An aerodynamics question (for experts only)

[GoneWest]

I've tried this one before and got nowhere - maybe because it's all cr*p.

I've been looking through the books again following attendance at an instructor seminar - and am getting really confused with the "forces acting on an aircraft in flight".

Let's go for the easy option (or at least what I believe to be the easy option), let's take a light aircraft, in straight and level flight, all balanced - no acceleration, no climbing, nothing spectacular - just cruising along, drilling a hole in the sky. (first wanabee to send me £10 can log the time :D )

Trevor Thom, et al, are telling me that there are four forces acting on the aircraft - lift, weight, thrust and drag. And I've been using this since my trial lesson back in 1973.

Now, my FIC renewal has brought out the comments that we have been using the terms of the four forces for the sake of convenience but that actually there are only three forces. Weight, thrust and total reaction (TR). This TR can be broken down, by the scientists, to show a horizontal vector (not a force) which we call drag - and a vertical vector which we call lift.

I'm getting numerous arguments in favour of both camps. Anybody (who actually knows what they are talking about) care to comment? Please!

['%MAC']

"If anyone tries to tell you something about an aeroplane which is so damn complicated that you can't understand it you can take it from me it's all balls."

- R.J. Mitchell (designer of the Supermarine Spitfire)

[MasterGreen]

If you were a study of Richard Feynman et al, you would understand this concept. What it is saying is that weight and thrust are relatively constant and independant variables acting on the body.

However there is a resultant and balancing vector that is comprised of (classically) Lift and Drag.

However Drag is partially and variably dependant on Lift so it makes sense to resolve these two items as a single vector that is continuously opposing Weight and Thrust (in unaccelerated motion). Simply resolve the vector (at the instant in time that you are interested in) and you have the two (more conventional) items - Lift and Drag.

It is another way of looking at the same thing. But mathematically a single vector of these two co-dependant and variable effects is easier to manage.

Don't sweat this complex math stuff (unless you enjoy the challenge) when there is lots more to understand well before you come to this. From a human / physical perception there is no difference.

MG

[Checkboard]

Fine - tell 'em if they want to play that way that there are still four fources - weight, thrust, "total reaction" from the wing and drag from the airframe.:rolleyes:

Or tell 'em that you have resolved the forces of thrust and weight into one force, call it the "speed independant" force, and the lift and drag into another, so that there are only "two real forces" acting on the airframe... :rolleyes:

There was a discussion about this topic a while ago on the Instructors forum, but I can't find it now.

[MasterGreen]

Exactly - just being clever cloggs by and bye. What are we teaching ? Maths or Flight

MG

[mutt]

GoneWest,

Mr Boeing states that for level unaccelerated flight, thrust and drag must be equal and opposite, and the lift and weight must be equal and opposite according to the laws of motion, you then get:

T=D
L=W

He then goes on to justify it mathematically, resulting in……

L/ ä = 1481.351 CL M2 S
D/ ä = 1481.351 CD M2 S
W/ä = 1481.351 CL M2 S
T/ ä = 1481.351 CD M2 S


Mutt.

[BEagle]

An aerofoil presenting an angle of attack to the relative airflow will experience a single force.

That force is conveniently resolved in 2 mutually perpendicular directions; normal to the relative airflow to give lift and parallel to the relative airflow to give lift dependant drag.

Similarly, a propeller is also an aerofoil. It also experiences a force due to the angle of attack presented to the relative airflow resulting this time from both propeller rotation and forward motion of the aircraft. This force is also conveniently resolved into 2 directions, that normal to the plane of rotation gives thrust, that parallel to the plane of roatation gives 'prop drag' which is the force the engine must work against to maintain the rotational velocity.

[bookworm]

Is there any testable difference in validity between the 3-force model and the 4-force model?

You can slice and dice forces any way you want for convenience. It's convenient to differentiate thrust from the "total reaction", but they're both contributions to the aerodynamic force on the aircraft -- so should it be a 2-force model?

Which model you pick depends on what problem you're tring to solve at the time.

[OzExpat]

If we're getting pragmatic here, why don't you concentrate on the number of forces that a P of F exam will ask about? Isn't this the final arbitrator?

[GoneWest]

Some very useful replies - (and a couple of useless, :p ), thanks guys.

MasterGreen - I like your explanation. Made it a lot clearer.

Mutt - I think it was said by many, long before Mr Boeing came along.

BEagle - you have quoted the CFS style approach to the question ......

........which really brings us to the question posed by a few of you "Does it matter?"

A different subject - but (for OZExpat, and others of the same frame of mind) one of the downsides of flight instruction is that every instructor is convinced that he/she is correct. This then brings us to the point that the definition of correct will likely be a matter of opinion - between the various instructors who state their own ideal as being the only correct one. You can read any number of threads in these forums and everybody argues that their posting is the facts of the case - and that anybody who argues is just plain wrong.

So, if that P of F exam mentioned by OzExpat, is a verbal one - administered by an examiner who may decide on the future of your position in the aviation industry - then I would think it quite important to be able to give him/her the answer that he/she wants. If it is a question of researching the individual examiners, to learn their favoured answer, then this really highlights the problem that BEagle brought up about instructor renewals being done by means of a mandatory flight and ground test with an examiner.

In which case, %MAC, MasterGreen, Checkboard, bookworm and OzExpat - it certainly does matter.

What I don't agree with is that I've spent 20 years of listening to instructors and reading Trevor Thom etc - all who tell me there are four forces and now - the people that MATTER (as far as this discussion goes) are telling me that there are only three.

Man has been flying for just short of 100 years now - surely somebody has sussed it out.

[bookworm]

So, if that P of F exam mentioned by OzExpat, is a verbal one - administered by an examiner who may decide on the future of your position in the aviation industry - then I would think it quite important to be able to give him/her the answer that he/she wants...

In which case, %MAC, MasterGreen, Checkboard, bookworm and OzExpat - it certainly does matter.

If the point that you're making is that examiners sometimes don't have a full grasp of the physics behind the questions they ask and that you sometimes have to learn an 'official' simple answer to a much more complex question, then perhaps I'd agree with you. But if that's the case here, your question is mis-titled: you mean "for examiners only", not "for experts only". :)

Most of the good examiners I've met (most of them) are not so insecure about their ability in physics as to be frightened by a little debate about a complex issue.

[GoneWest]

bookworm - agree entirely.....but what I's saying is that it's the examiners who are now suggesting that they more than fully understand the physics, and that it's their three forces is correct and that the four forces stuff that we've all grown up with is "convenient - but wrong".

The last guy I flew with was talking about this very subject and said "There is a whole generation of pilots out there that have no idea what is making the aeroplane fly"

....and, as far as the examiners are concerned, they are the ones with the full understanding and the likes of (presumably) you and I - and all the others that subscribed to the four forces theory - are in that generation......and that it is us who do not understand the physics.

Postings such as BEagles (who I greatly respect [on most topics]) confirm that there are people who agree on the three forces. I just don't understand why it is suddenly an issue - but it is fast becoming one.

[twistedenginestarter]

An aerofoil presenting an angle of attack to the relative airflow will experience a single force.

Of course there are in fact millions of forces acting on an aircraft in flight.

What Physics does is to say can we pretend there is only one, or two or three etc and get the correct predictions (-ish) whilst making the mathematics simpler.

So the answer to your question depends on what you are trying to do. As an example if you want to explain the effects of propellor wash you are going to have to consider two lateral forces on the airframe, perhaps one at the centre of mass and one at the centre of the fin.

[Old Aero Guy]

If you are going to the trouble to talk about resultant forces on the airplane, the answer is there are none. The sum of all forces has to be zero or the airplane wouldn't be flying straight and level at constant speed.

[BEagle]

If all forces are in balance, the aeroplane will not accelerate. But it would continue to maintain both horizontal and vertical velocity until influenced by another force.

[andrewc]

There is obviously only one single reaction force T to
the movement of the plane, however it is convenient
to split it into its two equivalent components Tx & Ty
which are acting at right angles to one another and
hence can be placed in opposition to the Thrust &
Weight components so as to balance in straight and
level unaccelerated flight.

To think that the three force model or four force
models are describing different things is to completely
miss the point.

-- Andrew

[chrisN]

Sorry to be pedantic, and at risk of being flamed, but the "obviously only one single reaction force" is not, ISTM, correct. As pointed out by an earlier posting, there are really zillions of little aerodynamic forces acting all over the aeroplane - but for many purposes they can be treated mathematically as though they were a single reaction force, i.e. their "resultant", or they can resolved into two forces at rightangles, by convention called lift and drag. Neither is more right or wrong than the other, it is more a question (as yet another pointed out) of which one suits a particular purpose at the time.

Incidentally, there are also couples (twisting or turning forces) acting upon an aeroplane. Couples cannot be resolved into a single force, if I recall my maths. and engineering training correctly. For basic explanations of P of F these are usually ignored. (Any unbelievers, consider e.g. trailing edge vortices - created by apply a couple to the airflow - Newton's third law leads to the deduction that there is an equal and opposite couple experienced by the aircraft. A single prop creates another couple, in a different plane.)

[OzExpat]

When I was instructing, I had to face examiners for rating renewal and upgrades. There was always a contentious issue that happened to be that particular examiner's pet hobby horse. They would deliberately set that subject for the "long briefing".

When I hit the area of contention and addressed it in my own way, a "full and frank exchange of views" would occur. The fact is that, so long as you can explain - justify - why you've made a certain statement, you won't be failed. Well, not in my experience in Oz, anyway.

So, okay, you've been brought up to believe in four forces. Fine, so was I. I can still argue my case for that now. I can see how someone might want to believe in just three forces and that's also fine by me, so long as they can justify it. I think this is where most examiners are "coming from"... they want to know that you have an understanding of the subject and are not merely doing it parrot fashion.

I think our students have a right to expect that we can explain the particular statement or subject in any of several different ways. I'm not sure that a consideration of three forces makes it any easier for a student to understand because it's still necessary to explain how they are derived. In doing so, one has little choice other than to talk about FOUR forces.

In any event, a verbal P of F exam, which I admittedly hadn't considered in my first post, provides scope to justify your belief, whatever it happens to be.

[ShyTorque]

The RAF (CFS) teaching for rotary winged aircraft uses a vector diagram for the forces acting on a rotor blade. This shows the resultant of lift and drag on a blade segment as a the "Total Reaction". This is then broken down into "Rotor thrust" acting vertically up the rotor mast and "Rotor drag" acting perpendicularly to it. Rotor thrust lifts and propels the aircraft. Rotor drag is opposed by the application of engine Torque.

Similar thing to the prop theory given by Beags. It has to be done for rotating things because the lift and drag on a rotor blade don't act coincidentally with the rotor axis. I don't think it's necessary to do it for plank wings as it over-complicates the issue.

[andrewc]

chrisN, the situation being discussed is the analysis in two
dimensions of the forces operating on an aircraft. The
intent is to simplify as far as possible while still being
able to generate useful predictions of the actual system
behavior.

So yes, the reaction force T, is the result of the sum of
the set of reaction forces operating on each element of
the aircraft...however that is a complication that doesn't
help us understand what is happening in the system.

Its easy to make things more complicated, the trick is
simplifying while keeping the nub of the problem intact,

-- Andrew

[twistedenginestarter]

The intent is to simplify as far as possible while still being able to generate useful predictions of the actual system behavior.

Spot on!

There are not one, two , three, four or whatever forces. There is whatever you choose to best explain what is going on.

(If anyone else doesn't understand this just try putting your hand out of each door and window in turn during the cruise. Then write back and tell us there are no forces...)

[cwatters]

For the plane to be in S&L flight TWO things must be true...

1). The sum of all the TRANSLATIONAL (straight line) forces must be zero

AND

2). The sum of all the TORQUES/ROTATIONAL forces must be zero.

If you define LIFT(cog), THRUST(cog), WEIGHT(cog) and DRAG(cog) as the orthogonal (right angles) translational components of all the forces that act through the center of gravity then it's true that they balance to zero.

But then you have to look at the other rotational forces. If the true center of drag is a bit above the CoG then that causes a rotational pitch-up. A high thrust line creates a pitch down. The wings pitching moment causes a pitch down etc etc. All these rotational forces must also balance to zero or the plane will be rotating about it's CoG - as it flys along S&L.

Consider what happens when you put the gear down. The translational component of drag increases so you must apply more power to maintain speed. However the center of drag also moves downwards. This causes a rotational force (pitch down) which may require a trim change. The trim change adds up elevator restoring the balance between rotational forces.

The TWO types of force must balance or the plane will either accelerate in some direction or rotate.

[StrateandLevel]

GoneWest why don't you take thrust out of the equation and consider a glider. It has a downward force due to its Mass and Gravity.

Newtons third law of motion states that every force has an equal and opposite force known as the reaction.

So now we have a glider with weight acting vertically down and the reaction acting vertically upwards. 2 Forces.

For convenience, we resolve the reaction into 2 "man made" forces called lift and drag to explain how the glider flies.

If we add a third force Thrust, there will be an equal and opposite reaction to that force however, the two reactions do not coincide with the established definitions of Lift and Drag. If we combine our reactions into a total reaction, there are now 3 forces acting on our aeroplane, one of which can be resolved into the two forces we require for the purpose of explanation thus giving us the conventional 4 forces.

The reference to 3 forces is to make you think about the forces acting and realise that Lift and Drag are not equal and opposite reactions.

[bookworm]

GoneWest why don't you take thrust out of the equation and consider a glider. It has a downward force due to its Mass and Gravity.

Newtons third law of motion states that every force has an equal and opposite force known as the reaction.

So now we have a glider with weight acting vertically down and the reaction acting vertically upwards. 2 Forces.

No, that's an abuse of Newton's third law. The 'equal and opposite reaction' phrasing of the law is concise, but leads to this sort of confusion. Newton's third law says that if Body A exerts a force on Body B, Body B exerts and equal and opposite force on Body A.

In the case you cite, the 'reaction' is the gravitational force that the glider applies to the earth, equal to the weight of the glider and applied vertically upward -- it has nothing whatsoever to do with the aerodynamic forces.

[RadarContact]

I think the tree-force-model has come from the aircraft and wings designing department.
Since both drag and lift are equally subject to airspeed and angle of attack for a specific wing or entire aircraft, it does make sense to calculate with one resultant vector. Otherwise, you would have to calculate lift and drag seperately and the put them together again. For all the graphs it's much easier with one resulting force.

But in the end, both models still describe the same effects, don't they? (Yes, it does fly - and yes, you need some energy to keep it doing so)

[OzExpat]

We're going round and round saying the same things - and this is from folks who understand the situation. Maybe the first real test might be to see how well a trainee grasps the concept. Perhaps its time we heard from them?

I suspect that we might be getting away from the "KIS" principle that has served us all so well, ever since the Wright Bros flew at Kittyhawk.

I have always referred to a "Total Reaction" force in a climb, for example, but this is a completely different situation to S+L flight. It's quite easy to introduce that concept at that stage because the trainee already has a grasp of the fundamentals that apply in S+L. I'm just not sure it's such a good idea to introduce the TR force in S+L, but I'm happy to accept that others can make it work without bamboozling a trainee.

[I Fly]

Does a 3 force aircraft fly faster or slower than a 4 force aircraft???????????
Look in you student's, or examiner's face. If you see understanding - you explained it the correct way. If not - try something different.

[bluskis]

On the subject of drag, I believe it manifests itself in three somewhat unrelated forms, one only of which is related to angle of attack/lift.

It cannot be analytically useful to consider only three forces, thrust, weight and was it TR?, but it is useful to consider thrust overcoming drag, producing speed through the air, thus acting on the airfoil and producing lift to overcome the weight, while unfortunately generating various drag forces which in turn reduce the resultant force available to produce speed through the air.

Equally unfortunatey none of these effects is linear so the equations relating to the above simplified consideration of flight are both multivariable and non linear.

[Capt Pit Bull]

I'm with bookworm on this one.

Add them, subtract them, do what ever you want. Just choose the method that explains most easily the behaviour you are trying to explain.

When teaching PofF I always used to avoid starting a discussion of the forces by looking at straight and level. Its too much of a special case. People get the wrong idea really easily about what orientation the forces are defined in, then you are really in trouble.

CPB

[acheo]

By definition a vector is a force with a direction (magnitude, sens and orientation).

Drag as always been a force (see defintion) in every aerodynamic books. Don't trust flight instructors but rather engineers for those issues.

cheers

[StrateandLevel]

Bookworm,

you say: "that's an abuse of Newton's third law"
-------------------------------------------------

NASA don't seem to be too upset by such abuse:

Quote from:
http://www.lerc.nasa.gov/WWW/K-12/airplane/newton.html

"The third law states that for every action (force) in nature there is an equal and opposite reaction. In other words, if object A exerts a force on object B, then object B also exerts an equal force on object A. Notice that the forces are exerted on different objects. The third law can be used to explain the generation of lift by a wing"

[B2N2]

I start most of my technical groundschool by saying:..' now this is not 100% technically correct but it is the easiest way to understand..'
Usually I get the point across.
I also keep a book from the University of Delft(Holland) from the division of aeronautics as a backup for the occasional smart-ass student....'but what if..'
In that case I get the book out,show them all the formulas and tell them they are only required to know the simple version.(Right?)
One vector split in two doesn't really change the right of existence of the original one now does it?
Our whole system of mathematics is man-made and a model that exists by the grace of not been disproved....yet.
Thrust and weight can also be conbined in a single vector resulting in a single angled force.Counteracted by the lift/drag combo,back to 2 vectors.
Question: Why do we need more power in slow flight,in fact almost as much as in cruise? (Single engine GA)


Most frequent answer: high angle of attack resulting in high lift therefore higher induced drag....


How about this one: as lift works perpendicular to the wing right?
In a turn we can split the lift into a horizontal comp. and a vertical comp.
The horizontal comp. makes the airplane turn and the vert. keeps the plane in the air.
Now for the shocker....the same applies in slow flight.
The horizontal component works in the same direction as drag in the 4-forces model,therefore increasing it.Hey presto more thrust needed.
Never even thought about this till the FAA examiner for my CFI mentioned this.I will forever be grateful.......
Taught most of my students the same thing.Makes great sense after you've experienced slow flight.
Now for the big question...
Does the 1st answer disprove or overrule the 2nd one?
No ....they co-exist.
Some people tend to think their way is the only way
it's not..
I never claim I'm a 100% right,I'm just less wrong than the the people I teach and I hope the'll end up being equally right.
Not more,that's embarrassing.

To finish: Confucius say: ..many thoughts lead to a cluttered brain....

:D :D :D

[B2N2]

Almost forgot :
The only true 4 forces acting are
lift
trust drag
faith

[bookworm]

StrateandLevel

The quote from NASA is correct. But if you read the pages linked to from the one you cite, you'll find that at no point do they use Newton's Third Law to assert that lift, drag, or total aerodynamic force is a 'reaction' to gravity, which is what you implied.

B2N2

Now for the big question...
Does the 1st answer disprove or overrule the 2nd one?
No ....they co-exist.

They don't coexist at all. The assertion that the direction of lift (or total aerodynamic force) is perpendicular to the chord line is simply wrong, and you can see that it's wrong by either:

a) inspecting any drag polar and noting that D/L is not equal to the angle of attack

b) thinking about how that rule might work when you know that changing the wing span changes the drag coefficient for th same AOA.

The problem is not that it's a little inaccurate or not 100% correct. It's not even close.

If you want to claim that your rule of thumb is a good aide-memoire for those who don't understand mechanics then fine, go ahead and teach it. But it doesn't make the physics correct, or more importantly it doesn't make the physics as correct as a proper explanation of induced drag!

[cwatters]

> How about this one: as lift works perpendicular to the wing right?

Well err no. There is no such thing as a single lift vector..

Sum ALL the force vectors acting on the wing and you get ONE vector pointing upwards and backwards. You can split this into two orthogonal (right angles) vectors pointing in any direction you like...

But for convenience it makes sense to choose one pointing upwards at right angles to the airflow because weight normally acts downwards, also at right angles to the airflow (unless you happen to be wave or ridge soaring etc).

You could define your lift vector to act perpendicular to the wing but you don't have to.

[Capt Pit Bull]

Again, I'm with Bookworm.

It really is depressing how much people misunderstand Newton.

Even the RAF have it wrong (I used to teach air cadets, and their manual has the same mix up!).

The equal and opposite the the gravitional force of the earth on the aircraft is:

The gravitational force of the aircraft acting on the earth.

Think about the moon. Not only is it pulled towards the earth by gravity, but the Earth is pulled towards the moon.

Good thing its in Orbit really (Itchy and Scratchy).

If you jump out of an aircraft, not only do you fall towards the Earth, but the Earth falls towards you. The gravitational force is equal and opposite.

Thing is, Earth has more mass, you have relatively little. Newton also tells us F=MA, so you do most of the accelerating.

Eat lots of crewfood till you weigh the same as the Earth. Then, if you jump out, you'll meet the Earth half way.


Acheo.

Vector is an amount and a direction, not a Force and a direction.

A force is a vector, but there are many other vector quantities.

Velocity. Acceleration. Momentum.
All the rotational equivalents, to name a few.


CPB

[Romeo Romeo]

The NASA quote...The third law states that for every action (force) in nature there is an equal and opposite reaction. In other words, if object A exerts a force on object B, then object B also exerts an equal force on object A. Notice that the forces are exerted on different objects. The third law can be used to explain the generation of lift by a wing...can be used to explain lift by a wing. The wing accelerates air downwards and forwards (by applying a downwards and forwards force to it), causing the wing to experience an upward (lift) and backward (lift drag) force. The more you accelerate it (by increasing the angle of attack), the greater the forces are.

What is actually going on when an aeroplane flies is enormously complicated. In order to understand it, we build simplified models. There are no 'right' or 'wrong' models; however some models explain certain aspects of flight better than others. Therefore, it is sometimes easier to think of an aeroplane as having thrust, drag, lift and weight, and sometimes it better to think of it as other forces (for example splitting the lift force into a horizontal and vertical component when the aeroplane is turning).

I think you should be comfortable with both the 3 and the 4 force model (and perhaps models with a few more forces as well to handle turning and spins and so on), but don't think that one is correct and the other is somehow wrong.

A good model should:-
Be simple as possible
Be able to explain observer results
Be able to predict future results

[Konkordski]

All forces acting on a body at any instant can be resolved into a single force vector.

If the aircraft is at constant altitude then the weight and lift forces cancel out.

If the aircraft is flying 'straight and level' then the lateral forces must also cancel.

And if it is travelling at constant speed -- and thus not accelerating -- then the longitudinal forces cancel.

Therefore there is no overall force acting on the aircraft. Perfectly in line with Newton's first and second laws.

QED :D :D :D

[bluskis]

Konkordski

Perfectly true and perfectly useless. I won't quote the usual joke.

Your statement applies equally to a stone resting on the ground.

The objective is to understand the forces acting on an airplane, taking a simplified approach at first and eventually a more refined and complex approach in order, in the case of a pilot, to understand the interreaction of the controls available to the pilot in order to equip the pilot with the ability to fly proficiently and safely.

The four force explanation of straight and level flight carries out the first steps of this path to full understanding .

A resting stone analysis fails to do this.

[cwatters]

It seems to me we're all taking the same language. The original post was about...

><snip> actually there are only three forces. Weight, thrust and
>total reaction (TR). This TR can be broken down, by the
>scientists, to show a horizontal vector (not a force) which we
>call drag - and a vertical vector which we call lift.

Actually their are an infinite number of forces acting on a plane that sum to zero in S&L constant velocity fight - but we prefer to resolve them (break them down into) components that act in specific directions.

You could say...

The sum of all forces = 0

or

W + T + D + L = 0

or

W + T + L + TR = 0

or

W + everything else = 0

It's all means the same thing.

The important bit is that the sum = zero. If it doesn't then there
is a residual force acting on the plane that will cause it to depart
from S&L constant velocity flight.

[OzExpat]

The analogy of the resting stone isn't entirely incorrect. It is still doing something and it can be argued that all the forces acting on it are in equilibrium (good word huh! :D ). That will continue to be the case until it is affected by an acceleration - thrust created by someone kicking the stone, perhaps?

Anyway, it doesn't adequately illustrate the point but the reality is that in S+L flight, all the forces acting on an aeroplane are similarly in equilibrium until something is changed. I've deliberately avoided any reference to 3 or 4 force (or whatever) models because, even if there are a thousand forces acting on the aeroplane in S+L flight, they are ALL in equilibrium.

[B2N2]

OK gloves are off this time...
How many actually get the idea?
The idea that we use mathematical models to make difficult things easier to understand.
Not to make things more complicated for people that know less about it.The definition of a model is a simplified version of what's really going on right?Please don't argue on this one.
For those of you that are NOT flight instructors,here's the shocker:
We train people to be a safe proficient PRIVATE PILOT,not a rocket scientist.:p
That's one of the biggest problems in aviation by the way.
All the smoke & mirrors to impress the non-initiated.
Go to an airport on a weekend day and watch all the know-it-alls
in their regalia(there's a thread on this one) pretend they're Gods gift to aviation as they try to impress their non-flying victims.
I'll admit I'll lose a discussion with an aeronautical engineer but the point is wether it is of any use in actual flying.Designing yes,real life no.
How many of you know what's actually happening to your car as you drive,let alone in your engine.Does'nt stop anybody from driving though does it?

3-forces...4-forces peace to all of you..I've said my thing.

Keep the pointy end in front and the wheels down..you'll be awright...

:D :D :D

[OzExpat]

Yes, I used to be an instructor - and a Chief Flying Instructor at one stage too. That was a real looooong time ago but, back then, it was common to talk about forces being in "equilibrium" in S+L flight. The only way this could be shown, during a blackboard/whyteboard briefing was via the 4-forces model.

Lift and Weight were equal to one another, so it readily made sense that they cancelled each other out. Thrust and Drag were clearly equal to one another and it therefore made sense that they, too, cancelled each other out.

Every student I trained in this concept understood the concept straight off. Indeed, after a few very carefully chosen words at the start of the discussion, the trainees would be telling ME how it all happened! I always love it when a plan comes together!

That provided the foundation upon which to build the briefing on "climbing", where you MUST have a "Total Resultant" vector - otherwise there is NO climb. Again, I could start the discussion with a few carefully chosen words and stand back to let the trainees tell me how it all happened.

And, ya know, the thing is that I'd learned it all in much the same way. I did smplify it. I really was easy to understand and, above all, it was also very easy to demonstrate in flight!

Well, that's my hand grenade into the topic, but it all worked very well over the 10 years that I spent as an instructor.

[Romeo Romeo]

The 'level' bit isn't necessary for the forces to be in equilibrium. All that is required is that the vertical and horizontal velocities are constant. If you're going straight and climbing at 500ft/min then you are in equilibrium. You are only out of equilibrium when you are accelerating (either horizontally or vertically) e.g. when transitioning from straight flight to a climb.

SoThat provided the foundation upon which to build the briefing on "climbing", where you MUST have a "Total Resultant" vector - otherwise there is NO climb. isn't correct. (And I say that pulling my forelock because I dare say you're a far better pilot that I will ever be!)

B2B2: Bang-on right about the model (and why it doesn’t matter that much!)

[OzExpat]

RR ... I wouldn't have thought there was any reason for a cynical comment, so I've now learned something from you. In any event, I intend to overlook it and merely ask how one achieves a climb, from S+L, in the first instance, without disturbing the aircraft's equilibrium - as when transitioning "from straight flight to a climb"?

[TeeS]

Oh well, just cos I enjoy an argument - "In straight and level flight at constant speed, ie flight at constant altitude and speed, all the forces cancel out". I would suggest - almost, but not quite!! If you set off in straight and level flight as above (and keep going!) you will end up back where you started having flown a circular path around the earth. There must be a resultant centripetal force acting towards the centre of the earth, ie in the same direction as the weight vector.
Forgive me for being flippant but it is a very quiet Sunday morning.

[Romeo Romeo]

Hi OzExpat, it wasn't meant to be a cynical comment! As B2N2 said, these models are just about helping you understand the basics - it doesn't help you fly. All I was trying to say was that flying's about getting from A to B safely and that is far more about experience and training than technical discussions about forces. I'm a 200 hour PPL and I KNOW I have a huge amount to learn. Sorry if it appeared cynical.

You're right about the forces being out of equilibrium during the vertical acceleration stage between level and climbing at a constant rate. During the constant climb rate stage they are back in equilibrium.

[bluskis]

During a climb forces are certainly not in equilibrium.

Upward force exceeds gravitational force, hence climb.

If they were in equilibrium, as in straight and level flight, there would be no climb.

If you are looking for the acceleration part of F=(W/g)xa, the a is the gravitational acceleration you are overcoming.

[TeeS]

Sorry Bluskis, I can not bring myself to agree to that one -
"Upward force exceeds gravitational force, hence acceleration upwards!!"
When they are in equilibrium your vertical speed will be constant.

[Oktas8]

Bluskis-

In a steady state climb (constant direction, rate of climb and airspeed) the forces acting on an aircraft are in equilibrium. Furthermore, lift is less than weight in this state. The NASA site referenced earlier in this thread (p 2 I think) is an excellent resource for this kind of issue. So is AC Kermode's primer on Aerodynamics - I can't remember the exact title. It's up to edition 10 now I think.

Your use of the equation F = W/gxa is out of context. In a steady state climb the aircraft is not accelerating - all speeds, rates of climb, directions are constant and unchanging in direction and magnitude. Thus "a" is zero. Thus net (total, sum, add-em-all-up-and-what-do-you-get) force F is zero, as many preceding posts have stated.

Yes, lift really is less than weight and forces are in equilibrium.

Really!

happy flying,
O8 :)

[bluskis]

I will rethink this one. Unfortunately without Kermode,as my copy is many miles away at this time.

[GoneWest]

Bluskis - and friends..

A C Kermode. Mechanics of Flight. Edition 10. Page 229. "Climbing".

First, I need you to imagine a picture - let's, for the sake of continuity, call it "Figure 7.1".

Imagine a picture of an aircraft - side view. Draw on it the four arrows that we are all familiar with. Identify them with the the famous four - L, D, T and, W. For those with bad memories - L, represents Lift and is drawn vertically upwards, W represents Weight and is drawn vertically downwards, T represents Thrust and is drawn horizontally (pointing to the left), D represents drag and is drawn pointing to the right. All the above assumes you have drawn the aircraft the right way up and flying from right to left.

Now - leave the W arrow still pointing vertically downwards, but rotate everything else in the picture - to represent a climb. The amount that you have rotated the picture, measured in degrees, will represent the "climb angle", and will be called "a". Thats a small letter "a" (presumably "alpha").

Imagine now, your thrust arrow is inclined upwards at angle "a", your drag arrow is inclined downwards at angle "a", and the lift arrow is inclined backwards at angle "a". Weight, still vertically downwards. Got it??

Onwards to Kermode....

Climbing

During level flight the power of the engine must produce, via the propeller, jet or rocket, a thrust equal to the drag of the aeroplane at that particular speed of flight. If now the engine has some reserve of power in hand, and if the throttle is further opened, either -

(a) The pilot can put the nose down slightly, and maintain level flight at an increased speed and decreased angle of attack, or

(b) The aeroplane will commence to climb

A consideration of the forces which act upon an aeroplane during a climb is interesting, but slightly more complicated than the other cases which we have considered.

Assuming that the path actually travelled by the aeroplane is in the same direction as the thrust, then the forces will be as shown in Figure 7.1

If "a" is the angle of climb, and if we resolve the forces parallel and at right angles to the direction of flight, we obtain two equations -

(1) T = D + W sin "a"

(2) L = W cos "a"

Translated into non-mathematical language, the first of these equations tells us that during a climb the thrust needed is greater than the drag and increases with the steepness of the climb. This is what we would expect. If a vertical climb were possible, "a" would be 90° and therefore sin "a" would be 1, so the first equation would become T = D + W, which is obviously true because in such an extreme case the thrust would have the opposition of both the weight and the drag. Similarly if "a" = 0° (i.e. there is no climb), sin "a" = 0. Therefore W sin "a" = 0. Therefore T = D, the condition which we have already established for straight and level flight.

The second equation tells us that the lift is less than the weight, which is rather interesting because one often hears it said that an aeroplane climbs when the lift is greater than the weight! One must admit, however, that the misunderstanding is largely due to the rather curious definition which we have assigned to the word 'lift'. Let us consider the second equation under extreme conditions. If the climb were vertical, cos 90° = 0. Therefore L = 0. So that in a vertical climb we have no lift. This simply means that all the real lift is provided by the thrust, the wings doing nothing at all to help. If, on the other hand, "a" = 0, cos "a" = 1, and therefore L = W, which we already know to be the condition of straight and level flight.

(Phew)

[OzExpat]

GW ... I couldn't have said it better meself! Really!! But, seriously, thanks for that - I'm in the same situation as bluskis, in that my now very aged copy of Kermode's wisdom is many miles away from me right now... :(

RR ... the written word can often be subject to misinterpretation, except perhaps in Kermode's turn of phrase. We can't all be as good at it as he is.

Geez, I wonder what edition of his book I have? :eek:

[LeadSled]

All,
When I look at this circular discussion ( if you take it all to the nth degree, using your handy backyard Cray, you will only ever wind up with an approximation of the answer) I am reminded of an Australian cartoon character, Ding Duck, in the Swamp series.

A cartoon, which was enlarged and graced the door of the then Dept. of Aviation ( or whatever the airstapo was then called) examination hall in Sydney showed the question: What are the four forces acting on an aircraft in flight.

The happless Ding's answer was: Lift, Weight, Thrust and the Department of Aviation.

Tootle pip !!

[bluskis]

GW
Thanks for the extract.

I was confusing the increase in upward force required for the climb compared with straight and level flight ,with equilibrium of the forces which Kermode clearly shows exists in a stable climb.

[Capt Pit Bull]

The basic problem here is that all sorts of people do not have even the most basic understanding of mechanics.

If you want to talk about forces at all, then the student has got to have a proper understanding of the basic laws of motion. Otherwise the best that you can hope for is that they will remember individual 'sound bites' of your explanation.

Generally, instructors tend to assume knowledge.

The average person on the street does not understand friction, so their everyday experience tell them that when you stop pushing something, its slows down and stops. As a result, they don't really understand what a force is, and hence understanding Newton is just a no hoper.

The other problem is that straight and level flight is the obvious first thing to talk about. Sadly, its not the best way to do it. In straight and level, the standard 4 forces are conveniently 90 degrees apart. So the student gets the idea (subconsciously, even though you haven't said it) that all the forces are defined in those orientations.

I have found it useful to just draw introductory diagrams of different flight phases to show how the forces are defined (whilst stating any simplifications). Then you've got a basis for teaching S&L without causing miscomprehension.

Then, to teach the detail of different flight phases, you have to do the same sort of thing as the air exercise.

e.g. Climbing. To teach, you show student how to start, maintain, and finish. Basically 3 different exercises.

To teach the theory, you need to differentiate between steady flight paths, and changing flight paths. In the former, the forces must be IN balance, in the latter they must not. To enter climb, must have more lift to alter flight path. To maintain ..(bleh, can't be bothered to spell it out). To return to level flight, must reduce to level off, then increase back to match weight again to avoid descent.

But fundamentally, most students have forgotten (if they ever knew) their basic science, and the pre-flight briefing just isn't long enough to rectify that knowledge gap.

So basically I'm saying that if one is going to use force diagrams to teach, then I feel it is essential to get them right. On the other hand, what training value is actually derived from so doing?

In some ways I'm with B2N2 as far as the necessity of theory goes. But if someone wants to know, then get it right or leave well alone!

CPB

[OzExpat]

CPB ... okay, if nobody else is going to take issue with what you've just said, I'll kick the ball into the air. First of all, I resent your implication that no other instructor gets the basics right with briefings for trainees. Do you really believe that every other instructor, or former instructor, around the world, is so completely idiotic as to overlook such a fundamental aspect as the way we show the forces in a (massively) simplified diagram?

What you say might well be the case, in your experience, in your part of the world, but it certainly is not the case in my experience, in my part of the world. So, please, no lofty generalisations.

Next, you've ignored one of the fundamentals of aerodynamics in saying that "lift" is increased in the climb. I invite you to read back over the quote from AC Kermode, made by GoneWest, and then tell us about lift in the climb.

[Capt Pit Bull]

Relax Ozexpat.

I said "generally, instructors..."

not "all flying instructors..."

Which means, off course, exactly what you said. I.E. generally, but not always, in my experience. And I also mean 'instructors including myself', not 'other instructors because I'm such a genius'. I've certainly made the mistake of assuming knowledge on the part of the student. And as a student, not just in flying, I've had the reverse situation.e.g.

"We're going to start by looking at M32"
"What's an M32?" :confused:
"Number 32 in the catalog of Messier Objects"
"What's a Messier Object?" :(

Having said that, I don't think most flying instructors are that hot on the basics of mechanics. But I don't think they need to be, in most cases, so why worry? I on the other hand spent several years teaching commercial groundschool, but am by contrast, at best a mediocre flying instructor.


As far as your last paragraph is concerned, I did not say ' "lift" is increased in the climb'. I said that lift is increased to enter a climb. For most average powered aircraft in mundane flight phases this is true.

Kermode says that the aircraft climbs because power is increased. That excerpt however doesn't tell us the mechanism by which the climb is established. His aircraft just suddenly, instantly, starts climbing and miraculously points its thrust vector along its new flight path.

In other words, in this excerpt, Kermode is not even attempting to discuss the transition from level flight to climb. He is instead interested in studying the forces once established in the climb.

Hopefully I've made myself a bit clearer and you can see where I'm coming from. If you actually want a discussion of the forces involved in entering, maintaining, and leaving the climb I'm quite happy to continue. I got the impression however that you were just pouncing on a perceived error on my part though, so I don't propose to continue unless you actually wish it.

CPB

[GoneWest]

Well, I wish it!! (does that count??)

[OzExpat]

CPB ... yes, I probably could've phrased my previous post a little less stridently. I did, however, believe that you were saying something that was a tad too generalised. I accept that not all instructors (ground and flight) are good instructors and, indeed, everyone can have an "off-day" once in a while.

After several thousand hours instructing, I found it was still possible to assume a certain amount of knowledge on the part of the trainee, sometimes. I suspect there's probably some "human factors" stuff at work there.

The passage that GW quoted from Kermode does, however briefly, touch on the forces at work in the initiation of a climb. That bit about what happens when you have reserve engine power in S+L flight. Using that additional thrust, without a change in angle of attack, will initiate the climb.

Thus, thrust contributes to lift to create the "total reaction" vector, which is how this came up in one of my earlier posts. This is the basis I've used in briefings to explain to a trainee why, in practice, thust needs to be applied to initiate the climb. Of course, such briefings are necessarily tailored to the typical type of trainng aeroplane which is not overly endowed with thrust and, naturally, that point must also be made in the briefing.

This works out in practice too because, while a climb can often be initiated simply by converting excess speed to height, such a climb cannot be sustained without an increase in thrust. This is pretty easily demonstrated in flight. One must then touch on the fact that, when you run of out reserve thrust, you're playing with a stall to attempt further climb.

When I was trained as an instructor, it was emphasised that we need to keep briefings as short and simple as possible, without bypassing the essentials. Also, that it was entirely appropriate to give a briefing that covered all the essentials for that exercise alone.

Thus, the S+L briefing was intended to teach the student why it is necessary to have particular combinations of power and attitude to maintain S+L flight. If the basics of this are covered well at that stage, it is much easier for the student to grasp the concepts related to initiation of a climb. And, of course. to the subsequent maintenance of the climb.

It is certainly useful to start a briefing on climbing by reviewing our defined forces in S+L. Indeed, it is always a good idea to start any new topic with a review of what the student should already know. This is where shortcuts can often occur, because an instructor can be easily persuaded to assume a certain level of knowledge by the student.

All things considered, I suspect we've been talking more about instructional technique than the material itself. It had not been my intention to hijack the debate along those lines. Still, its clear that we all need to exhange views of such matters periodically, as it helps to keep us all up to speed and, thereby, to make us all better instructors.

There's one helluva thin dividing line between giving the student as much as is needed to understand what is about to be taught in the aeroplane, and complicating the whole thing with too much theory. This is the line that each instructor walks every day because we all need to know more about the subject than we'll normally need to teach. This was another point that came up in my own training because you never knew when a student was going to ask one of them really complex questions!

[Capt Pit Bull]

Ozexpat,

Well, thinking about it, a certain ammount of knowledge has to be assumed. The assumption is that, in a properly structured course, the student has the knowledge required to be sitting in front of you for their briefing. After all, you can't start every briefing by "In the beginning, there was darkness....";)

But, as you say, enough about intructional technique and back to the topic at hand, which I see as a deeper look at PofF for interest, not necessity.

So, for Gonewest and anyone else interested:

Lets look at Kermodes model.

His aircraft is in S&L.
Thurst is increased.
The Pilot does not select a lower nose attitude, so the aircraft commences a climb.
He makes the simplification, stated later, that "Assuming that the path actually travelled by the aeroplane is in the same direction as the thrust..."

On that basis, it is not possible to enter a climb without seeing an increase in Lift to initiate it.

Argument:
Kermodes thrust is acting along the flight path. Therefore, when in S&L, Thrust is horizontal (to the world).

a) If all other forces were in balance, and
b) The pilot maintains the same attitude (granted, this requires control inputs),
c) assumption: ignore prop effects (imagine its a jet)

then initially only one thing will happen: the aircraft will accelerate horizontally. Hence speed will increase.

As speed has increases, this will effect lift production.

Thus far:

Pitch attitude unchanged.
Flight path unchanged.
therefore AOA is constant, along with everything else in the lift formula apart from speed (which has increased).
Hence lift is increased.

[Incidentally, drag will have increase as well, exactly in proportion, so the Total Reaction has also increased if you are using a 3 force model, but that is not of direct interest. The key point is that an extra upwards force component is now present that will start the aircraft climbing.]

Therfore, to initiate a climb in this manner, Lift has increased.


Yet Kermode clearly shows how, once established in a steady climb, Lift is actually less than it used to be in S&L. No arguement from here!

How do we get to a from a stage where Lift is greater than weight to establish a climb to being less than weight to sustain one?

The answer lies in looking at the Angle Of Attack.

We've said the pilot is maintaining the same attitude, however the flight path is now sloping upwards. Hence AOA will be decreasing. The higher the climb rate, the more the AOA decreases until that effect on lift production overwhelms the effect of the increased speed. Eventually, lift will decrease (in line with Kermode) until the forces all balance out. At which point the flight path (along with AOA) remains constant.

Summary:

Conditions:
- Attitude constant.
- Thrust increased.

Results in:
- Speed increased.
- Lift increased.
- Climb Rate.
- AOA reduction leading to new equilibrium, as Kermode, with L<W


Alternatively, it is perfectly possible to climb, sustainably, without increasing thrust. (proviso - unless you are already at ceiling, or already on the back of the drag curve, in which case a stall will result). The key thing is the necessity to get to a regime of greater excess power. This is not the same as saying more power.

Select a higher nose attitude by using the elevators. This increases AOA. At that moment, speed is unchanged, but the AOA increase makes more lift. Aircraft has unbalanced upwards force, so accelerates upwards. Since there is currently no excess power, kinetic energy must initially be traded to gain gravitational potential energy, so aircraft slows down.

But the speed decrease has the effect of reducing power required, so we now have some excess power with which to sustain climb, albeit with a lower airspeed.

Meanwhile, the AOA decrease described earlier, plus the speed recuction, combine via the lift formula to end up with a net reduction in lift to eventually be less than weight, i.a.w. Kermode steady climb analysis.

Summary:

Conditions:
- Attitude increased.
- Thrust constant.

Results in.
- Lift increase.
- Climb rate (initially zoom).
- Speed reduction.
- Power required decreased.
- Sustainable climb.
- AOA (& speed) reduction leading to new equilibrium, as Kermode, with L<W.


So we can see two methods by which a climb can be initiated, either increase thrust or increase pitch. Both of them involve an initial increase in lift to establish the climb, followed by a decrease as the new equilibrium is established.

Q.E.D.

In reality, we tend to do both for sustained climbs, i.e. increase thrust and select a higher pitch attitude. For temporary climbs (or descents), say to correct our cruising altitude, we tend to make a small power increase (decrease) and let everything else come out in the wash. The mechanism is, however, as shown.


Footnotes:

(1) Of course, a steady climb isn't really an equilibrium. If you fly a constant air speed in a climb, then your TAS is steadily increasing, so you are not technically in a state of uniform motion i.a.w. Newton. It may not seem like a big deal, but it can make a big difference in certain circumstances.
(2) As mentioned earlier, S&L isn't really a uniform state of motion either (due earth curvature), but it doesn't make much difference unless you fly very fast. Feeling keen? Try doing some sums for Concorde or an SR-71.
(3) In spite of the argument above, it is possible to show how a a climb can be established without a transient lift increase. it just isn't the way average aircraft usually do it. Can anyone see the hole in Kermodes assumptions?

CPB

[B2N2]

:D As usual it's always better to let somebody else do the fighting:D
Cheers ExOZPAT I'm right behind you on this one...:D :D

[OzExpat]

Looks behind, hoping that B2N2 isn't toooo close behind... :D

CPB ... I don't dispute any of what you say, with the possible exeption of that last bit. I'm not so sure that it's a case of there being any holes in what Kermode says. Remember that he's not trying to teach anyone to fly, merely to understand what is happening. For this, he needs to break things down, to make the various aspects of each topic easier for people to read and grasp.

Surely it is the flying instructors' job to put all of that in the context of how it is applied to the type of aeroplane being used for the training?

Indeed, in your own explanation, you simplified some of it by referring to a jet, rather than a prop. A wholly reasonable simplification, I might add, but isn't this pretty much the same as what Kermode was trying to do?

B2N2 ... feel free to jump in whenever you like. I won't be able to keep posting for much longer anyway, as I'm heading off shortly and epect to be away for at least a month. I hope that won't mean that this most interesting thread dies in the meantime.

[twistedenginestarter]

You need to understand an aeroplane operates in a force field so is not like a spacecraft. A spacecraft will accelerate if a net force is applied, otherwise it will continue in a straight line.

An aeroplane behaves similarly in the two dimensions parallel to the Earth's surface but not when altitude is concerned. Here the force field generated by gravity comes into play. As you climb, you move through this field. Although you gain no momentum at a steady rate of climb, you do gain potential energy by changing your position in the gravity field. This requires an input of work/energy. And that comes from a vertical vector greater than the weight of the aircraft.

Newton's 3rd Law only really brings anything to the party when you try to explain how propellors and turbines work. People don't normally explain lift this way because a wing is completely different to a thing which merely deflects air downwards.

[OzExpat]

I was on my way out to dinner earlier this evening, when a thought struck me. And, before you ask - yes, thanks, it WAS a painful experience! :D

Anyway, it occurred to me that we need to be a bit careful when talking about lift increasing in the initiation of a climb, from S+L flight. The sum of all lift is considered to act thru the Centre of Pressure, at a right angle to the relative airflow. This gets awkward to describe without the aid of diagrams but, if worst comes to worst, I'll make some and post them.

Anyway, the simple act of applying some or all of the reserve thrust, increases the magnitude of the thrust force. In a diagram, this is represented by lengthening the thrust vector by a representative amount. The additional thrust creates the change in AofA, and the increased angle creates a movement of the CofP and, in our blackboard/whyteboard diagram, we see that the lift vector line is inclined toward the tail of the aircraft, in relation to the horizon. This is because it is still acting at a right angle to the relative airflow.

The actual physical length of the lift vector line is still the same but its inclination, relative to the horizon has effectively reduced it's magnitude in comparison with the assumed weight vector line, which is still acting vertically downward thru the CofG. This is why we've said, for years, that lift is less than weight in a climb and it matters not at all whether it's at climb initiation, or in the ideally envisaged steady-state climb.

The "total reaction" is now of greater interest to us as this is what determines our climb. To find the location and magnitude of the vector for total reaction, we resort back to our imaginary diagram and draw a line from the top of the lift line, parallel to the thrust line. This line obviously moves forward. Stop drawing this line when a tangential line will meet the forward end of the thrust vector line. Connect this point to the CofP by a straight line and you've found the vector for total reaction.

This is the vector line that we compare to the weight vector line, to find a new equilibrium that allows the aeroplane to stabilise in the climb. Thus, it really isn't true to say that lift is either equal to weight or greater than weight. This cannot be true because we have upset those forces in initiating the climb.

There are other factors that come into consideration, such as :

1. the effect of prop torque, which I think the Americans call P-factor; and
2. how to sustain the climb.

I'm ignoring those because the basic topic is already complex enough.

Who wants to go next? :D

[bluskis]

I hesitate to post again having made a pigs ear once, and still a long way away from my Kermode, however

In a climb, during initiation and also once stable, if forces are resolved vertically and horizontally with respect to earth rather than the aircraft this should isolate the vertical forces responsible for the vertical component of the climb.

Amongst these will be a vertical component of the now inclined thrust, and acting in the opposite direction ,that is in the same direction as the weight ( or mass) of the aircraft, will be a vertical component of the now inclined drag force.

This will hopefully explain why although aerodynamic lift may be reduced in a climb, the combination of the vertical component of this and the vertical component of the thrust have to be greater than the S/L vertical force required, in order to balance the weight and the added downward component of the drag forces.

[Capt Pit Bull]

Hmm. Sorry Ozexpat, but you are way off base in several areas in your post.

Your Para 3:

The additional thrust creates the change in AofA...

[Whilst accepting that local AOA changes can be caused by thrust changes for prop aircraft, both of us have elected to ignore prop effects for the time being. Certainly a Jet discussion should suffice for considering climbing - the mechanics are not going to be fundementally different by powerplant.]

Changing thrust will not directly influence AOA. Granted, it may do indirectly by causing a pitching moment, but we are assuming a Piloted aircraft, i.e. the pilot controls that attitude of the aircraft.

The flight path has not yet changed, and the attitude is unchanged. Therefore the AOA is constant. there must be some other mechanism for commencing the climb. That mechanism is either a speed increase leading to more lift, or a pitch increase leading to more lift, in order to generate an upwards force to initiate the climb.


Your para 4:

The actual physical length of the lift vector line is still the same ...

No it isn't. We're talking steady state climb here, and both you, I , and Kermode agree that lift is less than weight at that point. The length of the line represents the magnitude of the force, so if it were the same, lift would be the same.

Look at the Kermode excerpt. He specifically states that he is resolving forces parallel and at right angles to the flight path. He is not talking about just the vertical component of lift, he is talking lift, all of it.


Your Para 5:

You are adding vectors here. No problems with that - vectors can always be added. But the Total Reaction is the addition of Lift and Drag. What you describe however is the addition of Lift and Thrust, and hence is not the 'Total Reaction'.

Incidentally, I think maybe you ment to make that an 'Excess' Thrust line, because then at least your diagram would have accounted for all the forces, even if the 'total reaction' was misidentifed.

What your diagram basically shows is that excess thrusy must be present in order for a climb to be maintained. Thats fine, correct, and I'm not arguing with it.

But it does not show how the aircraft gets into the climb!

To show that once an aircraft is climbing, a vertical component of thrust exists, is not an arguement to show how that condition came to be.

I reiterate: During the cruise, thrust and drag are horizontal, and lift and weight are vertical. If the aircraft is displaced vertically upwards in can only be because lift has increased, or weight has decreased. If something has not been dropped, fired, kicked or fallen off of the aircraft, then the only remaining possibility is that Lift has been increased. [note 1]

Let me show it another way. If you enter a climb, you feel G. Maybe not much, perhaps sub-threshold, but we all know it is there.

What is G? Answer = Load Factor (N)

What is Load Factor? Answer = Lift / Weight.

Therefore, if you are feeling G, Lift exceeds weight.

[G relative to aircraft Normal Axis. For progressively steeper climb angles, although you still feel your own weight, more and more of it will be on your back rather than the seat of your pants. In a sustained vertical climb, you'd have zero lift, and zero load factor, although you obviously would still feel your weight by virtue of the seat back pushing you up]


Note 1: Or our model is wrong. Which it is, in fact. Both Kermode and I have so far assumed that Thrust acts along the aircrafts flight path. This is not true, but rarely much of a real issue. But when you see highly agile aircraft performing high AOA manoeuvres with a large excess thrust, it becomes significant. Consider a slow fly past by a modern fighter. It will have a big AOA - maybe very big. Hence its thrust line would already be pointing up. If it then applies max burner you can easily see that a climb will result without lift necessaraly having been increased. But that isn't really the kind of thing we are talking about here. As I said earlier - mundane aircraft, routine manoeuvres.


Also, to reiterate, I agree totally that this is way deeper than most pilots need. Nevertheless, people have expressed an interest, and in mechanics terms it is not that deep - just a free body diagram and a few forces.

Cheers.

CPB

[OzExpat]

CPB ... I knew it was a mistake not to have a drawing, or series of them, to show how it all develops. But in any event, I suspect that we'll have to agree to differ on this subject, as I no longer have the time available to pursue it, due to an up-coming trip away.

However, just a thing to consider...

Changing thrust will not directly influence AOA. Granted, it may do indirectly by causing a pitching moment, but we are assuming a Piloted aircraft, i.e. the pilot controls that attitude of the aircraft.

The flight path has not yet changed, and the attitude is unchanged.
We started this idea with the premise, from Kermode, that if the pilot did nothing to the flight controls after increasing power (thrust), the aircraft would enter a climb. If it's a single engine, prop-driven aircraft, it'll yaw too tho we've agreed not to bother with that aspect for now.

Anyway, because the aeroplane has entered a climb, the flight path HAD changed.

But the Total Reaction is the addition of Lift and Drag. What you describe however is the addition of Lift and Thrust, and hence is not the 'Total Reaction'.
Okay, maybe the more correct terminology is the "Vertical Component of Lift", but that's just a word game anyway because the Total Reaction is the result of the acceleration that has led to the climb being initiated by simply increasing thust. At this point I am, of course, staying away from the principles associated with maintaning a climb.

But it does not show how the aircraft gets into the climb!
With respect, I thought that's precisely what I showed.

What is G? Answer = Load Factor (N)
I suspect that some care is necessary with this statement. "G" is actually acceleration, which can be positive or negative. It can also be zero, of course. To illustrate the point, in S+L flight, the Load Factor is 1 and G is zero. I agree that Load Factor will increase in any of the normal manoeuvres (I'm trying to stay away from the forces in the pull-out from a steep dive, of course, or abrupt levelling after a very steep climb), but the increase isn't especially noticeable in normal manoeuvres with the possible exception of a turn with a moderate to steep bank angle.

G will increase as well, any time that any form of acceleration (by the straight physics definition of it) is applied. But the increase is really quite miniscule in the initiation and maintenance of a climb, unless you're talking about jet fighters, space rockets, etc. This would seem to be out of the range of the topic as it was originally started.

Anyway, I would be very cautious about even mentioning such a complication as "G" in a climb.

...in mechanics terms it is not that deep - just a free body diagram and a few forces.
Makes me wonder how we managed to make it more complicated then, eh? :D

I'll come back to this thread whenever I can, but my rapidly aging copy of Kermode will be no closer to me than it is right now... :(

[Capt Pit Bull]

Ozexpat,

Well - certainly agree about a diagram - picture versus a thousand words and all that.

We started this idea with the premise, from Kermode, that if the pilot did nothing to the flight controls after increasing power (thrust), the aircraft would enter a climb.

Well, actually, thats an inference about what Kermode says. What he actually says is that unless a lower pitch attitude is selected, the aircraft will climb. That is not the same thing as saying this is a stick free situation.

However, now that you've said that, I think I can see what you are thinking about. Unless I am mistaken, you seem to be saying:

Increasing power makes the aircraft pitch up, and thus causes a climb directly, with there being no need for an increase in lift to deflect the aircraft from its flight path.

My position is this - that for an aircraft to be deflected upwards there must be an unbalanced force, acting upwards. With the exception of high thrust high AOA situations as discussed towards the end of my last post, that force must be lift.

I have shown that if a pilot maintains attitude, a thrust increase will cause a speed increase. More Speed, Same AOA = More lift.
I approached the problem this way because it seemed simpler to choose a circumstance that removed the need to consider pitching moments.

Nevertheless, you want to consider Pitching moments as well. Fair enough - I accept that in many aircraft, particularly primary trainers, application of thrust causes a nose up pitching effect. Surely you can see therefore, that at as the nose pitches up, AOA is increasing. In fact you've said it yourself:-

The additional thrust creates the change in AofA, and the increased angle creates a movement of the CofP and, in our blackboard/whyteboard diagram, we see that the lift vector line is inclined toward the tail of the aircraft, in relation to the horizon. This is because it is still acting at a right angle to the relative airflow.

If AOA has increased, lift has increased since nothing else has changed in the lift formula (speed decay has yet to occur).

So again, whether the aircraft pitches up as a result of pitching moments, or because the pilot intentionally pitches up to enter the climb, (or even holds the same attitude and accelerates), whichever way you cut it the mechanism that deflects the FLIGHT PATH is a lift increase, which was my original contention that you took issue with.

And I think this is the root of what is wrong with your view - you seem to be mixing up rotational effects with translational effects.

I know you feel that you have a good simple model that explains climbing, but I'm afraid that its wrong. I don't mean oversimplied, I mean 1+1=3 sort of wrong, because it misuses laws of mechanics.


Moving on:

"G" is actually acceleration, which can be positive or negative. It can also be zero, of course. To illustrate the point, in S+L flight, the Load Factor is 1 and G is zero.

Well, I'll admit that its been a while since I flew an aircraft with a G meter. But let me ask you a question - in a 60 Degree Banked turn how many G are you pulling? I reckon its 1 more than straight and level, i.e. 2. By your reckoning it would be 1.

But thats by the by.

I was just trying to provide another way for you to see that a lift increase is present.


Let me close by wishing you well on your trip. I'm off down route for a few days too.

CPB

[Capt Pit Bull]

Bluskis,

Yes, agreed.

Although conventionally, the climb analysis is carried out parallel and perpendicular to the flight path, that is just because it is convenient for resolving the forces.

There is no mechanics reason why you can not choose to use any set of axes that meets your purposes.

What you are saying is certainly true.

[edited for brain fade]

CPB

[OzExpat]

CPB ... just a quick one before I blast off...

Okay, yes, fine, lift increases. But it does so - at least initially - because of the extra thrust. The wings don't start contributing to lift immediately. The difference is the thrust, which is why we teach the trainee to apply both a power increase and an AoA increase together. This means the aircraft enters the climb in a more smoothly controlled manner and then maintains the climb (yes, okay, up to a certain point).

But, clearly, as was also said earlier, in a vertical climb, the wings produce zero lift. So there's a point at which the wings cease to generate that force. The upward force ("lift" seems to be the wrong term in this context) is then based solely on thrust.

I believe its important for a trainee to understand where the lift comes from.

But let me ask you a question - in a 60 Degree Banked turn how many G are you pulling? I reckon its 1 more than straight and level, i.e. 2. By your reckoning it would be 1.
You've just lost me here. Any turn at all can hardly be defined as S+L flight. Or maybe I should've qualified that condition by using the full terminology :- "straight and level unaccelerated flight"?

Happy trails mate!