Load factor = Lift / Weight

[gerardflyagain]

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

[Jhieminga]

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.

[Maoraigh1]

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.

[jonkster]

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.

[Crash one]

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” !

[rarelyathome]

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.

[BEagle]

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.

[PDR1]

No. You need to have another look at the vectors in wings level and banked flight.


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.

No - in turning flight the wing is producing more lift so it must be producing more drag, and more power is required to oppose that drag if you want to maintain a constant airspeed (you don't have to - that's the driver's choice). It's not compensating for the bank; it's just addressing the drag rise.

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...

PDR

[Crash one]

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.

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.

[rarelyathome]

............... OK.

[Right Hand Thread]

You can't just increase power* to change the lift...

You can actually.

Increase thrust and you increase airflow over any section on the wing within the propwash. Voila! An increase in lift! 😜

[jonkster]

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.

[Jim59]

Quote:Originally Posted by Crash one https://www.pprune.org/images/buttons/viewpost.gif (https://www.pprune.org/private-flying/615006-load-factor-lift-weight.html#post10300493)

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.

Not necessarily in my glider.

[Dutystude]

Not necessarily in my glider.

I think the discussion is on aerodynamic lift rather than thermal lift.

[Jim59]

I think the discussion is on aerodynamic lift rather than thermal lift.
Precisely. That's what keeps both aeroplanes and gliders in the air. Jongster was considering the thrust vector when climbing under engine power and the thrust is greater than the drag and is perfectly correct. In a glide (and aeroplanes can also glide even if not quite as well) one Is always flying 'downhill' with no engine thrust and only climbs due to the air it is in rising faster than it is sinking so it's a combination of aerodynamic lift and drag. Perhaps Jongster's point is also valid for a glide and weight and lift are equal only in true level flight.

[PDR1]

Precisely. That's what keeps both aeroplanes and gliders in the air. Jongster was considering the thrust vector when climbing under engine power and the thrust is greater than the drag and is perfectly correct.

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

[Jim59]

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.

[Dutystude]

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

Surely in a constant IAS climb Thrust is greater than Drag.

As you climb at a constant IAS, TAS increases. How does this happen if there is no resultant between Thrust and Drag?

[PDR1]

Surely in a constant IAS climb Thrust is greater than Drag.


Which is why I said:

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.


Yes, it does, but this is related to the change in air density as you climb, not the act of climbing per se and so isn't relevant to the discussion. I didn't mention IAS or TAS in my piece to avoid unnecessary complications to what is just a matter of vector arithmetic.


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"

PDR

[custardpsc]

Originally Posted by Dutystude https://www.pprune.org/images/buttons/viewpost.gif (https://www.pprune.org/private-flying/615006-load-factor-lift-weight.html#post10301633)Surely in a constant IAS climb Thrust is greater than Drag.
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.

[India Four Two]

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?

Certainly can, but I wouldn't want to fly it! The Convair XFY-1 Pogo:

https://cimg7.ibsrv.net/gimg/pprune.org-vbulletin/756x1024/8519611522_be281418de_b_68b5148719ec04147fa4cb7d987b22804bee ecd3.jpg

[Dutystude]

Originally Posted by DutystudeSurely in a constant IAS climb Thrust is greater than Drag.
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.

I am no aerodynamicist but I do have a passing acquaintance with Newton’s laws (as modified by Einstein), Vector Diagrams, the composition of the atmosphere and, what we used to call the Principles of Flight.

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

[Dutystude]

Certainly can, but I wouldn't want to fly it! The Convair XFY-1 Pogo:



Not sure I would either.

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.

[Jhieminga]

I don't know, there are six inclined planes at the front of this thing, all of them producing lift by accelerating a fluid medium.... ;)

‘Aeroplane’ means an engine-driven fixed-wing aircraft heavier than air, that is supported in flight by the dynamic reaction of the air against its wings.
Oh all right, the EASA definition agrees with you, I'll get my coat then...

[Dutystude]

I don't know, there are six inclined planes at the front of this thing, all of them producing lift by accelerating a fluid medium.... ;)


Oh all right, the EASA definition agrees with you, I'll get my coat then...

Ah, nice one ‘minga.

You are right. Just like a Rotary Aircraft those propeller blades (Inclined planes) are producing a force at right angels to the Relative Airflow over the blades: Lift by standard definition.

Indeed in any propeller driven aircraft the propeller produces a force at Rt Angles to the Relative Airflow over the propeller blades. But I will continue, for clarity at least, to call this force thrust rather than lift.

For me Rotary Aircraft are Rotary Aircraft not Aeroplanes. But I don’t mind what the general vernacular is.

[PDR1]

A picture tells a thousand words, so three pictures tells a short novel! I've had time to do some sketches to show what I'm saying:

https://cimg3.ibsrv.net/gimg/pprune.org-vbulletin/284x280/s_l_d65b08abdd0788b1ed8c2b6c5033dd8c56ed4624.png

https://cimg6.ibsrv.net/gimg/pprune.org-vbulletin/463x315/slow_0a092d788fd8ca803836d90eab898afb67f7aab9.png


https://cimg5.ibsrv.net/gimg/pprune.org-vbulletin/374x368/climb_823a14b9f30188de2b003b17387c8925ab35983d.png


I'm slightly mystified as to why Dutystude introduced IAS/TAS issues into this - unless I'm missing something it's irrelevant to the subject. I can't think of any rule of law that I've infringed. If I have please point me at it!

PDR

[PDR1]

Just to complete the set, here's the one for gliding:

https://cimg8.ibsrv.net/gimg/pprune.org-vbulletin/515x360/glide_03bde4a25c73bd66d3bad2c06f1c242c38eb7d78.png

PDR

[Dutystude]

Hello PDR,

First, if I might steal a line from Shakespeare’s Mark Anthony: “I speak not to disprove what PDR has spoke but to say what I do know.” I am enjoying the intellectual exchange and not looking for a fight.

What I know from many years of observation is this: if I fly my ac at sea level at 70 KIAS my TAS will be 70KTAS. If I apply full power and maintain 70KIAS in the climb when I pass 10000 ft (as if) my TAS will be 80KTAS or thereabouts.

This is what I know. The why bit has been of no practical use to me as a pilot but, perforce, I have been a flying instructor a number of times and been required to offer a plausible explanation to the moe inquisitive Tyros that I have had the pleasure to teach.

You ask why I bang on about IAS/TAS.

Well, to begin with Indicated Air Speed is not actually a Speed except at sea level in ISA conditions. Better to remember that what you see on the ASI is Dynamic Pressure delivered via the pitot tube. The ASI could equally be gauged in inches of Mercury or Hecto Pascals. But, I’m quite pleased that they chose MPH/Kts.

Dynamic Pressure, in simple terms, is composed of the speed of the air and the density of the air. This speed is indeed a speed. It is the speed of the air in our ‘frame of reference’ – Planet Earth. It is the speed that the ac is moving in the surrounding air and, by convention, we call it True Air Speed (TAS). It is the speed used in Kinetic Energy, Momentum, acceleration and Newton’s Laws.

So for we pilots IAS (dynamic pressure) depends on TAS (speed through the air) and air density. If the density reduces (unavoidable in a climb) IAS will reduce.

But it doesn’t, we know that. But what we forget (because, for most it is of no practical value) while we maintain IAS, TAS is increasing (The ac is accelerating ) to balance the drop in air density.

The ac is not accelerating under the influence of air density. In accordance with Newton’s second Law the ac can only accelerate if acted upon by an external force – density is not a force.

As to vector diagrams:

In level balanced un-accelerated flight, Forces on the ac are not overcoming weight (gravity) they are balancing weight.

In a constant IAS climb the Forces on the ac are not balancing weight, they are overcoming weight, otherwise the ac would not move away from the centre of the Earth – climb that is. And, the Forces on the ac are accelerating the ac in the surrounding air – TAS increasing.

Some force is raising the ac in relation to the Earth’s surface; some force is accelerating the ac. If, a chosen vector diagram does not show this then I guess, as many suspect, it is just FM - Flipping Magic.

Which, when I have been in the mood, I have, sometimes, given as an explanation to many of the mysteries of flight.

[jonkster]

there must be an initial acceleration when the aircraft enters a climb from S&L.

To cause that change requires unbalanced forces. It could be done by increasing power or changing angle of attack (or a combination).

This unbalance changes the velocity of the aircraft (ie causes an acceleration) from a level direction to one angled upwards.

In the 'steady' climb that (hopefully) results* after this change all the forces will again end up balanced, with the magnitude of the lift force being less than weight, hence my original statement, lift can be less than weight in some modes of flight).

*ok, in real life we cannot sustain a steady climb forever but for small climbs of a few thousand feet, we should be able to approximate reasonably closely a steady climb.

[PDR1]

with the magnitude of the lift force being less than weight

The second vector diagram shows that it isn't, but it's certainly a commonly-held misunderstanding.

PDR

[keith williams]

PDR1

In your second diagram the aircraft is climbing at constant speed and constant roc, so the forces must be in equilibrium. You show the thrust and drag as being in equilibrium, but we need to look a bit more closely at lift and weight

Let us start by sliding the lift arrow down and to the left until it's point touches the tail of the weight arrow. We now have two sides of a triangle of forces. For these forces to be in equilibrium, there must be a third force arrow joining the point of the weight arrow to the tail of the lift arrow.

You have shown lift being greater than weight, which means that lift is the hypotenuse and the third force must be at right angles to the weight. This means that we must have a horizontal force pointing to the left. What is this force?

If you are wrong and lift is less than weight, then the weight is the hypotenuse. In this case the third force arrow is acting up the flight path and meets the lift arrow at right angles. In conventional texts this is the excess thrust.

Now let's see where your arguments lead us if we have a very high-powered aircraft such as a modern fighter jet. If your diagram is correct it means that as we gradually increase the climb angle, the lift force must increase. When the climb angle is 90 degrees in a vertical climb, the lift force must be infinite. But no wing can produce infinite lift, so logically if your diagram is correct, no aircraft could ever achieve a vertical climb.

So, is it impossible to achieve a vertical climb, or are you mistaken in your arguments?

[Dutystude]

https://cimg8.ibsrv.net/gimg/pprune.org-vbulletin/2000x1379/adfa49e7_e53b_4b7e_beba_6219fb48d3fb_de35851273cf9963bf65fe0 c90f5f3176a2a6cf8.jpeg


In this diagram weight has been resolved into 2 component vectors at right angles; the long side equal and opposite to the Lift vector. You will notice that in this vector triangle Weight is the hypotenuse of the triangle and therefore, by observation, is longer (so greater) than the vector that is equal and opposite to the Lit vector.

QED: In a climb Lift is less than weight.

[jonkster]

The second vector diagram shows that it isn't, but it's certainly a commonly-held misunderstanding.

PDR

Your diagram #2 states T=D In that case there would be no net force along the flightpath coming from thrust or drag, so those 2 effectively cancel each other out - correct?.

If so we then only need to ensure lift and weight cancel each other for a steady motion.

W is shown with a rearwards component to the flightpath direction - is that correct?

L is also shown acting with a rearwards component to the flightpath - is that correct?

In that case don't we have an unbalanced set of forces with a net force acting rearwards to the flightpath?

This then cannot be a steady motion could it? - your diagram is showing a situation where the aircraft would be de-accelerating.

ie diagram 2 shows an unbalanced set of forces where there would be an acceleration slowing the aircraft, not a steady, unaccelerated climb.



I have a sneaking suspicion you know this and are wanting to see how far you can pull the wool :)

[Crash one]

I don’t think these vector diagrams are correct when they have the weight always pointing straight down. That situation is only possible when the weight is stationary.
if the aircraft is moving the weight is acting against that movement, along with the aerodynamic drag. (Modified by gravity of course)
so the W vector should be angled rearward by the addition of an inertia vector.
Taking all these vectors as they are usually displayed, as if the aircraft were stationary, Then remove the lift, thrust and drag, but leave the fact that the aircraft is moving, such as would happen if it left the atmosphere, then the aircraft would not just fall vertically down the W vector, it would continue forward/downward on a curved path as the inertia vector decayed to zero and replaced by W.

[PDR1]

PDR1

In your second diagram the aircraft is climbing at constant speed and constant roc, so the forces must be in equilibrium. You show the thrust and drag as being in equilibrium, but we need to look a bit more closely at lift and weight [...]

This made me pause and think, and you're quite right. I fell into the trap of doing a quick sketch containing the details I thought were relevant rather than doing an actual vector diagram which (as you point out) would need to show an equilibrium. If I had actually drawn the proper diagram I would have filled in the extra items which immediately show the obvious conclusion - that I've been talking out of my arse!

No excuses - saw something which made me jump to a particular conclusion but completely failed to do a proper analysis before spouting off. Please accept my apologies.

PDR

[jonkster]

no apologies required, I regularly talk out my arse so know the feeling :)

[India Four Two]

In the course of my life I have often had to eat my words, and I must confess that I have always found it a wholesome diet.

W.S. Churchill

PDR and Jonkster, you are in highly esteemed company! ;)

[Jim59]

Before reading this discussion I had rather assumed that the 1G stalling speed of an aircraft is the same (when it is in steady, straight, unaccelerated flight) whether it be level, climbing or descending. I now realise that the 1G stalling speed is at a maximum in level flight and reduces if the aircraft is either climbing or descending because the wing loading reduces in the climb or descent.