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4.9.2.6. Fighters will use the following minimum airspeeds below 5,000’ AGL:
4.9.2.6.1. F-15: 22 Units AOA
4.9.2.6.2. F-16 CAT I: 13 Units AOA
4.9.2.6.3. F-16 CAT III: 200 KIAS

3.2.8. The CAT III position of the Stores Configuration Switch will be selected when the aircraft is configured with a Category III loading IAW T.O. 1F-16-1-2.

5.3.3. All configurations are authorized for unlimited maneuvering as defined by AFI 11-214. Before conducting unlimited maneuvering in a CAT III configured aircraft, consider gross weight, drag, departure susceptibility and training requirements.

Translation of 5.3.3 Please don't screw my jet up ... please

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johnwill wrote:

Outlaw162,

So I'll say again, a cockpit g meter does not tell you acceleration, it tells you load factor, which is total lift divided by gross weight. A more simplistic case is sitting on the ramp, not moving. What does your g meter read? 1.0. What is your acceleration? 0.0.

Here is an equation relating acceleration and g meter reading:

accel (z axis) = g meter - cosine(pitch angle)

John

"Cockpit g meter does not tell you acceleration, it tells you load factor, which is total lift divided by gross weight":

On the ramp, not moving, we are generating NO aerodynamic lift, divided by gross weight, this would appear to equal zero on the cockpit g meter. But as you say the cockpit g meter reads 1.0.

This because the aircraft is still being accelerated by the earth's gravity (g).
Acceleration is the rate of change of the velocity vector not just the magnitude. We can't continue falling toward the center of the earth because the ramp is in the way.

If you do the F=ma equation, you can see on the ramp we are being subjected to an acceleration, otherwise all our precious fighter assets would float off into space.
The G meter reflects this.

regards, O.L.

I just somehow knew you would say something about lift being zero on the ramp. That's why I didn't say aerodynamic lift. Lift on the ramp is provided by the upward acting force on the three landing gear tires. So the forces acting on the airplane are gravity pulling the airplane down and the ground reaction pushing up on the three tires. The net force is zero, the acceleration is zero, the g meter reads 1.0.

Sorry, I can't say it any clearer than that.

By the way, your equation:

accel (z axis) = g meter - cosine(pitch angle) (should be flight path angle)

is only good in "lateral axis parallel to the horizontal plane" flight at a stabilized pitch attitude resulting in a constant FPA and is an excellent illustration of why the pilot is subjected to slightly less than 1 G on ILS approach final, & slightly more than 1 G in a normal climb.

And upon further review, I didn't understand my second paragraph either.

see ya, O.L.

Well, I said there was an error due to AoA. And the difference between pitch angle and flight path angle is AoA, so we are in agreement. Your "lateral axis parallel to the horizontal plane" cracks me up. Doesn't that mean "wings level"? heh heh!! I really think the equation applies to performing a loop (for example) as well as constant FPA.

Here is a trick question for you. Say you are at 20,000ft and perform a constant radius loop at a constant speed, with the g meter reading a constant 4g. What is your altitude when you complete the loop?

If you COULD perform a constant radius CIRCULAR loop, then by definition from a point anywhere on the circumference, eventually you should return to that same point in space on the circumference, assuming the air mass you’re working in doesn’t move vertically or horizontally (they do). You would also need the excess thrust to hold a constant true airspeed up the front side, and the necessary thrust reduction capability and drag devices to hold a constant TAS down the backside. In addition, you would also need the density altitude to remain constant throughout the altitude range used for the loop or you would have vary calibrated airspeed to maintain that constant TAS if there was a lapse rate involved.

In any case you couldn’t do it with a constant 4G on the G meter (I defer to your equation).

I believe a constant “4G on the G meter” type loop, in the real world, would kind of look like a “wide” script written small “L” from an external vantage point, probably not even symmetrical. The altitude you end up at on the bottom side then depends on “specific power” (P sub s) for the 4G and altitude conditions involved. Assuming speed held constant, positive “P sub s” and you end up higher, negative “P sub s” vice-versa.

Also interesting are the accelerations/load factors involved in a square loop like some of the rich civilian aerobatic folks do in their YAK-52’s. Airsickness anyone?

It's also very possible, I've missed something.

regards, O.L.

johnwill wrote:

I just somehow knew you would say something about lift being zero on the ramp. That's why I didn't say aerodynamic lift. Lift on the ramp is provided by the upward acting force on the three landing gear tires. So the forces acting on the airplane are gravity pulling the airplane down and the ground reaction pushing up on the three tires. The net force is zero, the acceleration is zero, the g meter reads 1.0.

Sorry, I can't say it any clearer than that.

Uhh... I'm not sure about that. I'm an engineering student, and gravity is always accelerating anything on Earth down. When I took physics 1 we had to do problems on rotational forces, and one of the problems had a looping aircraft in it. The correct result, as indicated by our textbook, would indicate that the "g" felt by the passengers is nothing more than downward acceleration caused by the passenger's inertia wanting to keep moving along the velocity vector but being restrained by the rotational movement of the plane doing loops. Think of a weight at the end of a string. If you swing the string the weight will pull the string taut. Why? Because you accelerated the weight attached to the string, and because the weight is tied to the string it cannot keep going in the direction of the velocity vector. It must rotate about you at the length of the string. This causes an outward (from you) acceleration (which is 90* from the velocity vector). The outward acceleration pulls the string taut. In aircraft this is reffered to as "G".

Given this, it seems to me that the g-meter on an aircraft would tell you how much is the downward acceleration. This is fairly simple to measure compared to the lift/gross weight equation. You'd have to know how much lift the a/c is generating at any one time: how do you do that? It is far simpler to have some sort of mass system where the acceleration will make the mass pull on a calibrated sensor and this feeds the g information to the HUD. I searched "G meter" in wikipedia and google and nothing comes up, so I cannot tell you for sure if this is how they work, but it would seem to me that it's something similar.

The beauty of physics is such that it wouldn't surprise me if lift/gross weight = what I just said (downward acceleration). I just don't think that's what the g-meter MEASURES.

Senior member Joined: Dec 14, 2006 Posts: 472

The way I look at it is that gravity is not acceleration. However acceleration can simulate gravity.

If you are in a vacuum tube within the Earth’s atmosphere and you drop an object it will accelerate yet it will have no weight because the force of gravity is being nullified by the acceleration until a balance is achieved. The object will accelerate to maintain zero G. for practical matters.

The way I look at it the gravity itself ( the attraction by mass) is for the most part a constant on all of the Earth. However the indicated gravity on the Earth’s surface may be different due to centrifugal force. If I remember correctly on the Earth’s surface stationery relatively to the Earth surface a 100 pound object at the equator will weigh about 106 pounds at either north or south pole.

Like Einstein said it's all relative.

I think a correctly calibrated G. meter (mounted on the Earth’s surface) will read one G. at the North or South Pole and will read about .95G mounted on the Earth’s surface at the equator due to the Earth’s rotation.

The Earth is not truly spherical. The Earth is wider at the equator because of the centrifugal force, it is shorter at the polls.

The way I look at it gravity is the attraction of mass to other mass. Artificial gravity is acceleration acting on mass.

ATFS_Crash wrote:

The way I look at it is that gravity is not acceleration. However acceleration can simulate gravity.

You are correct; gravity is a FORCE which CAUSES acceleration. When you decelerate a car you experience a force which tries to pull you forward (hopefully you are wearing a seatlbelt). This force is due to your body, having been already accelerated by the car, wanting to keep moving forward as the car stops. If you bend at the waist you will experience sort of the same "force"; this time it's gravity. So yes, acceleration can simulate gravity because gravity (the force) causes acceleration.

The g readout in a HUD is measuring the centrifugal force acting on the objects that compose the plane. Since the wings are generating most of the lift (this would be the centripetal force - all forces come in pairs), you could say that the wings are pulling the rest of the plane "up", while the body of the plane and anything bolted on to the plane are pulling on the wings "down". This is oversimplified, but it explains why you can literally rip the wings off an aircraft if you over-g it.

Excellent posts, ATFS_Crash. Thanks for the interesting perspectives/insights.

johnwill wrote:

I just somehow knew you would say something about lift being zero on the ramp. That's why I didn't say aerodynamic lift. Lift on the ramp is provided by the upward acting force on the three landing gear tires. So the forces acting on the airplane are gravity pulling the airplane down and the ground reaction pushing up on the three tires. The net force is zero, the acceleration is zero, the g meter reads 1.0.

Sorry, I can't say it any clearer than that.

JW's completely correct, of course, but lemme take a stab at it, maybe a different way.

An aircraft accelerometer (the 'g-meter') is designed so that it measures (by observation, not by calculation) 'acceleration' along a specific axis (from now on we'll assume our g-meter is in the 'normal' axis, i.e. along a pilot's spine (in a non-reclined seat)). The reason I parenthesized 'acceleration' is that while it is (obviously) the 'a' in F=ma its not the total 'a' for the aircraft. And this is exactly the point that I think is getting missed by some. In the acclerometer's private little world, the 'F' (and therefore the 'a') is not the absolute total of all the forces acting on the aircraft. It's the total of all the forces except gravity. It may help you to think of this in the terms that (since the g-meter is attached to the aircraft) gravity (i.e. the Earth) always affects the aircraft and the g-meter exactly the same, and so those effects cancel each other out.

So, back to John's example of an aircraft sitting on the ground. We know that in the big picture, the total 'F' is zero (because the total 'a' must be zero, because the aircraft is holding constant velocity (in our Earth-fixed frame of reference)). We happen to know that the mass of the aircraft is about 621.6 slugs, so we (correctly) assume that the Earth is exerting a force of about 20,000 lbs 'down' and the ground is applying a countering force to the aircraft of 20,000 lbs 'up'. In the aircraft accelerometer's world, though, the gravity force disappears and it only sees the 20,000 'up' force from the ground. Or, more precisely, it sees only the acceleration resulting from the ground force, which according to a=F/m, is 32.2ft/sec2, or '1g'.

So, now let's flip our poor F-16 over onto its back (still on the ground). In the 'big' world we know that the total 'F' is still zero. The difference is that the force applied from the ground to the aircraft (gravity ignored) is being applied as if someone was pushing 'down' on the aircraft, rather than 'up' on the gear, as before. The accelerometer now sees the same 32.2 ft/sec2 acceleration from the ground force, but it has been told that 'aircraft up' is positive, so it now reads '-1g'.

Similarly, if we balance the aircraft on its tail. Still, big picture, total 'F' is zero. In this case, though, the ground is not exerting any force in the axis that our accelerometer cares about. All the ground force is acting along the spine of the aircraft, not the spine of the pilot. So, our g-meter, seeing no force, reads '0g'.
Same aircraft, sitting quietly on the ground, with three very different g-meter readings, all dependent on where the forces (ignoring gravity) are coming from.

Obviously, this same description works for an aircraft in flight - the only difference is that the external forces on the aircraft are coming from lift/drag/thrust instead of from the ground.

As to the mystery quiz, john. I'm pretty sure you're gonna say the aircraft will wind up right back at 20,000' feet. I agree, with the caveat that it only works exactly if you could somehow get those four G's with zero AOA. The minor complication is that the instantaneous 'center' of the loop is on the stability axis lift vector, whereas our g-meter is (likely) sensing body-axis - but a minor nit.

Senior member Joined: Dec 14, 2006 Posts: 472

As far as acceleration vrs gravity. I have a foot in both camps. However if I understand what johnwill is saying, I would lean toward his definition/understanding/description as being more scientifically accurate.

Remember it’s called a G. meter, not an accelerometer.

I think the arguments that many of you are making against john will; is very similar to the arguments that people make wanting to call a strain gauge; a stress gauge. The strain gauge, measures the amount of strain, indirectly it is fairly easy and accurately to calculate/deduce the stress if all the properties of the structure that the strain gauge is mounted on is known. However technically it would be improper to call it a stress gauge. Though in some ways it may simplify the thinking to think of it as a stress gauge.

I think it’s much like the G.meter argument wanting to say it is measuring acceleration. I think it’s more accurate to say that the G.meter measures the amount of gravity ( the sum of true gravity and artificial gravity). Sure the artificial gravity is related to acceleration, thusly it is easy to think of it (a G. meter) as measuring acceleration, and indirectly it can be used to measure acceleration.

Though I have a foot in both of the camps line of thinking, I feel I must reiterate what I said before.

Remember it’s called a G. meter, not an accelerometer.

johnwill wrote:

Gums, the reduced roll rate in Cat 3 was not needed to protect the structure from "bending". In fact, the entire FSD external stores structural flight test was flown in Cat 1, 'cause there was no Cat 3 at the time. We flew full rate 360 degree rolls with the most severe stores on 3 & 7 - GBU-8 and 3 AGM-65 on LAU-88. Same story for 4 & 6. Worst stores were 370 tanks, 6 MK-82 on MER, and 4 CBU on MER. All flown with Cat 1, without any structural problems.

Hi John

This is surprising but very impressive nonetheless Shocked

Obi

ATFS_Crash wrote:

Remember it’s called a G. meter, not an accelerometer.

http://www.aircraftspruce.com/menus/in/ ... eters.html

http://www.chiefaircraft.com/cgi-bin/ai ... eters.html

http://www.duotechservices.com/gmeter.html

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