Negative G Question
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When an F-16 pilot is pushing forward on the control stick and gets exactly -1.0 G on the G-meter, will he have the same amount of drag as he would with no pressure on the stick, which would be positive 1.0 G? Or will there be some additional drag when pushing to -1.0 (pretending you could sustain -1.0G in level flight)?
Thanks,
Starfighter
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Huh, interesting, I have no answer but am interested in one.
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if you want to accelerate quickly, go zero G
that’s where you make a difference. -
By the very nature of gravity and aerodynamics, I’d say no, but I’m no expert on the matter, so I might be wrong…
For all I know, any (pilot controlled) diversion from 1G – regardless of whether it’s positive / negative G, or on which axis you’re changing the G-force – would always require moving at least 1 control surface from its central position. In doing so, they are positioned differently against the airflow, which would AFAIK automatically increase drag. That does assume aircraft are generally designed to have near-lowest drag situations in level 1G flight, however, which I’m not sure is the case for the F-16.
Furthermore, to maintain -1G in level flight, your AoA would be different from that in +1G level flight, also changing your drag factor.
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It depends on several things. Mostly airspeed and if it is a sustained -1.0 G descent. The short answer is yes because lift creates drag. If the speed is constant and the only thing that changes is G’s then yes, there is less drag. It can be easily understood by flying with a constant speed and constant thrust and trying to go into a -1.0 G descent. You will either gain speed or have to decrease thrust. Gravity is a factor and it can be confusing because we are going towards the pull of gravity, so we can think that is the reason for increased speed. But, that is the nature of a G.
If we want to talk about an inverted -1.0 G flight vs an inverted 1.0 G flight then it would be the opposite. There would be less drag at 1.0 G than at -1.0 G. It depends on lift vector.
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If it is a question about drag, one must not think in terms of Gs but in terms of AoA.
-1G is not less than 0G. ; -1G is the same than 1G … it is only a question of referential.
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Yes. G force is just a part of what effects drag. Think of this extreme. A harrier coming into a hover. That equation shows that thrust vector plays a big role, and in simple terms gravity can be seen as thrust. It is a force applied to an object much like thrust. Overcoming gravity can be done a few ways; thrust, lift, inertia… although inertia doesn’t really apply too much here because it’s close relationship with G’s. The part inertia plays gets larger the greater the delta is between a major gravitational object, such as a planet or moon. They all are pieces of the puzzle, as well as what Dee-Jay points out AOA and how an A/C’s flight control surfaces are tailored for certain densities/speeds/AOA’s.
Basically an object in flight has a few forces keeping it in the air, or even water; any fluid either aerodynamic/hydrodynamic.
Specific density is a factor as well. We can see the relationship between densities and drag with altitude.
It is actually a complicated and varied system of relationships and we look at a narrow spectrum which is A/C in air.
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1.0 (positive 1 G, no pressure on control stick)
0.9 (less drag)
0.8 (less drag)
0.7 (less drag)
0.6 (less drag)
0.5 (less drag)
0.4 (less drag)
0.3 (less drag)
0.2 (less drag)
0.1 (less drag)
0.0 (zero G)
-0.1 (increased drag)
-0.2 (increased drag)
-0.3 (increased drag)
-0.4 (increased drag)
-0.5 (increased drag)
-0.6 (increased drag)
-0.7 (increased drag)
-0.8 (increased drag)
-0.9 (increased drag)
-1.0 (negative 1 G)As you start pushing forward on the control stick, you go to 0.9, 0.8, 0.7, etc. As you do drag decreases. Once a fighter gets to zero-G, it is free from the the drag caused by producing lift with its wings. As you continue to apply forward pressure, you begin going -0.1, -0.2, -0.3, etc. As you do, drag increases. So I was wondering, once you reach -1.0, will you have just about the same amount of drag that you had when you were at 1.0.
Starfighter
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It depends on many other things. We have a case of chaos theory here, and would have to use something called poisson statistics. Too many variables. But in theory without accounting for the relationship between the mass creating gravity (Earth), other forces such as force and it’s vector (i.e. thrust) it can be seen as equal, because the variables you left out are the variables which in reality would make them not equal. If in an controled STP experiment with a symmetrical A/C the drag would be the same. In reality they are most often not the same for many reasons. The reason it is even a question is because of all the variables we are not taking into account here.
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For +1 or -1 acceleration there must be an equal magnitude force. This force is lift and the lift comes from AOA. AOA has an associated increase in drag. The AOA for +1 and -1 is probably not exactly equal as the craft isn’t physical symmetrical but first-order it should be close. The drag-per-unit-AOA is probably not identical either, close. I hazard a very tentative guess that there’s more drag per unit in the negative direction but that might not be so. So the drag at +1 and -1 are functions of their respective required AOAs and how much each of those AOAs induce drag. Those 4 numbers are probably all different.
Qualitatively the graph of drag v load factor would be a U-shape between +1 and -1 with a minimum at 0.
It’s commonly said that 0g gives the best acceleration and this is not strictly true long term. 0g suggests minimum induced drag but it neglects potential energy effects. Pointing the nose toward the floor also matters. Between 0 and -90 pitch you get a whole 1G longitudinal acceleration from gravity. Thus it might be worth it to suffer from induced drag for negative G in order to rotate the pitch down to enhance the gravitational accelerative effects.
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Frederf,
Thanks for your reply, and thanks to everyone else who has replied to my post.
Speaking of pointing the nose straight down toward the earth, should a fighter jet be able to go inverted while at zero-G, precisely at 0.0? Or should that not be possible, to push up against gravity and go inverted at 0.0? Would the fighter be stuck in a 90 degree nose-down attitude? Would the pilot have to push some negative G, -0.2, -0.3, in order to get the fighter to start going inverted?
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Speaking of pointing the nose straight down toward the earth, should a fighter jet be able to go inverted while at zero-G, precisely at 0.0? Or should that not be possible, to push up against gravity and go inverted at 0.0? Would the fighter be stuck in a 90 degree nose-down attitude? Would the pilot have to push some negative G, -0.2, -0.3, in order to get the fighter to start going inverted?
Yes, with 0g, the aircaft behaves ballistically, so no going inverted from wings level.
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This post is deleted! -
I’m not sure I understand the last question, or Cruse’s response…
Are you talking physically or theoretically? When you say “inverted” do you mean can you apply rotational force to the roll axis? 0G and 90 degree nose down have nothing to do with each other per se, the previous example was just an example for illustration. You can get 0G without 90 degrees nose down, and vice versa. So I’m a little confused about your question.
Are you asking A: Can you maneuver the aircraft in a roll at 0G, or B: Can you roll the aircraft when you’re 90 degrees nose down. The answer to both is yes, however, you won’t “technically” maintain a 0G state when you start to induce roll into the entire system. You have to remember that “G” is a term relevant to Acceleration, not speed - which means technically you can achieve a “0G State” at any speed or attitude, so long as the acceleration toward the ground is the same as the acceleration of gravity… And lift/drag are both functions of Speed. You can create the required force to effect the control surfaces at any “G”, so long as that “G” doesn’t occur in a vacuum.
Roll? Not sure why you inject it in that thread.
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Frederf, that makes sense. It made me think about how initiating a -G condition likely creates more drag from the control surfaces because the A/C is designed to pitch up more efficiently. I guess it would be a bad design to make an A/C more efficient at -1.0 G than at 1.0 G because the A/C is at positive G’s much more. So, I see the practical answer being that if you want to pitch away from gravity you of course want to pull back on the stick, and if you want to pitch towards gravity there is likely a point at which going inverted would be best. Of course you can only hold negative g’s for so long unless you are inverted. Eventually you would end up inverted going the other direction. If I have that correct.
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Frederf,
Thanks for your reply, and thanks to everyone else who has replied to my post.
Speaking of pointing the nose straight down toward the earth, should a fighter jet be able to go inverted while at zero-G, precisely at 0.0? Or should that not be possible, to push up against gravity and go inverted at 0.0? Would the fighter be stuck in a 90 degree nose-down attitude? Would the pilot have to push some negative G, -0.2, -0.3, in order to get the fighter to start going inverted?
If I understand your question, in the condition of -90 pitch is it possible to continue a negative load and pass -90 pitch? Yes, it is possible to complete any amount of “nose down” pitch change even so far as to arrive at the opposite horizon inverted and beyond. A complete loop (or more) with only nose down pitch rate is called an outside loop.
Frederf, that makes sense. It made me think about how initiating a -G condition likely creates more drag from the control surfaces because the A/C is designed to pitch up more efficiently. I guess it would be a bad design to make an A/C more efficient at -1.0 G than at 1.0 G because the A/C is at positive G’s much more. So, I see the practical answer being that if you want to pitch away from gravity you of course want to pull back on the stick, and if you want to pitch towards gravity there is likely a point at which going inverted would be best. Of course you can only hold negative g’s for so long unless you are inverted. Eventually you would end up inverted going the other direction. If I have that correct.
It’s more obvious if you consider a world which a mass of air but there’s no gravity. The zero-lift (ballistic) path is obvious: a straight line. To go down you’d have to apply negative lift and to go up positive lift. On Earth with gravity the ballistic path is a downward curve and all paths with more downward curvature require negative lift while all curves with less downward curvature require positive lift.
“Negative G” is not a direction in relation to gravity. It’s simply load factor expressed in units of G. If a weight on a spring in the cockpit stretches toward the floor that’s positive load factor and if it stretches toward the canopy it’s negative load factor. It’s theoretically possible to do endless loops of negative load factor.
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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