Inverted, unloading and reducing angle of attack on Modern Jet Aircraft

[bohpilot]

We are having a real active discussion at work on upset recovery training on a jet and in particular High angle of attack when inverted. My understanding has always been that even when you are inverted with an high angle of attach on the wing you have to unload the wing by "pushing".

The argument comes back that if you "push" when you are inverted then you are going to make the situation worse.

Would welcome some advice.

[111boy]

Unload is reducing the G load, so inverted or anything else, it means pushing less or pulling less, if you look up vg diagrams, it explains nicely. But basically if you reduce the load factor you reduce your stall speed.

[Dufo]

As long as you have AOA to provide sufficient lift, rules are the same for all attitudes.

[H Peacock]

I assume you mean inverted with a positive AoA, ie similar to the top half of a loop. If the nose is still above the horizon you could unload then roll wings level, but you'll invariably be decelerating throughout - so you'd need to have plenty of speed first to avoid running out (stalling). Better to wait until the nose is just below the horizon prior to gently unloading and then rolling wings level. In the latter case the speed will either gently increase or even remain stable. Of course if the nose is already well below the horizon (speed rapidly increasing and g load increasing) you'll need to be more aggressive with the unload prior to rolling wings level. Selecting idle power and perhaps a tad of spoiler might help control the speed, but it's now a careful balance between pulling too hard and overstressing the aircraft or not pulling enough and overspeeding the aircraft. Of course the latter assumes you're well clear of the ground so not a factor. Do you really need to pull 3.5g when say 2.5g may suffice?

Depending upon the trim settings, the unloading may be a relaxation of the pull or indeed a positive - but gentle - push.

[Goldenrivett]

if you "push" when you are inverted then you are going to make the situation worse.

It depends.

To unload the wing whilst inverted in level flight, you would have to "pull" (i.e. descend to reduce the g.)
If you were flying a vertical loop and pulling g over the top whilst inverted, then you would "push" to reduce the g and reduce the stalling speed.

[bohpilot]

Originally Posted by Goldenrivett

It depends.

To unload the wing whilst inverted in level flight, you would have to "pull" (i.e. descend to reduce the g.)
If you were flying a vertical loop and pulling g over the top whilst inverted, then you would "push" to reduce the g and reduce the stalling speed.

We have this slide on our briefing



If the wing only knows AOA, even upside down would not pulling in level inverted flight increase the AOA and thereby decrease the stall margin

[111boy]

If the question is " level inverted flight " then that implies minus 1 g. I.E a " push "

unloading would mean " stop pushing " .....

[bohpilot]

Why does level inverted flight mean -1G?

[Goldenrivett]

Hi bohpilot,

I can see why you are confused - the diagram has an error.

The inverted aeroplane's velocity vector would be more towards earth relative to the aircraft axis. (diagram shows the velocity vector impossibly giving lift in that attitude)

For level inverted flight, the aircraft axis would be have to be above the velocity vector axis.

[bohpilot]

In level inverted flight, the wing is still producing lift so why would the aircraft axis be above the velocity, would they not be the same ?

[aerolearner]

As long as we are not assuming that the aircraft is in level inverted flight, the diagram would be correct in my opinion. To me, the picture is correctly depicting the descending part of a looping with some substantial "pull" (e.g., following video at 1:06-1:08).
https://www.youtube.com/watch?v=gcfFTZmq7wM#t=1m05

[Goldenrivett]

In level inverted flight, the wing is still producing lift so why would the aircraft axis be above the velocity, would they not be the same ?

When inverted, the lift direction would have to be towards the belly (not the roof).

[111boy]

Originally Posted by bohpilot

Why does level inverted flight mean -1G?

Level " right way up flight is 1 g "......
in order to sustain level inverted flight you need the same force in the opposite direction .....-1 g.........

[PDR1]

Originally Posted by Goldenrivett

When inverted, the lift direction would have to be towards the belly (not the roof).

If you are in straight & level inverted flight, yes. But if you are pulling through (say, from the top of a loop) then you will have a *positive* AoA and pulling harder will lead you into buffet and even potentially flicking out of the loop.

If you do a round loop in a low-powered aeroplane (Tiger Moth, Stampe etc) then as the speed decays over the top of the loop you will be backing off the pull and may well get zero-G (or close to it). You will certainly get "reduced G", and that means that the stall-speed reduces accordingly. I've been in unstalled flight at 18kts (according to the ASI) through the top of a loop in a Stampe, but I wouldn'd want to try that as an approach speed...

[Goldenrivett]

Hi PDR1,

We agree. See post #5

[QuagmireAirlines]

Upset recovery: Unload (reduce alpha to near zero) first, followed quickly by a roll to get normal non-inverted flight. Then pull to get level flight.

If you don't reduce alpha to near zero first, before rolling, then your alpha becomes sideslip, which hits you with a lot of yaw.

If you're flying level upside down, you do pull, not push, to unload.

Maybe the quick (training) rule of thumb is to always pitch your nose towards your velocity vector, period, before rolling to the next orientation you want.

[Zaphod Beblebrox]

Originally Posted by bohpilot

We are having a real active discussion at work on upset recovery training on a jet and in particular High angle of attack when inverted. My understanding has always been that even when you are inverted with an high angle of attach on the wing you have to unload the wing by "pushing".

The argument comes back that if you "push" when you are inverted then you are going to make the situation worse.

Would welcome some advice.

In addition to lowering stall speed, unloading while upright or inverted increases roll rate and effectiveness of ailerons. As remarked by the NTSB in the US Airways 427 accident in Pittsburgh PA, the hard-over rudder accident in the B737-300, the recovery was partly stymied by the crews attempt to pull out. The resultant high angle of attack situation this caused greatly reduced the effectiveness of the ailerons to stop and correct the roll.

[Vessbot]

Originally Posted by bohpilot

We are having a real active discussion at work on upset recovery training on a jet and in particular High angle of attack when inverted. My understanding has always been that even when you are inverted with an high angle of attach on the wing you have to unload the wing by "pushing".

The argument comes back that if you "push" when you are inverted then you are going to make the situation worse.

Would welcome some advice.

As in any airplane in any attitude, if you want to unload the wing (and move away from a stall) you have to move the stick toward the middle, in order to reduce the AOA. If the stick is back and the AOA is high, that means moving it forward. In the high attitude inverted situation, this will allow the (presumably) low-speed critical airplane to pitch below the horizon (via the natural, longitudinal stability, or "lawn darting" if you will) with an AOA close to zero, and therefore a stall speed close to zero. Just the way you want it.

In any technical discussion by non-engineers, a lot of (even basic) detail can get lost in assumptions and mistellings. You didn't say what your friend says will be made worse, but I'll take a stab. He probably took "push" to mean more than moving the stick forward toward neutral, but rather past neutral and even more forward into negative AOA and therefore negative G. That could then prevent the nose from coming below the horizon, and force the speed to decay even more (not to mention what will happen inside a transport plane at negative G).

[Vessbot]

Originally Posted by Goldenrivett

Hi bohpilot,

I can see why you are confused - the diagram has an error.

The inverted aeroplane's velocity vector would be more towards earth relative to the aircraft axis. (diagram shows the velocity vector impossibly giving lift in that attitude)

For level inverted flight, the aircraft axis would be have to be above the velocity vector axis.

There is nothing wrong with the diagram. All of these configurations are possible, momentarily. The inverted airplane shows what is happening about 2/3 of the way through a loop.

You're assuming it's trying to show sustained inverted flight, but it's not. If it were, then yes it would need a negative AOA (what's shown is positive) and therefore aircraft axis above the velocity vector, as you say.

[Vessbot]

Originally Posted by bohpilot

In level inverted flight, the wing is still producing lift so why would the aircraft axis be above the velocity, would they not be the same ?

In level inverted flight, the wing has to produce negative lift, i.e., the lift vector is pointing out the belly instead of the roof. This requires negative AOA, which means the aircraft axis is above the velocity (seen by an upright observer)

Your last phrase "would they not be the same" belies a deeper confusion, which is enabled by the efficient cambered wings on our airplanes, that make lift at zero AOA at high speed. Imagine for a few minutes that all wings are symmetrical, and really inefficient at that [edit: earlier I typoed and wrote the opposite]. That means that to make any lift, it has to fly at a definite, high AOA. Zero AOA = zero lift. High AOA = high lift.

Even in normal cruise (level right side up flight), then, the aircraft axis has to be above the velocity. If they were aligned, then the AOA would be zero and no lift could be created. In level inverted flight, the exact same thing is happening. The aircraft axis has to be above the velocity, (or below, if this is being considered by someone sitting in the plane. Put it another way, in both cases the velocity is aligned exactly with the horizon, and the leading edge of the wing is pointed a bit at the sky.

When considering large scale effects of all-attitude flight, this simplification is more than close enough to use, and should hopefully make things clearer.