Help me understand wing loading

dmspilot said:
The load factor is equal to 1.41 in a coordinated 45° bank without vertical acceleration. No caveats about frames of references, whatever that is supposed to mean.
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Wrong, with a caveat. If you do not pull back on the elevator, the plane will continue to increase its descent and the load will not be 1.41. You can’t get something for nothing.

You simply changed “level” to “without vertical acceleration”, but that’s what was meant by level in the first place, and everyone knows that.
 
Salty said:
You simply changed “level” to “without vertical acceleration”, but that’s what was meant by level in the first place, and everyone knows that.
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No buddy, I did not interchange "level" with "without vertical acceleration", YOU did, and YOU are making it the key crux of your argument; a crux that is completely wrong and demonstrates a misunderstanding of "basic rules of physics".

They are not interchangeable, which is the crux of my argument. You are accusing me of making an argument that is actually the opposite of the one I am making!

What do you measure acceleration in? And is (for example) "-10,000 fpm" a measurement of acceleration?
 
dmspilot said:
No buddy, I did not interchange "level" with "without vertical acceleration", YOU did, and YOU are making it the key crux of your argument; a crux that is completely wrong and demonstrates a misunderstanding of "basic rules of physics".

They are not interchangable.

What do you measure acceleration in? And is (for example) "-10,000 fpm" a measurement of acceleration?
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Sure. You restated the same thing with different words, but I’m wrong. Congratulations.
 
A 45° banked turn in a steady-rate descent will still have a load factor of 1.41. Turning does not eliminate the load factor rules that apply to steady climbs or steady descents in straight-ahead flight. The only time you see a reduced load factor is when you are transitionmg from level flight to descending flight, and the only way to maintain that lower load factor is to keep increasing the rate of descent.

Too many people just don't get this stuff, and it kills some of them. The guy that buzzes the runway or some friend's house and pulls up hard is raising the load factor alarmingly as the airplane changes direction upward, and he might get the stall/spin he asked for.
 
Wing loading 101. Compare your plane's performance at gross to running empty with minimum fuel. Available power is the same but performance is very different. Now take that plane and extend the wings. Better at some things, like STOL ops, worse for others, like crosswinds and turbulence.
 
Back to wing loading: the weight of the airplane divided by the wing area (with the wing being assumed to pass through the fuselage and that area included) is the wing loading. It's somewhat misleading, since each square foot of the wing is not taking the same load as all the other square feet. Lift is highest just aft of the leading edge, and toward the trailing edge a lot less lift is generated over a large area, with some airfoils actually producing negative lift there. The pressure patterns shift somewhat as AOA changes.


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Power loading is weight divided by horsepower. Most of us need more of that.
 
Salty said:
the plane will continue to increase its descent and the load will not be 1.41. You can’t get something for nothing.
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If the plane "continue(s) to increase its descent" the "increasing descent" is vertical acceleration.

Examples of zero vertical acceleration include:

1. Level flight
2. A constant-rate climb
3. A constant-rate descent

Vertical acceleration occurs when the rate of climb/descent is changing.

A constant-rate climb, or descent, is a 1g manuver.
 
A banked turn will increase the wing loading because the lift generated isn't directly opposing graviity. It uses some of it to accelerate the aircraft into the turn. Yes it indeed is determined by the angle of bank (e.g. 45 degrees is 1.414... times what the unbanked loading would be)
 
Let's do a little math.

Let's say you roll into a 45° coordinated turn from level flight and allow the airplane to descend.

Let's say you pull back on the elevator enough to produce a significant increase in lift, but still less than the 1.414 factor required to maintain level flight. Let's say it's 90% of that required. That sounds like a significant increase doesn't it? (I'm being generous to try to do your argument a favor.)

So the vertical component of lift is 90% of the aircraft's weight. The aircraft thus accelerates downward at only 10% the acceleration of gravity.

With gravity being equal to about 32 feet per second per second, 10% of that would be 3.2 feet per second per second, or 192fpm per second.

After 10 seconds the aircraft is descending at 1,920fpm (10 x 192). Close to pegging most GA VSI instruments.

Thus I reiterate my posts 10 ("either the forces balance or you create an impact crater") and 30 ("my intention was to discuss an airplane in controlled flight, not one that is out of control").
 
Ryanb said:
ok cool thanks
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Means my head hurts trying to follow all y’all, lol.

By the way, sure, the wings hold up the mass, but all of it? Does the fuselage create any appreciable lift?
 
455 Bravo Uniform said:
By the way, sure, the wings hold up the mass, but all of it? Does the fuselage create any appreciable lift?
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The wings actually hold up a little more than the weight of the airplane.

A conventional tail produces a downward force to balance the nose-down pitching moment created because the center-of-lift is somewhat aft of the center-of-gravity. That downward force must be countered by additional lift from the wing to produce unaccelerated (vertically) flight. That's a chief advantage to a canard design. Both the stabilizer and wing are producing lift so total lift required is less.
 
I’m probably going to regret entering this. But load factor is not a measure of “g”(although it is often expressed that way for a misleading “clarity” reason) or acceleration

it’s a ratio based on vector math of forces involved in banking airplane. It has no unit. As such there is the same load factor(but not the same acceleration) in a coordinated turn of a specific bank angle regardless of level, climb, descent, or any vertical acceleration. Just because you have more or less than 1g vertical, doesn’t change the fact that wing experiences 1.4 times(for 45 deg) of whatever vertical acceleration (and weight) you have at that moment. It’s simply a matter of your bank angle. As long as there is any positive weight.

Say you are accelerating your descent in a straight flight. Your weight will decrease. Say it went from 2000lb to 1500lb(“.75g”). You put your plane into 45 degree bank and continue accelerating descent at the same rate, your vertical acceleration doesn’t change, but the wing will now carry 1.4x1500lb=2100lb. And a pilot will feel 1.05g, but load factor remains 1.4(1/(sin(45))
 
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455 Bravo Uniform said:
Does the fuselage create any appreciable lift?
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There is probably some up/down force, but a fuselage is a very inefficient airfoil shape (very, very low aspect ratio if nothing else) and would create a lot of drag relative to the lift. So you would want to minimize that factor by pretty much lining up the fuselage with the airflow.

Look at the glide angle of lifting body aircraft...
 
Salty said:
Sure. You restated the same thing with different words, but I’m wrong. Congratulations.
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I hope you now understand the error in your thinking. Your thinking that the turn must be level to achieve 1.41G in a 45 degree bank is no different from thinking that you can only achieve 1G with the wings level if you are not descending. I hope you understand that if you are in a stable descent with the wings level, you are at 1G. If you then do a 45 degree banked turn, still at a stable descent, you are at 1.41G. All you have to do is go fly an airplane with a G meter to prove this.
 
RoscoeT said:
I hope you now understand the error in your thinking. Your thinking that the turn must be level to achieve 1.41G in a 45 degree bank is no different from thinking that you can only achieve 1G with the wings level if you are not descending. I hope you understand that if you are in a stable descent with the wings level, you are at 1G. If you then do a 45 degree banked turn, still at a stable descent, you are at 1.41G. All you have to do is go fly an airplane with a G meter to prove this.
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Yes, I think I’ve made it clear I understand that. But if you are in level flight and then bank 45 degrees and then change to aN increasing descent, that is NOT what you just described. And that is what will happen if you don’t pull back on the yoke.
 
Salty said:
Yes, I think I’ve made it clear I understand that. But if you are in level flight and then bank 45 degrees and then change to aN increasing descent, that is NOT what you just described. And that is what will happen if you don’t pull back on the yoke.
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From any bank angle, you can also hold the yoke in a position that will load the airplane at exactly zero G until the wings come off or you hit the ground. We can list all the variations we want, but this all started by your stating that the turn must be level, which it doesn't. Glad you realize that.
 
Flying at any angle increases the load on your bank.

Well, unless your parents are paying for it.
 
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Salty said:
Yes, I think I’ve made it clear I understand that. But if you are in level flight and then bank 45 degrees and then change to aN increasing descent, that is NOT what you just described. And that is what will happen if you don’t pull back on the yoke.
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Moving the goalpost. Completely different from saying the turn HAS to be level to have the load factor of 1.41, which is incorrect.
 
455 Bravo Uniform said:
Means my head hurts trying to follow all y’all, lol.

By the way, sure, the wings hold up the mass, but all of it? Does the fuselage create any appreciable lift?
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It would be hard to test. Switching from wheels to floats or wheels to skis? Floats and skis definitely contribute to lift. I remember doing stalls with floats and having to be mindful of the floats masking not having the wings fully recovered.
 
What if you were in a 5/3 Bank, loaded, with wings?

Probably getting the cops called on ya, but you're having a good time.
 
Wing loading demo. Take a Skywagon up next to a mountain for face-bender training. Slow to 70mph, roll into a 50* bank toward the mountain, apply full power, and yank full flaps. Hard not to feel the Gs doing that. It works in Cubs, too. :)
 
RoscoeT said:
I hope you now understand the error in your thinking. Your thinking that the turn must be level to achieve 1.41G in a 45 degree bank is no different from thinking that you can only achieve 1G with the wings level if you are not descending. I hope you understand that if you are in a stable descent with the wings level, you are at 1G. If you then do a 45 degree banked turn, still at a stable descent, you are at 1.41G. All you have to do is go fly an airplane with a G meter to prove this.
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No, the original error was stating without qualification something that is not always true, namely that a 45* bank will generate 1.41g. That statement was corrected to a true statement, albeit one that did not give all of the cases that would make it true. The fact that the corrected statement did not state all of the cases that make the statement true does not change the fact that the original statement is not true as stated without qualification.

A 45* bank does not always result in 1.41g. To say so is wrong. A 45* banked level turn will result in 1.41g, so to say so is not wrong. There are other cases (e.g., stable 45* descending or climbing turn) that will also result in 1.41g. That doesn't change the preceding facts.
 
Oh, and as the FAA points out, these numbers are accurate for coordinated turns. YMMV.
 
genna said:
I’m probably going to regret entering this.
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Me too.
genna said:
Say you are accelerating your descent in a straight flight. Your weight will decrease. Say it went from 2000lb to 1500lb(“.75g”). You put your plane into 45 degree bank and continue accelerating descent at the same rate, your vertical acceleration doesn’t change, but the wing will now carry 1.4x1500lb=2100lb. And a pilot will feel 1.05g, but load factor remains 1.4(1/(sin(45))
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Say you went from 2000lb to zero lbs. The pilot feels zero 'g'. The wing produces no 'lift'. The airplane doesn't turn. There is no 'load factor'. Whatever the sin of theta says. IMO.
 
I think much of the argument in this thread illustrates how some things are better explained with a few equations and perhaps a diagram than a whole bunch of words.
 
dtuuri said:
Me too.

Say you went from 2000lb to zero lbs. The pilot feels zero 'g'. The wing produces no 'lift'. The airplane doesn't turn. There is no 'load factor'. Whatever the sin of theta says. IMO.
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This is why I said “as long as there is any weight”. Once you go weightless or negative, things get way more interesting
 
dmspilot said:
Moving the goalpost. Completely different from saying the turn HAS to be level to have the load factor of 1.41, which is incorrect.
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You created goalposts out of thin air. If you started out level, which is What the scenario was, then you must remain level in order to have the load factor be 1.41. Nobody would assume you were starting out descending and then banked and continued to descend at the same rate. That’s nonsense that you made up so you can say you were right. The premise is that you started level. If you don’t stay level and bank, and allow the plane to descend naturally, it will increase its descent rate and will Never have a load factor of 1.41.
 
Lindberg said:
No, the original error was stating without qualification something that is not always true, namely that a 45* bank will generate 1.41g. That statement was corrected to a true statement, albeit one that did not give all of the cases that would make it true. The fact that the corrected statement did not state all of the cases that make the statement true does not change the fact that the original statement is not true as stated without qualification.

A 45* bank does not always result in 1.41g. To say so is wrong. A 45* banked level turn will result in 1.41g, so to say so is not wrong. There are other cases (e.g., stable 45* descending or climbing turn) that will also result in 1.41g. That doesn't change the preceding facts.
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Thank you for stating it much more clearly than I was.
 
Salty said:
You created goalposts out of thin air. If you started out level, which is What the scenario was, then you must remain level in order to have the load factor be 1.41. Nobody would assume you were starting out descending and then banked and continued to descend at the same rate. That’s nonsense that you made up so you can say you were right. The premise is that you started level. If you don’t stay level and bank, and allow the plane to descend naturally, it will increase its descent rate and will Never have a load factor of 1.41.
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There was no scenario. I made a statement that a turn does not have to be level in order for the load factor bank angle relationship to hold true. You are the one that fabricated a scenario out of thin air and moved the goalpost by saying we were only talking about turns that start level and end up in an increasing descent. You were not even part of the discussion when I made my statement (Post 4).
 
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PeterNSteinmetz said:
I think much of the argument in this thread illustrates how some things are better explained with a few equations and perhaps a diagram than a whole bunch of words.
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Yes pictures are always good, but a basic understanding what those diagrams and equations mean is important too. To someone who has no or little experience with diagrams like that, they have little meaning. It's a tough subject to learn and understand. And even when you do think you have it, sometimes it gets screwed up or you forget something and bingo, you are wrong. Add gusty winds and/or strong updrafts and downdrafts and things get more difficult to describe.
 
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