legoguy
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Hey there. I have just started Aero Engineering at UNi and love it, despite the fact we have an assignment due already. It involves breifly describing 15 aircaft of choice etc etc and putting them in a table balh blah.

However one of the calculations I am required to work out is the wing loading of an aircraft. I have the formula

Wing loading = (MTOM x accel due to gravity) / wing area

I hope anyone can make sense out of that.

I am confused by several things so I thought I could try here before asking my teacher as I have all weekend to figure it out.

When finding the Maximum Take of Mass of an aircraft, usually it says Weight instead of Mass and is in Kg. (I thought weight was in newtons not Kg). So is there any difference between MTOW and MTOM?

Here is an example of a wing loading I have worked out for an a330-300

MTOW = 121970 Kg

Wing aera = 363.1 metres squared

Does this sound correct?

Also lastly, is there a rough average of the usual wing loading of an aircraft?

Any help would be appreciated.

Regards,

Dave
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KELPkid
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I would have thought it would have been much simpler than that (although I'm not an aeronautical engineer )-I would have thought that wing loading would simply be: MTOW/(area of wings).

I wonder where the moment comes into the equation you give? Although, in thinking about it, maybe the moment would account for tail lift vs. wing lift...just surmising here (and not necessarliy correctly!).

One thing physics and statics taught me in college: when in doubt, do a simple unit conversion Sometimes this reveals the logic behind equations...

EDIT: Ooops! My bad. I would have thought that MTOM is maximum takeoff moment, but in reading your post, it's mass!

Well, gravity works as an accelerating force on a given mass...and the wings must produce a distributed force equal to gravity so that the sum of lift and the sum of gravity is zero.

[Edited 2006-09-29 23:41:47]

[Edited 2006-09-29 23:44:19]
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dakota123
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No practical difference between weight and mass on earth. Minute differences depending where on earth the plane is, but I imagine you can ignore that for this exercise. Take the plane into space, though, and all bets are off...

legoguy
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Right I have looked on Wikipedia and found that the typical wing loading is 390 to 585 kg/m² for high-speed designs like modern fighter aircraft. So that seem it would make the wing loading around 4000 to 6000 N/m².... I think. I admit my maths is terrible but it's slowly improving

We were giving that formula by the teacher. If there is a difference between MTOW and MTOM then that is robally where my problem is, if I have one
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KELPkid
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Did you include the units in your calculation? Newtons (unit of force) per square meter (unit of area) is correct, I'm at work so I can't look at the units conversion on paper
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legoguy
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 Quoting KELPkid (Reply 4): Quoting Legoguy (Thread starter): Therefore wingloading = (121970 x 9.81) / 363.1 = 3295.31 N/m^-2 Did you include the units in your calculation? Newtons (unit of force) per square meter (unit of area) is correct, I'm at work so I can't look at the units conversion on paper

Right so the formula must be Homogenious to be correct. So units wise the formula is

(Kg x m s^-2) / m^2 = N/m^-2

From W = mg.... N = Kg m s^-2

so.... (Kg x m s^-2) / m^2 = Kg m s^-2 m^-2

Uh oh. Thats not right is it? Or did I make a mistake somewhere

EDIT... wohoo I just did it out on paper (much easier) and got homogenious units on each side of the equation.

Kg m^-1 s^-2 = Kg m^-1 s^-2

So the equation must be right?

[Edited 2006-09-30 00:06:28]
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dakota123
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I'm confused as to what your end result is supposed to be. Are you trying to find out wing loading by mass or by weight? If by weight, you already have the weight, no?

If by mass, wouldn't you divide the weight by gravity force (9.81m/s^2) to arrive at mass?

It looks to me like in your original example of 121970 kg you shouldn't be multiplying it by 9.81; That factor is already included in the 121970 if I read you correctly.

vikkyvik
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 Quoting Dakota123 (Reply 2):No practical difference between weight and mass on earth. Minute differences depending where on earth the plane is, but I imagine you can ignore that for this exercise. Take the plane into space, though, and all bets are off...

Well, if you're talking pounds-mass versus pounds-force, then yeah, the number will be the same. But there is definitely a practical difference.

Weight (being a force) is in Newtons in the metric system, and mass is kilograms. Those numbers will not be the same.

People use kilograms for both mass and weight, even though the unit is technically a mass (hence MTOW is often given in kg, or alternatively, tonnes). But since you want your end result to be a force, you have to convert to Newtons.

is correct.

(MTOM * accel due to gravity) will convert your mass to the force due to gravity. Dividing that by the wing area will give you your wing loading.

 Quoting Dakota123 (Reply 6):It looks to me like in your original example of 121970 kg you shouldn't be multiplying it by 9.81; That factor is already included in the 121970 if I read you correctly.

That factor is not included in the 121970 kg. Even though people might call it the airplane's weight, it is in fact the airplane's mass, and needs to be multiplied by the acceleration due to gravity to arrive at the weight in Newtons.

Hope I didn't confuse you more.

~Vik
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legoguy
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 Quoting Dakota123 (Reply 6):If by mass, wouldn't you divide the weight by gravity force (9.81m/s^2) to arrive at mass?

When looking up an aircrafts Maximum Take Off Mass .... it is always listed as Maximum Take Off Weight and is listed in Kg (not Newtons... adding to the confusion)

So if the units of the final sum is N m^-s... I think that to get the Kg into Newtons, I must multiply it by the 9.81 (accel due to gravity)

This is confusing!
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legoguy
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Ahhh confirmation! Yay. Thanks all you guys for your help! Much appreciated!

Just to check, does the wing loading of 3295.31 N/m^-2 sound correct enough for a commercial airliner?
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KELPkid
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 Quoting Legoguy (Reply 8):So if the units of the final sum is N m^-s... I think that to get the Kg into Newtons, I must multiply it by the 9.81 (accel due to gravity) This is confusing!

LOL-this is one place where American Engish units help us-we measure things in weight directly (pounds)   The Slug is the Engish system measure of mass, and I don't think I've ever seen that one used anywhere.

Maybe you should do your calculation in English units and convert it to metric when you're done   Wouldn't your professor love that!
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474218
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You all are making it too complicated, "Wing Loading" is nothing more than:

"Maximum Takeoff Weight" divided by "Wing Area".

vikkyvik
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 Quoting KELPkid (Reply 10): Maybe you should do your calculation in English units and convert it to metric when you're done

I frequently do exactly the opposite when confronted with a problem in English units!

Honestly, I actually think the situation in English units is more confusing. With pound-mass and pound-force having the same numerical value on Earth, one is automatically dissuaded from appreciating the difference between a force and a mass. Or rather, there ceases to be a need to understand it.

 Quoting 474218 (Reply 11):You all are making it too complicated, "Wing Loading" is nothing more than: "Maximum Takeoff Weight" divided by "Wing Area".

So what does one do when presented with a wing area of, say, 100 m^2, and an MTOW of 200,000 kg, and asked to find the wing loading?

The answer is not 2000 kg/m^2.

Unless they use a unit of kilogram-force. Which I've never seen anywhere.
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KELPkid
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 Quoting 474218 (Reply 11):You all are making it too complicated, "Wing Loading" is nothing more than: "Maximum Takeoff Weight" divided by "Wing Area".

However, Kilograms are a measure of mass, not weight...and hence the (slightly) complicated formula that we have at the beginning of the thread
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474218
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 Quoting Vikkyvik (Reply 12):So what does one do when presented with a wing area of, say, 100 m^2, and an MTOW of 200,000 kg, and asked to find the wing loading? The answer is not 2000 kg/m^2.

Example:

DC-10 Wing Area 367.7 Square Meters. Maximum Takeoff Weight 263,000 Kilograms. 263,000 divided by 376.7 equals a Wing Loading 715 Kilograms per Square Meter.

Or DC-10 Wing Area 3957 Square Feet. Maximum Takeoff Weight 580,000 Pounds. 580,000 divided by 3958 equals a Wing Loading of 146 Pounds per Square Foot

474218
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 Quoting KELPkid (Reply 13):However, Kilograms are a measure of mass, not weight...and hence the (slightly) complicated formula that we have at the beginning of the thread

One Kilogram equals 2.2 Pounds. Mass does not matter. A Kilogram of aluminum and 2.2 Pounds of aluminum, are exactly the same mass.

vikkyvik
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 Quoting 474218 (Reply 14):Example: DC-10 Wing Area 367.7 Square Meters. Maximum Takeoff Weight 263,000 Kilograms. 263,000 divided by 376.7 equals a Wing Loading 715 Kilograms per Square Meter. Or DC-10 Wing Area 3957 Square Feet. Maximum Takeoff Weight 580,000 Pounds. 580,000 divided by 3958 equals a Wing Loading of 146 Pounds per Square Foot

Oh well. Is that what is typically used in the industry?

I don't feel like getting into a whole discussion about units and what makes sense and doesn't, so I'll just accept this answer and shut up.

Legoguy, I would be interested to know what your professor was expecting for the answer, though. I'm positive that he'd have expected an answer in Newtons/m^2 in my classes.
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legoguy
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 Quoting KELPkid (Reply 10):Maybe you should do your calculation in English units and convert it to metric when you're done Wouldn't your professor love that!

LOL

 Quoting 474218 (Reply 11):"Maximum Takeoff Weight" divided by "Wing Area".

 Quoting Vikkyvik (Reply 16):Legoguy, I would be interested to know what your professor was expecting for the answer, though. I'm positive that he'd have expected an answer in Newtons/m^2 in my classes.

Yes the final units had to be in N/m^2. I must admit I had absolutely no clue about it but now I have used the formula and worked out that the fact it is indeed homogenious (same units each side of the formula)... it seems a whole lot easier now.

Again thanks for your help guys and will get back to you with what my teacher says on monday.
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sovietjet
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Wow you guys sure made this complicated lol. He has it right in Newtons. To get it by mass divide by 9.81. End of problem.

texfly101
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 Quoting Legoguy (Reply 5):Right so the formula must be Homogenious to be correct. So units wise the formula is

Very correct and a very important check on any calculation. There is a concept in engineering and physics called dimensional similitude. Basically, it says that the units on one side of an equation must be similiar (read equal) to the units on the other side of the equation. So in matters concerning weight and mass, the actual units will tell you what constants to use, or not use. Dimensions in the English system are at times very confusing with things like lbs-mass, lbs-force, and not being on a base 10 system like metrics. The gravitational constant g is usually needed in any calculation done on Earth. Since it is so basic, sometimes it gets applied wrong, usually in the units. To skip past this invites disaster as the Mars Probes in the late 90's pointed out. That's where the Lockheed team gave constants based on lbs-force to a NASA team that was operating the probe and was based on the metric system and newtons. So the probes got erroneous data that would have been very easy to have caught had they checked the dimensional units....just a point of order, but one that has saved my butt quite a bit over my career. Good original question and a great topic for a thread. Have fun with your studies.

BAE146QT
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Oh look - yet another university course using its own terminology (MTOM vs. MTOW) to screw up its graduates.

I wonder sometimes whether they do it deliberately so you can tell a graduate from a journeyman.

I regularly get CVs from people who have some qualification in 'NIS' or some such.

"Oh, you mean IT"
"Er, well yeah."
"So you have a degree but no experience and yet expect to look after £70M of this bank's kit and demand an £80K salary?"
"But...but...but... I have an NIS degree!"

Actually, there's nothing to complain about here. Makes 'em easy to identify.

Carry on.

///EDIT///

Before I get any wisenheimers telling me that mass =/= weight, I know that. I was talking about the fact that colleges often teach what they want with *no* reference to the real world.

Look - it might be true that you shouldn't start a sentence with a preposition, but in the real world people do. And I will too, apparently. See what I did there?

[Edited 2006-09-30 22:09:35]

[Edited 2006-09-30 22:13:44]
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vikkyvik
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 Quoting BAe146QT (Reply 20): Look - it might be true that you shouldn't start a sentence with a preposition, but in the real world people do. And I will too, apparently. See what I did there?

Actually, you're not supposed to end a sentence with a preposition.

 Quoting BAe146QT (Reply 20):Before I get any wisenheimers telling me that mass =/= weight, I know that. I was talking about the fact that colleges often teach what they want with *no* reference to the real world.

You are correct about that to a certain extent. From my experience, colleges teach everything the proper way - the way things really should be done in the real world (but obviously aren't).

In my job, where I deal with rather small-scale stuff, I'm continuously converting microns to mils and vice versa. It's annoying as all hell, and goes thoroughly against my deeply ingrained sense of keeping units straight.

Does anyone know if they do in fact use kg/m^2 in for wing loading in the industry? Thanks....

~Vik
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BAE146QT
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 Quoting Vikkyvik (Reply 21):Actually, you're not supposed to end a sentence with a preposition.

::LOL:: That is precisely the sort of grammatical pedantry up with wich I will not put! W. Churchill

 Quoting Vikkyvik (Reply 21):You are correct about that to a certain extent.

I suppose really I was griping about the fact that universities, (or colleges) send their charges out to the real world with all the confidence and qualifications they could ask for, but with no idea about how the world really works. I swear if I have to interview another supposed MCSE who doesn't know what RAID5 really means, I'll go postal. To wit;

"Yeah, that means, like, all your data is safe 'cause you have, like, five disks, you know?"

Gah! And these are sort of people who are trained to design aircraft?! I feel sorry for the "graduate" who corrects his boss thusly;

Boss: "And of course, you can't fly this plane with full cargo and full fuel, 'cause you'll exceed MTOW"

Much violence ensues.
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474218
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 Quoting Vikkyvik (Reply 21):Does anyone know if they do in fact use kg/m^2 in for wing loading in the industry? Thanks....

It is really not complicated MTOW divided by Wing Area = Wing Loading.

FredT
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3201
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For related discussion (i.e. where wing loading is a means rather than an end), see this thread, responses 19 and 23 mostly.

Careful...

Lego, definitely make sure on your work you clarify that it's MAXIMUM wing loading you're calculating. The actual wing loading is varying all the time as you're burning off fuel.
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KELPkid
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 Quoting 474218 (Reply 15):One Kilogram equals 2.2 Pounds. Mass does not matter. A Kilogram of aluminum and 2.2 Pounds of aluminum, are exactly the same mass.

I would still assert, with all that is in me, that wing loading is not measured in units of mass per unit of area, but I'm not going to create a huge stir in this forum over that
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vikkyvik
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 Quoting KELPkid (Reply 26):I would still assert, with all that is in me, that wing loading is not measured in units of mass per unit of area, but I'm not going to create a huge stir in this forum over that

Same thing I said here:

 Quoting Vikkyvik (Reply 16):I don't feel like getting into a whole discussion about units and what makes sense and doesn't, so I'll just accept this answer and shut up.

It does grate against all ingrained conventions though, doesn't it?

However, in actuality, measuring wing loading in mass per unit area can and does work, as long as one is staying within the Earth's atmosphere - or more specifically, close enough to the Earth that there is no noticeable change in gravitational acceleration.

So, you know, just watch out for those gravitational fluctuations!
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legoguy
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Thanks, seeing the list of aircraft with different maximum wing loadings helped! Also it made me realise that the example I used at the start of the thread (a330-300), I had accidently used the aircrafts minimum take off weight, so I have now corrected that in my work by using the maximum take off weight

Again, thanks. I never realised it would be changing. I have learnt alot from this thread so to everyone who replied, many thanks!
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keta
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Even at the risk of being too pedant, I will add that the maximum wing loading is more than that one At level flight, the force of the wing is the same as the weight, but there are phases of flight were the airplane needs an upwards acceleration, for example at takeoff. The lift will be greater than the MTOW, and the wing loading will increase too.

About the units, IMO the correct form is N/m^2, since the interesting info is force divided by area, that is, pressure. However, it's not hard to believe that kg/m^2 is used, it's just the same as when we say that we weigh x kg, we actually weigh x kiloponds.

BTW, why is wing loading important? I mean, what is it used for? I have heard that it gives some kind of information on how the aircraft behaves? But it only gives an idea of the mean pressure the wing is bearing with, not the local pressure anywhere in the wing.

[Edited 2006-10-04 20:54:36]
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KELPkid
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 Quoting Keta (Reply 29):BTW, why is wing loading important? I mean, what is it used for? I have heard that it gives some kind of information on how the aircraft behaves? But it only gives an idea of the mean pressure the wing is bearing with, not the local pressure anywhere in the wing.

Well, for one example, the amount of material used in the aircraft's spar and wing ribs is dictated by the amount of force the said parts need to withstand, and the strength of the material being used for construction. Planes can be "overengineered" like mad, but the downside to this is that effeciency will suffer (more mass than is necessary will be being carried around). Of course, any good engineer will try and optimise available construction materials vs. cost vs. fuel effeciency.
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3201
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It shows up in all sorts of performance calculations, gust load calculations, etc. Generally you want a high wing loading if you want your airplane to efficiently fly far and fast, and a low one if you want it to be maneuvarable and be able to fly slowly, and of course it's all a big tradeoff since you want your far/fast airplane to also be able to fly slowly and handle well for takeoff and landing.
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keta
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 Quoting KELPkid (Reply 30):Well, for one example, the amount of material used in the aircraft's spar and wing ribs is dictated by the amount of force the said parts need to withstand, and the strength of the material being used for construction

That's what I mean, to properly design a wing you need to know the actual distribution of forces, not just the average pressure all over the wing.

 Quoting 3201 (Reply 31):Generally you want a high wing loading if you want your airplane to efficiently fly far and fast, and a low one if you want it to be maneuvarable and be able to fly slowly

May I ask why is that?
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3201
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Well, I'm not really good at this, but I'll give it a shot. If someone like OldAeroGuy or AeroWeanie stumbles onto this thread, they'll probably do better. Basically for me it's one of those things that made sense through all the maths many years ago, and I hard-coded it into my brain, but my physical feel for it is very holistic, not too rigorous.

BTW speaking of that holistic physical feel, you can mess around with wing loading and paper airplanes. Think about taking two identical sheets of paper (thus the same weight) and making two different airplanes, one with much bigger wings than the other. Or actually do it.   Check out their different flight characteristics. One will be a lot smoother at higher speeds and long straight distances.

Aircraft that want to be good a cruising a long way:

Better gust response (as mentioned in the other referenced thread), because the change in lift due to a gust is proportional to the wing area, so as you increase wing area, the ratio of the change in lift (due to local variations in the air) to the weight of the aircraft is lower, so the ride is better, which makes it both more comfortable and more efficient and lets you fly faster without strengthening the structure more. Also the optimum lift coefficient (excluding compressibility drag) is proportional to wing loading and inversely proportional to the square of Mach, so a given lift coefficient will be optimal at a higher Mach number if you raise wing loading -- in other words, you can fly faster at a given lift coefficient and have it still be optimum if the wing loading is higher. There's a limit there, since too high a lift coefficient can give you problems with compressibility drag, although as airfoil design has gotten better, people have been able to keep compressibility drag down at higher cruise lift coefficients.

Aircraft that want to fly slow:

Stall speed is lower as wing loading goes down (for a given lift coefficient and atmospheric conditions, speed is proportional to square root of wing loading) so you can fly slower (which you want for field performance). Turn radius is also proportional to wing loading, for a given lift coefficient, so a lower wing loading lets you make tighter turns. This all comes from the fact that lift is proportional to wing area, and both stall speed and turn radius involve the ratio of lift to weight.
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lehpron
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 Quoting Legoguy (Thread starter):Wing loading = (MTOM x accel due to gravity) / wing area I hope anyone can make sense out of that. I am confused by several things so I thought I could try here before asking my teacher as I have all weekend to figure it out.

• Always remember your units and try not to mix English with metric. Doing so unfortunately costed NASA a Mars mission back in 1999 when someone didn't notice that feet was in meters (or vice versa) and a rover slammed into the planet. In fact I think they lost two mission that year.

• weight is a force in newtons in the metric system and varies mainly with gravity.

• In engineering, we will either refer to an aircraft's weight as 25,000 lbs or an aircraft's mass as 12,400 kg; there is no maximum takeoff mass, per se. Your professor may even say the plane's weight is 100,000 newtons at sea level.

• Here is an example for your problem:

Say we have a plane with weight of 600,000 lbs and wing area of 4000 sq ft.

In the English system, pound mass (lbm) and pound force (lbf) are very similar, and are usually mixed. lbm x g's = lbf, so 600,000 pound-mass x 1 gravity of acceleration = 600000 pounds of weight. Wing loading this way could be 600,000 lbf divided by 4000 sq ft = 150 lbf/ft2, gravity is already in the lbf.

You will have many professors that like to mix units to test you or some will only like metric or english units, so you have to learn to convert.

Book mark this: http://www.sciencelab.com/data/conversion.shtml
and this for Calculus stuff: http://www.calc101.com/webMathematica/MSP/Calc101/WalkD

600,000 lbf / (2.2046 lbf per kg at seal level) = 272,160 kg. Airbus calls the A380's MTOW in kilograms even though it isn't a weight, MTOW is just a title.

gravity is generally 9.8 m/s2 in metric system, multiply by mass in kg and we get 2667000 newtons of weight or force. 4000 sq ft x (1ft per 0.3048 meters)2 = 372 m2

metric system wing loading is now 2667000 N / 372 m2 = 7177 newtons / m2 or, divide by gravity 732 kg /m2

• If you're confused that is okay, I'm guessing you're just starting and are in some kind of engineering 101 survey course? I'm in my 3rd-to-last semester, hopefully. Though I have a bit of advice: Try to stay away from Wikipedia for engineering related research purposes. Many of our esteemed faculty have found gross errors in engineering subjects found there, like over-simiplifications, exaggerations and misinformation. That website is aimed at layfolks, not us technical people. We are told we will get a zeros on papers that have cited that website.
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3201
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 Quoting Lehpron (Reply 34):In engineering, we will either refer to an aircraft's weight as 25,000 lbs or an aircraft's mass as 12,400 kg; there is no maximum takeoff mass, per se. Your professor may even say the plane's weight is 100,000 newtons at sea level.

Working in industry, I have never heard anyone talk about an atmospheric (or surface) vehicle's weight in newtons. That doesn't mean it doesn't happen, but that sounds suspiciously like something that applies "In engineering classes at school" rather than "In engineering" to me.

Certainly the operators of existing real-life aluminum-and-carbon aircraft talk about "weights" in kilograms. The only time they don't do that is when they're talking about them in pounds. Is it strictly technically wrong? Yeah OK I guess so.

 Quoting Lehpron (Reply 34):Though I have a bit of advice: Try to stay away from Wikipedia for engineering related research purposes. Many of our esteemed faculty have found gross errors in engineering subjects found there, like over-simiplifications, exaggerations and misinformation. That website is aimed at layfolks, not us technical people.

That's good advice, but where does a.net fit in?
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FredT
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 Quoting Keta (Reply 29):Even at the risk of being too pedant, I will add that the maximum wing loading is more than that one At level flight, the force of the wing is the same as the weight, but there are phases of flight were the airplane needs an upwards acceleration, for example at takeoff. The lift will be greater than the MTOW, and the wing loading will increase too.

 Quoting Keta (Reply 29):About the units, IMO the correct form is N/m^2, since the interesting info is force divided by area, that is, pressure. However, it's not hard to believe that kg/m^2 is used, it's just the same as when we say that we weigh x kg, we actually weigh x kiloponds.

If misunderstandings are likely to occur, you can use kgf/kgm and lbf/lbm to separate masses and weights.

It is all more than a bit muddled up, so unless common sense and a bit of sensitivity to the context is applied it will rarely make sense. MTOW/MTOM is used interchangeably, weights (forces) and masses are given in kg, lb, N and pretty much every unit there is. It generally works out in the end.

 Quoting 3201 (Reply 35):That's good advice, but where does a.net fit in? innocent

Unless you apply a lot of common sense and know which people to listen to, any internet forum comes in well below Wikipedia in the ranking of sources to trust, IMO. This, unfortunately, applies to a.net tech/ops as well - even if the standard is higher than quite a few other forums.

Rgds,
/Fred
I thought I was doing good trying to avoid those airport hotels... and look at me now.

3201
Posts: 813
Joined: Sun Oct 10, 2004 4:16 pm

 Quoting FredT (Reply 36): Unless you apply a lot of common sense and know which people to listen to, any internet forum comes in well below Wikipedia in the ranking of sources to trust, IMO. This, unfortunately, applies to a.net tech/ops as well - even if the standard is higher than quite a few other forums.

Exactly my point. (Sorry, forgot about my American location and thus the assumption that the question was sincere rather than rhetorical!)

Learning whether to trust something is a valuable skill, though. At the graduate level, that's a skill you're actually supposed to learn, and it applies all the way up to refereed journal articles, not just free-for-alls like web forums.

[Edited 2006-10-05 23:12:50]
7 hours aint long-haul

YYZAeroEng
Posts: 160
Joined: Wed Jun 01, 2005 10:39 am

 Quoting Dakota123 (Reply 2):No practical difference between weight and mass on earth

Not quite true. If you mix up weight and mass in calculations you're going to have problems.

I've just completed my B.Eng in Aerospace Engineering. One thing I've learned is, you need to be comfortable using USC and Metric units and be able to switch between them.

The hardest part is trying to remember what the constants change to when you switch units.

Force = mass * acceleration

It's just applying one of Newton's laws of motion on a per area basis.
During the course of your studies you'll become very familiar with F=ma in all it's wonderful forms.
Mind that Bus! What bus? *Splat!*

louA340
Posts: 323
Joined: Mon Oct 03, 2005 2:19 pm

Slightly off topic a bit, but I'm also doing aerospace engineering and have done courses where we calculate things like, the max range, best climb rate, altitude that gives minimum fuel consumption, fuel consumption depending on weight, altitude, etc, and many other things. I presume pilots also have to know for each flight. In my work, however, these calculations are a bit long and tedious. I was wondering how pilots do this before their flights, do they have special equipment made for calculating such things?
RyEng

FredT
Posts: 2166
Joined: Thu Feb 07, 2002 9:51 pm

 Quoting LouA340 (Reply 39):I was wondering how pilots do this before their flights, do they have special equipment made for calculating such things?

Yes. Charts and tables. Page upon page upon page of them.

That's the AFM.

Electronic gadgetry is taking over though, little by little.

Rgds,
/Fred
I thought I was doing good trying to avoid those airport hotels... and look at me now.

3201
Posts: 813
Joined: Sun Oct 10, 2004 4:16 pm

 Quoting LouA340 (Reply 39):Slightly off topic a bit, but I'm also doing aerospace engineering and have done courses where we calculate things like, the max range, best climb rate, altitude that gives minimum fuel consumption, fuel consumption depending on weight, altitude, etc, and many other things. I presume pilots also have to know for each flight. In my work, however, these calculations are a bit long and tedious. I was wondering how pilots do this before their flights, do they have special equipment made for calculating such things?

Some of those things are done by the FMS, some are done by the flight-planning software, and some (especially for older aircraft) they do some things that are not optimized for each flight -- like for example they'll use a single climb schedule (constant IAS up to some altitude, then constant Mach) that's best for the aircraft over a range of operating conditions and not necessarily exactly optimal that day. In general, there are standard climb and descent schedules, and flight-planning software finds the best altitudes and speeds and the amount of fuel to load on the aircraft (including all kinds of contingency fuel), then the FMS flies the aircraft at the climb schedule, cruise speed, and altitudes requested by the crew (usually entered directly from the flight planning system via one of several kinds of electronic data link, otherwise entered manually by the crew). The altitudes aren't chosen just for optimal fuel -- there may be operational restrictions, lower altitudes mean faster flight and burning more fuel may be worth it, etc. Plus remember that only certain altitudes are available.

If there's not an FMS or computer flight planning system, a pilot would normally not do any of the calculations you mention -- there's a handbook that comes with the airplane that has plots of things like fuel flow rates for each weight, temperature, and altitude, or maybe a linear correction for one of the above, and the pilots or flight planners would do some quick arithmetic but nothing as detailed as what you learn. Your equations might instead be used to generate those plots.

The biggest difference between the real world and all the calculations you do in Aero Eng education, in my experience, is that in the real world there is a temperature variation from standard, and there is wind.

BTW, one of the calculations you mentioned is very commonly done wrong, even in the real world.
7 hours aint long-haul

YYZAeroEng
Posts: 160
Joined: Wed Jun 01, 2005 10:39 am

 Quoting 3201 (Reply 41): The biggest difference between the real world and all the calculations you do in Aero Eng education, in my experience, is that in the real world there is a temperature variation from standard, and there is wind

That is both a burden and a boon.
Living in "Physics World" makes calculations a lot easier, but it's not very accurate at times.

The more industry experience I gain, the more I realize how different the two worlds are.
Mind that Bus! What bus? *Splat!*

flybyguy
Posts: 1416
Joined: Sun Jun 27, 2004 12:52 pm

 Quoting Vikkyvik (Reply 12):I frequently do exactly the opposite when confronted with a problem in English units! Honestly, I actually think the situation in English units is more confusing. With pound-mass and pound-force having the same numerical value on Earth, one is automatically dissuaded from appreciating the difference between a force and a mass. Or rather, there ceases to be a need to understand it.

My sentiments exactly... when will America convert to the metric system... it is only American engineers and scientists that find metric far more practical?
"Are you a pretender... or a thoroughbred?!" - Professor Matt Miller

BAE146QT
Posts: 981
Joined: Sun Sep 24, 2006 4:58 am

 Quoting Flybyguy (Reply 43):when will America convert to the metric system...

Apropos of that, my other half reckons that mathematical skills in this country have declined since we embraced decimalisation*. She thinks that because our currency forced people to think in base 16, (and base 20 for shillings), children grew up with less fear of maths.

Maybe the Americans have a point, (even if it's unintentional).

* Whenever we have this conversation it usually ends up in violence because I ask her what a groat looked like and how many sheckels you would get for one.
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