Lift question

[overstress]

At what stage of flight is lift the greatest?. .. .(With regard to the CCDE final app q?)

[Genghis the Engineer]

The flare I'd think.. .. .G

[Silvertop]

I think that since the A/C was heavier on take off, the lift required would be greatest on rotation.

[Tinstaafl]

Ignoring accelerated flight, then I'd say S&L when done very soon after take-off.

It would depend on time spent climbing & climb gradient achieved.

My reasoning is that L<W in a climb or descent but L=W in S&L flight.

Maximum weight is at T/O but that's not necessarily when max lift occurs since the vertical component of T is partially supporting the a/c.

Next highest weight is S&L sometime after take-off. The shorter the period climbing the higher weight during the subsequent S&L so the greater the L being produced.

[oxford blue]

If you want a really informed and expert academic opinion on your question, I'd say post it <a href="http://www.oxfordaviation.net/forum/forum.asp?FORUM_ID=15" target="_blank">here</a> .. .. .You'll get the opinion of one of the ground school experts from one of the top schools in the country - or, indeed, the world.. .. .Give it a go and see.

[Skye Pilot]

INDEED! What are you like?

[low n' slow]

How about in a power off stall with all high lift devices deployed fully?. .If you consider situations where ground effect is included I would say flare at max landing mass as there is no thrust vector to give a helping hand.. ./lns

[Tinstaafl]

I think a distinction has to be made between accelerated & unaccelerated flight.

Accel. flight lets you bump up the quantity by whatever load factor you apply.

Unaccel. flight resolves about balanced forces ie a steady state.

If the flare - one example of accel. flight - is allowed then I'll bid for a simple, high 'g' manouevre eg high pitch rate or the steepest turn the a/c'd Thrust will allow. :D

[FlyingForFun]

I think a distinction needs to be made between lift, and coefficient of lift.. .. .Lift is greatest just after take-off. The main purpose of lift is to counteract weight, and weight is greatest just after take-off when there's maximum fuel. Any extra lift over and above the weight causes the aircraft to climb, which is exactly what it does just after take-off.. .. .Coefficient of lift is greatest at the point of the stall - this is a basic property of aerofoils. With regard to what phase of flight this relates to, I'd guess (although I'm not certain) that it would be during the flare, just prior to landing.. .. .FFF. .----------

[Tinstaafl]

Not correct, FFF.

An a/c climbs because it has thrust in excess to that being used to counter drag.

In a climb, T > D and L < W

[FlyingForFun]

Tinstaafl,. .. .You are quite right, I'm talking crap. I'm going back to study this some more.... .. .Sorry,. .. .FFF. .----------

[747FOCAL]

Lift is greatest at low altitudes.

[FlyingForFun]

Ok, I admit it, I've completely confused myself with the crap I was talking earlier. Can someone please tell me where I'm going wrong???. .. .I'm flying along, straight and level. Lift=weight, and thrust=drag, so according to Newton's 1st law, I'm not accelerating in any direction, which is great.. .. .Now, I add power (or thrust), and wait for everything to settle down. I'm now in a steady climb, at (approximately) the same speed as before - not accelerating in any direction.. .. .According to Newton, since I'm still not accelerating, lift=weight and thrust=drag (at least approximately - this won't be exact because the lift, thrust and drag will be at a small angle). But I know that thrust has increased (I increased it!), so that means that drag must have been increased too, right?. .. .But drag is 1/2*rho*V-squared*Coefficient-of-Drag. Rho hasn't changed. V-squared hasn't changed (very much). I haven't re-trimmed, so I'm still at the same angle-of-attack, so the coefficient of drag is the same. So it seems like the drag hasn't increased.. .. .But in this case, the forces wouldn't balance out, and I'd be accelerating.. .. .What have I got wrong here???. .. .I used to know this, but I've obviously forgotten! <img border="0" title="" alt="[Big Grin]" src="biggrin.gif" /> . .. .Thanks. .. .FFF. .---------

[dolly737]

@FFF:. .. .1st Error:. .“...I'm flying along, straight and level. Lift=weight, and thrust=drag...”. .This is a widespread myth. Because of wing & fuselage moments plus CG location effects, a downward elevator-force is required to achieve some kind of stability. Thus lift in s&l unaccelerated flight is always greater than weight.. .. .We have to differentiate quite precisely between the force created by the wing (lift?) and all the other forces acting upward and/or perpendicular to flight path, the sum of which support the vertical component of the total flight vector.. .. .Furthermore, it is this “small angle”, that makes a whole lot of difference in climbs and descents.. .. .To answer your confusion: The only thing you forgot in your calculation ref. drag is weight, which, in a climb, acts partially rearward, thus increasing the total force acting “against” the flight path.. .. .To be precise, during power-off descent, wing-generated lift is greater than during climb.... .. .happy landings

[Whipping Boy's SATCO]

FFF, if you study Newton too much I think you will find that you are always accelerating. Weight is a factor of mass and gravity with gravity being measured as a rate of change of velocity (acceleration). Therefore, to stay at the same altitude, you must have a constant acceleration towords the Earth's centre. Consequently, if you are in a steady climb or descent, your acceleration towards the Earth's centre will be difficult. Confused? . .. .My vote is with the simple explanation whereupon, unless in S&L flight, the forces of thrust & drag will always contribute towards 'lift' and 'weight'.. . . . <small>[ 26 March 2002, 20:14: Message edited by: Whipping Boy's SATCO ]</small>

[FlyingForFun]

Thanks dolly - spot on. In case anyone else is interested, I also found an excellent diagram of this in Trevor Thom, volume 4, in the chapter on climbs.. .. .I want to apologise to everyone for the huge amount of drivel I've been spouting on this thread. One of the great things about PPRuNe is that you can talk about any subject you like, safe in the knowledge that if you talk crap someone will pick you up on it before everyone else starts believe it! <img border="0" title="" alt="[Eek!]" src="eek.gif" /> I'll try to restrict myself to commenting on things that I know something about in future... but no guarantees! <img border="0" title="" alt="[Big Grin]" src="biggrin.gif" /> <img border="0" title="" alt="[Big Grin]" src="biggrin.gif" /> <img border="0" title="" alt="[Big Grin]" src="biggrin.gif" /> . .. .FFF

[Miles Magister]

This all seems a bit in depth for us more simple pilots. Being classically trained on the JP fun jet, I would have answered at 15deg AoA!

[overstress]

I've found the answer. Thanks for all the entertaining replies everyone!

Looking back in my ATPL groundschool & CFS notes (and a littlle bird reminded me) I find that:

Rotation:

Due to body angle and wheel contact, (lift generation begins at nosewheel unstick), thrust assists lift here.

Climb:

As at unstick, there is always a component of thrust 'assisting' lift. Therefore lift cannot be greatest in the climb.

Descent:

There is always a component of gravity present, therefore lift cannot be at a max here either.

Manoeuvre:

There is always a component of thrust or gravity present here (eg a loop)

This leaves the cruise, assumed to be S&L.

In S&L flight, lift=weight. (as thrust=drag)

Weight in the cruise is always at a maximum at the beginning of this phase, due to fuel burn (assuming no Air-Air refueling)

Therefore lift is at a maximum at the instant you level off in the cruise.

Tinstaafl - you were spot on thanks

[englishal]

lift is greatest at the wings max angle of attack just before the stall, regardless of speed, attitude, altitude, etc. You could be decending but at a high AoA therefore the wing is generating maximum lift possible. Equally you could be doing 150 kts, whack the stick back and stall the aircraft, in which case there is no lift being generated by the wing.........

Therefore it is 'most likely' that max lift will be generated at rotation (and subsequent climb out), and during the final flare, when the wings AoA will be at maximum (or should be in the ideal world).

Rgds
EA

[rubik101]

At Rotation. The thrust vector at this point is accelerating the aircraft forward, the only thing getting you airborne at rotation is lift.

[FlyingForFun]

EnglishAl,

At the risk of being completely wrong again... I don't agree with your last post. Lift isn't necessarilly greatest just before the stall, but coefficient of lift is greatest. If you enter the stall by reducing power and attempting to maintain altitude, the lift will be equal to the weight just prior to the stall, as for straight+level. If you enter the stall by pulling back very sharply from cruise speed, then yes, lift might be very high.

Also, during the stall, the wing still produces lift - it just can't produce as much. If there was no lift at all, a power off stall would effectively be a 0g dive, and you'd be weightless.

I've already demonstrated that this is an area in which I don't know as much as I should, though - so I'll leave it for others to either expand on my points, or correct them! :D

FFF
----------

[englishal]

If you slow down and keep pulling back maintaining altitude, you are increasing the AoA of the wing and hence the lift produced by the wing. Remember that the actual force-of-lift in Kg or N will not be as great becasue the reletive airflow is slower, but still just prior to stall the wing is producing maximum lift that it can (for that airspeed). So it is better to say that at whatever airspeed, the wing creates maximum lift at AoA just prior to stalling (roughly 16°). If the airflow is moving faster, the actual amount of lift in newtons or Kg will be more...but then this will be a pointless question as the 'amount of lift (in Kg)' will increase with airspeed indeffinitely (theoretically)

If you enter the stall by reducing power and attempting to maintain altitude, the lift will be equal to the weight just prior to the stall

This statement is true, but the wing is still producing the maximum possible lift based upon reletive airspeed.

It all boils down to AoA, or reletive airflow over the foil. If the yoke is yanked back at high speed, assuming the wings don't fall off then the AoA of the wing may exceed max, and the wing can stall even though the airspeed is high....This is why there is a manoeuvering speed or max abrupt control speed. Below this speed the wing will stall prior to causing structural damage, above this speed nothing is guaranteed !

Remember the aircraft can stall at any airspeed, and in any attitude.

Cheers
EA

[overstress]

Englishal

The question was when is LIFT greatest, not LIFT COEFFICIENT (CL) which you are getting confused with

Also you say <but then this will be a pointless question as the 'amount of lift (in Kg)' will increase with airspeed indeffinitely (theoretically)>

In straight&level flight, lift always=weight. As speed increases, the pilot decreases the aoa therefore decreasing CL - lift stays the same unless the weight changes

Lift=1/2*CL*density*vsquared



rubik 101

At rotation, you are still on the ground as the nosewheel lifts.
Therefore the weight is supported by the wheels. Think of a see-saw. Lift is not doing anything, in fact due to tailplane downforce, there can be more force on the wheels during rotate than when the nosewheel is on the ground.

At unstick, if that is what you mean, there is a considerable vertical component of thrust assisting lift, as I explained above, therefore lift is not at the maximum for that weight.

(eg if you rotate to 20degrees and assume the thrust is parallel to the a/c axis, the vertical component of thrust is sin20 which according to my trusty casio is 0.342 - so over a third of the upward force is provided by the engines!)

The correct answer stands as per Tinstaafl's post.

[englishal]

I think you're getting confused with the Coefficient of lift and a numerical value. Are you talking about the maximum upward force of lift in Kg? LIFT (available) is always greatest at Max AoA of the wing. Sure you decrease AoA with airspeed to stay SaL, but the question is 'When is the lift greatest'. With reduced AoA the Lift produced at cruising airspeed is the same as with a high AoA at slower airspeed (still SaL). Therfore the lift force (in Kg) will be greatest at high AoA and high airspeed.

Imagine you are cruising at 2000' at 60 kts, AoA is at 16° and the LIFT produced is say 1100Kg (equal to aircraft weight whatever that is, so you're flying SaL). Now you accellerate to 100Kts but keep AoA still at 16°, now the LIFT produced is say 2000Kg, and you'll climb. Now accellerate to 120 Kts, and the LIFT produced could be 2300 Kg and you'll climb faster.

Therefore the wing will ALWAYS produce most lift force (in Kg) that it can (at that altitude and that airspeed) at max AoA. If you travel faster it produces more lift for the same AoA, if you travel slower it produces less lift for same AoA. This is the same for all altitudes, air densities,temperatures etc. Of course at higher altitude the wing will not be capable of producing as much lift force (in Kg).

Lift=1/2*CL*density*vsquared

This formula should read 'Max possible lift force in Kg generated by the wing' and if you look at it, it says that lift (force in Kg) will be greatest with max coefficient of lift (ie max AoA, around 16°), in most dense air, and at highest velocity. This phase of flight is normally straight after rotation, (ignoring ground effect) when you're climing out at best rate of climb speed. You are attaining max forward speed that you can (for that AoA) and AoA is very near max, with density also at a max.


Cheers
EA

[overstress]

Sorry mate & at the risk of starting to sound boring, the question I asked was (at the start of this thread) when is lift greatest.

Not CL!

Cl varies with aoa, not lift

Lift = weight in S&L flight

Forget max aoa etc, it's a red herring

Please don't question the lift equation I quoted because a chap called Bernoulli (who was cleverer than I will ever be) invented it.

In your example you forget that as soon as the ac starts to climb, there is a vertical component of thrust which is helping it go up

May I suggest the excellent description at the Nasa website which shows the diagrams I am referring to

here (http://www.lerc.nasa.gov/WWW/K-12/airplane/climb.html)

You will see that as you incline the aircraft in its climb, the lift vector rotates away from the vertical and it actually becomes LESS than the weight.

The ultimate example of what I mean is the rocket - it goes up but there ain't a whole lot of lift generated initially!

The answer isn't as simple as you think!

PS -please no more replies I've had enough

[bookworm]

"PS -please no more replies I've had enough"

Oh but that's not how this game is played! :)

I think your answer is still open to considerable debate. You dismiss rotation on the basis of assistance from a component of thrust. That may or may not be correct depending on the rate at which climb is established.

More importantly, how about re-establishing climb after a level off early in the flight? That takes an acceleration upwards, and the the thrust component is minimal.

[rubik101]

An A/C will generate lift as soon as the wing is rotated to a positive angle, so nose wheel lift off is the moment of first lift generation. Seconds later the aircraft leaves the ground at its highest weight for the flight. If you then closed the throttles the aircraft would be generating its maximum lift. Re-applying the power makes no difference to the total lift generated at the moment of unstick. Think about it!
All the pwer does is accelerate and contribute but it doesn't affect the maximum lift generated by the wing.

[englishal]

Hmm, I don't want to sound boring but.....

I'm not questioning Mr Bernoulli, I know he's right, I'm just trying to clarify what he's actually saying.

Think you're getting mixed up with a 'Vertical force' and lift. A wing on a Pitts in a vertical dive is generating lift, which acts in a horizontal direction (unless the pilot puts the nose a few degrees beyond vertical to reduce AoA to zero), a glider pulling out of a dive generates Lift (in excess of the weight). A glider climbing in a thermal generates lift in excess of weight due to increase in AoA due to change in reletive airflow over the wing. A rocket doesn't rely on Bernoulli's principles but Newtons 3rd laws of motion, in that "every action has an equal and opposite reaction", so strickly speaking doesn't generate any lift at all, but pumps gas out of its arse at high speed which send the rocket up and away.

CL is proportional to AoA, and as Mr B pointed out, Lift is directly proportional to CL. Therefore, when CL is at max (when AoA is max) then Lift will be at max. The only way to increase lift now is to increase airspeed (may not be possible) or density (impossible).

Therefore LIFT will be greatest when CL is at max and hence when AoA is at max.

"Combined vertical acting force" will be greatest when the aircraft is climbing at the fastest rate of climb that it can. In normal opertations this would be best rate of climb speed (and hence BROC AoA...probably not far from max).

Cheers
EA
:)

[overstress]

Englishal

Best roc speed is dependent on many factors but the principal one is specific excess power SEP=(T-D)*V

The curves look vastly different for a piston (which I suspect is the type you have flown) and a jet.

In a jet, there is a broad range of speeds in which max roc is obtained. (In a piston, there is rarely much SEP!)

These jet speeds are nowhere near Cl max!

I restate the original question, which you have chosen not to answer:

When is lift greatest in flight (not in theory, or in a wind tunnel test)
Oh btw your reply - if you re-read what you have put, you will see that you have agreed with me (and NASA)! Tricky subject, isn't it?


Bookworm

Welcome to the debate

<You dismiss rotation on the basis of assistance from a component of thrust. That may or may not be correct depending on the rate at which climb is established>

During rotation and until unstick, although there is lift being generated, it is less than weight

At UNSTICK, the ac has already pitched-up a few degrees and therefore the lift vector is not the only thing contributing due to simple geometry. The rate at which this happens is irrelevant. So the lift is not at a maximum here.

In s & L flight, L=W. No matter what speed you are going.

So L is a max when W is a max, ie at the start of this phase.

<More importantly, how about re-establishing climb after a level off early in the flight? That takes an acceleration upwards, and the the thrust component is minimal.>

After a level-off, you will continue burning fuel, diminishing weight & therefore lift required. To establish a further climb requires an increase in SEP from zero. As the ac is rotated into the climb FROM THE CRUISE the thrust vector inclines as before and geometry takes its part again. The acceleration upwards is provided partly by the rotated lift vector and partly by a component of thrust.

To maintain IAS, there must be an increase in SEP, provided by an increase in thrust.

[overstress]

Englishal

I've just re-read your theory on the Pitts in a dive

You neglect to consider the influence of gravity

A Pitts, Hunter or MiG29 in a dive may well be experiencing slightly less than 1g. If weight reduces, so does lift. At zero g, weight is zero and so is lift. Fighter pilots use this effect to 'bug-out' from a fight,as there is no lift at this point, there is no lift-induced drag, hence faster acceleration

A lot of textbooks will tell you this!

[englishal]

Hello BIK_116.80,

I would be grateful if you could tell me why my theory is so flawed?

Overstress....

Look, I'll make this my last post as you clearly believe I am wrong :p .....I have looked at the NASA website, and I can see where you are coming from and I can also see what I am trying to say. Ignoring any additional forces from engines etc, Lift is the force generated by the wing due to Bernoulli's principal, it acts at 90° to the reletive airflow. This is what lift is.

If a Pitts is in a vertical dive, ie. nose pointing straight at the ground and accellerating, the wings are still at a few degrees AoA, with airflow over the wing. Therefore Lift will be generated at 90° to the reletive airflow, which in this case will be in a horizontal direction. Remember reletive airflow is defined as the angle the airflow makes with the chord of the wing. This will cause the aircraft to track horizontally across the ground. To overcome this the pilot will put the nose a few degrees beyond the vertical to reduce AoA of the wing to about -4°, which casues CL to be zero and hence lift to be zero. Now the aircraft is falling (with style!) due to the effects of gravity and its engine. The only thing that will stop it accellerating forever is the effect of induced drag (and imminent contact with the earth).

When is lift greatest in flight (not in theory, or in a wind tunnel test)

To answer your ambiguous question, Anytime the AoA of the wing is near max, be it in a decent, a climb, S&L.....the wing is then generating the maximum possible lift that it can

Its not just me who believes this, as Trevor Thom puts it:-

An aerofoil has the greatest lifting ability at high AoA, just prior to the stalling AoA....Lift is the total reaction at right angles to the reletive airflow....

Cheers
EA:)

[bookworm]

<More importantly, how about re-establishing climb after a level off early in the flight? That takes an acceleration upwards, and the the thrust component is minimal.>

After a level-off, you will continue burning fuel, diminishing weight & therefore lift required. To establish a further climb requires an increase in SEP from zero. As the ac is rotated into the climb FROM THE CRUISE the thrust vector inclines as before and geometry takes its part again. The acceleration upwards is provided partly by the rotated lift vector and partly by a component of thrust.


I agree with that as far as it goes but look at the magnitudes involved. If you enter a 6 degree climb, your lift requirement in the climb reduces by about 0.005 g (= 1 - cos 6 deg). By contrast, if you establish a 1800 fpm climb over 10 seconds, the average acceleration in that manoeuvre is 0.1 g, so you need a lift of 1.1 g. That's 20 times more significant than the change enforced by geometry.

The maximum lift is likely to be at the moment when the rate of climb is increasing fastest. The geometry isn't very important.

[overstress]

Englishal: Thank you for your kind reminders re the geometry of lift, it saves me from recalling it!

<To answer your ambiguous question, Anytime the AoA of the wing is near max, be it in a decent, a climb, S&L.....the wing is then generating the maximum possible lift that it can >

In S & L Lift = weight

Sorry to upset your theory, but lift is the same in S&L at the stall or at max speed. You constantly confuse CL with lift, they are not the same thing.

CL is at a max just prior to the stall in a conventional wing. I do not dispute that! The subtlety of the question has evaded you!

Bookworm:

Not sure I've heard it described this way before, ie referring such factors to 'g'. The tailplane rotates the ac about its c of g into a climbing attitude whereupon the lift vector has been rotated 'backwards' by the same amount. Sin6 degrees is about 0.1 so 10% of the upwards force is due to thrust, 90% due to lift

[bookworm]

Not sure I've heard it described this way before, ie referring such factors to 'g'. The tailplane rotates the ac about its c of g into a climbing attitude whereupon the lift vector has been rotated 'backwards' by the same amount. Sin6 degrees is about 0.1 so 10% of the upwards force is due to thrust, 90% due to lift


If you don't like g you can do the sums in pounds. Just multiply by the mass of the aircraft!

Your picture is not quite correct. The upward component of thrust is 10% of its magnitude. Don't forget that drag also tilts downwards. The easiest way to resolve the forces is parallel to and perpendicular to the flight path, in which case only the weight rotates.

[overstress]

Bookworm:

Yes I know drag rotates but I was trying to keep drag out of it. Pounds, g or whatever, we're not interested in numerical values. This explanation of yours makes no sense to me.

You can resolve forces however you like. However, it is more instructive in this argument to resolve them horizontally & vertically (as englishal has so ably pointed out, lift is perpendicular to the flightpath). If you don't resolve them this way, you will never be able to see how those forces come about. Have another look at the NASA link to see what I mean.
Specifically,
Nasa climb diag (http://www.grc.nasa.gov/WWW/K-12/airplane/climb.html)

[Flash2001]

Bik

Force is never measured in kilograms. It can be Newtons, pounds, even poundals, sometimes kips some but never in units of mass.

[Flash2001]

Delete "some" from my previous.