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Calculating An Aircrafts Wing Loading  
User currently offlineLegoguy From United Kingdom, joined Jun 2006, 3313 posts, RR: 39
Posted (8 years 3 days 22 hours ago) and read 12074 times:

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

Therefore wingloading = (121970 x 9.81) / 363.1 = 3295.31 N/m^-2

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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44 replies: All unread, showing first 25:
 
User currently offlineKELPkid From United States of America, joined Nov 2005, 6398 posts, RR: 3
Reply 1, posted (8 years 3 days 22 hours ago) and read 12075 times:

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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User currently offlineDakota123 From United States of America, joined Aug 2006, 116 posts, RR: 0
Reply 2, posted (8 years 3 days 22 hours ago) and read 12062 times:

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...

User currently offlineLegoguy From United Kingdom, joined Jun 2006, 3313 posts, RR: 39
Reply 3, posted (8 years 3 days 22 hours ago) and read 12059 times:

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  Smile

Quoting KELPkid (Reply 1):

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 Big grin



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User currently offlineKELPkid From United States of America, joined Nov 2005, 6398 posts, RR: 3
Reply 4, posted (8 years 3 days 22 hours ago) and read 12046 times:

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  Wink



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User currently offlineLegoguy From United Kingdom, joined Jun 2006, 3313 posts, RR: 39
Reply 5, posted (8 years 3 days 22 hours ago) and read 12040 times:

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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User currently offlineDakota123 From United States of America, joined Aug 2006, 116 posts, RR: 0
Reply 6, posted (8 years 3 days 22 hours ago) and read 12031 times:

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.


User currently offlineVikkyvik From United States of America, joined Jul 2003, 10048 posts, RR: 26
Reply 7, posted (8 years 3 days 21 hours ago) and read 12024 times:
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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.

Anyway, wing loading (a force per area) is typically MTOW/wing area.

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.

So your formula in the thread starter:

Quoting Legoguy (Thread starter):
Wing loading = (MTOM x accel due to gravity) / wing area

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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User currently offlineLegoguy From United Kingdom, joined Jun 2006, 3313 posts, RR: 39
Reply 8, posted (8 years 3 days 21 hours ago) and read 12019 times:

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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User currently offlineLegoguy From United Kingdom, joined Jun 2006, 3313 posts, RR: 39
Reply 9, posted (8 years 3 days 21 hours ago) and read 12013 times:

Quoting Vikkyvik (Reply 7):

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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User currently offlineKELPkid From United States of America, joined Nov 2005, 6398 posts, RR: 3
Reply 10, posted (8 years 3 days 21 hours ago) and read 12008 times:

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)  Wink 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  Wink Wouldn't your professor love that!



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User currently offline474218 From United States of America, joined Oct 2005, 6340 posts, RR: 9
Reply 11, posted (8 years 3 days 20 hours ago) and read 11996 times:

You all are making it too complicated, "Wing Loading" is nothing more than:

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


User currently offlineVikkyvik From United States of America, joined Jul 2003, 10048 posts, RR: 26
Reply 12, posted (8 years 3 days 20 hours ago) and read 11994 times:
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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

 rotfl 

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.



"Two and a Half Men" was filmed in front of a live ostrich.
User currently offlineKELPkid From United States of America, joined Nov 2005, 6398 posts, RR: 3
Reply 13, posted (8 years 3 days 20 hours ago) and read 11988 times:

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  scratchchin 



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User currently offline474218 From United States of America, joined Oct 2005, 6340 posts, RR: 9
Reply 14, posted (8 years 3 days 19 hours ago) and read 11980 times:

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


User currently offline474218 From United States of America, joined Oct 2005, 6340 posts, RR: 9
Reply 15, posted (8 years 3 days 19 hours ago) and read 11976 times:

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.


User currently offlineVikkyvik From United States of America, joined Jul 2003, 10048 posts, RR: 26
Reply 16, posted (8 years 3 days 19 hours ago) and read 11972 times:
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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.  Smile

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.



"Two and a Half Men" was filmed in front of a live ostrich.
User currently offlineLegoguy From United Kingdom, joined Jun 2006, 3313 posts, RR: 39
Reply 17, posted (8 years 3 days 7 hours ago) and read 11930 times:

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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User currently offlineSovietjet From Bulgaria, joined Mar 2003, 2614 posts, RR: 17
Reply 18, posted (8 years 3 days ago) and read 11910 times:
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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.

User currently offlineTexfly101 From United States of America, joined Jan 2006, 351 posts, RR: 0
Reply 19, posted (8 years 3 days ago) and read 11907 times:

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.


User currently offlineBAe146QT From United Kingdom, joined Sep 2006, 996 posts, RR: 0
Reply 20, posted (8 years 3 days ago) and read 11902 times:

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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User currently offlineVikkyvik From United States of America, joined Jul 2003, 10048 posts, RR: 26
Reply 21, posted (8 years 2 days 22 hours ago) and read 11884 times:
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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.

 Wink

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



"Two and a Half Men" was filmed in front of a live ostrich.
User currently offlineBAe146QT From United Kingdom, joined Sep 2006, 996 posts, RR: 0
Reply 22, posted (8 years 2 days 22 hours ago) and read 11879 times:

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"
Grad: "Surely you mean MTOM!"

Much violence ensues.



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User currently offline474218 From United States of America, joined Oct 2005, 6340 posts, RR: 9
Reply 23, posted (8 years 2 days 21 hours ago) and read 11873 times:

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.

http://www.answers.com/topic/wing-loading


User currently offlineFredT From United Kingdom, joined Feb 2002, 2185 posts, RR: 26
Reply 24, posted (8 years 2 days 7 hours ago) and read 11849 times:

Or gross weight by wing area to get wing loading rather than maximum wing loading... ?  Wink


I thought I was doing good trying to avoid those airport hotels... and look at me now.
25 Post contains links and images 3201 : 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, definite
26 Post contains links and images KELPkid : Okay, you are possibly vindicated here: http://en.wikipedia.org/wiki/SpaceShipOne lists the wing loading as "Wing loading: 240 kg/m²" I would still
27 Post contains images Vikkyvik : Same thing I said here: It does grate against all ingrained conventions though, doesn't it? However, in actuality, measuring wing loading in mass per
28 Post contains images Legoguy : 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
29 Post contains images Keta : 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 sa
30 KELPkid : 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 wit
31 3201 : 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 t
32 Post contains images Keta : 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. Ma
33 Post contains images 3201 : 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 b
34 Post contains links Lehpron : 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
35 Post contains images 3201 : 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 happe
36 FredT : When talking about wing loading in this sense, it is the 1 g wing loading. If misunderstandings are likely to occur, you can use kgf/kgm and lbf/lbm
37 Post contains images 3201 : Exactly my point. (Sorry, forgot about my American location and thus the assumption that the question was sincere rather than rhetorical!) Learning w
38 YYZAeroEng : 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. O
39 LouA340 : 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 rat
40 Post contains images FredT : 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
41 3201 : 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
42 YYZAeroEng : 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 e
43 Flybyguy : My sentiments exactly... when will America convert to the metric system... it is only American engineers and scientists that find metric far more pra
44 BAe146QT : Apropos of that, my other half reckons that mathematical skills in this country have declined since we embraced decimalisation*. She thinks that beca
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