Wing loading question
Thread Starter
Join Date: Aug 2005
Location: Glens o' Angus by way of LA
Age: 60
Posts: 1,975
Likes: 0
Received 0 Likes
on
0 Posts
Wing loading question
I understand that in a 60 degree banked level turn the wing loading reaches 2Gs, is there a calc or rule of thumb for the wing loading in a decsending turn at different bank angles ? I assume the wing loading would be less in the descent than in a straight and level turn.
The g force, or wing loading required to maintain steady flight (ie: not accelerating upward or downward) can be calculated by inverting the sine of the bank value. Sine 60 = one half. One over zero point 5 equals two, so load factor two. At seventy five degrees the sine value is a quarter. So, load factor is four.
It works out that in a 30 degree bank the load factor is about 1.4. As you can see it increases exponentially with increasing bank angle.
If the aircraft is descending, but in a steady state of descent, the load factor will remain the same. Ditto climbing (if you have an aircraft powerful enough to achieve much climb at a sixty degree bank.)
The next logical learning to this formula is to realize that the stall speed will increase by virtue of the increased wing loading. The revised stall speed is always the normal speed multiplied by the square root of the wing loading. So at a 2 g loading, stall speed is approx 1.4 normal. At a 4g loading, stall speed doubles.
I don't think descending changes it, except for when actually initiating the descent, which slightly changes the required load factor for the duration the aircraft is accelerating downward.
It works out that in a 30 degree bank the load factor is about 1.4. As you can see it increases exponentially with increasing bank angle.
If the aircraft is descending, but in a steady state of descent, the load factor will remain the same. Ditto climbing (if you have an aircraft powerful enough to achieve much climb at a sixty degree bank.)
The next logical learning to this formula is to realize that the stall speed will increase by virtue of the increased wing loading. The revised stall speed is always the normal speed multiplied by the square root of the wing loading. So at a 2 g loading, stall speed is approx 1.4 normal. At a 4g loading, stall speed doubles.
I don't think descending changes it, except for when actually initiating the descent, which slightly changes the required load factor for the duration the aircraft is accelerating downward.
Join Date: Nov 2006
Location: Down south
Age: 69
Posts: 363
Likes: 0
Received 0 Likes
on
0 Posts
At 60 degrees of bank, the G load is 2. But only if you pull on the stick to maintain level flight. In other words, it is the pulling (Or pushing) on the stick which determines the load factor negative or positive.
So if you pull during the descending turn or if you just reduce power on a trimmed a/c and let it descend without pulling you have 2 different scenarios.
So if you pull during the descending turn or if you just reduce power on a trimmed a/c and let it descend without pulling you have 2 different scenarios.
Imagine you are in level flight and decide to conduct a 60* level turn.
You roll on the bank, balancing with rudder, and apply back-pressure to maintain altitude.
Immediately after you are established in the perfect 60* level turn, holding 2g, the stall warner operates. (A)
You could decide to maintain this scenario, or:
You relax some of the back-pressure, reducing g slightly, but maintain 60* bank.
The stall warner ceases, because of the reduced g (wing loading).
What is happening to the altitude? It is reducing of course.
So, you are now in a 60* bank, balanced, descending turn with less than 2g. (B)
If you return to scenario (A) and wish to climb, maintaining a balanced 60* bank turn, how is this achieved?
By increasing the lift.
This is achieved by increasing back pressure, g and wing loading.
The aircraft will now climb, with the stall warner operating and buffet possibly being felt.
So you are now in a 60* bank, balanced, climbing turn with more than 2g. (C)
There are therefore three scenarios (A), (B) and (C), in all of which you are in a 60* bank, balanced turn but in which you have different amounts of g and wing loading.
I have deliberately not mentioned power/thrust changes. Let's assume that the power/thrust is constant and sufficient to achieve the results described.
Yes, I agree but don't know the 'rule of thumb', other than the obvious relationship that; the less the wing loading, the greater the rate of descent.
Not sure that I agree with that.
Exactly! See above.
You roll on the bank, balancing with rudder, and apply back-pressure to maintain altitude.
Immediately after you are established in the perfect 60* level turn, holding 2g, the stall warner operates. (A)
You could decide to maintain this scenario, or:
You relax some of the back-pressure, reducing g slightly, but maintain 60* bank.
The stall warner ceases, because of the reduced g (wing loading).
What is happening to the altitude? It is reducing of course.
So, you are now in a 60* bank, balanced, descending turn with less than 2g. (B)
If you return to scenario (A) and wish to climb, maintaining a balanced 60* bank turn, how is this achieved?
By increasing the lift.
This is achieved by increasing back pressure, g and wing loading.
The aircraft will now climb, with the stall warner operating and buffet possibly being felt.
So you are now in a 60* bank, balanced, climbing turn with more than 2g. (C)
There are therefore three scenarios (A), (B) and (C), in all of which you are in a 60* bank, balanced turn but in which you have different amounts of g and wing loading.
I have deliberately not mentioned power/thrust changes. Let's assume that the power/thrust is constant and sufficient to achieve the results described.
I understand that in a 60 degree banked level turn the wing loading reaches 2Gs, is there a calc or rule of thumb for the wing loading in a decsending turn at different bank angles ? I assume the wing loading would be less in the descent than in a straight and level turn.
If the aircraft is descending, but in a steady state of descent, the load factor will remain the same. Ditto climbing (if you have an aircraft powerful enough to achieve much climb at a sixty degree bank.)
At 60 degrees of bank, the G load is 2. But only if you pull on the stick to maintain level flight. In other words, it is the pulling (Or pushing) on the stick which determines the load factor negative or positive.
So if you pull during the descending turn or if you just reduce power on a trimmed a/c and let it descend without pulling you have 2 different scenarios.
So if you pull during the descending turn or if you just reduce power on a trimmed a/c and let it descend without pulling you have 2 different scenarios.
Join Date: Sep 2011
Location: Mare Imbrium
Posts: 638
Likes: 0
Received 0 Likes
on
0 Posts
Sorry Eckhard, you're not differentialting between steady state climb or descent and the transition from level to climbing or descending. In steady state climb or descent the vertical component of lift is always 1g - you only need more or less than that to initiate the climb or descent.
As has been said already
As has been said already
I don't think descending changes it, except for when actually initiating the descent, which slightly changes the required load factor for the duration the aircraft is accelerating downward.
In steady state climb or descent the vertical component of lift is always 1g - you only need more or less than that to initiate the climb or descent.
But we are talking about a turning descent, during which there must be an unbalanced force causing the aircraft to accelerate towards the centre of the turn. This force is provided by a horizontal component of the lift. So while the vertical component of lift (together with the vertical component of drag) is balancing the weight, the extra lift produced by 'pulling g' is used to accelerate the aircraft around the turn. Therefore, the total lift vector is more than the weight (or more than 1g).
Or have I got this all wrong somewhere?
Join Date: Sep 2011
Location: Mare Imbrium
Posts: 638
Likes: 0
Received 0 Likes
on
0 Posts
Thats right now. But that doesn't agree with what you posted earlier where you said the load factor was less in a descending turn than a climbing turn.
If it was, what is the extra force doing? (Forces create acceleration)
If it was, what is the extra force doing? (Forces create acceleration)
I think that there is a little ambiguity here about the definition of wing loading. I had always understood this to be weight/wing area, and what is being discussed here is load factor i.e. lift/weight and what a cockpit mounted normal accelerometer will read.
The way to think about this is to consider the 4 forces on an aircraft (lift, weight, drag and thrust) and then resolve weight (and thrust although that is not relevant here) with respect to lift and drag vectors. In a level turn, load factor = 1/cosine bank angle (Tarq57, almost right! Sin 60 = Cos30 = 0.8661, sin30 = 0.5). In a wings level climb or descent, load factor = 1/cosine flightpath angle. For example, in a 45 deg climb or descent the aircraft will have a load factor of 0.7g. If we now consider an aircraft in a 45 deg bank turn and in a simultaneous 45 deg descent, load factor = 1/cos45 x 1/cos45 = 0.5g.
The trick with this is always to resolve the forces with respect to lift and drag and not with respect to the weight vector because that really confuses me!
The way to think about this is to consider the 4 forces on an aircraft (lift, weight, drag and thrust) and then resolve weight (and thrust although that is not relevant here) with respect to lift and drag vectors. In a level turn, load factor = 1/cosine bank angle (Tarq57, almost right! Sin 60 = Cos30 = 0.8661, sin30 = 0.5). In a wings level climb or descent, load factor = 1/cosine flightpath angle. For example, in a 45 deg climb or descent the aircraft will have a load factor of 0.7g. If we now consider an aircraft in a 45 deg bank turn and in a simultaneous 45 deg descent, load factor = 1/cos45 x 1/cos45 = 0.5g.
The trick with this is always to resolve the forces with respect to lift and drag and not with respect to the weight vector because that really confuses me!
Last edited by LOMCEVAK; 7th Feb 2016 at 16:23.
