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Wilfred
19th Apr 2001, 13:17
The books say that the stall speed increases with altitude. Have I got the reason right?

1/2RhoV2SCl. For a stall in a given configuration, S will remain the same, and Cl(a combination of wing shape and incidence) will be the same. Stall always occurs at a certain incidence. So if air density (1/2Rho) decreases with altitude, then the airspeed at the stall would have to be higher to balance the equation.

Please can you put me straight folks.

Smurfjet
19th Apr 2001, 14:37
Its the same reasoning I came to, regarding stall at increased weight. I'd be interested in more input for both cases.

SJ

[This message has been edited by Smurfjet (edited 19 April 2001).]

supermunk
19th Apr 2001, 16:10
An aircraft will stall at the same IAS irrespective of height (more or less). As you go higher, for a given IAS, TAS increases due to the reduction in air density. In extreme cases (the TR1 is one of them) you can have only a very small speed range to aviate in between stall amd mach buffet.

Smurfjet
19th Apr 2001, 17:24
Ah! True but irrelevant ;)You're forgetting the basics, Lift = weight and air density :)

kabz
19th Apr 2001, 17:53
Not so fast smurfjet, supermunk has it right. IAS (indicated airspeed) will show a similar speed approaching the stall, to that at lower altitudes, simply because the amount of air the wing sees near the stall and the amount the airspeed indicator sees are both similar to lower altitudes. Higher TRUE airspeed takes care of the lower air density...

Lots of books cover this stuff, and go into the real detail of angle of attack, load factor and why planes suddenly start to feel really unsteady towards a stall.

Here's a real good reference:
http://www.monmouth.com/~jsd/how/

Smurfjet
19th Apr 2001, 19:12
Right, I mixed my TAS and IAS again, arrgh :mad:
Now thats irrelevant ;)

SJ

fireflybob
19th Apr 2001, 21:14
Aren't you forgetting compressibility effect?
If you stall at (very) high altitude then although the EAS (Equivalent AirSpeed) will be the same at Sea Level, the IAS will be somewhat higher - least that's what I seem to recall from a long time ago!

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Golf-Kilo Victor
19th Apr 2001, 22:34
I thought Wings stalled at the same Angle of Attack, not Airspeed..

Of course your all going to laugh at me now, cause you are all aeronautical engineers....at leaste that's what it sounds like...

kabz
19th Apr 2001, 22:42
GKV you're dead right. I think people above were assuming a gradual pull back of the stick with power off whilst maintaining level flight, in standard student manner.

If you rip the controls back, yup, it'll stall at the critical angle of attack, no ifs no buts.

I keep meaning to try it, but you should be able to get a recognisable stall out of a kite, if you jerk the string(s) hard enough.

... ahh "Peter Powell Stunt Kite, where are you now ??"

Smurfjet
19th Apr 2001, 23:08
At the critical angle the aerofoil (wing) is stalled.

Just before the critical angle I have maxLift. For a given weight = maxLift, I'm still flying level.

Now this maxLift is = 1/2 rho S V^2 Cl

Increase my weight by 20% gives me weight &gt; maxLift.
Brings me to the conclusion that unless I increase IAS (1/2 rho V^2) I will not be able to maintain level flight. I'd say as a pilot I would call this a stall? Bearing in mind I have no power left to increase my IAS.

Yank the stick backwards and you are beyond the critical angle and the aerofoil (wing) is stalled.

So from this I will try to answer the original question...

Well maybe not, there is still a missing link somwehre :)

Ball in your court gentlemen, time to dig mechanics of flight ;)

fly4fud
20th Apr 2001, 01:47
and as a practical example, when flying heavy at hi level (say 410) with the A-310, I remember having seen the coffin corner as small as 7 kts! You then just pray for calm air...

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... cut my wings and I'll die ...

static
20th Apr 2001, 09:59
Fly4fud,
These 7 knots are between your "yellow bands", so that`s not taking into account your 1.3 margin over Vstall and machbuffet.
The coffin corner will actually be much, much bigger.
But in these circumstances I`m also hoping for still air.

Wilfred
20th Apr 2001, 12:24
GKV nope, just a pilot!

DP Davies writes in Handling the Big Jets (or the Bible)that the indicated stall speed increases for two reasons.

1. Due to compessibility effects causing a larger difference between EAS and IAS. That I can cope with. But...

2. "The actual EAS stall speed increases due to Mach No. effect on the wing. At very high altitude the EAS stall speed occurs at a significant Mach No.(180 knots = 0.61 Mach No., for example); the pressure pattern is disturbed and a higher stall speed results"

Could someone explain, in easy words writ large for pilots, the second one please? Ta.

Also, was my opening assumption incorrect; nobody actually said for sure.

BOAC
21st Apr 2001, 01:46
Wilfred, you know that a stall develops when the airflow can no longer flow smoothly over the wing upper surface. Remember the airflow speeds up over this part. As you climb, the increased speed of the air gets towards the point where compressibility starts to have an effect, and once shock waves start forming the airflow begins to separate from the wing = stall. Used to be known as a 'shock stall'.
It becomes easy to stall a wing in a turn in these conditions, as the airflow speed further increases due to the increased lift generated by the wing.

John Farley
21st Apr 2001, 22:05
Wilfred

Another way to look at your para 2 point is to say that in general compressibility effects are seldom good news. They reduce aerodynamic efficiency in various ways. In this particular case the stalling angle of attack of a wing in compressible flow is a little less than the same wing in incompressible flow. This lesser angle is naturally reached at a slightly higher speed.

JF

alapt
22nd Apr 2001, 12:50
BOAC, I do not think Compressibility, unless you are at the critical mach number will enter the equation here. (Remember that is why we have SWEEP to delay the onset of compressibility) At high and low altitudes, the pressure distribution will in effect move the Aerodynamic center on pressure forward. (Just past the max Lift/Drag curve) At that critical point, flow separation will occur and you will eventually stall. The main thing to know is the difference between IAS, TAS, and EAS. Very simple I know, but the details would take three pages.

Wilfred
22nd Apr 2001, 13:06
BOAC and John Farley, thanks, but are you talking about the high speed stall situation? The quote from HTBJ in my last post here refers to 0.61 Mach; would sinificant shock waves be developing at such a low Mach No.?

I was under the impression that the plain old "there ain't enough air flowing over me wings" stall speed actually increases with altitude.

I get the feeling now that I am missing something so bl***y obvious. Don't lose patience with me please. Ta.

Wilf.

BOAC
22nd Apr 2001, 20:06
Let's not get into sweep or we'll be here all day talking about thickening wing-tip b-layers/tip stall/pitch-up etc!

Wilfred, cannot answer the question about 0.61M - it depends on how much the air accelerates over a particular thin or thick section. It only requires LOCALLY sonic conditions to cause separation. Mach 1 is always REACHED somewhere in the airflow around an aircraft when the actual physical Mach is below 1.0. Pulling 'g' aggravates the compressibilty effect. Ask any FJ pilot who has dropped a boom in combat while 'subsonic'! Try the B737 at 0.8 Mach and a steep turn (in the sim, of course!) - you'll get shock buffet.

John Farley
22nd Apr 2001, 20:59
Hi Wilfred

If BOAC has sorted you please ignore everything after this!

I think your comment about “high speed stall” might have given me a clue as to what is bothering you.

In my book a “high speed stall” is simply one that happens at more than 1g. It can still be wings level (say at the bottom of a loop) or it could be at the normal cruise speed for the type and involve a lot of bank on and pulling hard. It has nothing to do with “high speed” as in compressibility or Mach number effects. It just means the pilot made the aircraft stall at a speed that is higher than the wings level 1g case. Indeed the proper term for what we all refer to as a “high speed stall” is an accelerated stall.

So doing an ordinary stall (wings level with speed reducing in level flight) at say 40K feet does not mean you have experienced a high speed stall even though the Machmeter may well be some way round the dial at the time you stalled.

Just in case the above does not unlock the issue for you, I will spell out my understanding of the subject from the top (forgive the sucking eggs bit but I don’t want to miss out any step just in case that is the one giving problems.)

There are two types of mathematical approach to aerodynamics, one treats the air as if it is an incompressible gas and the other assumes (correctly) that it is compressible. This latter correct approach has the disadvantage that it is much more complicated.

The difference in numerical results between the simple theory and the complex one is very small indeed at low speeds - say below 150 kts - and is usually ignored in normal life because the scatter in results from a whole host of other factors tends to be greater than the compressibility effect. (By these “other factors” I mean inaccuracies in flying, effects of turbulence, instrument and sensor calibrations and errors and so on)

The fact that air IS compressible does effect some aspects of the simple (incompressible) theory more than others. What we are interested in here, namely the loss of the maximum available lift coefficient available from a wing (i.e. the amount of lift we can get before the flow breakdown that we call the “stall” happens) is probably affected as much as anything.

What I am talking about is the breakdown in the simple relationship between stalling speed and applied G. Simple (incompressible) consideration of the lift equation (L = Cl x ½ x density x wing area x speed squared) tells us that if we have a wings level stalling speed of say 100 kts, then at 141.4 kts (or whatever the square root of 2 is) we should be able to pull 2g, ie just manage a 60 deg bank steady level turn at the stall.

But somehow one can never quite achieve this and the shortfall at 173.2 kts is even greater (root 3 when we would expect 3g from simple theory).

Now I know we are taught that a wing always stalls at the same angle of attack (and nobody goes around saying THAT more than me) but it is actually only a true statement if the Mach number at the stall is roughly constant. Numbers for the Harrier in conventional flight are locked in even to my failing memory, so forgive me quoting type specific numbers to make what is a general point:

In a metal wing Harrier at low level the angle of attack you see in a 1g stall wings level is around 12 deg (clean configuration, flaps up) That number will not change even if you double the aircraft weight (although the IAS at which you reach it will be much higher) and equally if you turn too hard you will stall at the same 12 AOA regardless of weight or bank angle in use.

BUT, if you go up to 20K feet and spiral down at say .8 Mach in a high G turn, pulling harder and harder until it stalls, you will probably not see more than 9-10 degrees on the AOA gauge. This reduction in the max lift available has happened because of compressibility effects.

Hope that helps, if not get back to me as there is no such thing as a bad student only a poor instructor (or as my doctor says - there is no such thing as an impotent man only and incompetent woman - advice which I find very comforting)

JF

BOAC
23rd Apr 2001, 00:13
Nice to see you still quoting the 'bona jet', JF.
PS Have you got your doctor's phone number?
Mike

John Farley
23rd Apr 2001, 14:02
Sorry Mike

In my europhoria on the way home I lost it

JF

john_tullamarine
24th Apr 2001, 10:04
are we forgetting that the ASI is not a speed indicator at all .... only a differential pressure gauge ?... some assumptions and fancy designing labels the dial in speed instead of pressure ... if the conditions right here and now replicate the calibration assumptions then the gauge gives a correct speed .... at all other times it is wrong to a greater or lesser extent ... hence the various engineering tricks of different airspeeds to make the thing tractable for analysis .... worth reading the relevant sections of an engineering undergrad text on aerodynamics and the thing becomes a little easier to accept ..

Wilfred
24th Apr 2001, 13:00
BOAC - agreed, it a localised acceleration of the airflow to M 1.0, but at 0.61 surely that is unlikely. Standard B757 cruise is0.80 with MMo at 0.84, and we still experience no high speed buffet. However, I do not think this about pulling G at altitude.

JF. Excuse my apparent ignorance, but if you are pulling harder and harder in you r 0.80 spiral dive, are you not increasing your wing loading. Stall speed, being directly related to effective weight, will increase with increased weight. How far adrift am I?

Unfortunately, I am still baffled by the HTBJ reference I quoted earlier which states that the EAS stall speed increases with altitude. EAS means nothing to do with intrument corrections, position error, or compressibility error; this is the stuff that is important to the aerodynamicists - or at least that's what me Mum told me! So...why does it increase with altitude?

Guys, if you have already explained it and I have not cottoned on please excuse me and send me to back of the class , but I really am curious about this one.

john_tullamarine
25th Apr 2001, 01:33
Wilfred has been confusing EAS and TAS ... EAS removes density as a variable by definition and, on a simplistic view would suggest that stall EAS is independent of altitude. However, CL also is influenced by Re with the result that CLmax will be reduced at higher altitudes. Solving for EAS stall speed at CLmax then yields a value in excess of the SL figure. Shock separation is not relevant to the discussion.

john_tullamarine
25th Apr 2001, 02:28
Considering Wilfred's concern about his first assumption, it is quite correct to relate density's decreasing with height to an increase in stall TAS. Recall, though, that EAS relates to SL density to remove the height variation in density from the calculations .. ie

local rho TAS squared= SL rho EAS squared

Turbofan
25th Apr 2001, 18:30
Gentlemen, I believe we're over-complicating the original question with talk of compressibility and EAS (although I am finding it quite interesting).

Wilfred - if we consider only TAS and IAS, then TAS Vs will increase with altitude, and IAS will ramain the same, for the reasons you outlined. Remember that, taken from the lift formula, '1/2rhoV* = IAS'

And I do believe EAS Vs will increase slightly with altitude. I might have to verify with the 17 or 18 text books I've got up in the shelves though.

Vapour Trail
25th Apr 2001, 20:25
Could the increase in stall speed be linked to Reynold's number. Which states

R=(Rho*V*L) / viscosity

Where V=TAS, and L=chord length

Thus;

A reduction in density at higher altitude (all other factors the same as at low alt) will reduce R and hence reduce Cl max.

Therefore if you maintain the same weight (hypertheticaly) you would need a higher AoA at the same speed, thus less margin over the stall AoA, therfore you would stall at a higher speed????

I might be on the right track, but then again, it is 2:30 in the morning........ :)

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Eat My Vapour Trail

Wilfred
25th Apr 2001, 22:21
John Tullamarine, Noooo. I am not knowingly confusing TAS with EAS. As I said above, EAS corrects for instrument, position and compressibility, and for a given TAS at altitude, this is the equivalent flow of air over the airframe, ie what your hand would feel out of the window, allowing for compression.

It is EAS that is important in 1/2RhoV2SCl questions, as that is what the wing feels.

Keep it coming folks!

john_tullamarine
26th Apr 2001, 02:37
Wilfred.. you are getting there .. as I suggested earlier and vapour trail reiterated .. the problem relates to how you simplify things .. CL depends on a number of things .. including Re (Reynolds Number).. at altitude CLmax is reduced ... solving for stall speed then gives an increased value for Vs .. don't get too hung up on EAS .. it is basically something we engineer chappies need to make the maths a little easier to process ... basically EAS is a "standardised" TAS .. all the other "speeds" are the system's way of getting around the fact that the ASI's calibration lets us down anywhere other than at standard sea level

Hot End
26th Apr 2001, 09:52
Wilfred,

I'll make my pitch & jump in here. Probably a little off but, here goes.

As the boundary layer travels over the top aft surface of the wing, it begins to decelerate and reaches the aft stagnation point at the trailing edge. Normal flow separation in a stall occurs when the lower levels of the boundary layer don't have insufficient energy to overcome the adverst aft pressure gradient on the wing. They prematurely stagnate, ie, before the trailing edge, and separate.

Perhaps the "mach no." effect is the beginning of the onset of the shock wave forming on the wing as a result of higher speed or altitude flight. Depending on wing laminar flow designs, this may happen a quite low mach number but since the drag ( read fuel flow ) effect was minimal, it was disregarded. However, it may have a noticeable effect on boundary layer energy levels. Hence the reason for the stall speed increase.

Is that what you were asking? Phew, I need another red!

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Goddamit! Burnt another one

Hot End
26th Apr 2001, 09:59
sorry about the speeling! I think there was a double negative somewhere in there

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Goddamit! Burnt another one

john_tullamarine
27th Apr 2001, 04:56
This thread has meandered somewhat .. the answer to the original question is found in any basic engineering text on fluid flow .. either aerodynamics or hydrodynamics ... fluid forces depend on

(a) velocity
(b) mass density
(c) a measure of size
(d) viscosity
(e) compressibility

and no other variables ....

Running through the usual dimensional analysis tricks ... these come down to the following terms ...

(a) mass density
(b) a characteristic area
(c) velocity squared
(d) Reynolds Number (Re)
(e) Mach No (M)

and no others ...

Re is important at low speeds, M is important at high speeds .. and so we tend to ignore one or the other as appropriate, and, in general, ignore Re for most CL considerations.

Re depends on density and viscosity and viscosity is roughly proportional to the square root of the temperature.

At height the end result is that the CLmax reduces .... solving for speed at the (1g) stall gives

Vs squared is proportional to 1/(CLmax x rho)... end result is that the stall speed goes up at height ..... measured as TAS if you are dealing with actual rho .. or measured as EAS if you are using standard SL rho ...... and that is the story ....

John Farley
28th Apr 2001, 21:52
John Tullamarine

Its really nice to go away for a few days and on return find somebody has sorted out your problems. Hope you have now convinced Wilfred!

I had hoped that my attempt at convincing W that the Cl max would reduce with height (thanks to the stalling alpha reducing with increasing speed and or height, due to compressibility) but I only seemed to confuse him further – judging by his comments on wind up turns.

JF

john_tullamarine
29th Apr 2001, 07:32
you have me a little perplexed .. the stall speed increases a little due to Re changes .... compressibility is irrelevant at stall speed values ... I think you are thinking of the compressibility increase in EAS as Hp increases for a normal cruise value airspeed ..... the typical Vmo envelope shows this sort of effect ...

John Farley
29th Apr 2001, 18:52
JM

No, what I was thinking of was the effect I quoted in an early post repeated here:

What I am talking about is the breakdown in the simple relationship between stalling speed and applied G. Simple (incompressible) consideration of the lift equation (L = Cl x ½ x density x wing area x speed squared) tells us that if we have a wings level stalling speed of say 100 kts, then at 141.4 kts (or whatever the square root of 2 is) we should be able to pull 2g, ie just manage a 60 deg bank steady level turn at the stall. But somehow one can never quite achieve this and the shortfall at 173.2 kts is even greater (root 3 when we would expect 3g from simple theory).

Every aircraft I have flown demonstrates this point even at circuit height. The designers I have whinged to always blame compressibility. But you feel it is more Re?

JF

john_tullamarine
30th Apr 2001, 13:49
This is not something I have come across before .... when you have been looking at a banked stall, what were the conditions ie ..smooth air ?.. rate of change of bank ? ..rate of change of airspeed ? etc ... might give a clue to the answer ... I would have expected that, in a steady banked turn .. with the airspeed constrained to reduce slowly, the stall speed would be predictable. PEC ought not to be the problem .. Re is not varying ..and the mach number is too low to have any significance .... interesting ...

Checkboard
1st May 2001, 11:58
You might find that it is an instrument error at the high angle of attack, although I would expect the IAS to underread, not overread.

john_tullamarine
1st May 2001, 16:06
one wouldn't expect a PEC problem if the manoeuvre is relatively steady ... PEC certification covers the range back to the stall ... unlikely to be any sort of instrument problem ...

John Farley
2nd May 2001, 01:04
JT

You asked about conditions at the stall. The manoeuvre I am thinking of is the wind up turn (WUT). It is a very common and bog standard flight test technique used in all development and certification programmes to establish the Manoeuvre Boundary of any aircraft (how much g can it pull at what speed)

The technique is to start at the test height and slowly increase alpha at the required speed until the stall. Once full power is not enough to maintain speed (ie at the thrust boundary) you overbank and lose height to maintain the speed steady.

The aim is to have quasi steady state alpha and speed (albeit with changing height) at the stall (a bit like the must be less than 2kts per sec speed reduction used for 1g certification stalls) Once the height loss is significant you start the manoeuvre above the test height and (if you are lucky) have it on condition as you pass the height.

As I say this is bread and butter of flight test but always produces points that sit on a curve that is a tad less than V squared. And that despite the beneficial effect of high power which has to have a small component assisting lift (indeed not so small at high alpha)

It is this gradual reduction in Cl max as speed increases that I have always been told is down to compressibility effects.

JF

john_tullamarine
2nd May 2001, 05:07
sorry .... misinterpreted the thrust of your previous post ... the majority of my FT (not overly extensive) experience has been on performance not handling (other than basic work). The phenomenon to which you refer, I suggest, is either going to be a result of one or more of

(a) unsteady flow affecting the PEC depending on the manner in which the stall is approached
(b) compressibility if the aircraft concerned are of sufficiently high performance with attendent high stall speeds
(c) Re if the data be not reduced to SL

Wilfred
8th May 2001, 13:27
What a fantastic response. Thanks to all who posted here, and I think that I have finally begun to understand the concept properly. Especial thanks to John Tullamarine - you really cracked it for me.

Now if I can regurgitate all this at the Cx interview, they may give me a job!!

I've said it before, but this really is a great forum!

john_tullamarine
9th May 2001, 13:11
Best of luck with the interview .. if you get the job, do ask if I can have one as well ........