Help me understand wing loading

RyanB

RyanB

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I’m pretty rusty on this subject. Need some clarifying and education on it.

part of the POH for the Archer II says :

B64B2993-1F53-4E92-B5C9-0E488AF06C36.jpeg

So what are these numbers actually saying?
As for wing loading, what is 15.0lb/ft^2 meaning?

if I’m flying straight and level at altitude, how much weight are my wings supporting? If I roll the airplane into a 45deg bank, how much weight are my wings supporting then?

thanks all
 
Two different topics really.

Wing loading = maximum gross weight divided by wing area.

Load factor in a 45 degree bank turn is 1.41, meaning the wings are creating lift equal to 1.41 times aircraft weight.
 
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The turn doesn't have to be level.
 
math math math numbers numbers math.
 
Higher wing loading will ride better in turbulence but need faster takeoff and landing speed which produces long takeoff/landing rolls.

Lower wing loading will ride worse in turbulence but TO/LND slower/shorter.
 
dmspilot said:
The turn doesn't have to be level.
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It does to have the 1.41 load factor. The load factor only occurs if you remain level. If you allow the nose to drop as much as it wants to naturally, then load factor is 1, but you’ll be in a pretty brisk decent. If you allow the nose to drop but not all the way, you still won’t get 1.41 load factor, but something between 1 and 1.41.
 
Salty said:
The load factor only occurs if you remain level. If you allow the nose to drop as much as it wants to naturally, then load factor is 1, but you’ll be in a pretty brisk decent. If you allow the nose to drop but not all the way, you still won’t get 1.41 load factor, but something between 1 and 1.41.
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Wrong. When the vertical component of lift is less than one, weight will exceed the vertical component of lift which will cause downward acceleration. Such a condition doesn't last long (either the forces balance or you create an impact crater).
 
dmspilot said:
Wrong. When the vertical component of lift is less than one, weight will exceed the vertical component of lift which will cause downward acceleration.
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I have no idea what that means, but banking doesn’t magically make your plane weigh more, nor does it cause more lift to be created.
 
dmspilot said:
It's a good thing that I said neither of those things.
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So, if you’re not remaining level (or ascending), where is the Increased load coming from?
 
maybe you need more of a colloquial explanation
 
Salty said:
So, if you’re not remaining level (or ascending), where is the Increased load coming from?
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If you are at 1g, you're at 1g. Doesn't matter if you're climbing, level, or descending. Add a bank angle without accelerating vertically and you're at 1g plus the increased load factor due to the bank.
 
Larry in TN said:
If you are at 1g, you're at 1g. Doesn't matter if you're climbing, level, or descending. Add a bank angle without accelerating vertically and you're at 1g plus the increased load factor due to the bank.
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A turn is acceleration.
 
Salty said:
So, if you’re not remaining level (or ascending), where is the Increased load coming from?
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When the plane is banked, the wing still produces lift perpendicular to itself. So for explanation sake, let's say you are at a 45 degree bank, you have 0.707 of 1 g pointed straight up and 0.707 of 1 g pointed in the direction of the turn. If you leave the plane in this situation, 0.707 g of lift will not balance out 1g of gravity, you will descend. So you need to increase the lift of your wing, in this case, (if I'm doing the math right, I'm sure I'll be corrected if not) you need about 1.4 g of lift from the wing which will give you 1g in the up component, counteracting gravity and maintaining level flight. You will feel the extra gs in your seat.
 
PaulS said:
When the plane is banked, the wing still produces lift perpendicular to itself. So for explanation sake, let's say you are at a 45 degree bank, you have 0.707 of 1 g pointed straight up and 0.707 of 1 g pointed in the direction of the turn. If you leave the plane in this situation, 0.707 g of lift will not balance out 1g of gravity, you will descend. So you need to increase the lift of your wing, in this case, (if I'm doing the math right, I'm sure I'll be corrected if not) you need about 1.4 g of lift from the wing which will give you 1g in the up component, counteracting gravity and maintaining level flight. You will feel the extra gs in your seat.
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That’s correct. But that can’t happen if you allow the plane to descend. You’ll still be at 1 g if you allow it to descend naturally.
 
red4golf said:
please define
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It’s a basic rule of physics.

Any change in the velocity of an object results in an acceleration: increasing speed (what people usually mean when they say acceleration), decreasing speed (also called decelerationor retardation), or changing direction (called centripetal acceleration). Yes, that's right, a change in the direction of motion results in an acceleration even if the moving object neither sped up nor slowed down.
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https://physics.info/acceleration/
 
Salty said:
A turn is acceleration.
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He said "without accelerating vertically"

A turn is not a vertical acceleration, at least not the kind of turn I thought pilots normally talk about.
 
Salty said:
That’s correct. But that can’t happen if you allow the plane to descend. You’ll still be at 1 g if you allow it to descend naturally.
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Descending does not reduce load factor. What is the load factor during straight flight in a 10,000 fpm descent?
 
dmspilot said:
Descending does not reduce load factor. What is the load factor during straight flight in a 10,000 fpm descent?
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1. However, if you continue to hold bank without touching the elevator you will continue to accelerate your descent. The “death spiral” if you will. Your load factor will not go up unless you pull on the elevator.
 
Salty said:
1. However, if you continue to hold bank without touching the elevator you will continue to accelerate your descent. The “death spiral” if you will. Your load factor will not go up unless you pull on the elevator.
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Maybe completely true in an airplane with no stability, but only partially true for most airplanes. Airspeed will increase which does what to lift?
 
Salty said:
That’s correct. But that can’t happen if you allow the plane to descend. You’ll still be at 1 g if you allow it to descend naturally.
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Yes sir, until the ground rises up and smites thee.
 
dmspilot said:
Maybe completely true in an airplane with no stability, but only partially true for most airplanes. Airspeed will increase which does what to lift?
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True, there will be some increase in load. Good point. But it won’t be 1.41.
 
Salty said:
True, there will be some increase in load. Good point. But it won’t be 1.41.
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Regardless, my intention was to discuss an airplane in controlled flight, not one that is out of control.
 
dmspilot said:
Regardless, my intention was to discuss an airplane in controlled flight, not one that is out of control.
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Your statement that the turn doesn’t have to be level is wrong. For the load to be 1.41, the turn has to be level.
 
Salty said:
Your statement that the turn doesn’t have to be level is wrong. For the load to be 1.41, the turn has to be level.
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My statement is not wrong. Your statement is wrong. The turn does not have to be level.
 
dmspilot said:
My statement is not wrong. Your statement is wrong. The turn does not have to be level.
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We are both wrong and both right depending on the frame of reference. The only difference is that I’m willing to admit where you were correct.
 
Salty said:
We are both wrong and both right depending on the frame of reference. The only difference is that I’m willing to admit where you were correct.
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That's doubletalk and you haven't admitted anything.
 
Salty said:
True, there will be some increase in load. Good point. But it won’t be 1.41.
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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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