Crash landings/ditchings

[wsherif1]

Faced with a crash landing/ditching situation, why would a pilot make a ground/water contact at the full momentum, and actual weight of the aircraft? The momentum and actual weight of the aircraft are completely controllable by the pilot! The pilot is able to dissipate the aircraft’s excess kinetic energy, (momentum), in a ground effect flare maneuver. Maintaining a level flight attitude, out of the flare, in a flight path tangent to the ground/water, the second zero aircraft weight transition occurs, (the first occurs automatically on take-off), and ground/water contact is made at zero aircraft weight and reduced momentum!

Pilot flight training has never included this procedure in the training syllabus!

[CBA_caption]

There would be an awful lot of aircraft at the bottom of the ogsplosh if they started doing that!

Seriously though, is isn't that what most pilots do in order to arrive at the gate in one piece? I can think of one Nimrod crew a few years ago that managed to do it a few years ago without too many problems.

CBA

[ecj]

The answer lies in the flight manual of the specific aircraft.

Use the technique which the manufacturer recommends.

[Farmer 1]

Fine words. What do they mean?

In particular, could you please explain to me, in English, how:

a. I can completely control the actual weight and momentum of the aircraft?

b. ground/water contact is made at zero aircraft weight and minimum momentum?

Momentum = mass x velocity (learnt that at school). Therefore either the mass or the velocity (or both) must be zero to produce zero momentum. The only way for the mass to be zero is for the aircraft not to exist. So, the velocity must equal zero. How?

Pilot flight training has never included this procedure in the training syllabus!

I wonder why.

[ChrisVJ]

It may well be that some aircraft do not ditch best at minimum 'landing' speed.

I fly an seaplane amphibian and the one thing you do not do for a successful water landing is let the nose get above the optimum attitude. Jemima stalls at 39 knots with flaps at 22 deg and settles nicely on all three wheels on tarmac. Get near that on water and you have an immediate crisis. One has to fly her on between 45 and 50 knots in her 'on step'attitude and with plenty of elevator authority left to stop the taildropping suddenly when you touch causing you to play kangaroo. ( I've been there and it is quite frightening!).

I realise of course, that my two place 1400 lb amphib has little similarity to an airliner but it does serve as an illustration that the most obvious answer is not always the one you want.

When an airline lands on a runway the fulcrum is just behind the centre of gravity and the nose stays up until the pilot lowers it or the elevator runs out of authority. If you approached a ditching the same way when the aft fuselage hit the water the nose would probably come down quickly and uncontrollably turning it into a submarine. I suspect the proper answer is somewhere in between and according to Manufacturer's SOPdeveloped from computer simulations etc.

Certainly I am just an amateur but I am just suggesting that the immediately obvious may perhaps not be so with further thought.

[wsherif1]

Farmer 1

Your comments,

How

a. I can completely control the actual weight and momentum of the aircraft?

b. ground/water contact is made at zero aircraft weight and minimum momentum?


First, remember the aircraft carrier (straight deck) landing technique? This technique modified! (dissipates excess kinetic energy.)

Second, There are two zero aircraft weight transitions, the first occurs automatically on take-off. The second occurs in a normal landing approach. (Not available in the present day mechanically controlled rate of descent procedure.)

In a crash landing/ditching situation combine the two, a nose down attitude powered glide into ground effect, and then a positive flare into a level flight attitude. Holding forward pitch control pressure to keep the nose from rising in ground effect produces a flight path at a tangent angle to the ground. The second zero aircraft weight transition is reached and ground contact is made, imperceptibly! (Landing gear scissor switch is required to "announce" ground contact)!

Used this technique in a Lockheed Electra, stopped on a "Dime"!
(Had to add power and taxi off the runway.)

Used this technique in the last two landings of my 25 hour check ride on the CV 990, embarrassed the check pilot.
,

[Farmer 1]

wsherif1,

Thank you for your most patient reply. I think we are looking at this situation from different viewpoints. I fly the type of aircraft with the propeller on top, and I’ve never landed on an aircraft carrier. (I was sorely tempted to make the last recorded landing on the Ark Royal while it was at Plymouth waiting to be towed away for razor blade conversion, but I chickened out. But that’s another story.)

I did spend some considerable time trying to explain my ditching procedures etc., but I eventually decided it would serve no purpose here, so that’s all deleted (I described the floats, problems with the size of waves, top-heavy aircraft not designed to float, etc.). I thought initially that you were making the classic mistake of confusing weight and mass. I now know better, please forgive me. Several of your phrases were new to me, but obviously not to others with more knowledge than I.

So, I will say no more on this, and hope you receive some helpful information from them as knows.

Well done on your check ride, by the way. I hope you acted with suitable nonchalance.

Farmer.

[keithl]

I know little of "zero weight transitions", but with reference to "tangent angle to the ground" in the ditching case I would point out that it is difficult to fly a tangent to a surface that is rising and falling, sometimes rather sharply.

[anartificialhorizon]

wsherif1,

It's all very good in theory but usually the reason for ditching an aircraft in water is usually due to dire circumstances.......!!!

ie the aircraft may only be partially controllable ( if at all) and landing tangentially to the water may not be possible due to waves/ turbulence etc.....

It's one of those "what if's" easily mulled over in an armchair but completely different if faced in real life... probably why they cannot / do not train for it.

[wsherif1]

crash landings/ditching

Keithl,

Your comment,

I know little of "zero weight transitions", but with reference to "tangent angle to the ground" in the ditching case I would point out that it is difficult to fly a tangent to a surface that is rising and falling, sometimes rather sharply.

Why?

[AirRabbit]

Maintaining a level flight attitude, out of the flare, in a flight path tangent to the ground/water, the second zero aircraft weight transition occurs, (the first occurs automatically on take-off), and ground/water contact is made at zero aircraft weight and minimum momentum!

Please feel free to show me where I may have the wrong idea, but … where in the name of sanity can you, with any degree of seriousness, talk about “zero aircraft weight?” Farmer indicated what he “learnt” at school – he and I must have gone to the same school. Anything that exists has mass, and as long as that mass is in any kind of gravitational field, it is going to have weight.

Obviously, you seem to believe that sometime during the takeoff the airplane passes through this (I don’t even know what to call it) phase (?) of flight – and, it does it automatically! Is this designed into the airplane? Is it a function of aerodynamics (of which most aerodynamicists are ignorant)? And, then the aircraft does it again, when landing.

Methinks you were involved in writing some of the scripts for the TV series and movies of StarTrek! That must be the way they “takeoff” and “land” those huge spacecraft vertically without any sort of “lift generating” device.

Ah … (the light bulb turns on above his head) … I see. You just turn “ON” the Anti-gravity switch and the airplane becomes weightless! What pure simplicity!

In light of this revelation, I agree, there should not be anyone, ever again, who should land an airplane at anything above “Zero Weight.” Think of the savings that could be made! All we have to do now is get the world’s aviation regulatory authorities to require the installation of “anti-gravity” devices in each airplane. Hmmm… that may take a bit of effort … you know … the stodginess of those regulator types and all … but, while we contemplate this next obstacle, could you spare some of that “happy juice” you’re gulping and one of those little wiggly cigarette thingies might be nice as well.

Used this technique in a Lockheed Electra, stopped on a "Dime"! (Had to add power and taxi off the runway.)
Used this technique in the last two landings of my 25 hour check ride on the CV 990, embarrassed the check pilot.

I guess so…would’ve embarrassed me as well!

[barit1]

flight path tangent to the ground/water

Tangent (adj) Touching; esp. (a straight line) meeting a curve or surface and not cutting it if extended.

Inasmuch as the sea's surface is a plane with disturbances (waves) I find it hard to accept "tangent" to describe what
wsherif1 proposes. I suspect he proposes arresting the rate of descent to zero just as water contact is made. Good luck!

I'd go with ecj's advice. The manufacturers have done a bit of work (model hydrodynamic tests, etc.) on the subject and the odds are probably with them.

[AirRabbit]

In asmuch as the sea's surface is a plane with disturbances (waves) I find it hard to accept "tangent" to describe what wsherif1 proposes. I suspect he proposes arresting the rate of descent to zero just as water contact is made. Good luck!

Actually, I think he’s talking about arresting the RoD to zero AND arresting the forward motion to zero, both simultaneously with touching the wheels to the runway or the belly in the water.

I’m not aware of any combination of aircraft attitude and thrust that you could achieve that would give you zero rate of descent and zero forward movement, particularly at the same instant, in conventional aircraft. Now, the military Osprey, V-22, CAN do these things, because it was designed that way. There are (or were – I don’t know where the development is at the moment) efforts underway to develop a fighter aircraft capable of vertical takeoff, transition to forward flight to achieve supersonic airspeeds, and return to vertical flight for landing. But, absent such aerodynamic marvels, we’re dealing with traditional airplanes; at least that’s the premise from which I’m speaking.

When a traditional airplane is off the ground, zero rate of descent is achieved by having the lift equal the weight. When the airplane is on the ground, zero rate of descent is achieved by having the ground keep you from descending. In each case you have the airplane mass being accelerated toward the center of the earth at approximately 32 ft/sec per sec. The airplane, any airplane, WILL accelerate toward the center of the earth unless some outside force acts on it. In the first case, lift, an applied force, acts on the airplane exactly opposite the pull of gravity – zero rate of descent is achieved. In the second case, the earth itself (OK, over-laid with concrete and stuff, but its still the earth) applies a normal force upward, against the airplane, again, equal and opposite to the force of gravity, to keep it where it is. In neither case is the mass of the airplane changed, nor therefore, is its “weight” changed.

Getting the forward motion of the airplane reduced to zero is a bit trickier; unless of course you put the airplane on the ground. At that point you would have the added force of friction to help retard forward motion. However, our task is to get the airplane to zero forward motion before it lands. Again, looking at traditional airplanes (not the Osprey or the VTOL fighter) aerodynamic drag and reverse thrust are the only sources available to retard the forward velocity. One problem springs to mind almost immediately. The only way that enough lift can be generated to adequately counter the force of gravity and yield zero RoD, is to move the wing through the air (or blow the air over the wing) with sufficient velocity to generate that lifting force. Of course, you can slow the airplane and maintain sufficient lift by increasing the angle of attack. But, increasing the AOA too much will, as we all know, wind up stalling the wing. At that point, gravity has a holiday and attracts the airplane un-hindered. As the AOA of the wing is increased toward this maximum lift generating point, at least some of the velocity of the rear-ward flow of air or exhaust gases from the engines (the presumed source of thrust) will have a downward component, yielding an upward thrust component, and can be used to help arrest our RoD. However, most airplanes have a maximum pitch attitude that can be achieved without striking the tail when in close proximity to the ground. And this maximum pitch attitude is in the same area necessary for a sufficient pitch attitude (sufficient AOA) to generate enough lift to counter the weight. As I said earlier, as the AOA is increased, the wing gets closer and closer to “stalling.” When it stalls, the airplane, still being attracted to the ground, and with the weight/lift equality becoming unequal, the airplane is going to fall. If we have been really careful and either very, very good or very, very lucky that stall AOA will be achieved as the wheels are very, very close to the ground – so that the distance the airplane will “fall” will only be a small fraction of an inch. This little “maneuver” is called a full stall landing. This is not generally thought of as a reasonable landing choice primarily for two reasons. First, with the nose of the airplane that far above what would be necessary if a normal approach and landing had been flown, the pilot has virtually no directional control of the airplane; it is much more susceptible to cross winds (in that the rudder is likely to have little effect) and nose wheel steering is of no use at all, and lateral control and pitch control are also very limited if available at all. Second, the definition of a stall is when the airplane “quits” flying. If the height above the ground is not the small fraction of an inch we had hoped for, the airplane is going to fall that distance. How far? Well, how far above the surface were the wheels when the stall occurred? Ever wonder what would happen if you dropped a B747 from 15 feet? How about a B737 from 3 feet? Well, you get the picture.

And, of course we’re still going to be moving forward at some pretty good clip, so the momentum is still there. How fast? Well, that depends on the weight of the airplane, but I would guess something on the order of 85 to 100 knots.

So … zero weight and zero momentum landings? Certainly not zero weight. There is no way that is going to happen. Zero momentum? Hardly. I cannot see how anyone could get an airliner slower than 85 to 100 knots of forward speed – even doing all we’ve described. But, arresting the RoD to zero just at touchdown? Mmmm … maybe, but I wouldn’t go beyond exceptional skill, lots of luck, and acceptable risk in consequences. I’ve seen Bob Hoover do power off loops and “dead stick” landings – some of which were full stall landings. But just how good WAS he? How lucky WAS he. And, if he piled it up, other than his family and friends, most folks would probably scratch their heads and mutter something about “dern fool acrobat pilot, showin’ off gets ya hurt.” But if it was XYZ Airlines, it would be all over the news within minutes of it happening. Thanks, but no thanks.

[wsherif1]

Crash landings/ditching

AirRabbit,

Your comments,

"When a traditional airplane is off the ground, zero rate of descent is achieved by having the lift equal the weight." Correct!

"Actually, I think he’s talking about arresting the RoD to zero AND arresting the forward motion to zero, both simultaneously with touching the wheels to the runway or the belly in the water."

I never mentioned arresting the forward motion to zero!

After the flare maneuver, (in ground effect), holding forward pitch control pressure to keep the nose from rising, (in ground effect), the lift is >< the aircraft weight, and the second zero aircraft weight transition occurs at reduced momentum, due to the kinetic energy dissipation in the flare maneuver, (in ground effect). The touchdown occurs at the instant the weight is greater than the available lift, imperceptibly!

Note: This technique can not be used in todays prescribed, stabilzed approach procedure. There isn't any zero aircraft weight transition available in this mechanically controlled rate of descent approach.

[AirRabbit]

Wsherif1:

I never mentioned arresting the forward motion to zero!

I went back and read the thread again. I apologize; your statement was “…at minimum momentum…” not zero momentum. However, to reach minimum momentum, forward speed would have to be reduced to a minimum as well; and I still believe that achieving minimum forward airspeed before landing is, at best, un-necessarily difficult and bears an even more un-necessary risk.

After the flare maneuver, (in ground effect), holding forward pitch control pressure to keep the nose from rising, (in ground effect), the lift is >< the aircraft weight, and the second zero aircraft weight transition occurs at reduced momentum, due to the kinetic energy dissipation in the flare maneuver, (in ground effect).

I’m assuming (a dangerous thing to do) that by your use of the symbol “><” you mean approximately equal (?). As long as the weight of the airplane is being supported by the lift being generated, where there is no climb or descent, the lift does, in fact, equal the weight.

I’m not sure why you describe holding forward pitch control pressure to keep the nose from rising (in ground effect)… because whether or not that will be true, at least initially, is going to be a matter of where the pitch trim has been set. The kinetic energy dissipated while still airborne is going to be directly proportional to the aggressiveness of the flare, the amount of aircraft “frontal area” presented to the relative airflow, and the length of time the airplane is kept airborne. Again, the only way I can see of reducing that kinetic energy to its lowest point while still airborne, would be to achieve an AOA that provides L/D max (which as you know is, or is just short of, the stalling AOA). In my opinion, inside ground effect is no place to have the nose that high; and I explained my reasoning for this in my earlier post.

Modern jet transports fly final approach at a no-wind speed that is approximately 30% above the stalling speed for that configuration, and with wind present an additive of ½ the steady-state wind and all of any gust factor. The generally accepted technique is to have the throttles at idle at the same time the MLG touch (where throttle reduction is initiated somewhere between threshold crossing and reaching the level flight attitude – and the speed at which the throttles are retarded is directly related to the initiation point), and that touchdown should occur in the same level flight attitude achieved at the conclusion of the flare. Ideally, the time between achieving level flight attitude at the end of the flare and MLG touchdown will not be very long – in fact, it should be simultaneous, but certainly not more than a few seconds. In my world, pulling the nose to a position higher than what is necessary to maintain the correct level flight attitude is the wrong thing to do (again, see my earlier post).

The touchdown occurs at the instant the weight is greater than the available lift, imperceptibly!

IF, the MLG are imperceptibly close to, but not yet on the ground, at the instant the weight becomes greater than the available lift (and, by that, I am, again, assuming you mean when an aerodynamic stall is achieved) the airplane will touch down about as softly as you can touch down a modern airliner and landing weight. But it will touchdown at a forward speed of something in the neighborhood of 85 to 100 knots (give or take a few knots, probably; and assuming the power at idle) and in a condition that is, as I have pointed out, un-necessarily nose high, which would be at least un-necessarily and relatively severely limited in controllability, and un-necessarily farther down the runway. In my book that is unacceptable for any of these reasons.

My counter is to fly the final approach at a speed of 1.3Vs (Vref), flare to level flight attitude so that level attitude will be close to touchdown, retard the throttles through the flare, and touchdown at a speed something below Vref, with the airplane in a level flight attitude and the power at idle. This provides touchdown at a speed close to, but comfortably above, the stalling speed, will allow the touchdown to occur at a reasonable RoD (smoothly firm or firmly smooth, your choice of descriptions), and that will, each time, yield safe airplane controllability and consistent touchdown within the touchdown zone or the first 1/3 of the runway, which ever is less.

Note: This technique can not be used in todays prescribed, stabilzed approach procedure. There isn't any zero aircraft weight transition available in this mechanically controlled rate of descent approach.

And I would contend that the reason there are prescribed, stabilized approach procedures is to prevent specifically and intentionally, attempts to perform “full stall” landings.

[wsherif1]

Crash landings/ditching.

AirRabbit,

Your comments,
quote:
___________________________________________________
"As long as the weight of the airplane is being supported by the lift being generated, where there is no climb or descent, the lift does, in fact, equal the weight."

"I’m not sure why you describe holding forward pitch control pressure to keep the nose from rising (in ground effect)."
____________________________________________________

When descending into ground effect the relative wind assumes a more vertical angle, increasing the effective lift. Forward pitch control pressure is required to maintain a level flight attitude.

quote:
____________________________________________________
"I went back and read the thread again. I apologize; your statement was “…at minimum momentum…” not zero momentum."
____________________________________________________
I actually meant reduced momentum, my error.

quote:
__________________________________________________
"In my opinion, inside ground effect is no place to have the nose that high"
____________________________________________________
Actually, the aircraft's nose is held continually in a level flight attitude, while descending in a tangent angle, to the ground, flight path. A landing gear scissor switch is required to "announce" ground contact.



…

[Capt Pit Bull]

wsherif1

I'm very sorry, but most of your posts are indescipherable to me. I don't wish to be offensive, but you clearly confuse many issues of basic mechanics, such as weight / kinetic energy / momentum etc.

You may well have practical experience of landing / ditching issues that means you have something very pertinent to say about it, but sadly you lack the scientific terminology to put it across. You need to put your concepts across in more normal scientific terminology to have any chance of being understood. Sorry.

cpb

[chornedsnorkack]

Terms

Well, it sounds like "weigth" has differing meanings in aviation. I think "lift" and "drag" also have different meanings.

"Zero weight transition" sounds like the point where the lift equals the force of gravity. Thus "weight" is force of gravity minus "lift" or force of gravity minus total upwards aerodynamic force.

An aircraft flying level has zero "weight". Depending on definition, I am not sure whether an aircraft descending at a constant rate or descent, or indeed climbing at a constant rate, has zero "weight" or not.

An aircraft might have zero "weight" while flying level very near ground/water and then touch down with a minimal movement (at a substantial speed along the surface, of course).
Is it what the OP advises?

Then again, an aircraft might before actually coming to contact with surface decrease the lift a lot. Once the lift becomes less than the force of gravity, the "weight" becomes positive and the aircraft begins to accelerate downwards. And touches the surface at a nonzero rate of descent.

Depending on how fast the aircraft decreases the lift, it might actually be possible to decrease the angle of attack to zero - at which time the aircraft produces no lift at all - and therefore the aircraft would have "full weight" by the time it touches surface.

Is it what the OP critizises?

Speculating further, it seems that an aircraft might actually touch down at more than its full weight. If a wing flies at a negative angle of attack, the lift is directed downwards. Also there are things like spoilers which are meant for generating downwards lift...

Also, an aircraft must be able to fly, and land, at negative "weight". If lift exceeds the force of gravity, the plane will accelerate upwards. And descending at a decreasing rate of descent is accelerating upwards. Thus, if an airplane pulls out of a glide slope, it has to increase the lift so that it exceeds the force of gravity and the "weight" is negative, so the aircraft descends at a decreasing rate of descent and eventually reaches level flight. Or the aircraft might meet the surface while still descending at a decreasing rate, so land at a negative "weight".

[dublinpilot]

Well, I don't understand pretty much anything that wsherif1 is saying (no offence intended), but one thing that I am pretty sure of is that the touchdown in a ditching (ie without power) will NEVER be imperceptable.

[AirRabbit]

Well, it sounds like "weight" has differing meanings in aviation. I think "lift" and "drag" also have different meanings.

There are probably some in the industry who don’t understand the terms used, but that doesn’t mean that the terms have, or should have, different meanings. In order to communicate effectively you have to use terms that are commonly understood. Aviation is no different. If we start assigning different definitions to previously understood terms, no one will be able to understand anyone. The terms “weight,” “lift,” and “drag,” are well understood terms. As my British friends are wont to say, “Let’s not muck around with them.”

"Zero weight transition" sounds like the point where the lift equals the force of gravity. Thus "weight" is force of gravity minus "lift" or force of gravity minus total upwards aerodynamic force.

Personally, I’ve never heard of “zero weight transition,” but, then again, I hadn’t heard the term “warp” or “beam me up, Scotty” until the original writers of StarTrek first brought them to my attention. Let’s not go around inventing new terms and leaving everyone the daunting task of trying to figure out what the term means and how it might be used. Again, even the original StarTrek authors had to develop concepts, regardless of their “believability,” and use them consistently or their ideas would have gone the way of the channel changer.

An aircraft flying level has zero "weight". Depending on definition, I am not sure whether an aircraft descending at a constant rate or descent, or indeed climbing at a constant rate, has zero "weight" or not.

Any time a gravitational force is attracting any thing that has mass that mass will have “weight.” The amount of “weight” will be determined by the amount of mass and the amount of attraction. Airplanes are no different – unless you know of a “mass-less” airplane. We know that the airplane changes mass; i.e., the airplane is burning fuel and turning some of the mass into energy. It is that energy that is used to adjust the “balance” of the forces acting on the airplane; and it is this adjustment in the balance of forces that result in the airplane climbing, descending, or maintaining level flight. No more, no less.

An aircraft might have zero "weight" while flying level very near ground/water and then touch down with a minimal movement (at a substantial speed along the surface, of course).

As I said, unless you can make all the mass “disappear,” the airplane will have “weight;” and it makes no difference as to where it is flying. It is true that due to the compressibility of air and its interaction with the surface of the earth (ground or water) the amount of “lift” necessary to balance “weight” when inside of this “ground effect” is a little different (somewhat less) than that required outside of this area. However, even considering ground effect, as I said earlier, theoretically, if you could control the force of lift in very fine, very small increments, and do it accurately and consistently, it might be possible to adjust that lift to have it just a bit less than the weight it is countering, allowing a rate of descent (RoD) to develop. But, that RoD would have to be so small that it would take a while to close the final distance between the wheels and the runway (or between the belly and the water’s surface) in order for the “touchdown” or “splash-down” to occur imperceptibly. To my knowledge, humans don’t have the ability, at least with the tools they have at their command at the present, to control the balance of forces acting on an airplane in flight to the degree that I’ve just described. Keep in mind that all of the lift being generated is directly related to the speed of the wing through the air and the Angle of attack (AoA) between that relative wind and the wing itself.

Is it what the OP advises?

The simple and direct answer is, NO. While it might be desirable to be able to touchdown with an imperceptible vertical speed, it is NOT desirable to do so when considering the trade-offs that would be necessary, even if it could be done repeatedly. Remember, the airplane is moving in 3-dimensional space. The rate of touchdown is only considering one of those dimensions. The performance in this one dimension is significantly related to and directly dependent upon the performance in at least one of the other dimensions.

Then again, an aircraft might before actually coming to contact with surface decrease the lift a lot. Once the lift becomes less than the force of gravity, the "weight" becomes positive and the aircraft begins to accelerate downwards. And touches the surface at a nonzero RoD.

Yes – if you always maintained a zero RoD, you’d never get to the ground. All landings occur at some RoD.

Depending on how fast the aircraft decreases the lift, it might actually be possible to decrease the angle of attack to zero - at which time the aircraft produces no lift at all - and therefore the aircraft would have "full weight" by the time it touches surface.

Again, “weight” does not change. You can rotate the airplane around the lateral axis all day, increasing and decreasing the AoA, but the “weight” of the airplane does not change (except of course, for the mass of fuel that is turned into energy). The “transition,” if we must use that word, occurs when the ground supports the aircraft and the lift no longer does. In order to get an “imperceptible” touchdown, this “transition” would have to occur slowly enough to be below the threshold of recognition of the occupants of the airplane. And, again, to my knowledge, humans don’t have the ability, or the tools, to control the balance of forces acting on an airplane to such a fine-tuned degree to achieve this set of circumstances, other than by pure fortune.

Speculating further, it seems that an aircraft might actually touch down at more than its full weight. If a wing flies at a negative angle of attack, the lift is directed downwards. Also there are things like spoilers which are meant for generating downwards lift...

I think I’ve mentioned that “weight” isn’t going to change. Yes, you can accelerate an airplane into the ground and it will impact with a greater force than it would impact if gravity were the only force acting on it. But I don’t think that’s what you are referencing – although, it wouldn’t be the first time I’ve miss-interpreted someone.

I hope you don’t take this next comment to be anything more than for what it is intended … spoilers don’t “generate” anything. The purpose of spoilers is just what the name implies – to “spoil” lift. Nothing more, nothing less.

Also, an aircraft must be able to fly, and land, at negative "weight". If lift exceeds the force of gravity, the plane will accelerate upwards. And descending at a decreasing rate of descent is accelerating upwards. Thus, if an airplane pulls out of a glide slope, it has to increase the lift so that it exceeds the force of gravity and the "weight" is negative, so the aircraft descends at a decreasing rate of descent and eventually reaches level flight. Or the aircraft might meet the surface while still descending at a decreasing rate, so land at a negative "weight”.

There is only one way to say the following: there is no such thing as “negative weight.” However, since acceleration is defined as a “change in velocity;” and velocity is defined as “speed in a direction” if you change the RoD you are technically “changing the speed while maintaining the direction.” You would be negatively accelerating downward, and, therefore, you are correct, you would be “accelerating upward.” You are also correct in that when you flare to land you are adjusting the AoA to increase lift, reducing the RoD. But, if you decrease the RoD to zero, and keep it there, you will never land. And, certainly if you establish an AoA that increases the lift to some value in excess of the weight of the airplane, you will establish a rate of climb (RoC) – and, unless you’re landing on a runway built on an increasing slope (increasing faster than your RoC), you’re still not going to land. To land, even from a very small distance above the runway, you’ll need to establish a RoD. The idea is to land (to touch down) at a RoD that is safe considering all of the factors acting on the airplane. And, for one final time, there is no such thing as a “negative weight.”