Gravity and AH's maximum g

[LoneStarBuckeye]

I am wondering how gravity interacts with AH's g-limit, which I understand to be 6 g. In particular, is the 6-g limit a limit on the g that a plane can generate or is a limit on the total g that affects a plane?    

To illustrate what I believe to be the difference, consider two different, related situations:  (1) a plane in level flight, with sufficient speed to allow it to pull maximum g, performs a non-oblique, maximum-g Immelman (i.e., straight "over-the-top" half-loop reversal); and (2) the same plane performs a maximum-g, non-oblique split S (i.e., straight "underneath" reversal).  If AH's g-limit is such that a plane can generate no more than 6 g, it seems to me that at the very beginning of the maneuvers, the plane performing the Immelman reversal will experience 5 g (i.e., the 6 g generated by the plane minus the 1 g due to gravity), but the plane performing the split-S will experience 7 g (i.e., the 6 g generated by the plane plus the 1 g due to gravity).  If, on the other hand, AH's g-limit is such that the total g affecting a plane can be no more than 6 g, at the very beginning of the maneuvers both planes would be affected by 6 g--the plane performing the Immelman will generate 7 g, but the plane performing the split-S will generate only 5 g.

If anyone has any insight or corrections, I would really appreciate reading them!  Thanks in advance.

- JNOV

[john9001]

the 6 g limit is on the pilot not the aircraft, it's a arbitary number by AH for pilot black out, in RL the pilot g limit varies by training , physical condition, body size, etc.

the g limit for the aircraft is where parts of your plane break do to the g force , IE wings and tail fall off. this will vary by aircraft model and type.

the force of gravity ( 1-G ) has nothing to do with the loads placed on the pilot/plane other than to be used as a referance number. the loads on pilot/plane are from linear acceleration, (trying to make the plane change course from stright flight) and are measured in reference to the force of gravity. 2-g, 3-g, etc

gravity does have a effect on vertical moves ( loop, immelman, split-s), on the way up gravity will pull the plane down making the veritical turn radius smaller and on the way down will make the turn radius larger, but gravity will not change the loads placed on the pilot/plane. it's kinda involved but i hope that helped.

44MAG
knight barbarian

[LoneStarBuckeye]

I don't think its correct to state that:

the force of gravity ( 1-G ) has nothing to do with the loads placed on the pilot/plane other than to be used as a referance number. the loads on pilot/plane are from linear acceleration, (trying to make the plane change course from stright flight) and are measured in reference to the force of gravity. 2-g, 3-g, etc.

I think I understand what centripetal acceleration is, and I also think I understand its relationship to velocity, turn radius, and the letter "g."  Namely, centripetal acceleration = velocity^2 / turn radius (for constant velocity and radius).  I also believe that the result of a centripetal (or any other) acceleration acting on a mass (e.g., a plane) is a force (force = mass * acceleration), just as the result of gravity's one-g acceleration acting on a mass is a force (in the case of you sitting at your desk, that force is your weight).  Finally, I believe that the net force acting on a plane is the product of its mass times the vector sum of its acceleration (centripetal or otherwise) and the one-g accerlation due to gravity.  Thus, I think that a plane's turning performance is affected by the orientation of its lift vector with respect to the vector direction of gravity's one-g acceleration.

My question is how AH's g-limiter takes gravity into account.  I read your post to suggest that the g limit is an arbitrary physiological limit on the pilot.  If so, I think that the 6-g limit is a limit on the total g-load (including the component due to gravity) that can act on the pilot.

Why doesn't gravity affect the load on a pilot?  It seems to me that the pilot has mass just like anything else, and the force exerted on him will always include a component due to "God's g."  For instance, if you've ever been on an amusement park ride that takes you around in a vertical circle at a constant angular velocity while keeping your head pointed toward the middle of the circle (e.g., a "witches wheel"), don't you feel something different at the top of the circle than you do at the bottom?  I think you do, and the reason is that the constant, downward force that gravity exerts on you adds to the centripetal force (always pointing toward the center of the loop) that you feel at the top of the loop but subtracts from the centripetal force at the bottom of the loop.

[john9001]

maybe i missunderstood the question, if you were asking if AH models in the force of gravity, the answer is yes

[LoneStarBuckeye]

Actually, John, I think that I have been thinking about this erroneously, and when you stated that "gravity will not change the loads placed on the pilot/plane" you were quite right.

In the case of of a turning airplane, centripetal acceleration is defined by v^2/r.  Thus, v^2/r defines the centripetal load on the pilot, regardless of the orientation of the plane's lift vector with respect to gravity.  My mistake was in reasoning that acceleration due to gravity is something in addition to centripetal accelration, rather than a component of it.

Thus, a pilot of a plane traveling at a constant velocity of v at the top of a loop of radius r will feel a force of

m*total centripetal acceleration = m*v^2/r = m*(centripetal acceleration due to the plane's lift PLUS the one-g acceleration of gravity),

where m is the pilot's mass.  The same pilot (in the same plane, traveling at the same contant velocity) at the bottom of the loop will feel the same force (albeit in the "up" direction rather than in the "down" direction):

m*v^2/r = m*(centripetal acceleration due to the plane's lift MINUS the one-g acceleration of gravity).  

Thus, the force or load on the pilot/plane is the same in both cases, but in the former case, the plane is generating less centripetal acceleration (in the form of lift) than in the latter case.  (And my "witches wheel" example from above is wrong, although the force the rider feels through the ride's seat is greater at the bottom than at the top, due to gravity.  That doesn't change the total force the rider feels, because that force is defined by v^2/r.  It does, however, change the comonent of that force contributed by the ride's seat pressing against the rider's backside.  That seat force is the analog of the plane's lift from the looping example above.)

So, I guess you've helped me to answer my own question.  AH limits instantaneous v^2/r so that it cannot exceed 6g, regardless of the orientation of a plane's lift vector with respect to gravity.

Thank you.

- JNOV

[bozon]

LoneStarBuckeye,
the "G" number is just how much lift the aircraft generates (aerodynamicly) in own-weight units (lift/weight).

by your logic, flying straight and level would be 0G, but this is considered one G - the lift is exacly the weight.

Bozon

[LoneStarBuckeye]

Bozon:

What do you mean by "your logic"?  I guess I am missing your point.  

The net vertical force (and acceleration) acting on a plane flying straight and level is zero.  A plane's g-meter reads 1.0 in this circumstance because the plane is generating just enough lift to offset the 1-g acceleration of gravity.  (I'm not sure what, if anything, this has to do with "weight units" or "lift/weight").  If you roll the plane inverted and maintain level flight, the g-meter should read -1.0 (because the plane is generating "negative" lift just sufficient to offset gravity), notwithstanding that the total vertical force (and acceleration) acting on the plane is still zero.  Thus, there is a difference between the net acceleration acting along a plane's lift vector and the indication on its g-meter.

If I understand the way that AH's g-limiter works, I believe that this means that the g-meter of a plane flying straight and level that pulls into a maximum-g loop will read 7.0 but the g-meter on a plane flying straight and level that rolls inverted and pulls into a maximum-g split-S will read only 5.0.  In each case, the magnitude of the net centripetal acceleration acting on the plane is the same (v^2/r=6g), but the amount of that acceleration geneated by the plane's lift differs due to the contribution of gravity.

- JNOV

[Lephturn]

No, you'll black out at 6G's either way.  You won't see 7G's on the G meter, and you won't black out at 5G's.  Wether you are fighting gravity or not, the aircraft must still generate all of that lift to - 6G's.  The only difference is the effect of the airplane generating 6G's of lift on your flight path.

When you pull up, initially part of that 6G's of lift is counter-acting 1G of gravity.  You still black out at 6Gs and the plane still generates 6G's of lift, you just change direction a bit faster going down, then when going up, at least to start.  It's not nearly that simple of course, because when you go up you are trading kinetic energy (speed) for potential energy (altitude), where as when you go down you do the opposite.  Since airplanes perform quite differently at different speeds, this changes the game big time.  The plane breaking upward likely won't be able to maintain 6G's for long simply because his speed will drop to a point where he can't generate enough lift to turn at 6G's.  The plane breaking downwards will be trading altitude for speed, and possibly gaining speed, making it easier to maintain 6G's, but at a higher speed than the fellow that went up.  The result being that the upward break will end up being a tighter radius, lower rate turn than the downward break, which will be larger radius, but higher rate turn.

[bozon]

What do you mean by "your logic"? I guess I am missing your point.

I meant by the way you are "counting" the G number.
G has nothing to do with the planes direction or position, it is just an indication of how much stress is on the frame (or how much lift is generated). this is the acceleration felt by a person attached to the plane's frame of reference.

for the acceleration as seen by an inertial viewer (lets say a man standing on the ground), you need to add the gravitational acceleration, as you described - this will give you the initial rate of turn (or turn radius or whatever). but you can't refer to "G" number.

my point is that the pilot blacks out at 6G by the plane's meter. it has nothing to do with directions.

Bozon

[LoneStarBuckeye]

Thanks for the replies -- now I'm confused again:)  (I am not trying to be obnoxious or confrontational, I am genuinely confused and want to figure this out.)

As I see it, there are two ways that AH's 6g limiter could work:  

1.  Planes are limited to generating lift force equivalent to a lift-line acceleration of 6g.

OR

2.  Planes (and their pilots) can experience a lift-line acceleration of no more than 6g (due, perhaps, to modeling the pilots' physiological threshold).

Unless I'm mistaken, those two possibilities are different.

If I understand all of you correctly (which I may not), John seems to believe that case 2 is correct, but Leph and Bozon believe that 1 is correct.

My limited imperical evidence seems to weigh in favor of possibility #2.  For example, I know that planes' g-meters can register, at least temporarily, readings higher than 6.0, because I have seen it.  There seems to be a delay in the black-out effect, and if you pull, very abruptly, into a climb, you can see the g-meter swing well past 6.0.  This is consistent with what I believe to be the fact that when entering a climb from straight-and-level, in order for the plane and pilot to be affected by a net lift-line acceleration of 6g, the plane must generate 7g to compensate for the 1g due to gravity.  (As Bozon correctly noted above, the g-meter registers 1.0 in straight-and-level, non-inverted flight, during which the plane and pilot experience zero net lift-line acceleration.)  This suggests to me that #1 above is not true, at least not absolutely.  

Leph and Bozon:  If I understand you correctly, you are saying that AH limits planes to 6G, as indicated on the g-meter, and it is that indicated 6G that corresponds to the pilot's blackout.  Is that right?

Shouldn't pilots blackout when they experience a particular centripetal acceleration (i.e., accleration along the plane's lift line and also roughly along the pilot's spine, from tailbone to head)?  It seems that whenever the pilot experiences 6g, regardless of his orientation, the same 6g acceleration is forcing the blood away from his head.  (I may well be wrong about this, but if you can explain why, I'd really appreciate it.)  

In an (imaginary) AH plane that continually loops in a constant-speed, constant-radius flight path, shouldn't the plane and pilot always experience the same centripetal acceleration, as defined by v^2/r=6g?  And, in that hypothetical case, shouldn't the plane's g-meter read 5 at the top of the loop and 7 at the bottom?  If not, why?  It seems to me that in order to maintain that constant-speed, constant-radius, vertical loop, the net acceration acting along the plane's lift vector must always be equal to v^2/r=6g.  If so, the plane generates 7g at the bottom, 5g at the top, and 6g when vertical (i.e., at 3:00 and 9:00).  The pilot is blacked out the whole time, but the plane's g-meter continually fluxuates between 5.0 and 7.0 (isn't the stress on the plane's wings different at the top of the loop than at the bottom, notwithstanding that the net lift-line acceleration affecting the plane and pilot is the same?).  If that is wrong (and it may well be) and you can explain why, I'd really like to read your explanation.  

Thanks again for the insight - JNOV

[Dwarf]

Don't know whether this will help, but think of the G-meter as an adding machine.  It automagically totals all accelerations acting on plane and pilot.

So, at the top of your loop, the aircraft and pilot experience 6G's of centripetal acceleration minus 1G of "God's gravity" and the G meter reads 5.

At the bottom of the loop, unless the pilot eases off on the stick, the aircraft and pilot will experience 6G's of centripetal acceleration plus 1 G of gravity, and the G meter would read 7.

Add in the velocity changes due to vertical flight, and you get a tactical egg rather than a round loop for a pilot who modulates stick pressure so as to always show 6 G's on the meter.

Dwarf

[Lephturn]

I just want to point out, that the pilot and plane are modelled differently in AH.  The pilot will black out at 6G's and lose vision, but that in no way limits the air frame.  It is entirely possible to pull more G's than the airplane can handle and rip off the wings... I'm just pointing out that no matter what you do, the virtual pilot will black out at 6G's and you'll lose vision, you will often pull more than 6G's in AH, it's just that you'll black out at 6.

AH doesn't limit how much G the airframe can generate arbitrarily... it just makes the pilto react to those G's at about 6 G's.  You can pull more than 6G's in AH... it's just that you won't be seeing anything past that point.  The airframes themselves, I believe, follow the real airplanes in terms of how much G they can stand before being damaged.  Try it sometime... dive a fast plane to 500 IAS and yank the stick back.  If the aircraft maintains enough control authority at that speed to generate more than 6G's, you'll black out.  Keep pulling, and although you won't see anything, if you can exceed the structural limitations of the airframe, you'll tear the wings off of it.

[LoneStarBuckeye]

Dwarf:

Thanks.  I think (but am not entirely sure ) that you and I are on the same page!

Leph:

Thanks for that information; it is very useful and, in fact, much more important than the answer to the question I originally posed.  Indeed, it renders that question moot.  (Perhaps the reason my question was hard to understand is that everyone but me appreciated the substance of your response!)

I had been assuming that AH did artificially limit a plane's ability to pull Gs.  If, as you state, that is not the case, a pilot that pulls well into the blackout stage should outturn (at the cost of energy) a pilot that tries to ride the edge of blackout (in some situations, being blacked out is a small price to pay for a temporary boost in turn performance).  I assumed that by riding the edge of blackout, I was turning at max performance.  Doesn't your response also imply that Badboy's EM plots (excellent, by the way), which use the 6g line as the "roof of the doghouse," are a little misleading?

Thanks again - JNOV

[J_A_B]

You don't want to pull "in the black" for any signifigant length of time because your blackout will rapidly turn into G-LOC (when this happens you will remain blacked out for awhile even if you start to fly straight and level--NOT a good thing and almost certain death in AH).

But yes for short periods of time you can turn "in the black" to get away from somebody and in fact such a tactic is quite common in AH although there are probably better evasive maneuvers available that don't waste so much "E".


J_A_B

[Lephturn]

Yes, you can pull more G's despite the black-out.  Were that not the case, you wouldn't be able to rip the wings off of a plane... and I assure you I've done it.  To get a better idea of what's happening, you should run some offline tests with, say, a Pony and film it.  Do some high-speed dive and recovery tests, try using manual trim to pull huge G's, or hit one of the auto-trim modes to execute a hands-off pull-out in excess of 6G's.  See if you can rip the wings off of one due to G's in a pull-out instead of speed.  Go back and view the film with the stand-alone film viewer and you will be able to see the G's pulled without the blackout.  It's interesting, and I think would help you see what's happening.

Badboy's EM charts are still very useful, in that they will give you a good idea of the comparitive performance envelope of the aircraft.  The 6G blackout line is an effective line to use for comparison, as anything beyond that removes vision, is dangerous on the airframe, and will be fleeting.  I think the 6G line will give us enough information to be able to effectively compare different planes, given that either pilot has the ability to temporarily exceed 6Gs in some situations.