Negative G Question
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It’s theoretically possible to do endless loops of negative load factor.
Actually, what I was saying is that you cannot pitch down towards Earth and stay at -1.0 G’s endlessly. Once you gain inertia in the direction of gravity you are stuck unless you deviate from -1.0 G. You will either hit the ground or you will have to pull more than -1.0 G in order to level off. Of course you can go in endless loops but with gravity once you get to the point you are trying to fly straight and level, you will have to not only overcome gravity but your inertia towards the ground, hence passing -1.0 G’s. I was talking about how you can endlessly do -1.0 G’s and exactly -1.0 G’s by flying inverted and pitching away from gravity, but you cannot do that when pitching towards the Earth, because eventually you will hit the ground or have to pull more than -1.0 G. Maybe I didn’t explain it well enough. You can do horizontal loops. But not vertical while staying at -1.0 G’s. The effect of gravity makes it impossible. It’s like if you bungee jump or sky dive. In order to decelerate while falling, you have to pull more than 1 G; or if you jump off a bench or something. It’s not the falling that kills you, it’s the sudden stop.
Negative G can be a direction in relation to Gravity. It is a metric of force. I think you misunderstood what I posted because I never posited anything about negative G being a direction. It is a force, and the Earth’s gravity is in the direction of Earth. When accelerating towards or away from Earth we experience G forces due to the direction we take in relation to the center of mass. Everything you have ever done in your entire life is in relation to the direction of gravity. Unless you have been out in space, you have never experienced gravity not having a direction. You just overcome it with an opposite force which creates G forces.
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Thanks very much to everyone who has replied to my post.
I just wanted to clarify my question.
If a jet fighter pilot was in a 90 degree nose-down attitude – heading straight down to the earth – if he pushed the stick forward just enough so that the G-meter read exactly 0.0 G, would he start going inverted? Or would he have to push the stick a little more to get at least, say, -0.3, -0.4, or -0.5 in order to begin going inverted?
Thanks,
Starfighter
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Thanks very much to everyone who has replied to my post.
I just wanted to clarify my question.
If a jet fighter pilot was in a 90 degree nose-down attitude – heading straight down to the earth – if he pushed the stick forward just enough so that the G-meter read exactly 0.0 G, would he start going inverted? Or would he have to push the stick a little more to get at least, say, -0.3, -0.4, or -0.5 in order to begin going inverted?
Thanks,
Starfighter
Going inverted means flying upside down, to me. As well clarify that ASAP
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Going inverted means flying upside down, to me. As well clarify that ASAP
Actually, in air/space, upside down is 360 degrees. The only difference is at which degree you are fighting more or less gravity!!!
C9
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@Cloud:
Actually, in air/space, upside down is 360 degrees. The only difference is at which degree you are fighting more or less gravity!!!
C9
Language problem now. OP should clarify, just MHO.
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Thanks very much to everyone who has replied to my post.
I just wanted to clarify my question.
If a jet fighter pilot was in a 90 degree nose-down attitude – heading straight down to the earth – if he pushed the stick forward just enough so that the G-meter read exactly 0.0 G, would he start going inverted? Or would he have to push the stick a little more to get at least, say, -0.3, -0.4, or -0.5 in order to begin going inverted?
Thanks,
Starfighter
I see what you mean. For any dive angle the g-meter reading which continues a straight path is the cosine of that dive angle. For a straight line -60 dive the g-meter reads 0.5. For a straight line -90 dive the g-meter reads 0.0. If the a airplane is at -90 and the g-meter is 0.0 then the airplane will continue in a straight line. A dart doesn’t pull G. It’s always at 0g and we know a dart will approach -90 pitch and never go beyond that.
Actually, what I was saying is that you cannot pitch down towards Earth and stay at -1.0 G’s endlessly. Once you gain inertia in the direction of gravity you are stuck unless you deviate from -1.0 G. You will either hit the ground or you will have to pull more than -1.0 G in order to level off. Of course you can go in endless loops but with gravity once you get to the point you are trying to fly straight and level, you will have to not only overcome gravity but your inertia towards the ground, hence passing -1.0 G’s. I was talking about how you can endlessly do -1.0 G’s and exactly -1.0 G’s by flying inverted and pitching away from gravity, but you cannot do that when pitching towards the Earth, because eventually you will hit the ground or have to pull more than -1.0 G. Maybe I didn’t explain it well enough. You can do horizontal loops. But not vertical while staying at -1.0 G’s. The effect of gravity makes it impossible. It’s like if you bungee jump or sky dive. In order to decelerate while falling, you have to pull more than 1 G; or if you jump off a bench or something. It’s not the falling that kills you, it’s the sudden stop.
The cockpit instrument only measures acceleration in the vertical (meaning top of airplane) direction. It does not measure lateral or longitudinal. An F-16 suspended from a crane with its nose pointing exactly up or down or even rolled 90 degrees either way would have its meter read 0.0 despite gravity being 1g toward the tail or nose or right or left wingtip. Given enough altitude a constant -1.0g load factor would have a radial acceleration of -2 from level to -1 pitched down and 0 approaching inverted level (the last result taking infinite time). A -1LF loop is not a closed maneuver because the radius of curvature would become infinite half way through it. But any LF more negative than -1 is a closed maneuver.
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So if you’re in level flight, and you push the stick forward until the G-meter says exactly 0.0, the nose of the fighter will start to pitch down. If you keep holding the stick at 0.0, the fighter’s nose will eventually be pointing straight down to earth, and will not go inverted until you push the stick forward some more, pushing some negative G.
Starfighter
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that is correct.
Its worth noting that the stick movements for this are so simple due entirely to the G-command system the FLCS uses in the F-16. Also worth noting is that you should technically never reach directly downward. As the nose drops, you will note the pitch rate decreases as the dive angle steepens, if you hold constant 0.0g in the dive. Three other factors affect the final dive angle; due to drag you can get essentially all the way down given enough time, then given sufficiently slow speed and high angels, you can get to a very nose low attitude - essentially 90 degree dive.
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I read the following from a publication called “Interceptor” by USAF ADC:
“Zero G is a ‘best exercise’ when you find you’re in a nose up, bleeding airspeed situation. But delaying initiation of a zero G recovery can place the airplane in a low speed situation so delicate that zero G cannot be maintained practically.”
What is the minimum airspeed required to push forward on the control stick and enter 0.0 G?
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@Red:
if you want to accelerate quickly, go zero G
that’s where you make a difference.I did a week of aerobatic flying in an Extra 300 with Patty Wagstaff down in Florida a couple of months ago and I asked her about this. Specifically, whether or not unloading and maintaining zero G would yield a greater speed build up versus nosing over with negative G and getting helped by gravity. She leaned toward negative G + gravity being a faster method, but was interested at the premise.
We were going to do some tests on our last flight… but ended up forgetting until the end and I wanted to spend the last bit of fuel in our acro tanks on seeing how high I could get the Gs in a loop entry. (7.2Gs! The two seat Extra is rated for +8/-8.)
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The cockpit instrument only measures acceleration in the vertical (meaning top of airplane) direction. It does not measure lateral or longitudinal. An F-16 suspended from a crane with its nose pointing exactly up or down or even rolled 90 degrees either way would have its meter read 0.0 despite gravity being 1g toward the tail or nose or right or left wingtip. Given enough altitude a constant -1.0g load factor would have a radial acceleration of -2 from level to -1 pitched down and 0 approaching inverted level (the last result taking infinite time). A -1LF loop is not a closed maneuver because the radius of curvature would become infinite half way through it. But any LF more negative than -1 is a closed maneuver.
This gives more clarity to the lazy way I was trying to inform how to best test the drag at -1.0 and 1.0 G. Doing a loop would not be an accurate test because of the relationship to the Earth’s mass and the limits of the A/C’s G sensor. The reason I even mentioned this is because testing drag would be best with no vertical speed and inverted for -1.0 G. The 1.0 G test could be done in standard level flight. My point was that going into a loop would be a bad way to test drag. I didn’t really make that clear when I mentioned it. I kind of let my thoughts spill out onto the post without clarifying.
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For a fighter aircraft to “unload and extend” to build speed really translates to reducing induced drag - drag due to lift, or that energy that gets wasted in span wise flow. There is also a fixed drag component due to configuration, but one can always kill induced drag by decreasing lift. This also manifests in that the more G you pull the higher the drag, because what you are really doing in pulling G is creating more lift - and bleed speed during a hard turn.
Just because the pilot puts the stick forward may not mean the aircraft will pitch down - at least, not if he does it right…it also depends on what he does with the throttle. AOA should decrease during an unload (hence reduced lift) and the jet will follow a more ballistic trajectory through the sky, but if the purpose of the unload is to gain speed the jet shouldn’t go nose low…or shouldn’t dive, anyway. For jets.
A prop plane is a different sort of case mainly because of how power is applied/transmitted…P-factor and torque get involved.
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That would depend on your flight path and speed and throttle setting. Pointed high and low you may or may not have the control authority to get to 0g. Even if you achieve 0g you may continue to lose speed on this trajectory to the point you lose control. For minor nose up-slow conditions a 0g trajectory would have a minimum airspeed that is still controllable. Probably should could get away with 130 knots entry at 30 degrees up. The speed would get awful low at the top.
For the F-16 refer to the HARTS maneuvers which are progressively more severe scenarios and how to best deal with them. Check section 3.13.1 http://falcon.blu3wolf.com/Docs/Basic-Employment-Manual-F-16C-RoKAF.pdf
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Depends on the aircraft - a Hornet can fly quite nicely to about 90 knots over the top, forex. The trick is to get to 0 G and not go negative - if you go negative you’re only producing lift out the other side and then you’re back into creating induced drag…not as much but more than the desired effect.
But I wasn’t talking about vertical maneuver - just an unload to gain speed. As long as you’re ballistic you may lose altitude but you’re not changing pitch - think of an arrow. It flies flat but it falls. For a jet if the thrust comes up it accelerates, but falls…which only aids the accel. But you have to do it right…I know I can’t do it right in a trainer, but the pros know how. And when…I’ve never been able to figure out which is most important. But I have developed the ability to know when I’m over-pulling and to relax.
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This gives more clarity to the lazy way I was trying to inform how to best test the drag at -1.0 and 1.0 G. Doing a loop would not be an accurate test because of the relationship to the Earth’s mass and the limits of the A/C’s G sensor. The reason I even mentioned this is because testing drag would be best with no vertical speed and inverted for -1.0 G. The 1.0 G test could be done in standard level flight. My point was that going into a loop would be a bad way to test drag. I didn’t really make that clear when I mentioned it. I kind of let my thoughts spill out onto the post without clarifying.
Absolutely. And just for fun I did exactly that (obvious in hindsight). Level flight at 0 bank takes about +2 AOA. Level flight at 180 bank takes about -7 AOA (was 380 IAS or so). Showing that the lift to maintain level flight is quite asymmetrical in the F-16.
@tpn:
I did a week of aerobatic flying in an Extra 300 with Patty Wagstaff down in Florida a couple of months ago and I asked her about this. Specifically, whether or not unloading and maintaining zero G would yield a greater speed build up versus nosing over with negative G and getting helped by gravity. She leaned toward negative G + gravity being a faster method, but was interested at the premise.
We were going to do some tests on our last flight… but ended up forgetting until the end and I wanted to spend the last bit of fuel in our acro tanks on seeing how high I could get the Gs in a loop entry. (7.2Gs! The two seat Extra is rated for +8/-8.)
It’s not even close. Getting the nose down quickly will be worth many knots and worth the drag penalty from the maneuver cost of the AOA. From level flight at 30kft 200KCAS and you are only allowed to select a constant load factor 1, 0, -1, -2, -3 to accelerate to say 500 KCAS I bet the winner would be quite a negative value for the minimum time (not exceeding -90 pitch of course). AOA has a drag penalty but getting a hefty fraction of 1 g longitudinal acceleration is a major factor. Small AOA has a pretty small drag penalty. Surely you want to dump a lot of very dragged high AOA to accelerate and the more the merrier but degrees it’s practically all parasite. Trading PE for KE is where the lion’s share comes from in that situation.
A 0g trajectory will change pitch attitude. 0g is 0AOA and so your nose position is your TVV and it will evolve downward over time. You’re a dart and follow a parabola. Given a level or slight climb initial condition the 0g trajectory doesn’t develop a low pitch very quickly but it’s always rating the nose downward.
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What is the minimum airspeed needed to push the sick forward and get the G-meter to read 0.0 G? Let’s say a jet fighter pilot was in a zoom climb, his engine flamed out, and he is bleeding speed rapidly at a very high altitude. If his airspeed dropped to, say, 50 KIAS, would the pilot be able to get the jet into zero G being at such a low airspeed? Or, if he pushed forward on the stick at 50 KIAS, would nothing happen?
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What is the minimum airspeed needed to push the sick forward and get the G-meter to read 0.0 G? Let’s say a jet fighter pilot was in a zoom climb, his engine flamed out, and he is bleeding speed rapidly at a very high altitude. If his airspeed dropped to, say, 50 KIAS, would the pilot be able to get the jet into zero G being at such a low airspeed? Or, if he pushed forward on the stick at 50 KIAS, would nothing happen?
The answer is, it depends. And this depends on a lot of design factors, specific to that aircraft. Rotating the nose in most aircraft is due to a lift imbalance between the horizontal tails and the wings. Most conventional aircraft, the wings have positive lift and the nose has negative lift, in straight and level flight. The F-16 reverses this, with both the tail and the wing providing positive lift, contributing to the lift the aircraft can generate (an important factor in determining how fast the aircraft can turn).
To push the nose over requires pitch authority. This authority comes from the ability (in a conventionally designed aircraft, including FBW fighters with conventional non TVC controls) to increase the lift generated by the tail(s), and in doing so induce a torqueing moment on the aircraft about its CoM. The ability of the aircraft to increase the lift on the tails is then an important factor in determining whether a pitch rate is attainable at all, and how much of a rate is possible. Lift is of course proportional to air density, wing area (of the tails in this case), the velocity squared, and the coefficient of lift (itself made up of a series of other frustrating factors). Of note is that at zero airspeed, there is zero lift, by definition. As altitude increases, air density decreases, so of course increasing altitude will mean less ability to rate the nose for a given airspeed. The question itself uses KIAS too, rather than KTAS, which means that altitude will affect everything else already mentioned, because true airspeed will vary with air pressure for a given KIAS.
Design factors can affect these factors, too. Airspeed over the tails affects the ability to generate lift, but this can be separate and distinct (although probably not independent) from the airspeed read in the cockpit, or by ground radar. Perhaps at the angle of attack in question, the airflow experiences separation of flow over the tails, such that the airspeed over the tails is zero even though the airspeed past the cockpit is 50 knots? This is something you cannot determine programmatically without an extreme amount of processing power and a LOT of input data.
So lift on the tail is important, what else? Well, the torqueing moment is opposed by the moment of inertia, which you can think of by comparing how hard it is (how much torque is required) to tilt a model C-130, versus tilting an actual C-130 through the same rotation (in the absence of drag, friction, etc). The same force applied at the same distance from the CoM for the same duration will add the same energy to the C-130 in question, but the model will spin MUCH faster than the actual C-130. Similarly, a larger, heavier plane (both size (technically distribution of density) AND mass are significant in determining moment of inertia, unlike its linear counterpart mass) will require more of a moment applied by its tail in order to generate the same nosedown rate.
Well, distance of the tail from the CoM will also affect the moment generated. A moment or torque is a force applied offset from a point of rotation (typically the CoM) which induces a rotation (or tries to). It is proportional both to the magnitude of the force (discussed above already) and also the magnitude of the offset. So, a tail which is further from the CoM will generate more of a torque than the same tail, with the same lift, which is closer to the CoM.
Ive perhaps missed additional factors which influence that question too. Certainly Ive glossed over the effects of airflow and how they will affect it - however, you can rest assured that those effects, like the others discussed above, will vary heavily, especially when comparing different aircraft. We can say it depends on the aircraft design in question, the air density (effectively, what the temperature and altitude is), and the location of the CoM (not in fact constant and something that evolves with fuel burn and stores offloading/release).
We have to limit the discussion to just conventionally controlled aircraft. At extremes of airspeed and pressure, we find ‘aircraft’ like the international space station, which has a lift response independent of airspeed (well, essentially), and thrust vectoring capability (well, the Reaction Control Systems constitute such, although they are not the primary control). Such outliers serve only to confuse discussion - similarly, for any sufficiently capable thrust vectoring system, there is essentially sustained zero airspeed maneuverability.
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Blu3wolf has it spot on…about all I can add is that the factor in the amount of tail authority is dynamic pressure - so it’s a function of speed, altitude, and configuration of the aircraft. Also, in a conventionally configured aircraft the stabilizing force on the tail is down - so for a statically stable aircraft a decrease in tail force (and not just forward stick alone) should result in the nose returning to the horizon; i.e. - a positive pitching moment. An F-16 is not statically stable, so yes - forward stick is required to get the pitching moment going.
And I’ll point out once again - just because the aircraft is at 0 AOA/0 G does not mean that the nose has to come down substantially - the ballistic path is parabolic. For this reason if you stay close to the center/origin of the parabola the less change in aircraft attitude is required to actually achieve the effect of unload. AOA will increase as rate of decent builds without adjusting attitude and it’s at this point that a pilot would have to actually transition to a diving attitude to maintain a low or near zero AOA…and it’s somewhere approaching this point that a pilot knows he’s reached the limit of his unloaded accel ability and starts thinking about maneuvering room and advantage generated (or lost) - I have yet to figure out instinctively how to spot this point for myself (requires change of focus on what I’m looking at in the HUD), but I do know that if I’m diving I’m past it and have to re-evaluate my options - like convert my maneuver to a low yo-yo, or something like that.
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Would a fighter’s V-N diagram tell us what the minimum airspeed would be for the fighter to enter zero G?
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nope. But you need more parameters anyway; You can definitely enter close to zero g in a cessna in a power on stall, for example. alpha wont be zero, though. In the F-16 its quite possible from 150 knots in the vertical. Its essentially impossible from 60 knots in the vertical with low initial pitch rate.
Its all about that ability of the tail to generate positive lift, and the moment of inertia of the aircraft. And the relationship between the CoM and the effective CoP. And the overall design of the aircraft. Its my suspicion that there is no analytical solution to your problem, no magic formula - although given sufficient data and a narrower problem set there might be a solution.
Sorry Im not real helpful on this one : its a horrible problem.