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I am convinced that fighter jets (every flying object) do stall. So from quora, Why don't fighter jets stall?

But most of the fighters have thrust:Weight ratio > 1 which means that they are not following aerodynamics at that time (during vertical climbs). You can consider them to be like rockets with propulsion system. It suggests fighter planes don't follow aerodynamics. That means stall can be prevented using the jet power. So what causes them to stall?

Edit: I realized my question is currently away from singularity. Based on comment, jamesqf, zeus and Robert maybe right. That quora answer might be wrong. It claims that fighter jets don't follow aerodynamics. I mean every flying object do follow it. So my question is if fighter jet has enormous power compared to their weight, will it prevent stall by a bit? My prediction is stall only depends upon AoA. Regardless of airspeed, if AoA exceeds critical angle, stall can't be prevented (even using high thrust propulsion). Only way to prevent stall is to decrease its AoA. Are my claims correct? What if flying object relies on propulsion engine for lift (lot like missile, rocket), they keep flying and nothing hinders their speed?

Stay Safe...

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  • $\begingroup$ So you know how stall-resistant fighters like F-16 are? $\endgroup$ – Abdullah May 19 at 11:48
  • $\begingroup$ what caused you to ask this question? telling that to us might help us answer the question better. Because nearly every airplane there is - fighter or otherwise - stalls. $\endgroup$ – Abdullah May 19 at 11:49
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    $\begingroup$ I think the question is based on a false premise: both fighter jets and commercial airlinere DO stall (and you can find accident reports to prove it), it's just that the pilots are trained to avoid stalling, and generally manage to do so. And of course fighter jets have a good deal of power to pull out of stalls - just as I can use increased power to fly my Cherokee right at the edge of stall. $\endgroup$ – jamesqf May 19 at 17:10
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    $\begingroup$ @jamesqf I think there is another misundertanding on definition of stall and answer should also clarify this.flying in a stall regime (above AoA) doesn't prevent from flying under control (either hovering (I'm thinking of the AV-8B) or in really high angle of attack (explored with the X-31) or more like a rocket/space shuttle on acsent) $\endgroup$ – Manu H May 19 at 17:16
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    $\begingroup$ "Most of the fighters have thrust:Weight ratio > 1" - not true. And even those that do, mostly have it at less than full weight. And even in such conditions, they rarely use it just to climb vertically: this is very inefficient. Fighters are still airplanes, and they do "follow aerodynamics". $\endgroup$ – Zeus May 20 at 1:03
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Every aircraft can be made to stall with a combination of low speed, rear c.g. location and high load factor.

Thrust acts in the airplane's lengthwise direction. Lift acts in a mostly perpendicular direction, so whether the aircraft stalls does not depend on its thrust when lift demand exceeds the available lift capacity.

As you observe, this will not be the case in a climb. We know that lift demand is actually lower in a climb than in steady flight. So we need to look somewhere else.

The obvious case is in turning flight when the aircraft has to counteract not only gravity but also a centrifugal force. Now lift demand can grow to multiples of that in steady flight and any aircraft will readily stall if mass times normal acceleration surpasses the lift capacity of its wings. More installed thrust will only reduce the rate of energy loss in a tight circle.

Only if pitch authority is not sufficient can stall be avoided, but then the control system will not allow to trim the full angle of attack range. This could conceivably happen with a very forward center of gravity location but should place this configuration outside of certifiable limits.

As a side note: Stall depends on more than just the angle of attack.

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  • $\begingroup$ Thanks @Peter kampf More thrust will only speed up plane in a circle?right? $\endgroup$ – Auberron May 20 at 5:54
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    $\begingroup$ "Thrust acts in the airplane's lengthwise direction". Yes, mostly, but thrust vectoring exists at least since the X-31 and is largely implemented on Sukhoi Su-27 derived fighters and on the F-22. $\endgroup$ – Manu H May 20 at 7:00
  • $\begingroup$ @ManuH: Since vectoring happens in the nozzles, this is meant for creating pitch moments, not as a substitute to lift. Only the Harrier, the F-35B and the Yak 141 would use direct lift. $\endgroup$ – Peter Kämpf May 20 at 7:55
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    $\begingroup$ @Auberron: Speed can also be gained by diving. Thrust helps to maintain altitude or even climb in a tight circle where others would only sink in order to compensate for the massive drag created by circling. $\endgroup$ – Peter Kämpf May 20 at 7:57
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The question perhaps may be less about the inherent characteristics of a given aircraft, and more about how the aircraft is being flown.

Some relevant points--

  • Given sufficient control power, a wing can be placed at the stall angle-of-attack at any airspeed. Even if thrust is greater than weight. Of course, in such a case, as the aircraft approached and then exceeded the stall angle-of-attack, the resulting flight path would not be anything like a straight line.

  • Some aircraft are designed to be maneuverable beyond the stall angle-of-attack -- e.g. the "Cobra" maneuver.

  • Given the constraint of a linear flight path, the steeper the climb angle, the lower the lift force. (For more, see Does lift equal weight in a climb?) This means that the steeper the climb angle, the lower the wing loading, if we define wing loading as lift force per unit wing area. This means that the steeper the climb angle, the lower the unaccelerated stall speed.1 At very steep climb angles, the unaccelerated stall speed would seem to drop so low that the concept of stalling is no longer very meaningful.2 If thrust is greater than weight, a sustained vertical climb is possible. Even if thrust is less than weight, a temporary vertical climb can be attained by exchanging kinetic energy for altitude. If the pilot applies control inputs as needed to maintain a purely vertical flight path, then we know that no longer how low the airspeed drops, the wing must still be held at the zero-lift angle-of-attack, so the wing cannot reach the stall angle-of-attack. Of course, if the airspeed drops to zero and then the aircraft starts to tailslide backwards, the angle-of-attack at that point is arguably far beyond the stall angle-of-attack.

  • The same "unloading" of the wing, and resulting decrease in unaccelerated stall speed, happens in diving flight as well. For example, if the pilot is making control inputs as needed to maintain a linear flight path, the aircraft could obviously never stall in a vertical dive, even if the pilot deployed a gigantic drogue chute that dropped the airspeed below the normal straight-and-level stall speed. But that doesn't mean the pilot can't stall the wing by pulling the stick back to "pull G's" to pull out of the dive. The stall speed under heavy G-loading will be much higher than the normal straight-and-level stall speed.

Setting all theory aside, to get a feel for how a high thrust-to-weight ratio can make possible some extraordinary maneuvers, some of which involve angles-of-attack far beyond the stall angle-of-attack, search the web for videos of model airplanes engaged in "3-D" flying, such as this one. Of course, some of the maneuvers that involve essentially zero airspeed, using the propwash to keep the control surfaces effective, would not be possible in (most if not all) jet fighter aircraft. Nonetheless, even with no propeller, thrust-vectoring can allow for some amount of maneuvering at zero airspeed, as demonstrated by a radio-controlled model airplane in this video.

Footnotes--

1 -- For the purpose of this answer, when an aircraft is flying along a linear (not necessarily horizontal) trajectory, we'll define the unaccelerated stall speed as the speed at which the wing will reach the stall angle-of-attack if the pilot makes control inputs as needed to maintain that linear trajectory while the airspeed is slowly decreasing. This may be an unconventional use of the term. It is certainly not the same thing as the 1-G stall speed, unless the flight path is horizontal.

2 -- The "other side of the coin" here is that if the airspeed does decay to a very low value during a steep climb, then a modest pull on the stick might put the wing at the stall angle-of-attack while creating only a mild increase in G-loading. In other words, if the airspeed drops very low during a steep climb, it may be easy for the pilot to accidentally cause an accelerated stall at some airspeed that is still well below the normal straight-and-level stall speed, but above the unaccelerated stall speed corresponding to the climbing linear trajectory.

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  • $\begingroup$ Disclaimer-- I've never flown a high-perfomance jet... $\endgroup$ – quiet flyer May 20 at 13:04
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A320, F-16, and paper airplane stall exactly the same way. They are going too slow to make enough lift to hold themselves up. They try to make more lift by increasing their angle of attack of the wing, but when the angle of attack is too much the airflow separates from the top of the wing, creating less lift and a lot more drag.

The drag makes it worse by slowing the plane down even more. The loss of lift makes the plane fall, which makes the angle of attack of the air on the wing even greater.

The solution for all 3 is to pitch the nose down to regain speed and re-establish the proper airflow over the wing.

The problem is the amount of altitude needed to recover. Paper airplane: about 4 inches. A320? Maybe 20,000 feet. The much greater mass of the A320 makes it much more difficult to recover. This is why airliners try so hard not to stall.

The F-16 is much lighter than the airliner, but still would need a lot more altitude to recover than a paper airplane, or a Piper Cub.

Edits for the editor:

most fighters have a thrust to weight ratio > 1, so they are more like rockets.

Any aircraft can stall if the wing AOA exceeds the stall limit. "Flying around like a rocket" burns up fuel very quickly and is very impractical. Simply adding power will not necessarily break a stall. Wing AOA must be reduced. Unfortunately for a giant airliner, this may involve putting it into an un-recoverable dive.

Fighter planes do not follow aerodynamics

Anything moving within the earths atmosphere follows aerodynamics. As soon as it moves, drag effects must be considered to fully understand all forces involved.

common cases of fighter jets and airliners stalling.

These events are very uncommon, and potentially disasterous. This is why we need reliable AOA sensors and adequate safety systems to prevent this from happening. In order to model the wing loading down to paper airplane scale, the paper airplane would have to be made of lead.

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  • $\begingroup$ that guy seems to have more rep and badges than someone who'd ask a stupid question. see the discussion below original post. $\endgroup$ – Abdullah May 19 at 11:57
  • $\begingroup$ The scale of the recovery made it interesting enough to answer, perhaps $wingloading$ would also apply. $\endgroup$ – Robert DiGiovanni May 19 at 12:27
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    $\begingroup$ I'm curious as to where this 20,000 foot figure comes from, it seems excessive... $\endgroup$ – Ron Beyer May 19 at 13:25
  • $\begingroup$ @Ron Beyer if you are falling at 150 mph how long does it take to go 4 miles? How do you stop 250 tons from falling without either exceeding G loads or re-stalling the aircraft? A buffet or near stall recovery (by man or computer) would certainly be less, but a full blown stall/sink with the wing loading of an airliner would not be on my aviation to do list at any altitude. $\endgroup$ – Robert DiGiovanni May 19 at 15:26
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    $\begingroup$ @RobertDiGiovanni Here's an account of a 737 simulation from FL380 that shows recovery in about 3500 feet. Here's a video of a full stall recovery in a 737 with an FAA approved aerodynamic model, they lost about 12,000 feet I believe. Second part shows a full stall at 15,000 feet, losing probably 3,000. $\endgroup$ – Ron Beyer May 19 at 16:26

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