7

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.

yippy_yay
  • 7,365
  • 14
  • 52
  • 67
  • 5
    Are you asking how a ~ 22 ton F-22 will affect a 100,000 ton USS Gerald R. Ford? – Farhan Nov 20 '14 at 18:08
  • 11
    Seems like a good question for http://physics.stackexchange.com . – egid Nov 20 '14 at 18:24
  • We seems to have rather strong disagreement between yeas and noes, and only experts on physics can decide. Or maybe we can agree that effect will be very small and not measurable (but not zero). – Peter M. - stands for Monica Nov 21 '14 at 00:32
  • I think you're asking whether there's a down-draft (force), onto the landing deck, that's equal to the weight of the plane. My guess is that the answer is "no": that what keeps the plane in the air isn't downdraft or at least isn't only downdraft (unless it's a VTOL jet or a rocket), rather it's a partial vacuum (i.e. a low-pressure system) on the upper surface of the wings. – ChrisW Nov 21 '14 at 02:43
  • 3
    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? – DeltaLima Nov 21 '14 at 14:59
  • @Ryno can you show the [meta.aviation.se] or [meta.se] discussion where reasoning validity is a requirement for commenting, pretty please? (If you can, I'll gladly delete this comment, if you cannot, your comment is rude, IMHO. – CGCampbell Nov 21 '14 at 15:25
  • @DavidRicherby - "IF the plane's vertical speed is decreasing..." - if it isn't decreasing, that doesn't make the reasoning wrong. And I maintain that it must be decreasing, or the plane will hit the deck with too much vertical speed. Whether all of the lift force will act on the deck depends on the height of the plane. I maintain that shortly before contact, it is close enough that all of the lift force IS acting on the deck. It will be spread over more than the footprint of the plane, sure, but still on the deck. – Ryno Nov 21 '14 at 18:08
  • 1
    @Ryno My understanding is that carrier landings are done pretty hard. At any point, some of the lift force is coming from work done against the viscosity of the air. At any point, not all of the lift force of the plane is transmitted through the column of air below it to the "ground" below. – David Richerby Nov 21 '14 at 21:36
  • +1 to DeltaLima for mentioning the upwards gravitational force on the boat from the airplane itself. Also, the increased density of air molecules near the deck would create even more upwards gravitational force. :) I tend to disagree that this question is really off-topic here. Sure, it's a physics question, but, definition of weight aside, the primary portion of the answer to that question is an explanation of how wings create lift, which seems decidedly on-topic for Aviation SE. It seems like we should have no shortage of aerodynamics experts at Aviation SE. – reirab Dec 31 '14 at 05:41
  • 1
    @Farhan ... I think an F-22 landing on the Ford would be there for a while ... their gear and landing hook are too weak and will break, and (last I checked) they have no catapult attachments. (Granted, they may have enough thrust to takeoff unaided.) – r2evans Jul 07 '17 at 23:45

7 Answers7

15

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.

Tyler Durden
  • 13,282
  • 4
  • 43
  • 67
  • I generally agree, but helicopters don't weigh 72,000. Hovering, a helicopter should displace it's weight in air. – Rhino Driver Nov 20 '14 at 23:35
  • 3
    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. – Peter M. - stands for Monica Nov 20 '14 at 23:46
  • 1
    @SHAF just because a helicopter weighs a certain amount does not mean that amount of force is applied to the GROUND. The PSI will show a gradient from the blades to the ground. At the blades it will be the weight of the helicopter, but then it will decrease from there. – Tyler Durden Nov 21 '14 at 00:09
  • @PeterMasiar You are right it is a physics question, but just claiming somebody is "obviously wrong" without giving any reason is against the principles of Stack Exchange. Consider that you have 0 reputation and no credentials whatsoever, and have offered no reasoning whatsoever, its kind of BS for you to go around talking about how wrong somebody is. – Tyler Durden Nov 21 '14 at 00:12
  • @SHAF Yes, I recomputed the PSI, appears to be only 1-2 PSI. – Tyler Durden Nov 21 '14 at 00:16
  • 4
    @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. – Rhino Driver Nov 21 '14 at 00:17
  • @Tyler Durden Downward force towards carrier before landing is slight but not zero. I might not have aviation credentials but remember my Newton laws. If you disagree, take it to physics experts to decide. – Peter M. - stands for Monica Nov 21 '14 at 00:20
  • 4
    @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. – Marcks Thomas Nov 21 '14 at 00:32
  • OK I can see that it is not "aircraft weight" what changes - but result (downward force on the aircraft) will behave exactly like increased weight (will push carrier down, and so slightly that it will be not measurable). So we need to define "before". :-) – Peter M. - stands for Monica Nov 21 '14 at 00:38
  • 3
    @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. – David Richerby Nov 21 '14 at 00:42
  • 2
    @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. – Jongosi Nov 21 '14 at 11:49
  • @Jongosi Actually, it does. It represents 100% of the aircraft's weight (directly underneath the blades, at least... the effect on the ground will be less.) The total force applied to the air under the rotors of a hovering helicopter will be exactly equal to the lift generated by the rotors, though. As with all forces, the forces are equal and opposite. – reirab Dec 31 '14 at 05:34
  • @reirab No, it does not. The primary force opposing the helicopter's weight is lift, not downward thrust. Google vindicates me; https://www.google.co.uk/search?&output=search&sclient=psy-ab&q=how%20does%20a%20helicopter%20work&oq=&gs_l=&pbx=1&bav=on.2,or.r_cp.r_qf.&bvm=bv.82001339,d.d24&ion=1&biw=1434&bih=855&ech=1&psi=0QusVLW0HMm4Ud-2g6gL.1420561348726.3&ei=1AusVInIIov4UtnygKAL&emsg=NCSR&noj=1 A helicopter pulls itself into the air, it does not push away from the earth. – Jongosi Jan 06 '15 at 16:22
  • @Jongosi Downward thrust is lift, at least for all practical purposes. You could argue that part (or most) of the lift is generated by 'pulling' air downward rather than 'pushing' it downward, but, either way, the downward force excerpted on the air is exactly equal in magnitude to the weight of the helicopter when it's not accelerating vertically. When moving horizontally, the airframe itself probably also generates some lift, but, again, it's generated by deflecting air downwards. Regardless of whether it's pushed or pulled, the change in momentum of the air will equal that of the aircraft – reirab Jan 06 '15 at 16:33
  • This really has nothing to do with the question, but for the record most of the vertical force of a helicopter blade is due to positive displacement of air dependent on the the blade pitch angle not airfoil lift. In fact, it is possible to make a propeller or helicopter rotor out of a completely flat blade that is not an airfoil at all. – Tyler Durden Jan 06 '15 at 16:57
11

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.

Community
  • 1
Peter Kämpf
  • 231,832
  • 17
  • 588
  • 929
  • 1
    Lift is primarily created by a pressure differential above and below the wing. – Keegan Nov 20 '14 at 19:04
  • 2
    @kjmccarx: Yes, but how is this pressure difference created? And what happens to the air once it has left the wing's surface? – Peter Kämpf Nov 20 '14 at 20:18
  • 3
    @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. – Jan Hudec Nov 20 '14 at 21:40
  • Yes I would say there's no net increase of pressure on the ship itself. This would be a fun experiment to do with RC aircraft though. – Jasmine Nov 20 '14 at 23:21
  • 1
    Newton or Bernoulli? – rbp Nov 20 '14 at 23:25
  • A Mythbusters episode was devoted to this, in this case to a truck full of flying pigeons. I suppose Youtube could be your friend. – Sanchises Nov 20 '14 at 23:37
  • 8
    I think you are confusing displacement and weight. – Rhino Driver Nov 21 '14 at 00:25
  • 8
    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. – David Richerby Nov 21 '14 at 00:40
  • @DavidRicherby: It depends on how you define weight. The mass of the carrier does not increase, but now it has something on its deck that was not there before and makes it sink into the water. If that is not weight, what is it? Pressure, yes, but that is only the means how this force acts on the carrier. What would you call the force, if not weight? Aren't you confusing mass and weight? – Peter Kämpf Nov 21 '14 at 07:28
  • 4
    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 http://hyperphysics.phy-astr.gsu.edu/hbase/mass.html http://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. – Federico Nov 21 '14 at 07:43
  • 2
    @PeterKämpf If you want to say that the weight of the carrier has increased after the plane has landed, because the plane now counts as part of the carrier, that's fine. Before the plane has landed, it does exert a downward force on the runway but that force is not the weight of the plane and it does not contribute to the weight of the carrier. It is just a downward force. And no, I'm not confusing mass (amount of matter) with weight (the force exerted on that mass by gravity). – David Richerby Nov 21 '14 at 08:33
4

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.

Jon Egerton
  • 141
  • 3
  • Ha! Nice piece of lateral thinking. – David Richerby Nov 21 '14 at 14:25
  • 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. :) – reirab Dec 31 '14 at 05:08
2

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.

Volker Siegel
  • 1,738
  • 1
  • 17
  • 31
1

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.

Meower68
  • 1,257
  • 11
  • 11
  • 1
    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. – Skip Miller Nov 21 '14 at 20:50
1

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.

Karl Stephen
  • 849
  • 6
  • 11
  • Correct: any discussion needs to include buoyancy, discussion of the "system" of the carrier and aircraft, etc. –  Nov 21 '14 at 15:42
0

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.

Jasmine
  • 188
  • 1
  • 10
  • 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... – Ryno Nov 20 '14 at 23:47
  • 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. – Jasmine Nov 21 '14 at 00:01
  • 2
    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. – David Richerby Nov 21 '14 at 00:45
  • That's what you said, and you were wrong. With the assumption (that most people have made) that by "weight" we mean "downward force on" - there IS an increase in downward force on the deck before it touches down. The plane pushes down on the air, and the air pushes down on the deck. Before the plane is on the deck, it's weight is being exerted downwards on the deck indirectly. That's what's incorrect about your answer. – Ryno Nov 21 '14 at 03:48
  • If we are strictly talking about weight (the effect of gravity on mass) then the boat's weight never changes. Ever. No matter what you do, or put on top of it. – Ryno Nov 21 '14 at 03:49
  • And I'm saying absolutely not. The airplane wing produces a local pressure gradient whose net force is zero. – Jasmine Nov 21 '14 at 16:04
  • @Jasmine : "The weight of the carrier + airplane system begins to increase ..." That's the incorrect part discussed here ;) The force applied by the system Boat+Aircraft on the water below is what you call "apparent weight" but is not weight. It's combined forces not all pointing down (AC lift is also produced by low pressure air above the wings - ground effect is a building up pressure) When decelerating, gears produces pressure due to momentum caused by tail hook braking. Anyway, a 4G impact isn't weight either. It's a contact force => series of landing load and recoils... – Karl Stephen Nov 21 '14 at 18:10
  • 1
    @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... – Karl Stephen Nov 21 '14 at 18:19
  • 1
    @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! – Ryno Nov 21 '14 at 18:23
  • Again, no. The ground effect does not create a force against the ground. That isn't even how it works in the first place. Ground effect is about disruption of the wing vortex, not about "air pushing on the ground" which would have no effect on the aircraft. http://en.wikipedia.org/wiki/Ground_effect_%28aerodynamics%29 – Jasmine Nov 21 '14 at 18:51
  • @Jasmine From paragraph 2 of the "Principle of Ground Effect" section of the wiki you linked: Flying close to a surface increases air pressure on the lower wing surface, known as the "ram" or "cushion" effect, and thereby improves the aircraft lift-to-drag ratio. The increased air pressure acts against both the lower wing surface and the ground surface. Ryno is correct that the net force acting on the ground is definitely not zero. Also, ground effect aside, part of the downwash from the wing acts on the ground. – reirab Dec 31 '14 at 05:22
  • The part about the weight gradually increasing isn't really true. The weight of the boat remains (mostly) constant (see above answers regarding minute differences due to gravitational forces between the boat the aircraft and slight decrease in distance from CG of Earth.) The weight of the aircraft remains constant (again, ignoring the aforementioned effects.) – reirab Dec 31 '14 at 05:26
  • The apparent weight of the carrier (i.e. its displacement) will first start increasing when the downwash acts on the deck, then will increase sharply when the wheels hit the deck (because the entire vertical speed of the aircraft has to be stopped over the distance of the maximum of the sum of tire compression + gear strut compression.) Once the vertical speed of the aircraft is stopped, downward force on the deck will sharply decrease briefly, then begin to increase again as the lift from the wings decreases. – reirab Dec 31 '14 at 05:29