Wing-Loading
 

I know this might sound like a silly question, but how do airliners manage to fly reasonably well with such horrendously high wing-loadings?

It is not a silly question but it is not easy to answer your post in a meaningful way without asking you for some clarification:

What is your aviation background?

What do you mean by 'fly reasonably well' (for example do you mean they have high top speeds,are controllable in turbulence, have good takeoff and landing handling characteristics etc etc)

What do you mean by 'horrendously high' (do you just mean many times that of a GA light aircraft? or something else?)

However I will offer that in general aircraft with a high wing loading will need greater distances to takeoff and land (because they will have a higher stall speed), will be more comfortable in turbulence and will not turn so tightly at low speeds as aircraft with a low wing loading.

And for an aircraft that is hover-capable - when in the hover there is little or no aerodynamic wing-loading? ;)

:)

Depends if you call a helicopter an aircraft i guess ;)

Okay, I meant (and should have specified!) hover-capable fixed-wing aircraft! ;) I have a feeling John F will know an awful lot about that.

I bet he does ;)

One can always ask the opposite question: Why do most small aircraft have such low wing loading? Their slow approach speed means that crosswind or tailwind landings are very dicey, and the large profile drag of a big wing limits their top speed. How profitable can a light transport aircraft be with its weather limitations and slow cruise speed? :8

John Farley,

Aviation buff. I'm not a pilot or an aerospace engineer

Mostly good takeoff performance, but not just that. The other issue has to do with turning performance and such, you'd expect a plane with the wing-loading figures typically seen on a commercial-airliner (727, DC-10, etc) to lose speed much quicker than it does for it's wing-loading in turns.

I have heard about numerous cases in which relatively high g-loads were pulled (not intentionally, by accident) and while the plane lost speed in the turns, you would think that for the thrust to weight ratios an airliner has, and the heavy wing loadings, the plane would have quickly lost all it's speed and just dropped out of the sky like a brick.

Well, as I understand it wing-loadings over 85 lbs/ft[sup]2[/sup].

The original 727-100 had a MTOW around 160,000 pounds and a wing-area of 1,650 square-feet, yielding a wing-loading of 96.97 lbs/ft[sup]2[/sup].

The DC-10-10 had a MTOW of 430,000 pounds and a wing-area of 3,550 square-feet, yielding a wing-loading of 121.13 lbs/ft[sup]2[/sup].

There are many more factors in aerodynamic design than just wing loading-- span loading (the flip side of aspect ratio), airfoil, finess ratios, thrust loading, drag, in all its components and installed high lift devices. A number of fighters have greater wing loadings than the DC-10 and perform fine, for their designed mission. Yes, under G loads they lose speed, but FAR 25 design is for 2.5 positive G and all of them possess the power to overcome the resultant induced drag.

Yes, airliners have undergone proof levels G-loadings, usually they have been associated with a deep nose-low attitude and no amount of G prior to breaking the airplane would result in a stall and "falling out of the sky".

galaxy flyer,

I assume by fineness-ratios you mean T/C ratio; by span-loading you mean the ratio of span to chord; and thrust loading you mean T/W ratio?

Like the F-105, it performed decently well in regards to sustained agility when flying faster than a bat out of hell at low altitudes when lightly to moderately loaded; at lower speeds it bled off speed way too easily, and at full-loads, even it's high-speed agility fell off.

They can all hold 2.5g without loss of airspeed?

I was largely talking about while in level flight, not in a steep-dive.

No, span loading is weight divided by span, related to aspect ratio which is what is what you are speaking of.

Yes, fineness is T/C and thrust loading is T/W

The performance under G loads is related to specific excess power, that is, wing loading causes loss of IAS, specific excess power is how the plane can overcome the increase in induced drag. The Thud didn't have a lot of Ps at high loadings or low speed.

I cannot speak for all transport category aircraft but mostl of them at reasonable altitudes can sustain 2.5 Gs for at least 180 degrees and need no more. At cruise levels, 1.2G or 1.3 G is about standard stall margins and sufficient.

Not many transport category planes have any need to exceed 2.5 G at any time and would absolutely need to only in an out of control dive recovery.

Wing loading is not a factor in any conceivable flight regime for the types of planes we are talking about. Transports don't engage in turning fights for survival or "scissor" with another plane to get landing priority.


Were it not for these "horrendously" high wing loadings they could not do their jobs as jet transports. Try designing a jet transport with a wing loading of, say, 50 lb/ sq ft.
GF

Jane-Doh,

not exactly,
Fineness ratio, is more about the physical cleanness of the wing it concerns boundary layer T/C is the thickness ratio

Span loading, is the weight divided by the span-a measure of how much load is distributed along the span the

T/C is the thickness ratio...important when chord changes as in flap deflection

The maximum wing load or the maximum weight to ever be carried by the wing is carried at the stall...and is dependent on how far above the stall speed you are if you are at Vs then only one g can be pulled at 2Vs then four times the load will be imparted at the stall at 3Vs 9g...etc...

I hope that helps...:)

Great Question:ok::}

I stand, actually sit, corrected by my learned colleague, PA

GF

PA and Jane DoH

Actually small planes have such low wing loadings due to FAR 23 limiting stall speeds to 61 knots due to crash survival considerations. Cost consideration rule out sophisticated high lift devices, hence large wing areas. Th Helio was an exception but long gone.

GF

GF
One learns to think differently about aviation everyday, thanks:)

galaxy flyer,

I'm sorry, I got that backwards... :}

Understood

That's the energy-maneuverability formula right?

It would need to have very thin wings that were very large for it to even remotely work, though I can think of at least one proposed commercial aircraft design that had a wing-loading of 62.6 pounds/square-foot. It was called the L-2000; it was built by Lockheed as a competitor in the SST program.

In a more serious note, I assume that planes designed to sustain high g-loads (fighter-planes) are built with wings that are excessively large for what is needed for optimum efficiency in level flight?


Pugilistic Animus,

Fascinating.

Yeah, I got span-loading and aspect-ratio backwards

You mean like fowler-flaps?

Jane
 

Thanks for responding to my request.

I will not answer it directly now as subsequent posts by yourself and others have moved things on a lot further.

You clearly are familiar with a lot of aspects of aircraft design and performance but I feel have got some of the basic building blocks a bit jumbled up in your head.

I will PM you with some of the basics of lift and drag which I think might help you straighten out some of your concepts.

However all other things being equal if you literally double the wing loading of an aircraft (by filling an airliner with fuel or covering a military aircraft with bombs) then you really won’t change much beyond

• The stall speed which will be 1.414 times what it was before (this will up your takeoff and landing speeds and distances)
  
• The aircraft will have much more inertia (this will make the aircraft appear ‘sluggish’ in response to you trying to change its flight path – especially in pitch compared to the light case)

When it comes to the effects of wing loading on manoeuvre there are two aspects called the manoeuvre boundary and the thrust boundary.

The manoeuvre boundary is a measure of how much g you can momentarily pull at a particular speed (this will reduce as wing loading goes up)

The thrust boundary is a measure of how much g you can sustain without loss of airspeed and is primarily affected by the thrust available. It will suffer as you put up the wing loading but likely less than the manoeuvre boundary.

JF

Low wing loading
 

My hang glider has 188 square feet and a wing loading less than 1.5 pound per square foot even carrying a porker like me.

It will take off at 18 mph, and will fly in a dive at 44mph. In all but the calmest conditions it requires active control inputs to avoid being thrown around like a leaf.

For a sport it's fine. As a means of transport it leaves a lot to be desired.

Reasonably????? Well?????

:)

Mine was a rogallo, and had a glide ratio of about 6:1. Its wing loading was ~ 1.75lbs/foot2, and yeah, it flew like a potato chip in a gale.

Two thoughts...

1) Air pressure at sea level is around 2000 Lbs/sqft so is 85 Lbs/sqft really "high" in percentage terms (just 4%)?

2) Reynolds number. Big fast wings are more efficient than small slow ones.

Low Reynolds Number Airfoil Design

FAR 23 limiting stall speeds to 61 knots

Originally 70mph prior to metrication.

From a discussion with an ancient engineer chap who was in the precedent organisation in the early days as a boy, 70mph was a figure plucked out of the air by the senior folk as being a reasonable starting point to put into the rule book .. and then it just stayed and was never altered.

J_T

Always wondered where 61 knots came from--rather odd sounding. But metrication? Perhaps, metrification or nauticalification?

GF

It's not really metrication, is it ? I ought to be a tad more precise at times .. have to blame the advancing decrepitude, I guess ...

(a) current rule

(b) old FAR 23 rule which changed to 61 kt at A/L 23-7 Sep69

(c) CAR 3 goes back to 1949 (refer 3.83). I don't have any Net references to earlier US regs but, no doubt, someone else in the sandpit will be able to cite a link to the real olden days .. 1949 - I was too young to have much of an interest in aeroplanes ..

Here endeth the useless bit of information for the day ...

Some pre-incarnation of CAR3 goes back to 1930s. Even racing aircraft of that era were designed to a Vso of 70 mph. :8

No wonder I can't fly an ILS; too much metriculation manoeuvreing conversion calculations !

Boy, am I sorry I brought this up!

GF

John Farley,

You're welcome

Understood

Yup, I got your message.

I assume that has to do with exponent in the the L = (CL)½(ρ)(S)(V^2) formula (1.414 and on is the square root of 2)?

Yeah, more mass always equals more inertia

The maximum instantaneous g-load you could pull without getting an accelerated stall?

Makes sense: There are planes that have good instantaneous agility and poor sustained agility.


cwatters,

I never knew low Reynolds numbers had any drawbacks -- I just thought they hurt you when you scaled a wing up without sharpening the leading-edge.


Willit Run,

I've usually found the metric conversions fairly easy. You just have to memorize all the conversion tables. Maybe that's easier said than done, but I have most of that memorized since 7th grade.

No. You will have to reach the stall otherwise how do you know you have reached the max g? DoH - sorry!

When briefing for tests to get this data the most important thing is the characteristics that the team choose to define the stall. (wing drop, buffet level, sideslip behaviour and so on)

John Farley,

I assume that data would have been derived from the wind-tunnel data and flight-testing...

The boundaries can only be fully established by flight test.

Tunnels are just one of the estimation tools available - but estimates are just that estimates.

.

John Farley,

Sorry about that...

galaxy flyer,

I'm sorry, I don't even know why I'm responding but I thought that was hilarious... :ok:

Knots (kts) = nautical miles per hour (nmph).
1 nautical mile = 6,000 feet.

1 statute (land) mile = 5,280 feet.
MPH = statute miles per hour

1 mile = .88 nmile

None of this has anything to do with the metric system.



KPH = kilometers per hour
1 kilometer ≈ 3,281 feet ≈ .62 mile ≈ .547 nmile

A nautical mile is 6076 feet not 6000

The nautical mile now days (since 1929) is actually defined in metric terms, being 1,852 metres, which in turn equates to 6,076.1155 feet.

Although it's conventional to think about the "metric system" in terms of metres (= 1/40,000,000 of the circumference of the Earth at the equator), today the metre is better defined, and it's all called SI (systeme internationale d'unites). I tend to think of ANY system as "a metric system".

And so the nautical mile is 1 minute of longitude at the equator, so = 1/(360x60) of the circumference of the Earth at the equator.

I'm afraid not barit.

The nautical mile was historically defined as a minute of arc along a meridian of the Earth, making a meridian exactly 180×60 = 10,800 historical nautical miles. It can therefore be used for approximate measures on a meridian as change of latitude on a nautical chart. The originally intended definition of the metre as 10−7 of a half-meridian arc makes the mean historical nautical mile exactly (2×107)/10,800 = 1,851.851851… historical metres. Based on the current IUGG meridian of 20,003,931.4585 (standard) metres the mean historical nautical mile is 1,852.216 m.

The historical definition differs from the length-based standard in that a minute of arc, and hence a nautical mile, is not a constant length at the surface of the Earth but gradually lengthens with increasing distance from the equator, as a corollary of the Earth's oblateness, hence the need for "mean" in the last sentence of the previous paragraph. This length equals about 1,861 metres at the poles and 1,843 metres at the Equator.

Other nations had different definitions of the nautical mile. This variety in combination with the complexity of angular measure described above along with the intrinsic uncertainty of geodetically derived units mitigated against the extant definitions in favor of a simple unit of pure length. International agreement was achieved in 1929 when the International Extraordinary Hydrographic Conference held in Monaco adopted a definition of one international nautical mile as being equal to 1,852 metres exactly, in excellent agreement (for an integer) with both the above-mentioned values of 1,851.851 historical metres and 1,852.216 standard metres.

Use of angle-based length was first suggested by E. Gunter (of Gunter's chain fame), reference: W. Waters, The Art of Navigation in England in Elizabethan and Stuart Times, ( London, 1958). During the 18th century, the relation of a mile of 6000 (geometric) feet, or a minute of arc on the earth surface had been advanced as a universal measure for land and sea. The metric Kilometre was selected to represent a centisimal minute of arc, on the same basis, with the circle divided into 400 degrees of 100 minutes.

Thanks for the upgrade; however MY navigation was never that precise! :O

When getting to the seventh significant figure, do we need to consider the effect of global warming on sea level and hence on the earth's circumference? Or the fact that aeronautical miles are longer because the earth's circumference is greater when at altitude?

mike-wsm,

Very good observation about the distances being greater at altitude. If you drew a line 30,000 feet up in the air at your takeoff and landing points you technically cover a little bit more distance than if you were at sea-level the whole time.