The same question goes for any runway, but I guess that for small aircraft carriers, one might actually be able to measure the exact point of weight increases (e.g. measure the submersion of the hull). For helicopters, it seems that this is the case, so I would presmue the same for fixed wing aircraft.


closed as off-topic by DeltaLima, Pondlife, CGCampbell, NathanG, Jay Carr Nov 21 '14 at 16:21

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question does not appear to be about aviation, within the scope defined in the help center." – Pondlife, NathanG, Jay Carr
If this question can be reworded to fit the rules in the help center, please edit the question.

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    $\begingroup$ Are you asking how a ~ 22 ton F-22 will affect a 100,000 ton USS Gerald R. Ford? $\endgroup$ – Farhan Nov 20 '14 at 18:08
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    $\begingroup$ Seems like a good question for physics.stackexchange.com . $\endgroup$ – egid Nov 20 '14 at 18:24
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    $\begingroup$ "My guess is" - your "guess" is not a valid reasoning for an answer (or comment) on Stackexchange. There is a correct answer to this question, it's not opinion based. So let the experts answer it. That answer is "Yes" the force (not the weight, but it's obvious what's being asked here) exerted on the boat before touchdown is equal to the weight of the plane (in fact, if the plane's vertical speed is decreasing, then the downward force is more than the plane's weight). $\endgroup$ – Ryno Nov 21 '14 at 3:54
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    $\begingroup$ @Ryno The downward force is less than the weight of the plane. A plane flying a constant glideslope is descending at a constant rate, so it is not accelerating upwards, so the net vertical force on the plane is zero (lift = weight). So the plane is exerting a downward force on the air equal to the plane's weight and the air will exert a downward force on the carrier deck. Now, not all of the lift force will act on the deck of the carrier while the plane is still in the air because the air column between the plane and the deck is viscous and turbulent. $\endgroup$ – David Richerby Nov 21 '14 at 8:39
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    $\begingroup$ Unclear. What do you mean by weight? The common definition is W=mg. The mass of the carrier doesn't change, nor does the gravity. Although one could argue the aircraft's gravity field will attract the carrier so the result is the carrier being lighter, everybody will agree that the effect is negligible. It seems you are looking for the effect of the aircraft's pressure field on the carrier's deck, @PeterKampf covered that one. For those arguing that the carrier's weight increases only after the aircraft touches down: why do forces subjected through rubber tyres count, but through air don't? $\endgroup$ – DeltaLima Nov 21 '14 at 14:59

No. Weight is defined as the effect of gravity on a body. Until the aircraft becomes part of the body of the carrier, it has no effect on the weight of the carrier.

I think you are confusing force with weight. For example, if wind shear drives wind against the top deck, this does not make the carrier have more weight, but it does cause the carrier to move down. Likewise any gust from an aircraft may apply a force to the carrier, but does not increase its weight. You might consider that force to increase the "apparent weight" of the carrier, but it cannot be said to increase the weight of the carrier itself.

Note also that the downward wind force of a landing fighter is relatively slight. For example, a person can easily walk under rotating helicopter blades and still stand up and walk. The force is probably only around 1-2 pounds per square inch, maybe 5,000-15,000 pounds total. For an aircraft it will be even less.

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    $\begingroup$ Not sure how this (IMHO) obviously wrong answer gained 3 upvotes. @Peter Kämpf is right. Maybe we should let physics exchange to decide this one. $\endgroup$ – Peter M. Nov 20 '14 at 23:46
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    $\begingroup$ @PeterMasiar Peter Kampf is incorrect, the weight of the carrier doesn't increase until the aircraft touches the deck. Displacement increases as a force is applied, but the weight remains the same. $\endgroup$ – Rhino Driver Nov 21 '14 at 0:17
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    $\begingroup$ @PeterMasiar: Perhaps you've misunderstood this answer. The air underneath a landing aircraft does push the carrier down, but 'weight' is not the correct name for this downward force; that name is reserved for the component caused by gravity. $\endgroup$ – Marcks Thomas Nov 21 '14 at 0:32
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    $\begingroup$ @PeterMasiar Weight is a force. Any other force is a force. In that sense, all forces "feel like weight". But that doesn't make them weight. $\endgroup$ – David Richerby Nov 21 '14 at 0:42
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    $\begingroup$ @TylerDurden You are mistaken; a helicopter does not "push" itself away from the ground, rather the rotor blades generate lift (they are mini-wings, essentially), "pulling" it upwards into the air. Obv we know that there is an opposite reaction of downward force from the blades, but it does not necessarily represent X proportion of the aircraft's weight. $\endgroup$ – Jongosi Nov 21 '14 at 11:49

Excellent question! Yes indeed, it does! At least in the sense that it is pushing the carrier down. Now we have an intense discussion going on if this is weight or not. If we avoid the nitpickers and call it a downforce, I think all will agree that the landing aircraft adds a downforce equivalent to its weight to the weight of the carrier, once it is flying above the deck. Before, this downforce was acting on the water surface (and on the ocean floor ...).

Lift is created by pushing air down continuously, and this air movement has to be stopped somehow. In the end, this stopping happens by friction as long as the aircraft is at altitude. The movement is dissipated by shear forces between the air molecules. Close to the ground the movement is not stopped by shear force, but by a pressure field. Below the aircraft is an area of higher pressure, and this pressure pushes the aircraft carrier down.

Even the high-flying aircraft creates an area of slightly higher pressure below it, but the area is so big that the pressure increase is extremely small. In the end, the mass of the whole earth does not change when an aircraft takes off or lands somewhere.

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    $\begingroup$ @kjmccarx: Yes, but how is this pressure difference created? And what happens to the air once it has left the wing's surface? $\endgroup$ – Peter Kämpf Nov 20 '14 at 20:18
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    $\begingroup$ @kjmccarx: Don't forget about the law of action and reaction. If the air applies upward force to the wing, the wing must reciprocally apply downward force on the air. $\endgroup$ – Jan Hudec Nov 20 '14 at 21:40
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    $\begingroup$ I think you are confusing displacement and weight. $\endgroup$ – Rhino Driver Nov 21 '14 at 0:25
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    $\begingroup$ You are right that the approaching plane exerts a downward force on the carrier. But that force is not weight. Weight is the force of gravity on a mass, not the force of air displaced by a plane on that mass. $\endgroup$ – David Richerby Nov 21 '14 at 0:40
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    $\begingroup$ The weight of an object is the force of gravity on the object and may be defined as the mass times the acceleration of gravity hyperphysics.phy-astr.gsu.edu/hbase/mass.html en.wikipedia.org/wiki/Weight#Gravitational_definition any other force is not weight, but an external force. It's not nitpicking, it's the official definition of Weight. $\endgroup$ – Federico Nov 21 '14 at 7:43

Actually, yes it does make it weight more, while the plane is still in the air, but over the carrier:

While the force the plane actually exerts downwards on the carrier does not directly increase the carrier's weight, that same force does lower the carrier (minutely) in the water as it displaces slightly more water.

Therefore, since the carrier is now less far from the center of the planet, the force excerpted by gravity on the mass of the carrier is now (minutely) stronger, so the carrier weighs more.

  • $\begingroup$ Ha! Nice piece of lateral thinking. $\endgroup$ – David Richerby Nov 21 '14 at 14:25
  • $\begingroup$ But how does this effect compare to the gravitational force that the plane itself (and the increased density of air under its wings) applies to the carrier? My guess would be that the plane and air's gravity would win because gravity follows an inverse square law. The arm from the CG of the carrier to the CG of the air and aircraft is decreasing by orders of magnitude while the arm from the CG of Earth to the CG of the carrier is decreasing by an extremely small amount. Of course, both forces are negligible, but +1 for pointing this out anyway. :) $\endgroup$ – reirab Dec 31 '14 at 5:08

As other answers explain, it's not the weight that increases, because the mass does not increase.

But there is a downward force that adds to the downward force of weight.

It is caused by the ground effect, which leads to an increased pressure under the aircraft, if it is flying lower than roughly its wingspan.

The added downward force is the force caused by the increased pressure over the area of carier surface.


Everything depends on what you meant by "weight" in

aircraft landing on a carrier increase the carrier weight

Don't confound

  • mass
  • gravity force applied on a mass by the Earth
  • and the collection of forces components pointing down to the center of earth including ground effect pressure (or induced air pressure produced by a fluid that has been moved by another object)

By the way, applying a downward pressure aft of a long floating thing will produce a momentum that induces lift on the front part of that thing (center of buoyancy) due to the principle of flotation which involves displaced water equivalent to the total mass of the thing.

Then comes a consideration about thing, ie, what system are you talking about ?

The aircraft carrier is an object floating on the water while the aircraft is a moving object that exerts forces on anything in its vicinity including air, ground, whatever... until this aircraft landed, taxied out, applied parking brakes (and tied to the flight deck) At that moment, the aircraft can be assumed as non moving part of the boat and the system mass becomes boat mass + aircraft mass => defined new (and non moving) center of gravity with a new weight.

So ?

Weight is the force applied on an object due to gravity. Assuming changes in mass, of the aircraft carrier as well as the aircraft can be neglected, and the two objects are close enough to have the same local gravitational magnitude, we can safely declare their weight doesn't change whatever the aircraft is doing. The weight of the system [aircraft carrier + aircraft] is also constant.

So what is changing.

Everything but weight !

  • Pressure (fluids)
  • Contact forces (impact, or much more complex at molecules level with moving air)
  • Friction (induces momentum)
  • Stress (at a neglectable scale, but exists anyway)

All this actually induce movement of the aircraft carrier due to the landing plane before, during and after its landing. Whether that movement is neglectable or not only depends on your taste, but it does occur whatever you may think.

In computing, just like trying to simulate a stellar system, even multithreading can't cope with continuously (analogic) varying forces simultaneously applied. But human brain can imagine it.

Everything becomes ultimately complex when you think of permanent oscillations, vibrations turbulences or viscosity. The models we are currently using doesn't allow you to get an exact answer allowing to know the precise vector defining the forces applied by a landing plane on a carrier. You can separate the logic in many small parts, but you'll always have to neglect one or several criterias to get as close as possible to what is really happening.

Measure the submersion ?
Ocean is not still water. And air is compressible (high pressure/low pressure) Due to everything I said above, unless you know every parameter at any given time, you can't measure submersion... and at that scale...

Try this :

Take a (precise and working) roberval balance, and a paper (A4). Hold the paper horizontally above one of the plates, like 6 or 8 inches above. Drop the paper. You'll notice the arrow telling the weight will move towards the plate on which the paper is landing, but will come back nearer to the center. Neither the weight of the paper or the balance has changed, but new forces came into action when you dropped the paper : Air pressure, contact forces, friction...

Want another example ?

Imagine a very wide window on your roof (to get a panoramic view of the sky at night) Then take an F14 or an F22 and make it fly supersonic some feets above your house... Weight ? No. Air pressure.

  • $\begingroup$ Correct: any discussion needs to include buoyancy, discussion of the "system" of the carrier and aircraft, etc. $\endgroup$ – user3305 Nov 21 '14 at 15:42

Insomuch as the aircraft is flying in ground effect, yes, it exerts a downward force on a runway or aircraft carrier before the wheels actually touch down. This is not weight, per se, but it has the same effect on the runway/carrier. This is true for fixed-wing aircraft and helicopters.

Helicopters are better known for having significant ground effect. Indeed, if you see a helicopter flying or hovering over water, and the surface of the water is visibly disturbed by the rotor-wash (downward moving air from the rotor), the helicopter is flying in ground effect. In which case, it's pretty easy to visualize. One of the more difficult tasks in flying a helicopter over uneven terrain is the fact that you may be going in and out of ground effect as you cross the terrain. This makes, at best, for a bumpy ride. At worst, crossing a narrow chasm can cause you to go out of ground effect, drop significantly, and collide with the opposite side of the chasm.

Do a YouTube search for "ekranoplan" (Russian term) or "Caspian Sea Monster" (western term for the same thing). This was a very large wing-in-ground effect vehicle the Soviets were developing during the Cold War. The aircraft was amphibious, meaning it took off and landed on water. Even when the aircraft is "flying," off the surface of the water, the water underneath is visibly disturbed by it.

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    $\begingroup$ The term "amphibious" does not mean the ability to take off and land on water. Amphibious means it can operate on BOTH land and water. And "Caspian Sea Monster" is a nickname for the ground-effect vehicle you describe. Sometimes these vehicles are described as WIG vehicles - Wing in Ground Effect. $\endgroup$ – Skip Miller Nov 21 '14 at 20:50

The weight of the carrier + airplane system begins to increase when the wheels touch and gradually reaches a maximum when the wings stop providing significant lift. It should be obvious that when the plane is in the air, it has no effect on the weight of the carrier, and equally obvious that when the plane is at rest, the weight of the carrier + plane system is exactly the full weight of both vessels added together.

The airplane only puts a momentary twist into this action. The apparent weight of an airplane at touchdown is very small, and increases as the plane settles. This is also obvious if you watch the gear suspension depress during the roll out.

You can do an easy experiment to verify this. Slowly lower a weight onto a scale. The measured weight will increase gradually as the 'lift' provided by your hand decreases to zero. Lateral motion and air pressure and the fact that an airplane is "self-lifting" can really be ignored here.

  • $\begingroup$ This answer is just plain(plane?) wrong. It's factually incorrect. The fact that an airplane is "self-lifting" cannot be ignored. When you pickup a weight, that extra weight is supported by your body, and that weight is exerted on the floor through your feet. Whatever is "lifting" a weight, the weight is exerted downwards somewhere... in this case, on the deck of the boat. The answer above is correct... $\endgroup$ – Ryno Nov 20 '14 at 23:47
  • $\begingroup$ Don't freak out dude, that's exactly what I said. The weight is on the deck of the boat. When the plane is in the air, it's not. The airplane pushes a small amount of air down to the deck and this may produce a force against the boat, but it does not increase the weight of the boat, and as mentioned above, the net force over the area is zero anyway. So, ignored. $\endgroup$ – Jasmine Nov 21 '14 at 0:01
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    $\begingroup$ No. The weight of a system is the force caused by gravity acting on its weight. The weight of a system is independent of whatever other forces might be acting on it. Scales only measure weight when they are in equilibrium, i.e., when the scale is the only thing supporting the object on it and nothing is moving. While you slowly lower an object onto a scale, the scales are not in equilibrium so the number they display is not the weight of anything. $\endgroup$ – David Richerby Nov 21 '14 at 0:45
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    $\begingroup$ @Jasmine. But what you said makes sense (in a way) and is much more "understandable" than a plain physical explanation. That's why I think it doesn't deserve a downvote nor extensive arguing (but it's just me) Just like Peter Kämpf said, it depends on what OP assumed to be weight... $\endgroup$ – Karl Stephen Nov 21 '14 at 18:19
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    $\begingroup$ @Jasmine I understand that that is what you are saying. It is incorrect. You don't need to keep repeating yourself, this is not a misunderstanding of your answer. The "net force over the area" is NOT zero! $\endgroup$ – Ryno Nov 21 '14 at 18:23

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