such as "thrust is parallel to aircraft axis", but referencing a flow magnitude (lift) to a random unrelated variable (gravity direction) doesn't make sense at all. How would you answer the lift statement question in my first comment? I really couldn't explain basic concepts such as "lift coefficient" or "increased stall speed in a turn" with your lift definition. Changing the definition depending on the question is for sure more confusing than just using the correct one. $\endgroup$ – Gypaets Jul 12 '17 at 17:43
  • $\begingroup$ @MSalters There are infinite directions (aka plane) perpendicular to the fluid flow and lift points in one of them. The good thing about having a consistent lift definition is that you don't have to change it depending on the flight attitude- it remains the same even in a nosedive in Mars. Admittedly, there are special situations where you need to define reference directions or quantities yourself, but it's not the case now. $\endgroup$ – Gypaets Jul 12 '17 at 17:57

Short Answer: No

Long Answer:

In real flight the equilibrium of forces in the Z axis is a little bit more complex. You should consider also the effect of the different angles and component of Thrust as indicates the next Figure: enter image description here

The equation should be:

enter image description here

If we consider static flight without normal acceleration and considering small values for the angles; the Lift Force would be $L = W-T*alpha$


Yes. Lift equals weight in level flight.

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    $\begingroup$ Answers should have a little background provided $\endgroup$ – Carl Witthoft Jul 11 '17 at 15:02
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    $\begingroup$ Perfect answer. $\endgroup$ – acpilot Jul 11 '17 at 17:08

A better way to say would be that in case of a steady level flight, the aerodynamic forces (including the propulsive force) and the body force are in equilibrium.

The aerodynamic force includes the lift created by the wing, fuselage, the down force created by stabilizer etc. So, in your case, the lift equals the weight plus the tail down-force.

  • $\begingroup$ So if the total weight of the aircraft is 100,000 lbs and the downward force from the tail is 10,000 lbs, can I legitimately say that the total lift the aircraft is producing in level flight is 110,000lbs? $\endgroup$ – lemonincider Jul 11 '17 at 12:00
  • $\begingroup$ @lemonincider No, the total lift would be 100,000 lbs since it is the sum of all positive and negative lift vectors. $\endgroup$ – SMS von der Tann Jul 11 '17 at 12:32
  • $\begingroup$ @SMS von der Tann Okay then let me put it this way. Let's say the total weight of the aircraft on the ground was 100,000 lbs, and when it took off and leveled off, the down force from the tail was 10,000 lbs. Then what's the total lift required to support the level flight? $\endgroup$ – lemonincider Jul 11 '17 at 13:10
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    $\begingroup$ The "down-force" the tail produces is lift - it just happens to be pointing downward. Calling it anything other than lift is confusing. In level flight, weight == net lift, where net lift == wing lift + body lift + tail lift. If the tail lift is negative (which isn't true for all aircraft), the wing lift or body lift must be more positive to make up for it. @lemonincider, in your example (wing lift + body lift) = weight - tail lift = (100k) - ( - 10k) = 110k lbs. $\endgroup$ – Carl Kevinson Jul 11 '17 at 14:44
  • $\begingroup$ @lemonincider: Can we agree that the lift of the wing-fuselage-combination is 110,000 lbs but the total lift is 100,000 lbs? $\endgroup$ – Peter Kämpf Jul 11 '17 at 15:28

You also forgot thrust. Particularly with vectored thrust at slow speed, you can get level flight with a high angle of attack/thrust vector and most of the weight is counteracted by thrust. e.g. this harrier transitioning to level flight

or this park flyer struggling into a head wind at an almost vertical angle of attack (hence little aerodynamic lift)
. Both of these are flying level with near to zero aerodynamic lift.

Net lift can be considered to be acting at 90 degrees to the velocity. Net lift includes contributions from wings, tail, canards, brakes, flaps and body, some of which may be acting upwards or downwards. Weight acts downwards. Thrust can acts at an angle to the velocity, either due to pitch or vectoring. So depending on pitch or vectoring, even if velocity is horizontal and net lift vertical, thrust can have a vertical component which acts against some of the weight.

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    $\begingroup$ I think the second video needs some kind of clarifying statement. As it is, "struggling into a headwind" statement followed by "near to zero aerodynamic lift" could lead the unwary to believe that wind doesn't contribute to aerodynamic lift. $\endgroup$ – Erin Anne Jul 11 '17 at 15:11

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