0

I have come across this question on Twitter. I have found a similar question here. The aircraft isn't flying but just doing Cobra Maneuver. Since the direction of the lift is in the same direction and the weight is the same, can I conclude the lift in both cases will also be the same given that it isn't climbing?

enter image description here

Auberron
  • 1,567
  • 1
  • 11
  • 34
  • 3
    The question you linked addresses the thrust: "[...] thrust provides an additional upwards component", so why haven't you included the vertical thrust component? –  Apr 07 '22 at 12:22
  • From one of the answers ( https://aviation.stackexchange.com/a/56476/34686 ) to the related ASE question cited in your question-- " For the purposes of the answer, we'll assume that Thrust acts parallel to the flight path through the airmass, although this clearly not always exactly true. This simplifying assumption leads to the following vector diagram:..." Obviously that assumption would be extremely invalid for the "Cobra" maneuver! You need to take into consideration the part of the Thrust vector that is acting perpendicular to the flight path. – quiet flyer Apr 07 '22 at 14:18
  • (Ctd) That is not normally considered part of the Lift vector. And that should give you the answer. – quiet flyer Apr 07 '22 at 14:20
  • Is the top aircraft in equilibrium, or climbing? (The cobra maneuver is dynamic, but the picture is just a snapshot in time.) – Michael Hall Apr 07 '22 at 14:47
  • @MichaelHall Its in equilibrium – Auberron Apr 07 '22 at 15:10
  • OK, so the red arrows must counter the weight as you point out, and they will be equal. In the Cobra more of that comes from the engine thrust, but I trust you realize that... – Michael Hall Apr 07 '22 at 16:35
  • @Auberron, The top aircraft is generally not in equilibrium,. When the cobra is performed, as Angle of Attack (AOA) increases, the aircraft lift is increased dramatically, until it reaches, and then exceeds, the critical AOA. But because the maneuver overshoots the critical AOA it places the aircraft AOA on the lift curve to the right of the critical AOA, where the Cl may actually be close to or equal to the Cl at a much lower AOA. The coefficient of drag (Cd) of course is much higher. But this is momentary, as the aircraft unloads back to a normal AOA again. – Charles Bretana Nov 16 '22 at 15:15

2 Answers2

4

...can I conclude the lift in both cases will also be the same given that it isn't climbing?

No you cannot. If the aeroplane is not climbing in both instances, it means the vertical forces are in balance. But with a horizontal flight path and the nose tilted way up, thrust contributes to the upwards vertical force - therefore the wing lift must be lower.

Note that even though the nose is lifted up, lift is still vertically upwards if the flight path is horizontal, the plane is just flying with a large angle of attack.

Koyovis
  • 61,680
  • 11
  • 169
  • 289
  • Can you differentiate the case between the plane climbing up and not climbing at all? In which case lift will be smaller? – Auberron Apr 07 '22 at 13:22
  • 3
    Direction of the lift vector is defined as being perpendicular to the flight path. So in a climb the lift vector is tilted backwards, reducing the vertical force compensating for gravity. Thrust will have to have a larger contribution in a climb - lift in a climb will be smaller. – Koyovis Apr 07 '22 at 13:56
4

This question is a favorite on the internet threads and forums, which can be easily resolved after a review of technical terms.

According to the NASA Glenn research center Lift is a force the acts perpendicular to the free stream, not the heading, of the aircraft.

This presents a very interesting quandary to our understanding of "lift" in that by the NASA definition, any force component acting in this manner can be considered to be lift.

From that point of view, "lift" of both aircraft can be considered equal.

However, the contribution of the thrust vector to this stalled yet still flying manuever, in opposition to gravity, leads to a better understanding of forces involved with the level flight (or climb) of the "Cobra" aircraft.

Robert DiGiovanni
  • 20,216
  • 2
  • 24
  • 73
  • Care to provide a link re the second paragraph? It would surprise me if they were defining the lift vector as the sum of all forces that act perpendicular to the free stream, including any such component from the motor/ engine. I.e. any such component in what we would normally call the "thrust" vector. But, you never know, I've been surprised before, could be an example either of sloppy writing (e.g. not making clear they were addressing a simplified case where thrustline is assumed to be parallel to flight path), or could be an unusual alternative definition. – quiet flyer Apr 07 '22 at 14:00
  • There it is, could not resist the perpendicular to "freestream" definition point of view. – Robert DiGiovanni Apr 07 '22 at 14:03
  • I'd say that's some arguably sloppy writing in that source-- no real basis for saying "Lift is the force that directly opposes the weight of an airplane " unless they clarify they are speaking of level flight-- yes they certainly do state that lift is "directed perpendicular to the flow direction"-- but I'd say that the caption in red in the upper left corner tends to suggest that they are not including any component of (what would normally call) the thrust vector which acts perpendicular to the flow direction. – quiet flyer Apr 07 '22 at 14:07
  • (Ctd) Anyway the intent of the page is not completely clear but I don't think it really should be cited to support the idea that a component in the thrust vector acting perpendicular to the free stream should be "counted" as part of the lift vector. – quiet flyer Apr 07 '22 at 14:09
  • @quietflyer Well, the argument in favor of our cherished ASE beliefs is the subsequent definition of lift as the interaction of a (wing) with a fluid. Therefor, thrust does not count as lift. It is just that the level aircraft presented a new twist to this old question. – Robert DiGiovanni Apr 07 '22 at 14:09
  • Now, applying that to helicopters gets interesting. – Robert DiGiovanni Apr 07 '22 at 14:11
  • Typical confusion resulting from using terms that have multiple definitions, with multiple conflicting assumptions, based on multiple different contexts/scenarios. Before we can start discussing something, we should decide amongst ourselves what the context/scenario is, and what we wish the terms to actually mean. Lift is especially vulnerable to mis-definition/confusion. Remember, ALL definitions are nothing more than arbitrary definitions, made to suit the problem/context we are working in. Defining Lift as the force opposing weight is useless for an inverted aircraft. – Charles Bretana Apr 07 '22 at 17:21
  • @CharlesBretana good comment but ... if the plane is inverted, is the wing (functionally)? But 100% agree on consistent definitions. This one is a little fun, and with vectored thrust, perhaps way better flown unstalled with thrust doing most of the "lifting" and forward propulsion, like a helicopter. A bit of a tease, would a fan jet be producing lift? – Robert DiGiovanni Apr 07 '22 at 17:44
  • @Robert, As you have probably garnered from my numerous posts on this topic, to me, Lift is just an arbitrary definition of some defined component of the TOTAL aerodynamic forces generated by the collision of fluid particles with the surface of an airfoil. We decide which component and which sections of the total airfoil to include, in order to suit the problem space we are attempting to analyze and work in. But as to a Fanjet, are you referencing the engine? like as opposed to a TurboJet? – Charles Bretana Apr 08 '22 at 01:46
  • I think the first and most basic, definition of Lift should be that Lift (if we are attempting to distinguish it from Drag, is that component of the TOTAL aerodynamic force which lies perpendicular to the free-stream airflow passing over the airfoil. Then the two components (Lift and Drag) will align with the frame of reference of the aircraft's motion through the fluid, and therefore Drag will affect only the magnitude of the Velocity, and Lift will only affect the Direction (angle) of the velocity. This makes the DEs easier to work with. – Charles Bretana Apr 08 '22 at 01:51
  • "This question is a favorite..." I beg to differ! It certainly isn't my favorite. :) – Michael Hall Apr 08 '22 at 03:16