Total drag questions

[300-600]

I have two drag related questions.

1) I have a vauge recollection that total drag in the climb is higher than in level flight due to the increase in lift associated induced wing tip drag. Anyone confirm this and / or got any idea as to what the percentage increase in drag is in a low rate (say 500 ft/min) climb approaching say 35,000 feet vs level flight at 35,000 feet.

The only figures I have are for a 737 - 800 @ 70 Tonnes FL 350 at minimum drag speed (0.74 mach).

For this condition the total drag is 8,750 pounds.

2) The same Boieng chart shows the total drag curve at all speed ranges. The level of drag at mach 0.8 is the same as mach 0.67 for the level flight 35,000 feet 70 Tonne 737).

The second question is can I assume the total drag in a climb at mach 0.8 is the same as the total drag in the climb at 0.67?

Thanks in advance

[hawk37]

I'll take a stab at this...RMC, I think you may have it a bit backwards...

Lets consider the aircraft at FL 350, at 220 kcas. If it's climbing then one can expect the thrust vector to be canted up slightly, hence providing some force in the vertical direction. Therefore, for a steady climb, the wing will be producing less lift than if the aircraft were in level flight at the same speed, and thus the climbing aircraft is at a lower Coeff of lift, Cl

Hence, since a component of total drag includes induced drag (Di), and Di is roughly porportional to Cl squared, the aircraft described above will have less drag in the climb than in level flight.

So, this is in fact the exact opposite to what you were wondering. Further, a portion of induced drag is in fact from wing tip vortices, and thus the decrease in induced drag in the climb, (versus the same CAS at that FL in cruise), means that the wing tip vortices drag will also be less for the climb situation.

Caveats apply.

Is there any way to post a picture of this graph you have of the drag of the 737? Perhaps upload it to a site such as photobucket (Image hosting, free photo sharing & video sharing at Photobucket (http://www.photobucket.com)), then link to there from a reply to this post?

Perhaps others (Mr Enicalyth?) will take glee in shooting down my analysis, or can take a stab at your second question.

TDV

[gearpins]

1.in level flight thrust=drag & lift= wt
2.any increase in pitch/angle of A will increase lift thus climb
3.increase in pitch/angle of A will also increase drag thus we loose speed
4.we compensate speed loss with thrust.
to sum up in climb lift+thrust> drag+wt
waiting to be shot down :)

[Dick Whittingham]

Dunno what happens in real life but thr theory says that in a steady climb lift is less than in level flight - Wcos(climb angle) instead of W. Thrust is up, at Wsin(Climb angle) plus drag. Drag is reducing as climb angle increases until in a vertical climb it is zero lift drag only, as lift is zero.

Dick

[ChristiaanJ]

Let me have a go...

Let me define lift and drag as follows:
Lift is the force acting on the aircraft at right angles to the air speed vector.
Drag is the force acting on the aircraft at right angles to the lift, hence along the air speed vector.

In steady and horizontal flight, that means the lift points straight up, and is equal but opposite to the weight, which points straight down.
The drag points horizontally backwards along the line of flight, and is equal but opposite to the thrust, which is pointing forwards *)

In other words, the classic diagram we all know.

Now, what happens in a steady climb at the same speed?
The weight still points straight down.
But lift, drag and thrust now all have rotated over the climb angle.

Since we are talking about a steady climb, the weight is now balanced, not by the lift L (as defined before), but by Lcos(climb angle).
And that means that L has increased, which in practice is achieved by an increase in angle of attack (AoA, alpha).
And an increase in L means an increase in induced drag.

CJ

*) I've ignored all the second-order effects, such as the thrust line not being exactly along the line of flight, etc. which are not part of the basic question.

[ft]

W = L*cos(climb angle) + T*sin(climb angle) - D*sin(climb angle)

Climbing means thrust will no longer equal drag in a steady state. Hence, part of the weight is being carried by the thrust of the engines and lift will in fact decrease.

It is easier looking at in in a coordinate system fixed to the velocity vector. Assuming thrust to remain parallell to the direction of flight, the steady state equation perpendicular to the direction of flight:

L = W*cos(climb angle)

And again, we see that lift decrases with increased climb angle.

Or, for a more intuitive approach, analyse the extremes. Assume the climb angle to approach the vertical. Lift will go to zero. It thus makes sense to assume that lift will in fact decrease with increasing angle of climb.

[keith smith]

I think ft is right, and the vertical flight case illustrates the point simply. It is called VTOL, trouble is you need thrust a bit larger than weight:)

[ChristiaanJ]

I see I'll have to do the rest of the diagrams and maths before I can convince anybody...

T*sin(climb angle) - D*sin(climb angle) pretty well cancel out, and are an order of magnitude smaller than L and W.

ft,
Stick with the original coordinate system.
Otherwise your "L = W*cos(climb angle)" is wrong for a steady-state climb condition.

If you convert to a coordinate system locked to the velocity vector, the weight of the aircraft is no longer W, but W / cos(climb angle).

CJ

PS: VTOL is not really relevant here, no?
No lift, no drag. Just T=W.

[ft]

ChristiaanJ,
I do believe you will have to show all the diagrams and math. I note with interest how changing the coordinate system will change the weight of the aircraft. Will the local gravitational acceleration be altered through changing the coordinate system? If not, I regret not having discovered this while I was still doing mass properties! ;)

As for the relevance of VTOL, I trust it you do not contest the fact that lift will be zero when climbing vertically? If we assume you are right and that lift does indeed initially decrease with increasing climb angle, at which climb angle will lift start decreasing with increasing climb angle?

Cheers,
/Fred

[hawk37]

ChristiaanJ, you have defined lift to be different from the rest of us. Here is what you said:

"Lift is the force acting on the aircraft at right angles to the air speed vector."

And since the airspeed vector is not level for a climb, your definition of lift has a backward cant that is equal to the climb angle. However, in fact, lift is typically defined as the vertical component of the force produced by the wing, ie in a direction straight up, regardless of the airspeed vector. The force produced by the wing will normally have a rearward component, which is equal to the induced drag, and which is included as part of the total Drag.

ft is correct, and Keith's case of taking the situation to it's extreme also provides a simple example that supports it.

Lift force is vertical, opposite to weight. And since we're talking about an equilibrium situation, and thrust has a vertical component, then lift must be less that weight by the vertical contribution of the thrust.

[Mad (Flt) Scientist]

I'm with ChristianJ on the lift vector direction. Lift is normal to the velocity vector, drag along the velocity vector. It's not fixed in earth axes.

[hawk37]

Yes, now I think about it a little more, lift is generally portrayed as perpendicular to the velocity vector, not to the Earth's axis. Disregard my post above with respect to that. Thanks, MFS, for not flaming me on that one.

[Mark1234]

Ditto. (MFS) It's far easier to build and understand the picture that way, but I don't think it matters what you take as your frame of reference, so long as you treat all the forces in a consistent manner:

If you work parallel and perpendicular to gravity, weight is unfactored, thrust, drag and lift all have angular terms (parallel and perpendicular components). if you work parallel and perpendicular to flight path, the thrust, drag, lift are unfactored and weight has an angular term.

Using vertical flight as a thought experiment, the lift requirement reduces as we pitch to vertical flight, which should reduce the induced drag (but not parasite/form drag). However, as much as that is happening, you are also introducing a new retarding component. It's easiset working parallel and perpendicular to the aircraft path to see this as some component of gravity working normal to the path, and some component working paralell (aft) to the path - it's not drag per-se, but it requires additional power to overcome it, just like drag (or at the same power setting you slow down, reducing drag until some point where everything reaches equilibrium).

If you insist on working with reference to the gravity plane, in the climb you have:
Vertically, positive up:
weight (-ve), a component of thrust (+), a component of drag (-) a component of lift (+ve)

Horizontally, positive left:
a component of thrust (+), a component of lift (-ve), a component of drag (-ve)

That's much more complex to understand (draw it!), has the same resultant - you're increasing the -ve components, and have to increast the thrust, (or allow the aircraft to slow until drag reduces to match thrust), in order to have everything balance.

[edit], post appeared while I was posting!

[Mark1234]

I'll also have a stab at 2), though I'm guessing somewhat here - happy to be set straight!

The only way you can have the same drag at 2 differing speeds is if they're either side of the minimum drag point, recall a L/D curve is some sort of U shape with: On the left high drag dominated by induced drag courtesy of excessive AOA. On the right high drag dominated by parasite drag due to excessive speed. The L/D curve is the result of combining the parasite drag curve, and the induced drag curve. Somewhere in the middle there's a low point where the combination hits a minimum.

given that, and what we already decided about lift in the climb being slightly reduced: you're going to move the induced drag curve a bit to the left (the wing is compelled to generate less lift at any given speed - effectively you have a lighter aircraft). that will move the min l/d left, and closer to the .67M figure.

So, I believe that transitioning from level flight to climb will favour the lower speed from a drag point of view. I have no idea of the magnitude of the change, or if it's significant. I would also expect it would only really figure at high altitude, as at any lower levels, the speed at which the low point occurs (best glide effectively) will be well to the left of where I'd expect anyone to be operating.

Does that seem reasonable?

[balsa model]

CJ: I think that you might have made a manipulation mistake with your vectors or the trig.
With your definitions (and references), I get:
L = W*cos(angleofclimb),
D = T - W*sin(angleofclimb)
This happens to satisfy all extreme cases of horizontal flight and vertical flight (if power setting allows such a steady state).

(1) Total drag D is what the poster asked about.
My answer:
Some assumptions are missing in the question.
If he doesn't touch the throttles as he levels off, the aircraft will accelerate and D will progressively increase to match T.
If he keeps his airspeed as he levels off, he will have to reduce the throttles. In this maneuver, D will be constant throughout.

(2) The second question is even more open.
You can't maintain speed and transition to climb without touching the throttles.
If you don't touch the throttles, we still need to know something about the vertical speed. Are they supposed to be the same? You will need less angle of climb for the same VS the faster you go.

I'm not waiting for the return fire - I'm going to bed - bm.

[300-600]

Hawk 37, You asked for a link to the graph...can't send you there directly but SmartCockpit - Airline training guides, Aviation, Operations, Safety (http://www.smartcockpit.com/search) type in search 737 & drag and it will come up with the Boeing presentation on high altitude flight. I think it is slide 13.

[Dick Whittingham]

Answer 1: Lift and drag in a climb are less than in level flight at the same speed.

Answer 2: Yes, with qualifications. The normal drag curves are for level flight and therefore based on one specific weight/lift. In a climb lift is less than in level flight so the speeds taken from the level flight curves are no longer accurate. At small angles of climb the error is small, but as climb angle increases from zero Vimd and best rate of climb speeds reduce, reaching zero in a vertical climb

Dick

Edit for typo

[Keith.Williams.]

Now here's a really counter intuitive one:

As an aircraft goes from climbing flight to level flight at constant IAS what happens to the angle of attack?

[Dick Whittingham]

Hi Keith,

I'll leave it to Christian to answer your question!

Dick

[balsa model]

Now here's a really counter intuitive one:

As an aircraft goes from climbing flight to level flight at constant IAS what happens to the angle of attack?


Ignoring changes in air density, at constant IAS, AoA is a fuction of required lift. Since we have shown (some disagree) that in climb, required lift is lower, the answer must be:
AoA will increase in level flight.

And yes, it is counter intuitive.
Personnaly, I started suspecting that aerodynamics will never be intuitive to me when I realized that people who say "you climb with power; accelerate with pitch" are right. Now if I'm in doubt, I draw them vector balance diagrams, and write power balance equations. If I can't prove it, then I must say that I don't know the answer.

[ChristiaanJ]

Hi all,
As an ancient, I'm enjoying this, so I'll have to do some scribbling and calculating and posting this weekend to compose a proper answer!

Rainboe, welcome!
"total lift balances total mass"
Sure, but you forget again to define "total lift".
Your "total lift" points straight up during a steady climb (or it wouldn't be steady).
So your wing will have to produce an additional force perpendicular to the airstream.... it doesn't "know" it's climbing.
And with the wing providing typically ten times as much lift as the engine thrust (ballpark figure, obviously) it's the wing which will provide the additional force to balance the aircraft, rather than the T*sin(climbangle).

Part of the problem of this discussion is that there is no such thing as "lift" and "drag".
Strictly speaking there are only three forces acting on an aircraft (in the context of this discussion).
Weight (or mass if you like), always pointing down.
Thrust, acting pretty well along the airspeed vector (other effects being second-order).
Aerodynamic forces, acting mostly on the wings, which we like to split out as "lift" and "drag", but which again is not a scalar but a vector tilted backwards from the airspeed vector.

Thanks to all for joining in!
I'll try to come up with something 'clear' this weekend!

CJ

[ChristiaanJ]

"you climb with power; accelerate with pitch"
Yeah, and I'm supposed to be an aeronautical engineer and I still had some problems to get my mind around that....
So don't worry, you're not the only one!

Since we have shown (some disagree) that in climb, required lift is lower... Nope....
The "lift", in the sense of the vertical force needed to counter gravity is exactly the same as the aircraft weight (we're talking a steady-state climb).
So the force provided by the wing, which force is now tilted back by the climb angle, now has to be larger (the upwards force provided by the thrust is an order of magnitude smaller).
To increase that force, the AoA has to be larger.

CJ

[ft]

Allright, let's do the full math in your preferred choice of coordinate system then.

Earth-fixed steady state.

ca = climb angle

Vertical force balance gives

W + D*sin(ca) = T*sin(ca) + L*cos(ca) (1)

Horizontal force balance gives

T*cos(ca) = D*cos(ca) + L*sin(ca) (2)

(2) gives

(T-D) = L*sin(ca)/cos(ca) (3)

(1) gives

(T-D) = (W-L*cos(ca)/sin(ca) (4)

(3) + (4)

L*sin(ca)/cos(ca) = (W-L*cos(ca))/sin(ca)

L*sin2(ca)/cos(ca) = W-L*cos(ca)

L*(sin2(ca)/cos(ca)+cos(ca)) = W

L = W/(sin2(ca)/cos(ca)+cos(ca)) = W/(1/cos(ca)*(sin2(ca)+cos2(ca))) =
= W/(1/cos(ca)*1) = W*cos(ca)

L = W*cos(ca)

Now, where did I hear that before... ? :)

The point where you go wrong is when you mistakenly assume that the thrust can safely be ignored in the vertical force balance. Take a look at the relative rates of change of the vertical component of thrust and the vertical component of lift with increasing climb angle. The longitudinal component of the weight of the aircraft increases with the sine of the climb angle and thrust has to increase with it to keep the aircraft from decelerating.

The vertical component of thrust also increases with the sine of the climb angle. The vertical component of lift, however, will only decrease with the cosine of the climb angle. In order to maintain equilibrium, the magnitude of the lift force has to decrease.

[balsa model]

http://img230.imagevenue.com/loc437/th_97275_lift-in-climb_122_437lo.jpg (http://img230.imagevenue.com/img.php?image=97275_lift-in-climb_122_437lo.jpg)

Geometrically, it would appear that in equilibrium:
L = W*cos(AoC)

[fullyspooled]

I can't even spell aerodinamics, but I'm with the lift reduces camp.

In any steady state climb the accelerometer indicates less than 1 G. The steeper the (steady) angle of climb, the less the G force, until we reach the vertical where the G force (and stalling speed) reach zero.

I can't explain why, but surely if the acceleration force is reduced it must follow that lift has reduced also!

Standing by to learn something new today.

[Dick Whittingham]

Here's one more misconception. An aircraft G meter measures CN, the coefficient of normal acceleration. "Normal" means on the aircraft normal axis. In a steady climb although the aircraft is experiencing 1g the element on the normal axis is less than 1g. In a vertical climb the G meter will register zero

Dick

[chornedsnorkack]

An aircraft G meter measures CN, the coefficient of normal acceleration. "Normal" means on the aircraft normal axis.

So, does it mean that on takeoff acceleration and landing braking deceleration, the G meter will register 1,0, because the aircraft vertical speed and acceleration are zero and the accelerations along fuselage are not counted? And that during steady horizontal cruise in a slightly nose-up attitude, the g meter will register less than 1,0?

[ft]

"So, does it mean that on takeoff acceleration and landing braking deceleration, the G meter will register 1,0, because the aircraft vertical speed and acceleration are zero and the accelerations along fuselage are not counted?"

Yes.

"And that during steady horizontal cruise in a slightly nose-up attitude, the g meter will register less than 1,0?"

Yes.

"Yes Dick, you can calibrate your g meter as you like. For convenience, subject to normal surface gravity, you can make it 1.0. Out floating in free space subject to no thrust, you can make it zero. Then you can take that meter and put it on the surface of any planet to get a good reading of gravity relative to the Earth's surface, or stick it on an aeroplane in Earth's atmosphere in a steady climb, cruise or descent, with no acceleration effects in which case it would read 1.0!"

If the local gravitational acceleration is 1g and, assuming as is the norm a one-axis g meter, if the g meters sensing axis is aligned with the local gravitational field.

In the case of an aircraft which is not level, the g meter will not be aligned with the local gravitational field and will thus only register the component of local gravity parallell with the g meters sensing axis. Hence it will read less than 1 g.

[ft]

Rainboe,
if that is indeed the case then your accelerometer

a) does not measure the normal load factor, which is the load factor we are usually interested in in aviation

b) differs from most other such accelerometers employed in aviation.

I'd say you are the one throwing in unnecessary complications by turning away from the industry norms without clearly specifying so.

[Dick Whittingham]

Rainboe, does your dangly g meter read 1 positve g in all steady flight angles up to the vertical? And does it read 1 positive g in inverted flight? And what does it read on acceleration on takeoff or deceleration on landing?

Dick

[ChristiaanJ]

Rainboe,
Is your "g meter" just a thought experiment, or does it exists, and if so, is it used in aircraft?

I can't quote recent aircraft, but in the big white pointy one, the 'g' indicator ran off a one-axis accelerometer installed with the sensitive axis along the z-axis of the aircraft, hence it indicated normal acceleration relative to the aircraft coordinate system.
The accelerometer was installed on frame 56, so as close to the "average" CG as possible, to minimise pitch angular acceleration effects.
As it was, there was nothing specific about the accelerometer... identical accelerometers were used to sense longitudinal and lateral acceleration (for other systems). They were just mounted differently.

Your "g meter", as you describe it, would be "contaminated" by longitudinal acceleration or deceleration.

CJ

[Dick Whittingham]

Rainboe - you're blathering

Dick

[krujje]

Bit of a digression here. Just thinking about this theoretical straight vertical climb...

Post #7Or, for a more intuitive approach, analyse the extremes. Assume the climb angle to approach the vertical. Lift will go to zero. It thus makes sense to assume that lift will in fact decrease with increasing angle of climb.

In a straight vertical climb, wing lift will definitely not got to zero. If the aircraft is climbing vertically, nose pointing up, then the wings see a relative airspeed, therefore they produce a force perpendicular to the relative velocity vector. In this case the velocity vector is perpendicular to the ground, as the aircraft is moving straight up. So the force produced by the wings is acting to push the aircraft away from the vertical. It's kind of counterintuitive to call it "lift", since it's not acting to lift the aircraft, but it is wing lift nonetheless. So, in a straight vertical climb, lift is not zero.

Furthermore, if the climb velocity is not that great, then there is a good chance that the wings will actually be stalled, which means there will be airflow separation over the wing surface, and that means there will be lots of drag, not no drag. Therefore, the thrust from the engines not only has to act against the weight of the aircraft, but also against a good amount of drag.

Just a thought. Sorry for the digression.

[Von Klinkerhoffen]

If something is in steady state , it is in balence i.e. all accelerations acting on it cancel out . So surely it's a constant climb rate , as opposed to a steady state climb ? For the aircraft too climb , the total upward vertical acceleration i.e the total lift made up of wing lift componant and thrust lift componant , must be greater then the total downward acceleration due to gravity i.e. weight , therefore it cannot be steady state ! Steady state can only be applied to cruise where the aircraft neither climbs or descends , speeds up or slows down.


Krujje , quote:

Furthermore, if the climb velocity is not that great, then there is a good chance that the wings will actually be stalled, which means there will be airflow separation over the wing surface, and that means there will be lots of drag, not no drag. Therefore, the thrust from the engines not only has to act against the weight of the aircraft, but also against a good amount of drag.


:= Sorry have to disagree , I think you are confusing a stalled wing due to excessive AoA with one that is not producing lift due to insufficient airspeed . In vertical flight , AoA will be such that the wing produces no lift and therefore induced drag will fall to zero , of course all other forms of drag will remain. All upward acceleration will be provided by engine thrust alone acting against weight and the remaining drag....as for a rocket .

[ft]

krujje,
sorry, no joy. :)

In a steady state vertical climb, the lift will most definitely be zero. As you say, the lift is perpendicular to the trajectory through the air. With no opposing force, the trajectory would not remain vertical very long if the wing indeed generated lift. With lift, a vertical climb is not steady state.

You will have to reduce AoA to the zero-lift AoA for the climb to remain vertical. At the zero-lift AoA, the lift coefficient is zero and... and here it comes... there's zero lift.

With no lift being generated, critical AoA cannot be reached (as this would mean generating lift) and a stall is impossible. The wing will not stall until you go into a tail slide, at which point the climb will not be a climb nor vertical. Of course, steady state means no deceleration and no risk of going into a tail slide either, so stalling is in fact not possible given those boundary conditions.

As long as there's forward airspeed, there will be drag.

T = W+D in a vertical climb.

[ft]

Klinkenhofen,
steady state effectively means "no acceleration".

For an aircraft to climb in a steady state, the forces striving to accelerate the aircraft upward have to equal the forces striving to accelerate the aircraft downwards. Force balance means no acceleration means steady state.

If the upward forces were in fact greater than the downward forces, as you suggest, the aircraft would be accelerating upwards and not in a steady state climb at all.

[fullyspooled]

I can state with complete certainty that all military and civilian aircraft that I have flown and displayed in bygone days used accelerometers of a type that Rainboe is not familiar with, and they do indeed read less than 1G while in any steady state of climb, and precisely zero G while climbing (or descending) in the vertical.

Rainboe, think of it this way. Imagine you have a set of scales under your bum while seated in whatever airplane you fly. While in straight and level flight the scales would read exactly your weight, ie 1G, but while climbing in the vertical plane all of your weight would be acting towards your seat back, and not on the scales. The scales would therefore read zero. While in a steady state climb, they would read somewhere between the two extremes. The accelerometer we use in aircraft act in a plane aligned exactly as the scales I have just described, and they are in no way used for measuring the acceleration forces one may observe while braking after landing, or while accelerating along the runway during take off roll. For the gague to do this it would have to be laid flat on the aircraft floor!

You do however correctly point out (in my lesson of the day) that the accelerometer will indicate temporary accelerations caused by pitching moments. It will also indicate the constant acceleration (greater than 1G indication) throughout any duration of a balanced erect turn - or less than -1G while in a balanced inverted turn. For display pilots the accelerometer is used not only to measure the stresses on an aircraft, but also as a guide to determine when the aircraft will stall.

That much I did not have any misconception about, but I am confused when it comes to the relationship of wing loading and lift. Surely if the G force is reduced, so must the wing loading. As we have not in anyway reduced the wing area, how is it possible that the lift cannot reduce also?

Over to the scientists....

[balsa model]

original question:
2) The same Boeing chart shows the total drag curve at all speed ranges. The level of drag at mach 0.8 is the same as mach 0.67 for the level flight 35,000 feet 70 Tonne 737).

The second question is: can I assume the total drag in a climb at mach 0.8 is the same as the total drag in the climb at 0.67?


My answer - assuming that we are comparing climbs of equal RoC - nope.
The parasitic drag components will remain unchanged (no change in speed). The induced drag of the slower case will go down farther, since the AoC will have to be greater to achieve the same RoC, and from our previous discussion re question 1, the greater is AoC, the smaller is required lift and thus the smaller is induced drag. On top of that, in the slower case, the induced drag component is a greater proportion of total drag to begin with, making the above reduction even more poignant.

What do I win? :O
bm

[Mark1234]

Nothing! I answered question 2, with explanation back on page 1 :p

[krujje]

Von Klinkerhoffen;

Sorry have to disagree , I think you are confusing a stalled wing due to excessive AoA with one that is not producing lift due to insufficient airspeed

Yes. My bad. There will be no wing stall. Stall comes as a result of too-high AoA, not insufficient velocity. In straight-and-level flight, the two are linked, as minimum amount of lift is required to balance weight. However, in the theorized straight vertical ascent, thrust is balancing weight, so wing-lift is what it is. When one examines non-traditional scenarios, one must also re-examine assumptions taken for granted.

ft;

In a steady state vertical climb, the lift will most definitely be zero. As you say, the lift is perpendicular to the trajectory through the air. With no opposing force, the trajectory would not remain vertical very long if the wing indeed generated lift. With lift, a vertical climb is not steady state.

You will have to reduce AoA to the zero-lift AoA for the climb to remain vertical. At the zero-lift AoA, the lift coefficient is zero and... and here it comes... there's zero lift.

Sorry. I have to disagree. It is true that the trajectory would not remain vertical very long if the wing generated lift. And it is true that you will have to reduce AoA to the zero-lift AoA in order to achieve this. However, unless you have an all-moving wing, how do you change your AoA without adjusting your climb angle? Remember the constraint on this theoretical problem: straight vertical climb. If you adjust your climb angle to other than 90 degrees, you then have the probem of a thrust component not in the vertical direction, which has to be balanced by a force perpendicular to your trajectory.

All I'm trying to say here is that for most applications, this straight vertical climb we're talking about does produce some wing lift. You could have zero-lift in the case where the wing zero-lift angle is parallel to the aircraft longitudinal axis, but very few aircraft are designed this way.

[ChristiaanJ]

krujje,
"...how do you change your AoA without adjusting your climb angle?"

You're confusing climb angle and pitch attitude!
The climb angle is 90°.
For a "steady-state" vertical climb, lift has to be zero or the climb wouldn't stay vertical, so you have to fly at zero lift AoA.
On most aircraft, that corresponds to a slightly nose-down pitch attitude.
So while your flight path is vertical, the nose of your aircraft is not pointing exactly to the zenith....

Oh, and that leads to a nice one....
In our theoretical perfect steady vertical climb..... the thrust line is unlikely to be perfectly aligned with the flight path. So there will be a small but noticeable component normal to the flight path.... which you then have to compensate for with a small but measurable amount of "lift".

fullyspooled, re accelerometers, I fully concur.
But you said...
You do however correctly point out (in my lesson of the day) that the accelerometer will indicate temporary accelerations caused by pitching moments.Negative. If the accelerometer is installed in the right location, pitching moments accelerations become second-order.
Of course.... if you install it under the pilot's bottom... that's a different story.

ft, re zero lift... so sorry to contradict you slightly, above, because your reasoning is basically correct!

CJ

[ft]

"However, unless you have an all-moving wing, how do you change your AoA without adjusting your climb angle?"

You are confusing deck angle with climb angle. The climb angle will be 90 degrees, but the pitch angle will be less.

I now see that Rainboe with some additional elegance pointed out the exact same thing in the next post. :)

"If you adjust your climb angle to other than 90 degrees, you then have the probem of a thrust component not in the vertical direction, which has to be balanced by a force perpendicular to your trajectory."

Yes, and this compensation will be some residual lift. However, in this discussion we have been careful to specify that we assume the thrust to remain parallell to the direction of flight, a reasonable simplification to make in a discussion of basic principles of flight.

"All I'm trying to say here is that for most applications, this straight vertical climb we're talking about does produce some wing lift."

Not unless you factor in other forces in the aircraft xz plane perpendicular to the direction of travel it won't.

"Nobody is saying the wing does not produce lift in a vertical climb."

Yes, I am. As long as we stick with the basic assumption above. :)

[ChristiaanJ]

ft,
Each time, we're talking about what are essentially four vectors in the vertical plane: W, L, T and D.
So let's not fully neglect the fact that they are not necessarily at right angles all the time.

In steady vertical flight, L and the normal component of T are obviously second-order. So, I garee with you, to explain the basic state of affairs, T=W+D makes perfect sense.


But to get back to the original question, I'm still convinced that in a "typical" aircraft in a "typical" steady climb, the extra force needed to compensate W/W*cos(climbangle) i.e., a slightly higher vertical force, comes for 90% from a slight increase in "lift", by means of a slightly higher AoA.
Only 10% will come from the vertical component of the increased thrust - which will have increased slightly to compensate for the increased drag from the slightly higher AoA (second order effect).
It will have changed by T*sin(climbangle), which will still be far less than the "lift" effect.

Yes, I know I'm sticking my neck out... especially since I promised to look at it this weekend but had other things to do...

Just to clarify, when I talk about a "typical" aircraft and a "typical" climb, I'm talking about a Cessna, or a Boeing, where lift/weight is about 10 times thrust/drag and a climb angle of something like 5° to 10° at the most.

Talking about vertical climbs confuses the issue, since by then you need engine power/thrust sufficient to maintain steady vertical flight, and the aerdynamics are no longer the same at all.

CJ

[Mark1234]

Q2:

No, the drag will not be the same at both M.8 and M.67 in the climb; The drag will be lower at .67 climbing than it will be at .8 climbing; i.e. climbing will favour the lower speed regime.

[balsa model]

Mark1234 said:
Nothing! I answered question 2, with explanation back on page 1

Ooops.. I see it now: post #14. Sorry, it didn't get through to me, at the time - I was too focused on question 1 controversy.
Honestly though, we might have to deduct some points for saying "you're going to move the induced drag curve a bit to the left". Under s.s. climb, induced drag curve shrinks down by a factor of cos(AoC). The portion of total drag curve that is dominated by induced drag will appear to move to the left.
(I'm really crying for attention here :O )

[Mark1234]

D*mmit! Bang to rights there... :)

[FE Hoppy]

I'm loving this thread! Thanks chaps.

Could I ask about a glider going by virtue of a thermal from a flight path angle of 0 to a flight plath angle of >0 at the same IAS?

More or less drag? Lift?

[Mark1234]

Sure :)

1) Transient increase in AOA while it's being accelerated, which corresponds to a nice feeling in the pilot's butt.. (probably a continued increase in AOA too due to beardy glider pilot slowing and cranking it onto a wingtip - but we'll ignore that)
2) ASI will surge (I have no idea why, that's empirical)
3) Once established in the theoretical climb, without circling, All will be as before. The glider is moving in exactly the same manner realtive to the airmass that it was, the 'deck angle', AOA, drag and lift etc., will all be as before, just that the whole lot's going up.

<also a glider pilot, without beard... and yes, I'm having a very bored day at work!>

[300-600]

Guys...thanks so much for the replies to this thread. Have learnt a lot. I hear a lot of non believers knocking PPRUNE as a source of technical info...the quality of your replies has been excellent.
Note to mod (please can I have that free personal title now):D

[ft]

But to get back to the original question, I'm still convinced that in a "typical" aircraft in a "typical" steady climb, the extra force needed to compensate W/W*cos(climbangle) i.e., a slightly higher vertical force, comes for 90% from a slight increase in "lift", by means of a slightly higher AoA.

a) The required net vertical force generated by T, D and L will still be the same as in level flight: W.

b) I don't know where you get W/W*cos(climb angle) from. It's the same thing as cos(climb angle) BTW... and why would you ever need to compensate a plain cosine?

c) You saw all the math above, giving you L = W*cos(climb angle). If you still think lift needs to increase, perhaps you should point out the error in the calculations... ?

[ft]

As for #2, I'll have a more analytical stab at it:

In both cases, total lift generated will decrease by the same amount as a climb is initiated. L = W*cos(climb angle), no speed dependency.

Parasite drag will essentially remain the same and can thus be ignored in this context.

As for induced drag,

L = S * rho/2 * V^2 * C_L

or, solving for the lift coefficient,

C_L = L/(S * rho/2 * V^2) = k1*L/V^2 (k1 is a constant)

The induced drag coefficient is proportional to the lift coefficient C_L squared,

CDi = k2 * C_L^2 = k2 * k1^2 * L^2 /V^4 = k3* L^2/V^4

(Induced drag is proportional to the induced drag coefficient)

The rate of change of the induced drag coefficient as the lift changes is calculated,

dCDi/dL = 2 * k3 * L / V^4

We see that V^4 remains. Conclusion: The rate of change of the induced speed as the amount of lift generated changes is highly dependant upon the true airspeed. The amount of change of total drag as you enter a climb is very unlikely to be the same at M.67 and M.8. At a high speed, the change in induced drag will be a lot smaller than at low speed.

It's late(ish). If I messed something up I'm sure I will be corrected. :)

[ChristiaanJ]

Rainboe,
Don't tell me....
I should print out this thread, sit down with pen and paper and a calculator, and try to get my ideas sorted out, rather than answering all these posts off-the-cuff.

By the time you talk about a really steep climb, you're talking about hanging on your prop, or sitting on your exhaust... i.e., a thrust-to-weight close to 1, and your wing normal force is becoming pretty well irrelevant in your steady climb.

CJ

And... oh shoot....
Suddenly saw I missed half a dozen posts, and it's getting late at UTC+2 (I think). See you tomorrow... ?

[john_tullamarine]

Note to mod (please can I have that free personal title now):D

.. too far down the food chain to be of much use to you there ..

... than answering all these posts off-the-cuff.

A lesson which just about all of us learn early in our careers .. in my case, being a bear of very little brain (as Milne would have it) .. I think it took me a tad longer than the average teddy bear to learn that lesson as a young engineer ... with several monumental clangers along the learning path.

[ChristiaanJ]

john,
Welcome to this thread!

I think you'll find two kinds of bears of little brain here....
The young withersnappers (sp?) that know it all, and the ancients (of which I am one) that still remember something, and don't mind getting it wrong, because they no longer have a boss looking over their shoulder.

Please, let's keep this one going, if only between us, until we all agree?

Christian

[john_tullamarine]

.. somewhere in the past I recall it being put in terms of "age and treachery will always win out over youth and enthusiasm .. ?" :}:}

[ChristiaanJ]

.. somewhere in the past I recall it being put in terms of "age and treachery will always win out over youth and enthusiasm .. ?"
John, PLEASE? This is a fun topic, even if we don't all agree...
Keep American politics out of it.....
We would like to keep this in "Tech Log", but this way it'll be relegated to "Jet Blast" in no time!

I think the current participants would like to continue kicking the subject around until we can all agree.

Christian

[krujje]

This discussion would be so much easier with a blackboard and chalk...

[Permanent Standby]

I like drag, anything in red or with sequins....

[Mark1234]

Christiaan,

I can see where you're coming from, and at face value it's kinda compelling, but I believe your simplifying assumptions are flat wrong, hence the problem - you've stated a couple of times you're ignoring '2nd order effects of thrust vector etc', yet you do pay attention to the 2nd order effect of the tilting lift vector. That's not cricket - if the 2nd order effect of 1 is significant, so is the other.

Staying well away from the vertical, I think/hope we all agree that in level flight, T=D, L=W (and we'll leave it at that).

Rotate to a 10degree climb, stabilise so we're unaccelerated.

Now, you rightly point out that the W vector is unchanged (I'm trying to stay in your earth parallel frame of reference), and the L vector is tilted slightly back. But the T vector is also tilted slightly up, and the D vector down; sure, it's small, but so is the backward tilt of L. Don't ignore it :)

So we now get that (vertical component of L)+(vertical component of T)-(vertical component of D)=W Nothing too radical so far.


Now the question is, 'does the component of T' make up for the loss of L, and vertical component of D?

If we climb at the same speed as we were in level flight, T must increase - we all know that intuitively, but the maths is harder in your frame of reference:
(H component of D)+(H component of L) = (H component of T).

I don't have any numbers, but I'd argue that that increase in T is what makes the (vertical component of T) term larger than the reduction in L due to the tilt of the lift vector.


As previously stated you're free to pick any frame of reference you choose, so long as all terms are treated equally to that frame - it's common practice in maths to choose a frame that simplifies the calculations:

It's much easier to visualise if you take the aircraft as the frame of reference, then the only thing that factors is W; Pitch the aircraft and whole frame of reference; T must increase to combat the component of W that is now acting rearward along the deck plane, and L must decrease because it is combating a reduced component of W.

And no, I don't know it all.. I'm working the maths as I go along. Quite happy to be proven wrong, and learn something.

[fullyspooled]

ChristiaanJ, you quoted me...

fullyspooled, re accelerometers, I fully concur.
But you said...
Quote:
You do however correctly point out (in my lesson of the day) that the accelerometer will indicate temporary accelerations caused by pitching moments.
Negative. If the accelerometer is installed in the right location, pitching moments accelerations become second-order.
Of course.... if you install it under the pilot's bottom... that's a different story. And I have pleasure in responding......

My Dear Sir, and I mean that most sincerely, but I think you are now guilty of splitting hairs, or dare I say it, having gotten slightly confused! (Should I say ooops now, I wonder?)

If the accelerometer is installed "in the right location," and by that I mean vertically on the instrument panel, it will indeed reflect accelerations occasioned by pitching moments! By the very nature of "pitching," assuming that one has sufficient airspeed to alter the trajectory of flight in the pitching plane, an acceleration takes place - and any such acceleration will be measured, and indicated on the accelerometer mounted as I have described.

The accelerometer, as has been previously mentioned, cares not at which attitude the airplane is, it simply measures acceleration in the "pitching" plane of an aircraft that is IN MOTION. If the aircraft is in motion and pitches, there is undoubtedly an acceleration, and I promise you that it WILL be measured by an accelerometer installed on the instrument panel!

However, if the aircraft is NOT in motion, you can pitch it all day long and no accelerometer in the World will notice a jot of change. An accelerometer does not in any way detect changes of pitch - that I know, and I can further testify that when correctly flown, several auto -rotational manoeuvres that involve massive pitch changes at VERY low airspeeds result in very small accelerometer deflections - due only to the fact that "directional" motion is limited to almost nothing. The lomcevak is a classic example.

Perhaps I was guilty of not qualifying my statement as thoroughly as you seem to do so naturally, but I trust that we now agree. If we do, I have another topic of discussion that I just know you will love to debate.........once this thread has finished.





.

[Dick Whittingham]

Karl Popper has pointed out that the more information a statement contains the less likely is it to be true. The statement " If the aircraft is in motion and pitches there is..an acceleration" is a sweeping generalisation and therefore unlikely to be true. A lot of qualification is needed here.

For example, if the accelerometer (or aircraft vertical axis acceleration sensor) is mounted at the center of rotation of the pitching movement there will be no record of the angular change in pitch. If the accelerometer is a simple spring and bob mounted on the instrument panel it may well indicate a small acceleration as the rotation in pitch is read as a linear vertical axis movement at the instrument location.

If the change of pitch occurs at some point in the flight regime where a change in pitch generates a change in lift then the pitch change will be followed by a linear acceleration on the vertical axis, read by either type of measurement system. However, if the pitch change does not result in a change of lift then there wil be no follow-up acceleration on the vertical axis.

So there is at least one set of factors that would result in a pitch change of an aircraft in motion recording no acceleration at all.

Where in the flight envelope would a change in pitch not reult in a change in lift? How about the flat bit of the CL/alpha curve around CLmax?

Going back, we never did fully explore the various linear and rotational forces that have to be brought to zero to sustain a true vetical flightpath. But that is for all you youngsters

Dick

Edited to correct "Z axis", which is an earth axis, to aircraft vertical axis.

[KristianNorway]

This was an interesting subject..

Initially my hunch was with ChristiaanJ, but as of now I am not sure if I completely agree.

Since there is much discussion about definitions in this thread I will state my own. I am certainly no aeronautical engineer, but I see these definitions as a prerequisite for the discussion.

I will use thrust (T), drag (D), lift (L) and weight (W). We have a vertical and horizontal axis on our aircraft through the center of gravity. Weight is always directed towards the center of the earth. Thrust and drag are for simplification aligned with the flight path, and lift is directed perpendicular to the flight path. I will also call the angle between the flight path and the horizontal through the center of gravity the climb angle (a).

To then sum up the positive vertical vector during climb we have:

T*sin (a) - D*sin (a) + L*cos (a) - W

This would give a fixed lift for any given speed regardless of pitch-angle and thus the induced drag will be the same as for that specific speed in horizontal flight.
The negative component of drag ( D*sin (a) ) and reduced vertical component of lift ( L*cos (a) ) must from what I see be countered by the increased positive component of thrust ( T*sin (a) ).

So.. neither increased nor reduced induced drag during climb.

As stated earlier I am only a layman on this subject, so any corrections making me wiser are very welcome.

[ft]

Kristian,
welcome to the discussion!

"T*sin (a) - D*sin (a) + L*cos (a) - W"

Almost correct. You included all the factors but forgot the condition, as specified by the requirement for a steady-state climb:

T*sin (a) - D*sin (a) + L*cos (a) - W = 0

"This would give a fixed lift for any given speed regardless of pitch-angle"

Sorry, it will not. T will increase while D will remain largely constant for small changes of a, especially at high airspeeds (see the answer to OP Q2!). W will, obviously, remain constant. L will reduce, and the condition of vertical force balance as specified by the equation will remain satisfied.

BTW, you defined a as climb angle and then proceeded to make conclusions about the effect of pitch angle from the (half... ;)) equation including a but no reference to pitch angle. Easy to mess up in this, innit? :)

[ChristiaanJ]

OK, hands up to whoever said the real issue is the lift vector being turned back over (climbangle) degrees.

Everything else is secondary, except that the thrust angle is turned up over the same angle.

If you feel like doing some scribbling, try with an aircraft of 2000 kg, a CL/CD of 10, and a climb angle of about 6° (sine = 0.1, cosine = 0.995, so practically =1).

I'll try and clean up my scribbles and post them, but here goes in the meantime.

In level flight, drag and thrust are 200 kgf.

Start a steady climb at abt 6° climb angle (nothing excessive, that's 2000 ft/min at 200 kts).

Now you have to drag 2000 kgf upstairs at W*sin(ca), i.e., 200 kgf more.
Aye, there's the rub.....

So T now becomes 400 kgf, and the vertical component = T*sin(ca) = 40 kgf.

Since we're talking steady state, the component from the lift will have to be reduced by the same 40 kgf. But.... L=W*cos(ca) is only about 4kgf less than W. So we have to reduce the lift by another 36kgf to get back to a steady state.


Since 2000 kg, CL/CD=10, and a climb at 6° are hardly excessive, it would seem Rainboe's intuition about looking at a steep climb was right after all....

Now to question 2.....

[ft]

ChristiaanJ,
welcome to the dark side!

However,

L = W*cos(ca)

does hold true. You forgot to factor in D*sin(ca) in the world-vertical force balance. Have a look at balsa model's diagram in post #26.

Thanks for the credit for post #6 and #58. ;)

[ChristiaanJ]

ChristiaanJ,
welcome to the dark side!LOL.... But I don't mind admitting I got it wrong.... Nothing beats pen and paper and thinking about it, rather than "off-the-cuff".
However, L = W*cos(ca) does hold true.Right. My mistaek was not looking up cos(ca) for something like 6° climb angle, which is pretty well negligeable compared to sin(ca), hence negligeable in the vertical force balance relative to T*sin(ca), even with T ten times less than W.
You forgot to factor in D*sin(ca) in the world-vertical force balance. Have a look at balsa model's diagram in post #26.Oops. Yes, I goofed there. But even in my quick-and-dirty example, T-D is still 200 kgf, so the vertical component is still 20 kgf, so the lift will still have to decrease by about 16 kgf.

Thanks all! I'm enjoying this.

CJ

[Mark1234]

In all fairness Rainboe, not everyones mind is wired to think that way / can do that - it's a visual 'I see' thing. Sometimes to get the message across you have to speak *their* language, not your own. Bit like someone trying to convince you or I with trigonometry when all they need is a decent picture.

[KristianNorway]

Hello ft, glad to be a part of the discussion!

You are quite right:

T*sin (a) - D*sin (a) + L*cos (a) - W = 0

That's the total equation.

And my intention was of course to refer to (a) as 'climb angle' throughout the post.

But still my point stands. If one climbs with the same speed and the same weight as a similar plane would for level flight the angle of attack would be the same in both situations. Therefore I haven't included it in the equation.
Assuming then the airspeed is the same the force perpendicular to the thrust is the same. So the lift vector is the same.

The vertical component of lift - L*cos (a) - will on the other hand be less. By increasing thrust and hence the vertical component of thrust - T*sin (a) - one compensates for the reduced vertical component of lift.

[Mark1234]

If one climbs with the same speed and the same weight as a similar plane would for level flight the angle of attack would be the same in both situations. Therefore I haven't included it in the equation.

And there's where you went wrong :) Humour me - consider the problem with the aircraft deck angle as your frame of reference. Forget factoring L, D, T, and look at what happens to W...

weight on the wings reduces, so the AOA will reduce. Still works in your frame of reference, but you need to plug in real numbers to see how it comes to pass.

[ft]

Allright, it is Friday. Time for some entertainment! I was stuck in a dull meeting and ended up scribbling a few illustrations for this thread. Here’s how most people seem to think about it at first:

http://img127.imageshack.us/img127/5281/earthreferencedsmalluv4.th.gif (http://img127.imageshack.us/my.php?image=earthreferencedsmalluv4.gif)http://img127.imageshack.us/images/thpix.gif (http://g.imageshack.us/thpix.php)

Same old reference system we grew up in (the antipodeans excepted), so I don’t suppose that is too surprising. The problem is that intuitive understanding of the issue at hand is not obvious and if you try to go the analytical route, you end up with the trigonometrical mess I posted a while ago (and yes, Rainboe, some of us do both the nice piccies and the analytical part behind closed eyelids while holding a nice cuppa – no promise not to spill though!).

If you instead think about the climb case in the aircraft reference frame, like this:

http://img255.imageshack.us/img255/6204/aircraftreferencedsmallju1.th.gif (http://img255.imageshack.us/my.php?image=aircraftreferencedsmallju1.gif)http://img255.imageshack.us/images/thpix.gif (http://g.imageshack.us/thpix.php)

everything becomes much clearer. Suddenly, the lift decrease (and thrust increase) is easy to understand intuitively. Even sorting it out analytically becomes a breeze. The only real problem is that the water will seem likely to spill out of the lake, but that’s a small sacrifice to make on the altar of aerodynamical understanding.

Yes, you can all have my autograph when I become famous for my art!

[ft]

But still my point stands. If one climbs with the same speed and the same weight as a similar plane would for level flight the angle of attack would be the same in both situations.

That's where you are trying to disguise an assumption as a fact. On what do you base the assumption that AoA would stay the same after entering a steady climb?

AoA will, under the specified conditions, have to be changed to whatever it needs to be to maintain equilibrium. And that does mean a decrease... :)

Again, doublecheck with the extremes and if it doesn't match up, you're probably doing something wrong. Will the AoA be the same in a vertical climb as in level flight?

[KristianNorway]

Indeed.. sounds right. Do you have any way of proving this? I would very much like to understand the fact better.
It would be interesting to see how much the climb angle would affect the AoA.
I would estimate the effect to be fairly small, but still.
In any other case the AoA would be directly connected to airspeed.

It's interesting because it tells us how much of an effect that will have on lift during climb.

[JABBARA]

Hi

I am on leave now.

At my first flight soon (A 330), I will check the AOA from ACMS page for same weight and IAS: Initially at level flight and immediately after this, at stabilized climb.

I hope this will put a full stop to this discussion...

Regards.

[ChristiaanJ]

Rainboe,
Been there, done that.
Initially I expected a something squared to jump out of the woodwork and bite...
I've now come over to the dark side :)

But....
JABBARA,
Can you give us a typical figure for the climb angle (or vertical speed and TAS) of your stabilized climb?

For instance, for a climb at 200 kts TAS and 2000 ft/min, the climb angle is about 6°. Since the cosine of 6° is about 0.995, the lift (normal to the flight path) at that sort of climb angle is reduced only by a very small amount (see all the earlier formulae...), and so is the AoA (we're talking a percent or so).
So unless you do some careful flying and 'measuring', you may not see the effect at all.

CJ

[ChristiaanJ]

Rainboe,
Don't over-simplify... the increase in alpha is not the same as the cosine(climbangle). But your basic intuition of decreasing alpha with climb angle is right (except possibly for something with L/D less than one.... I'll have to look into that....)..

To all,
OK, promises, promises...
Here's my scribble.
http://img.photobucket.com/albums/v324/ChristiaanJ/Total_drag.gif
Not quite so artistic, so let me explain.
What we are looking at, are the horizontal and vertical force balance (steady state, remember?).

I've been looking at the other extreme from Rainboe's example, i.e., the typical climb rather than the vertical climb.
With a climb at 6° (my earlier example) or even 15° (cos = 0.97, quoted by Rainboe)...

Horizontally:
- the drag component barely changes,
- thrust changes a lot, because suddenly the thrust has to "drag" L*sin(ac), the horizontal component of L, "upstairs".

Vertically:
- the weight doesn't change,
- the drag component is now tilted downward,
- the thrust component is tilted upwards, and thrust is now larger than the drag, so the net result is an upwards force.

To get back to a steady state, we'll have to reduce L*cos(ac).
After plugging in the rest the maths... that comes down to reducing L, hence the AoA.

So, conclusions?

- At the typical climb angles of most aircraft, be them Cessna or Airbus, you will barely notice the change in angle of attack (it's all cosine something, and the cosine is almost 1 (one) ).
- You will certainly notice the difference in thrust (or engine power - which is the same in this context) to maintain a steady climb under the same conditions as in horizontal flight.
- If you're a fighter jock with more puff behind you than your aeroplane weighs... have a go at all means to try the other extreme of our discussion. But remember what we said. You may need to pitch down a bit for that perfect vertical climb.

CJ

[JABBARA]

Chritiaanj,

I will try to record as many as paramaters when I do this test. But of course to be able to read a remarkable AOA difference in two cases, I will try do it as low altitude as possible (to get a significiant difference between cruise and climb thrusts) and as close as possible to Green Dot speed (to get a noticable body attitude during climb). Obviously I should comply with ATC requirement as well. I hope ATC instructions coincide with my goaled flight conditions.

I will try to record V/S and TAS as well, but in this issue, I do not consider they are so important as body attitude.

Anyway, I will record and you will decide.

Regards

[ft]

In a vertical climb, AoA for all intents and purposes would be zero,meaning the fuselage would have negative pitch (ie be pointing at about 86 degrees pitch whilst plane is climbing vertically)

Rainboe,
todays nitpicker comment: In a vertical climb most aircraft (which have cambered wings) will in fact have a zero-lift AoA which is negative rather than zero, giving you a few additional degrees nose down relative to vertical to add to those caused by the angle of incidence. ;)


Don't over-simplify... the increase in alpha is not the same as the cosine(climbangle). But your basic intuition of decreasing alpha with climb angle is right (except possibly for something with L/D less than one.... I'll have to look into that....)..

ChristiaanJ,
Rainboe is spot on with his assumption which is not at all an oversimplification. In the normal flight region, lift for most wings is more or less linearly connected with the AoA. Hence it is perfectly valid to assume AoA = cos(climb angle) when L = L_level_flight*cos(climb angle).

If the lift coefficient curve is highly non-linear, Rainboe's assumption won't be correct. However, for your typical wing in normal speed flight, that's not the case.

Tried tilting your diagram so that the lift vector is poing straight up and redoing the calculations yet, as in my very artistic sketches before?

Horizontally
Thrust equals drag plus the longitudinal component of weight

Vertically
Lift equals the flight path normal component of weight.

Much easier, methinks.

L/D isn't a factor at all in how much the lift will decrease. We know exactly how much lift will decrease. The thing L/D in the current flight conditions will tell us how much drag will decrease as a result.

Jabbara,
thanks for volunteering. However, as has been pointed out, it will be surprising if your AoA sensor can sense an AoA decrease of a few hundreths of a degree. Don't worry about the body attitude, it doesn't matter. Only climb angle does. However, slower will give a larger absolute change in angle of attack, everything else remaining equal.

Will the A330 tell you approximately how much thrust you are generating? That's a more interesting parameter to compare, as the change should be close to W*sin(climb angle).

TAS and V/S are interesting, as they will tell us the climb angle. Of course, any change of AoA will cause ASI errors... which again might or might not be compensated for by the ADC based on if it can sense the reduction in AoA or not. Not that I think these factors will be significant at the huge AoA changes we are considering here. :)

Oh, the joys of flight testing. Todays quiz: We once (well, more than once but bear with me) experienced climb performance data from a performance flight test which was way outside of the normal parameters. The aircraft just climbed as a bat out of h_ll for a segment of the recorded climb. What do you think happened, and how were we able to determine the cause and compensate?

[hawk37]

ft, increasing headwind will do this. Wouldn't most test aircraft have inertials that would record this?
Windshear recognition comes to mind as a similar result of changing air mass velocity.

[ChristiaanJ]

ft,
Re the quiz:
My guess would be either thermals, or a standing wave flying parallel to a mountain ridge.

A cunimb might also do the trick, but I doubt you would do perf test flying near a cunimb.

CJ

[ft]

hawk37 nailed it, the aircraft climbed through a layer of gradual wind shear.

No inertial sensor worth its salt nor any external tracking as it was a line aircraft performance flight test.

The solution to these problems is to always fly the same climb on the reciprocal course along the same track. Looking at the data from the two climbs together, it is then obvious what happened.

[Rainboe]

ft, did you not replicate with
Rainboe,
todays nitpicker comment: In a vertical climb most aircraft (which have cambered wings) will in fact have a zero-lift AoA which is negative rather than zero, giving you a few additional degrees nose down relative to vertical to add to those caused by the angle of incidence.
with what I wrote in the previous post:
Originally Posted by Rainboe
In a vertical climb, AoA for all intents and purposes would be zero,meaning the fuselage would have negative pitch (ie be pointing at about 86 degrees pitch whilst plane is climbing vertically)?

We can produce all sorts of fancy formulae (which really won't mean anything to anybody without a diagram!), but the answer will be- AoA will be in proportion to Cos climb angle. A 15 degree climb will have .97 of level flight lift- you know it, I know it- I feel it in me bones!

[ft]

Rainboe, yes, I was replying to you. AoA will typically be a few degrees negative in a vertical climb rather than zero, due to camber. Fuselage pitch relative to the airflow will be another few degrees more negative due to angle of incidence.

Of course this means I goofed earlier. AoA in a climb won't be given by

AoA_climb = AoA_level*L_climb/L_level

at all, even though the relationship is essentially linear.

(AoA_climb-AoA_zero_lift) = (AoA_level - AoA_zero_lift)*L_climb/L_level

should hold more or less true though!

As for diagrams, you have them earlier in this thread. In addition, this is rather obvious to a lot of people without diagrams so you're making a rather broad statement there. After all, this is only the basic T/W/D/L diagram.

[ChristiaanJ]

It seems we've now dealt with the original question, and we've all agreed lift and AoA decrease in a steady climb, even if only very little for small climb angles in a 'conventional' aircraft.

So we're now dealing with Rainboe's pet situation, a steady vertical climb.
Admittedly, at first sight we are just talking T=W+D+L, but since none of those vectors point in exactly the right direction, another diagram may be called for?

CJ