Talk:Wing loading

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effect on performance[edit]

http://en.wikipedia.org/wiki/Wing_loading#Effect_on_climb_rate_and_cruise_performance

the first paragraph is almost completely wrong.

  • A lighter loaded wing will have a superior rate of climb compared to a heavier loaded wing as less airspeed is required to generate the additional lift to increase altitude.

The climb performance depends on the amount of energy the motors can provide minus the amount of energy required to stay in the air divided by the weight of the plane. For example, a guy put a propellor and some bike pedals under his paraglider. Motor power (human power) was less than the power required to stay airborne. So he had very bad "climb performance" but still a low wingloading (as is normal with paragliders). So the wingloading DOES play a role, but it is not the main factor.

  • A lightly loaded wing has a more efficient cruising performance because less thrust is required to maintain lift for level flight.

This is completely wrong. The wingloading does not play a role at all. The amount of energy required to transport a weight over a distance is the weight * distance / glide ratio. Nowhere does the wing loading play a role.

Suppose I build an exact (1:1) copy of the 747 which is 100 times lighter. It will have a 100x lower wingloading, it will have 10x slower flightspeed, but the same glide ratio (about 14). To transport 1kg with either the real 747 or the copy takes the same amount of energy. Of course, to transport a ton of payload the original 747 will have to transport a whole 747 from A to B as well, so the lighter airplane will be more efficient. But that's because it's lighter and not because it has a lower wingloading. -- REW jan 31 2010. —Preceding unsigned comment added by 95.36.47.159 (talk) 16:09, 31 January 2010 (UTC)[reply]

On further reading, the rest of the chapter is bogus as well. "how rapidly the climb is established" is the most glaring problem. It simply comes down to the energy equations. So the energy of the system is m*g*h + 1/2 m v^2. Depleted is: m * g * Sink-rate and added is motor-thrust * airspeed.

If you're in level flight and suddenly point the nose up, you trade speed (1/2 m v^2) for height (m.g.h). The accelleration required will cause an AOA change sufficient to provide the neccesary force.

(if you suddenly point the nose upward while almost at stallspeed, you will stall if the momentary AOA is too high...) —Preceding unsigned comment added by 95.36.47.159 (talk) 17:22, 31 January 2010 (UTC)[reply]

An important aspect of wing loading that is not discussed in this article is that of the vertical acceleration of an aircraft as it encounters a vertical gust. When encountering vertical gusts, as the angle of attack and, therefore, the lift produced, changes, the aircraft with the higher wing loading will experience lower vertical acceleration due to its greater relative mass. The effect of this is that aircraft with high wing loadings ride more smoothly through turbulence. This is why modern jetliners handle turbulence so much more nicely than general aviation aircraft or older airliners. Perhaps this wording is not how it should appear in the actual article, but I think the article should mention this, since it's an effect most people will experience themselves, and a factor that is much more effected by wing loading than either cruising or climb performance. Harleygnuya (talk) 22:29, 8 August 2017 (UTC)[reply]

Units[edit]

I'm no engineer, but if "wing loading is the loaded weight of the aircraft divided by the area of the wing" then in metric shouldn't it be measured in N/m² instead of kg/m²? Blaise 21:08, July 19, 2005 (UTC)

Er... You're right, but strictly, N(Newton) is not just metric but SI. The "kg" in the article probably means "kgf" in fact...I'm sorry I was completely wrong. I didn't see the article well... See the discussions on talk:A/C project and its subpage for much more datails, if you're interested. - Marsian / talk 00:02, 20 July 2005 (UTC)[reply]
That loaded weight is a measurement of mass in kilograms or pounds, not of force in newtons or pounds-force or kilograms-force. That's the measurements we have. I'm less certain of what should be measured for the "wing loading" quantity, and whether that should have dimensions of force divided by area (the same dimensions as pressure or stress), or of mass divided by area. Gene Nygaard 00:40, 20 July 2005 (UTC)[reply]
At least one Japanese aeronautical engineer(Akira Azuma) uses N/m² for wing loading. Though in "Aircraft Design" by D.P.Raymer, lb/ft² and kg/m² are uesd (I'm not sure these are mass or force). - Marsian / talk 15:46, 20 July 2005 (UTC)[reply]
Using mass/area would define wing loading so that it is the same for a particular aircraft independent of its location, whereas using force/area would make the wing loading of a particular aircraft dependent on whether the aircraft is sitting on earth, the moon, or flying through Jupiter. I would think that mass/area would be preferable because of its constancy, but I'm not an expert so I won't commit the article to one or the other. Just a point. -Guest User 02:09, 16 February 2006
As a physicist, not an aerodynamicist It seems to me it doesn't much matter whether we use mass/ area (e.g. kg m-2) or force (weight)/area (e.g. N m-2) so long as we give the units. As Gene N says above, the figures we are usually given are masses (though we always call them weights!) in lb or kg. We have to remember that if we want to equate (in level flight, say) the lift (force)/m2 to the wing loading and the latter is in (say) kg m-2, we need to multiply it by g = 9.81 m s-2 to go from mass to weight, from kg to N.TSRL (talk) 17:59, 4 December 2008 (UTC)[reply]
This needs to be corrected. POUNDS ARE NOT UNITS OF MASS. Wake up and smell the science. 76.252.78.92 (talk) 21:15, 9 April 2009 (UTC)[reply]

With respect, and in lower case, the pound is a unit of mass. Any scientific dictionary will tell you this. I have the paperback "A Dictionary of Physical Sciences" ed Daintith pub Pan) in front of me and it defines the pound as "an imperial unit of mass equal to 0.453 592 37 kg". There are related imperial force units: the poundal (force need to accelerate a mass of 1 lb at 1 ft/sec/sec); the pound-weight (symbol lbwt, the force that would accelerate a mass of 1 lb at the local value of g); and the pound force (lbf, the force that would accelerate a mass of 1 lb at a fixed value (standard acceleration) of 32.174 ft/sec/sec, not far from the value of g anywhere near the Earth's surface).

Aircraft "weights" are given in mass units, lb or kg; engine thrusts in force units, lbf or N.

In pounds-weight the gravitational force on a body stays the same wherever you are, because the unit depends on the local g value; the force in pounds-force or poundals does vary slightly with position, with g. So of course does the force in Newtons. If we measure the wing loading as mass/area which is the commonest way (ie in lb/ft2 or kg/m2) then g is not involved. If anyone were to use pounds-weight/area, that too would be position independent. I don't recall seeing anyone using (pounds-force or poundals)/ft2 either, though they might; N/m2 is used. All of these give a loading which is position dependent.

Do we need to worry about such variations when calculating a number not usually quoted to better than 1% (think of variations in mass during a flight, or even from one take off to the next)? The value of g is highest at the pole, greater than at the equator by a factor about 1.0053, or 0.5%. g also falls with height and it is easy to calculate that an aircraft (we are talking wing loading) at 20 km will experience a fall in g by about a factor 0.994, or a fall of 0.6%. So these effects, whilst real enough are probably not of much consequence to engineers. In any case, as I said in the usual mass/area units (rather than force/area) wing loadings are independent of position.TSRL (talk) 20:49, 11 April 2009 (UTC)[reply]

Animal flight[edit]

I would really like to see some discussion of wing loading in flying animals. All the discussion ignores animals altogether, and perhaps the article should be reworded to reflect the fact that wing loading applies to more things than man-made aircraft. John.Conway 19:05, 23 December 2005 (UTC)[reply]

Yes maybe someone could start with the Eagle family. Enlil Ninlil (talk) 20:51, 4 April 2009 (UTC)[reply]

References[edit]

This page really needs some references, and sections would help it's readability. John.Conway 19:05, 23 December 2005 (UTC)[reply]

 Done I have had a go at tidying this article, it is not quite right but I believe that it is much better than it was. Cheers. Nimbus227 (talk) 17:28, 8 February 2008 (UTC)[reply]

WikiProject class rating[edit]

This article was automatically assessed because at least one WikiProject had rated the article as start, and the rating on other projects was brought up to start class. BetacommandBot 10:06, 10 November 2007 (UTC)[reply]

Range of values[edit]

This needs another look: there are many lower and higher values than quoted here and the impression is given, wrongly, that only fighters have high wing loadings. To pull out a couple of data, the little Cessna 152 has a WL of 51 kg m-2, and the Boeing 747 740 kg m-2. The data is on WikiP, but it would be nice to have a table: lowest (-ish) typical that, typical this - biggest (-ish) to give a better feel for the numbers. The intention of what we have is right but not the detail.

I think it might help, too, to spend a bit longer on wing loading and level fight, before getting into the manoeuverability issues. Needs a bit of thought and perhaps even a formula.TSRL (talk) 20:56, 4 December 2008 (UTC)[reply]

Lift to mass (L/M)[edit]

Maybe I am being dim, but I'm baffled by the second sentence from the start, which talks of a broad connection between wing loading and L/M. The former is a well defined property of an aircraft at a given mass. The latter, particularly the lift depends on speed, altitude and attitude: it's not a property of an aircraft, or at least only an instantaneous one. Can there be a significant relationship?TSRL (talk) 21:42, 4 December 2008 (UTC)[reply]

For the record, this sentence was "It [the wing loading] is broadly reflective of the aircraft's lift-to-mass ratio, which affects its rate of climb, load-carrying ability, and turn performance." I removed it. TSRL (talk) 14:52, 30 January 2009 (UTC)[reply]

In Stick and Rudder, Langewiesche talks about the technical (read 'accurate') definition of "lift." An aircraft's lift at any given moment is congruent to its weight; otherwise it would balloon upward or descend. Langewiesche ends up using the term "buoyancy" to describe what this article terms as "lift," but only after an explanation that may not be appropriate here. In the interest of keeping Wikipedia accurate- semantically and otherwise- can anyone come up with a better way to refer to "lift?" --99.180.72.12 (talk) 03:56, 30 January 2009 (UTC)[reply]

The standard definition of lift (it is in, e.g. Anderson's "History of Aerodynamics", but I'm sure turns up in most aerodynamical texts) is that it is the aerodynamic force at right angles to the free stream velocity. The latter is the air velocity (speed and direction especially) of the air seen from the aircraft but measured far enough away for the air to be undisturbed by the plane. A less formal definition, fine for aircraft if not for aerofoils in wind-tunnels is "the aerodynamic force at right angles to the flight-line". This is the definition by diagram, if not always by text in introductions to flying such as The Air Pilot's Manual.
The magnitude of the lift depends on speed and angle of attack. Balancing forces, so that eg vertical accelerations (ballooning) are minimised still requires care: an aircraft in level flight (so with lift vertical) but pitched up will not have lift equal weight, even though these are co-linear, as there will be an extra vertical upwards component of force from the engine thrust, so lift here is less than weight. Likewise, in the climb and assuming for simplicity the flight- and thrust lines are now co-linear, the component of weight at right angles to the flight-line is less than the weight. Surprisingly perhaps the lift gets smaller as the climb gets steeper: the extra force is provided by the vertical component of thrust which does get bigger. TSRL (talk) 13:09, 30 January 2009 (UTC)[reply]
I've just noticed that the definition above is essentially identical to that in lift.TSRL (talk) 15:00, 30 January 2009 (UTC)[reply]

How aircraft climb![edit]

Aircraft climb as a result of "excess power" not lift! —Preceding unsigned comment added by 134.129.221.252 (talk) 00:11, 16 April 2009 (UTC)[reply]

Indeed so. That was the point I was making in the last para of the 13:09, 30 January note above, but I got it wrong in the article (last line on climb) by forgetting about the motor! I've changed it now, with both upward components included.TSRL (talk) 20:02, 22 April 2009 (UTC)[reply]
I also had the feeling that it got confusing there. I'm no specialist, but what I've always been taught in flight school is that a low wing loading allows for a greater angle of climb, not rate of climb. Angle of climb is basically how much altitude an aircraft gains while covering a certain ground distance, and it depends largely on the lift/drag ratio. Climb rate is basically how much altitude an aircraft gains in a certain time, that is, how "fast" and aircraft climbs, and that is the one that benefits from excess power. Since it also benefits from lower drag, it's taught that airplanes with a small wing area relative to their mass, that is, with a high wing loading, will have a higher rate of climb, while airplanes with a large wing area and a low wing loading will have a higher angle of climb. Cpt Vidal (talk) 05:49, 20 June 2013 (UTC)[reply]

Wing area[edit]

Does the wing area given include that of the tailplane and/or other surfaces, e.g. canard wings? Thanks. --TraceyR (talk) 16:22, 4 June 2010 (UTC)[reply]

It wouldn't normally include the tailplane, canards I would think yes as they share the lifting load (in a canard with similar sized wings). Is that missing from the article or unclear? Have not looked myself. Cheers Nimbus (Cumulus nimbus floats by) 22:21, 4 June 2010 (UTC)[reply]
I can't see any discussion in the article of the tailplane as a lifting surface. I imagine that, in 'conventional' aircraft, the tailplane provides lift to 'keep the tail up' as well as e.g. controlling pitch via elevators. I'm currently looking at a 1910 aircraft which had a main wing area of 38 sq.ft. and a tailplane area of 5.4 sq.ft. - not inconsiderable. If a tailplane has a suitable aerofoil section, it must contribute lift, surely? --TraceyR (talk) 06:45, 5 June 2010 (UTC)[reply]
The tailplane may provide lift, depressing the nose, or it may raise the nose by providing negative lift. It is there partly to counter the net couple arising from the non-alignment of wing lift and weight, added to that from the non-alignment of thrust and drag. So the lift the tailplane generates depends on its incidence, chosen to provide a compensating couple as the aircraft is trimmed. The tailplane incidence is normally less than that of the wing, because then (for example) a drop of the nose makes the tailplane's incidence negative, providing a nose raising downforce on the tail.
So the area of the tailplane is interesting but not to be added to the wing area. Even with the tail out of the way, there are lots of ambiguities in the definition of wing area. Should we include aileron area? Flight's articles from the early days used to separate the two. Modern sources never do; I guess ailerons are included, but not sure. What about flaps etc? Even with the dangly bits accounted for, there are still different practices on what to do at the root. Include fillets? Include the projected area to the centre line, even though this includes non-lifting fuselage? Some do (there is a ref somewhere I'll try to dig out). There was a discussion about wing area for the geometrically simple Boulton Paul P.111, though the puzzle was not solved. Nightmare! Cheers, TSRL (talk) 09:13, 5 June 2010 (UTC)[reply]
See :http://adg.stanford.edu/aa241/wingdesign/winggeometry.html on definitions of wing area.TSRL (talk) 09:30, 5 June 2010 (UTC)[reply]
Thanks for the comments. I'll ignore the tailplane in this case. It's strange that there is apparently no universally accepted 'right' way to calculate wing area. This also means of course that "wing loading" is just as vague. --TraceyR (talk) 16:32, 5 June 2010 (UTC)[reply]
Most of the vagueness is removed when the manufacturer supplies the figures (i.e. they worked out what the wing area actually is or more likely they initially designed it to have a certain wing area and loading), which is probably the case for most of our aircraft type specification tables. Wing span is another tricky one, it's the distance between the wing tips but not necessarily the length of the wings!! I do have quite a few reference books showing how to calculate wing area but they are mainly for model aircraft (the same principle applies). I use an Excel programme for calculating model glider wing loading, moment arms, stability and centre of gravity numbers, the dimensions of a single wing (one side but might be multiple panels) have to be input so there at least the fuselage is not counted. Nimbus (Cumulus nimbus floats by) 22:06, 5 June 2010 (UTC)[reply]
Apropos of nothing, I remember a doumentary film which showed an F15 (or was it an F16?) which lost a wing in flight but which the pilot managed to land safely, because the fuselage and wing root provided enough lift! The pilot might even have been unaware of the extent of the damage. --TraceyR (talk) 22:43, 5 June 2010 (UTC)[reply]
It was an Israeli F-15 I think, mid-air collision. Certainly some aircraft are designed to use the fuselage as a lifting surface, there is a section on it in this article. Nimbus (Cumulus nimbus floats by) 22:47, 5 June 2010 (UTC)[reply]

Climb rate and cruise performance[edit]

The entire section titled Effect on climb rate and cruise performance is unsourced. In its entirety, it is technically unsound. It appears to have been written using nothing but intuition. For example, it implies that for an aircraft to climb requires that lift exceed weight. That is incorrect - climbing requires that thrust exceed drag. See Rate of climb and Angle of climb.

I am happy to re-write the section, based on reliable published sources. But first I will erase the entire section. It is worse than misleading. Dolphin (t) 05:10, 29 March 2018 (UTC)[reply]

I have erased the offending section. See my diff. Dolphin (t) 05:11, 29 March 2018 (UTC)[reply]