Load factor = Lift / Weight
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From: near an aeroplane
You can't just increase power to change the lift, unless you're in a helicopter of course. Think about the elements in the lift formula, that's what you've got to work with. Also, the weight at that point is a perceived 'weight' composed of your weight vector and the centrifugal force. Changing your power to increase speed for example, will also increase the centrifugal force, so you'll never get to one again.
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From: Moray,Scotland,U.K.
I don't understand the original statement.
Assuming you're in a 2g turn, the only way to reduce the g is to make it a descending turn,
If you're in a 2g level turn, the wings are producing 2*AUW lift.
To produce the extra lift, the angle of attack must increase compared to straight and level at that airspeed.
To maintain airspeed with that increased angle of attack will need more power.
Assuming you're in a 2g turn, the only way to reduce the g is to make it a descending turn,
If you're in a 2g level turn, the wings are producing 2*AUW lift.
To produce the extra lift, the angle of attack must increase compared to straight and level at that airspeed.
To maintain airspeed with that increased angle of attack will need more power.
Last edited by Maoraigh1; 2nd November 2018 at 20:23. Reason: Add
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From: Sydney
If lift = "1" and weight = "2" wouldn't the load factor be 0.5 not "2"? ie lift is less than weight so the aircraft would be be feeling 0.5G?
In a stable turn, the weight (say "1" unit) is obviously constant however the lift is now pulling at an angle to it.
The angled lift is being used to cause the aircraft to turn. If we maintain the same "1" level flight unit of total lift, angling it sideways means lift no longer balances the weight force.
To balance weight in a banked turn, we will need to increase the total lift from the wing so the component acting directly away from the earth (and so neutralising weight) equals "1" unit.
The wing may therefore need to be be producing say "2" units of total lift (depending how steeply we bank the aircraft) - part of this extra is acting to exactly neutralise weight, the rest is causing the aircraft to turn by accelerating the aircraft constantly towards the centre of the turn.
The only way you could get a 1G turn would be to turn without banking (skid it with rudder?) or somehow otherwise get some other force than lift to neutralise your weight
You in the cockpit feel the "2" unit lift force as an acceleration and you feel pushed into the seat with twice your normal weight.
In a stable turn, the weight (say "1" unit) is obviously constant however the lift is now pulling at an angle to it.
The angled lift is being used to cause the aircraft to turn. If we maintain the same "1" level flight unit of total lift, angling it sideways means lift no longer balances the weight force.
To balance weight in a banked turn, we will need to increase the total lift from the wing so the component acting directly away from the earth (and so neutralising weight) equals "1" unit.
The wing may therefore need to be be producing say "2" units of total lift (depending how steeply we bank the aircraft) - part of this extra is acting to exactly neutralise weight, the rest is causing the aircraft to turn by accelerating the aircraft constantly towards the centre of the turn.
The only way you could get a 1G turn would be to turn without banking (skid it with rudder?) or somehow otherwise get some other force than lift to neutralise your weight
You in the cockpit feel the "2" unit lift force as an acceleration and you feel pushed into the seat with twice your normal weight.
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From: Scotland
Shirley you cannot be in a level turn producing lift =1 and weight =2. In order to fly at all, lift must equal or exceed weight. Except perhaps in a ballistic situation though even then lift =0 and weight (g) =0 “temporarily” !
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From: Temporarily Unsure!
No. You need to have another look at the vectors in wings level and banked flight. With the lift vector tilted you need more thrust to compensate.
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From: Quite near 'An aerodrome somewhere in England'
In level flight, L=W
In a turn, the component of lift which balances W is (L cosAoB). This means that L must be increased; if TAS remains constant, then the only way of doing this is to increase the lift coefficient by increasing the angle of attack.
An increase in angle of attack also cause an increase in drag coefficient, which will try to decelerate the aircraft. To oppose this drag increase, thrust must be increased to maintain TAS at the desired value.
In a turn, the component of lift which balances W is (L cosAoB). This means that L must be increased; if TAS remains constant, then the only way of doing this is to increase the lift coefficient by increasing the angle of attack.
An increase in angle of attack also cause an increase in drag coefficient, which will try to decelerate the aircraft. To oppose this drag increase, thrust must be increased to maintain TAS at the desired value.
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From: Scotland
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From: UK
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From: Sydney
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From: Luton
Quote:
Originally Posted by Crash one 
No I wasn’t talking about turning, I said “in order to fly at all, lift must equal or exceed weight”. How you apply vectors and numbers has nothing to do with that statement.
No I wasn’t talking about turning, I said “in order to fly at all, lift must equal or exceed weight”. How you apply vectors and numbers has nothing to do with that statement.
lift is less than weight in a stable climb.
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From: Luton
I think the discussion is on aerodynamic lift rather than thermal lift.
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From: Luton
If we take it to the limit when an aeroplane can generate more thrust than its weight then can it can enter a stable vertical climb with zero lift? I guess it still has weight. As it approaches this limit it will be climbing mainly due to thrust but only need a small amount of lift. Thrust is not balancing drag it is greatly exceeding it.
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From: UK
Umm...no it isn't, because whilst in a climb there is a vertical component of thrust which opposes weight, there is also a vertical component of drag which adds to the weight. Thrust and drag are equal and opposite, so these two vertical components cancel out. I'm aware that the "lift is less in a stable climb" myth is taught to PPLs, but it's by no means the only bit of basic aeronautical theory taught to PPLs doesn't really stand scrutiny.
But even this simplistic view doesn't really cover it. The point becomes obvious if you draw out the force diagram. Weight always acts vertically downwards. Lift always acts at right angles to the airflow (it has to - it's a hydrostatic pressure effect) and drag always acts parallel to the airflow (because it's a dynamic pressure effect) so lift and drag always act at rightangles to eachother. Thrust acts in the direction which it is pointed - for simplicity lets assume that the designer made a decision to bolt the thrust-maker so that it's alighed with the direction of the drag vector at cruise speed. So in stable straight&level flight (with respect to the surrounding air mass) at cruise speed we find lift,drag, weight and thrust all at right-angles to eachother. This is the picture you'll find in all the PPL textbooks.
What happens if we slow down, adjusting power as required to remain straight& level? Obviously lift, drag and weight will still be mutually at right angles, but to maintain height we need to trim back, so the thrust vector angled upwards. So thrust now has a horizontal component equal to drag PLUS a vertical component. So in slow S&L flight "wing lift" will be less than it is a cruise speed, because total lift must equal weight.
OK, accelerate back to cruise and retrim for S&L. Now add power and climb at 10 degrees. The "air velocity vector" is now angle 10 degrees upwards, so the drag vector points 10 degrees downwards, adding to the weight, but this is cancelled out by the trust vector being pointed upwards. Weight is still acting vertically, but that is now angled 10 degrees backwards compared to the velocity vector so thrust and weight are no longer at right angles. Therefore you will need to add thrust to allow the engine to do work against gravity (or you could view this as the horizontal component of lift in the climb - same thing) as well as just against drag to restore the equilibrium. Last of all we have lift - lift acts at right angles to the airflow velocity vector, so it is angled 10 degrees backwards. That means that it is no longer parallel to weight, so only a COMPONENT of the lift is opposing gravity. Therefore the total "wing-lift" must be GREATER so that (lift*cos(climb_angle)) is equal to weight.
NALOPKT(&EFGAS),
PDR
But even this simplistic view doesn't really cover it. The point becomes obvious if you draw out the force diagram. Weight always acts vertically downwards. Lift always acts at right angles to the airflow (it has to - it's a hydrostatic pressure effect) and drag always acts parallel to the airflow (because it's a dynamic pressure effect) so lift and drag always act at rightangles to eachother. Thrust acts in the direction which it is pointed - for simplicity lets assume that the designer made a decision to bolt the thrust-maker so that it's alighed with the direction of the drag vector at cruise speed. So in stable straight&level flight (with respect to the surrounding air mass) at cruise speed we find lift,drag, weight and thrust all at right-angles to eachother. This is the picture you'll find in all the PPL textbooks.
What happens if we slow down, adjusting power as required to remain straight& level? Obviously lift, drag and weight will still be mutually at right angles, but to maintain height we need to trim back, so the thrust vector angled upwards. So thrust now has a horizontal component equal to drag PLUS a vertical component. So in slow S&L flight "wing lift" will be less than it is a cruise speed, because total lift must equal weight.
OK, accelerate back to cruise and retrim for S&L. Now add power and climb at 10 degrees. The "air velocity vector" is now angle 10 degrees upwards, so the drag vector points 10 degrees downwards, adding to the weight, but this is cancelled out by the trust vector being pointed upwards. Weight is still acting vertically, but that is now angled 10 degrees backwards compared to the velocity vector so thrust and weight are no longer at right angles. Therefore you will need to add thrust to allow the engine to do work against gravity (or you could view this as the horizontal component of lift in the climb - same thing) as well as just against drag to restore the equilibrium. Last of all we have lift - lift acts at right angles to the airflow velocity vector, so it is angled 10 degrees backwards. That means that it is no longer parallel to weight, so only a COMPONENT of the lift is opposing gravity. Therefore the total "wing-lift" must be GREATER so that (lift*cos(climb_angle)) is equal to weight.
NALOPKT(&EFGAS),
PDR
As you climb at a constant IAS, TAS increases. How does this happen if there is no resultant between Thrust and Drag?
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From: london
Originally Posted by Dutystude
No, Thrust will equal drag. Newtons first law. If at rest or constant speed, sum of forces must be balanced. If one exceeded the other there would be acceleration.
Surely in a constant IAS climb Thrust is greater than Drag.
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From: Nanaimo (CAC8)
If we take it to the limit when an aeroplane can generate more thrust than its weight then can it can enter a stable vertical climb with zero lift?
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From: UK
PDR
I may have misread your post if you were saying the the Vertcal Components of Thrust and Drag are equal rather than Thrust and Drag are equal.
Of course the relationship between IAS and TAS is a function of air density. But Density is not a Force and plays no part in Newton’s Laws. If we are going to use vector mechanics we need to stick to the Rule of Law.
Custard
You have the First Law right. But in a 70 kt IAS climb the aicraft is not in a steady state but is, indeed, accelerating.
IAS is a red herring. Our ‘frame of reference’ for Newton is the Earth and the appropriate velocity is TAS not IAS.
To an observer on the ground, as you climb at 70 kt IAS you will be seen to accelerate. TAS will increase.
As you maintain a, commendably steady 70 kt IAS climb, to an observer in the aircraft looking out of the window, you will be seen to accelerate in relation to the ground. TAS will increase.
This is is real life. And, according to Newton’s Second Law the forces on the aircraft must be unbalanced.
In our vector diagram for the climb Thrust must be greater than Drag.
Or, if you like, Drag must be less than Thrust
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From: UK
Of course, in vertical flight, while this contraption is an still an Aircraft it is not an Aeroplane/Airplane. Not producing lift by accelerating a fluid over an inclined Plane.
Plane (AKA wing) as in Bi-plane, Mono-plane, Tri-plane.