Load factor = Lift / Weight
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Load factor = Lift / Weight
Say I'm in a turn where lift = 1 and weight = 2. Load factor is therefore 2g. Can I decrease it back to 1g by increasing power until lift = 2? i.e 2 / 2 = 1g
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.
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 Nov 2018 at 20:23. Reason: Add
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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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” !
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.
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.
Yes, you do. Weight due to gravity remains the same, regardless of whether the aeroplane is flying level, turning, climbing or diving and it always acts vertically.
In turning flight this weight must be reacted by the vertical component of lift (l*cos(bank_angle)).
In turning flight an additional force is required to produce the turn (m*v^2/r where r is the radius of turn and v is the airspeed at a tangent to the turn). This force is provided by the horizontal component of lift (l*sin(bank_angle)).
These two forces are at right-angles to eachother so the "weight" experienced by the driver is the vector sum of the two - as calculated using archie mede's famous squaws and hydes relationship. That defines the amount of lift the wing must produce in doing it, and that is the "g" the driver will experience.
With the lift vector tilted you need more thrust to compensate.
There is a wrinkle here because the vertical component of thrust has an effect, but if we assume this is a low-powered light aeroplane we can decide it's too small to worry about and ignore it. This comes up in the smartarse PPL-level question "is the AoA higher or lower in a steady state climb than it is in S&L flight at the same airspeed" because the intuitive answer would be that AoA must be higher to climb when the questioner will claim it could be lower because the lift required is the same and the vertical component of thrust adds to the lift. It's a "smartarse" question because it's more complex than that. If you draw out the vector diagram you see that while weight remains vertical, lift, trust and drag rotate by the climb angle so drag adds to weight and the lift vector is no longer parallel to the weight vector, negating the whole argument. But I digress...
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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.
I think the discussion is on aerodynamic lift rather than thermal lift.
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
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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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?
Which is why I said:
Originally Posted by PDR1
Therefore you will need to add thrust to allow the engine to do work against gravity
As you climb at a constant IAS, TAS increases.
How does this happen if there is no resultant between Thrust and Drag?
It is nothing to do with "the resultant between thrust and drag"
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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.