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Calling Nick Lappos - Blade Stall

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Calling Nick Lappos - Blade Stall

Old 28th Sep 2016, 19:13
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Thanks Crab, I'll try not to let it keep me awake ...
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Old 29th Sep 2016, 02:19
  #22 (permalink)  
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( and your explanation of why u use false forces is a laugh
" In this case "false" forces become relevant and directly usable."
thats ok with me )
Even I, a non engineer or academic, am familiar with the term "false force" as used by academia and engineers, also referred to as "fictitious force". You might like to cogitate if gravity is a "real" or "fictitious (false)" force.
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Old 29th Sep 2016, 03:51
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And even whether it is a pushing force instead.....

Phil
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Old 29th Sep 2016, 04:29
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Anfi will take issue with centripetal and centrifugal forces being involved in any equation as why use two forces when statistically any equation with just one force is as good
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Old 29th Sep 2016, 06:40
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Bell Ringer....you are a very naughty boy!
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Old 29th Sep 2016, 12:22
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But very funny
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Old 4th Oct 2016, 18:05
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Coning angle is the result of the integration of some of the local forces on each element of a blade.

At each element the local forces that are relevant to coning are the moments about the flapping hinge. (Ie those forces in the plane normal to the flapping hinge at their respective distance from the flapping hinge.

They arise from two forces
They are
1 the apparent outward force (generated by the reaction of a mass to Centripetal Accn), call it Centripetal Force (Cf) if you like.
AND
2 the aerodynamic forces in that plane, traditionally thought of as ‘lift’ (although it’s not strictly true, nevertheless the aerodynamic forces are still a product of local v^2 notionally)

The first can be expressed thus
Moment = ∫(from 0 to R) mxω^2 xsin(κ) dx
m is the mass of a small length of blade (dx)
x is the position along the blade up to R the radius
ω is the angular rate in radians per second, ω≈RRPM/10
κ is the coning angle in radians
MCf is the Moment about the flapping hinge causing the blade to (not) cone up
That integral gives
MCf = -1/3 mR^3 ω^2 sin(κ)≈ 1/3mR^3 ω^2 κ (for small κ)


The second moment can be expressed like this
Moment = ∫(from 0 to R) 1/2ρA©x^2 ω^2 x dx
Where
ρ is the density of air
A is the Area of a dx sized piece of blade (ie Chord)
© is the ‘Coefficient’ of Aerodynamic force in the plane normal to the hinge (similar to CL ) but it also assumes an ideal twist (T, or ‘washout’) that reduces CL by an amount that ensures the Lift and Cf remain in proportion to minimize unnecessary longitudinal blade bending. So it embodies an x^-1 term, (so something like CL (R/(R-x))
ML this is the Moment caused by aerodynamics in the plane of the flapping hinge
x is the distance along the blade and ω is again the angular rate in Radians/second
so xω is the local v (speed of airflow, at that x)

That integral gives
ML = 1/6ρA©R^3 ω^2

Since ML + MCf = 0
Then ML= - MCf
1/3mR^3 ω^2 κ = - - 1/6ρA©R^3 ω^2
solve for κ (THE CONING ANGLE)
κ = ρA©/2m
SO regardless of how exact that rough working is you can see clearly that the coning angle is not dependant on RRPM (equivalent to ω)
What you can see is that coning angle depends on the ρ (density) , A (the chord), the shape and Angle of Attack (CL) and the mass of the blade.

Since there is a ©STALL then there is also a κSTALL and it does not depend on RRPM



It is irritating to have to pointlessly integrate your pointless formula, especially since you obviously barely understand it yourself

Since you have chosen to align yourself with the playground oaf I seem to have no choice but to perform the pointless maths you ask for.

I say pointless because the variable cancels out prior to having to perform any integration anyway.
So your (pointless) set of ridiculous assumptions (eg twist for constant Induced Flow, and I used a similar (but better) assumption) are unimportant since.

The whole thing can be stated far more simply (which is generally a good thing) like this

Κ = arctan(L/CF)
Which for small L and Large CF is (to 4 decimal places) the same as
Κ = L/CF
Well it should be obvious that RRPM cancels out of that but here it is
L = a times RRPM^2
CF = b times RRPM^2
So
Κ = L/ CF = a times RRPM^2 / b times RRPM^2
CANCELLING OUT you get Κ = a/b
(where a is the collection of other (unimportant here) factors (density of air that sort of thing. b is another collection of other (unimportant here) factors (mass of blade etc))

Don’t blame me for giving that answer, I was goaded into it by an over self satisfied aeronautical engineering degree holder and a playground oaf.
I’m sure there’ll be all sorts of nit picking over the detail of my maths, but I don’t have time to improve it so please don’t bother. I think it is enough to prove my point though
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Old 4th Oct 2016, 21:00
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I’m sure there’ll be all sorts of nit picking over the detail of my maths, but I don’t have time to improve it so please don’t bother. I think it is enough to prove my point though
Which was what, exactly? Sorry, I'm easily confused, but I didn't see any reference to how many engines were involved......
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Old 4th Oct 2016, 21:05
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If you're saying local speed at any blade point is distance out x angular rate, that ignores forward speed, which was one of the fundamental sticking points way back at the start.
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Old 4th Oct 2016, 21:06
  #30 (permalink)  
 
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Originally Posted by 212man
Which was what, exactly? Sorry, I'm easily confused, but I didn't see any reference to how many engines were involved......
Ironically there wasn't one point.
There were at least two which is, statistically, better than one - ok this is wearing a tad thin.

Over and out..
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Old 5th Oct 2016, 12:48
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AnFI, I think you forgot the QED at the end: Quod Erat Demonstrandumb
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Old 5th Oct 2016, 13:19
  #32 (permalink)  
 
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Oh..AnFI,,,you are so clever, so intellectually superior... how could any of us ever have doubted you???

I am sure there is a proper adult with more maths knowledge than my poor O level here on this forum who will blow big holes in your 'Proof' that you were right all along (as ever) and that we are all idiots who shouldn't even be allowed to breathe the same air as you

Not sure what all your pseudo-intellectual posturing is all about but it's certainly not winning friends and influencing people.
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Old 6th Oct 2016, 04:06
  #33 (permalink)  
 
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AnFI, you said:
you can see clearly that the coning angle is not dependant on RRPM
However, I can clearly see that a chopper doing an engine failure at the hover will have its coning angle increasing with decreasing RRPM, even though the lift is slowly decreasing.

But I suppose that, as the revs decay, the CL would be increasing with the pilot pulling in pitch to slow the descent and cushion on, which from your equation would increase the coning angle. Apparently the change in angle between low RRPM (pitch high, CL high) and high RRPM (pitch less, angle less) is in fact due solely to CL and not RRPM.

What a clever little secretary. Learn sumpfink every day. But yer average dum student can easily comprehend it if we (mistakenly) say that coning angle is the result of lift (upward force) and RRPM (outward force), the resultant vector of which lies neatly along the span of the blade.
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Old 6th Oct 2016, 13:10
  #34 (permalink)  
 
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AnFI I haven't aligned myself with anyone. But I did make one mistake, I assumed that you were someone who was here, like a great number of users of this forum, to expand your own knowledge by interaction with others. I therefore chose not to directly post a refutation of your analysis, but instead gave you an opportunity to look at the conditions and statements that I and others had posted and see if it revealed to you where the inconsistency lay through your own exploration of the physics. From the tone, attitude and content of your post it is obvious to me now that that isn’t the case, so I won’t waste too much time here.
We'll use your equation.

κ = ρA©/2m
© used here by you to denote coefficient of lift, is a normalised value of lift, for any given aerofoil. An aerofoil has a curve associated with it, often obtained empirically, that links © to α. Assume we are in the linear part of the lift curve and we'll call dClbydalpha "a" and substitute.

κ = ρ.A.α.a /2m.

Unlike a fixed wing, a rotary wing is rotary. Therefore the "v" that any aerofoil element experiences relies on rrpm and the element's distance from the hub let's say Ωr . However the downwash the element is seeing is independent let's call it u.
Therefore
α = arctan(u/ Ωr)
Assuming a small angle
α = (u/ Ωr)
And substitute

κ = ρ.A. (u/ Ωr).a /2m.

As you can now see, there is a term for rotor speed in your equation. Q.E.D.

Last edited by dClbydalpha; 6th Oct 2016 at 14:23.
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Old 7th Oct 2016, 09:15
  #35 (permalink)  
 
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Hi a

(we'll call dClbydalpha "a" and substitute)

I am afraid you are wrong
I am trying to get an important point across.

Please analyse your friend crab responses and see if he is helping get the point across. That is where the frustration comes from.

I think your assumption that downwash remains constant is incorrect. Really the downwash in your scenario would increase with RRPM so there's the RRPM term on top that you 'accidentally' left out.

ie it still cancels.

But that's not the point, the point is there is an interesting way to look at an Ultimate Coning Angle

Yes there is some bathwater (like Airspeed, which would reduce the UCA)
but it is useful to think of UCA being dependant on the terms left in my previous math

You can see it is useful because there is a frequent error made in thinking that coning of 'say' 45deg might be acheived.
In 'layspeak':
This 'can't' happen because if a blade is going slowly enough to make feeble Cf then it's going too slowly to make enough lift (and the attempt to make more lift by increasing AoA will have resulted in stall prior to that point (actually at (or close) to the UCA)

dCl/dA I hope that you can find the valuable point here, and help me explain it. (rather than showing that you learnt some maths once, it's nice that you did, why not use it for the forces of good, truth and the American way? Instead of encouraging that moronic oaf.)

AscendC quite right
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Old 7th Oct 2016, 11:30
  #36 (permalink)  
 
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As ever AnFI, it is never clear exactly what point you are trying to get across - and even less clear as to how it will improve a student's knowledge and make him fly a helicopter better.

Also, as usual, when someone who does know the subject (maths in this case) shows you the true path, you immediately refute it because it doesn't fit your argument (whatever that is).

Enjoy your posturing but, as the oaf in the playground, I do know who is rapidly becoming the big-headed bluffer in the classroom
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Old 7th Oct 2016, 16:00
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AnFI, the discussion of such concepts is important, but fundamentally the premise that for a given rotor system there is a "coning angle" that is the equivalent to a "stall angle" for a fixed wing does not stand up against the physics of the situation.

I am glad you acknowledge the nature of downwash. Downwash is the important factor that appears to be missing from your conceptual model, and where my musings began. The reason I suggest a uniform downwash, and a linear twist is that it makes the first calculations easier. If you can envisage the form of the integral that incorporates a linear twist and a uniform downwash then you can see how it can be extended to more exotic implementations.

So there are a number of forms of the equation below but it is pretty well recognisable

dL=1/2 ρ c dr a [ θ - u/Ωr - τ(r/R ) ]

c is the chord at that section
dr is the element of radial blade
a is the coefficient of lift of the aerofoil with α
θ is the original pitch of the blade
u/Ωr is the difference of α generated by the downwash with respect to the tangential velocity of the blade
τ(r/R) is the difference of α generated by the twist of the blade

If you choose not to have a uniform downwash, u, then you replace u with a function that represents the value of u at blade element r. Similarly for τ. But even in its simplest form it shows the importance of the induced downwash term when calculating lift.

Your suggestion of providing a blade design that matches lift to centrifugal force is intriguing, but can only really be considered as self-defeating to your hypothesis. I'll explain the steps why in words rather than equations.

Change CL such that it is proportional to the distance along the rotor.
Lift is now changed with proportion to the distance along the rotor, except for the fact that the change in lift has induced a change in the relationship between local downwash and tangential velocity i.e. it has changed the α and therefore the lift, so the lift is slightly off of the expected value.
You compensate this delta by tweaking the rate of change of CL along the blade and so now the lift is as you required.
What you find by doing this is that the "tweak" is only valid for a given rotor speed. You change the rotor speed and you need a different relationship between CL and r. Without adaptive aerodymanics this is impossible.


So what have we discovered?
The concept of relating coning angle to a "stall" condition can only be done for a known rotor system, with a specific relationship between CL and r, and at a particular rotor speed. Something that I hoped I had pointed to in my original posted reply. This is a long way from the concept of a generic, rotor speed independent relationship hypothesised.

This of course is totally academic as it has so far only touched on the rotor in hover. As soon as any cyclic commands are input, then the disc tilts and the axis of the cone along with it. Having done that we will start to translate, creating an asymmetry of lift. The whole concept of "stall" then involves flapping, advancing blade and retreating blade dynamics.

There is a relationship between aerodynamic and inertia forcings, called the Lock number, which is important to rotor design. But that is another topic.

So I'm genuinely sorry AnFI I don't see what this important topic is. I must be missing the scenario you are thinking of as there doesn't appear to me to be any reason to monitor coning angle as a critical flight parameter. To the pilots out there, have any of you flown with a coning angle indicator marked with a critical value?

Last edited by dClbydalpha; 7th Oct 2016 at 16:33.
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Old 7th Oct 2016, 18:59
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Yes, the indicator is called the Collective, and the critical mark is when it reaches your armpit.

The other indicators are "RPM LOW" and "beeeep beeeeep beeeep".
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Old 7th Oct 2016, 22:27
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dCl/Da
I did cover all those points didn't you bother reading?

and it is important because there are many accidents from attempting to pull more g resulting in less lift

elegantly the rotor actually physically performs the integration (not that it's neccessary), and as AC simply puts it coning is the ratio of 'lift' to Cf. you just can't get that ratio beyond a (fairly low) value even if you want to. (because if the Cf were low (although that's not the point, the normal RRPM case is) then the 'Lift' that can be made before stall isn't that great either)

yes, nobody flys with a 'critical cone indicator' (perhaps they should!) but it would help pilots understand why you can change pitch of the aircraft at a greater rate but should not neccessarily expect a greater TRT hence why they annoyingly hit the lake/snow/ground so often.

I alluded to higher speeds, the points dCl/Da makes are not relevant (especially Lock number)
really what happens when you pull more g in a helicopter at speed is parts of the disk stall but the effect is distributed, there's a load of vibration and the disk as a whole starts to make less TRT rather than more.
It's a little like the real stall in an aeroplane in as much as the whole wing (generally) does not stall at the same parameters, so it's a 'soft stall'. For a helicopter it is very like the phenomenon described in the manual extract by (i think it was) SASless, about 'mushing'.

valuable ? possibly would have been to the pilots with egg on their face and might help other pilots to avoid egg on thiers

Crab nothing valuable to add as ever, you really are an oaf, and nowt wrong with my maths either thank you (dCl/da neatly avoided the fact that there's an RPM term on top missing in his simplification !) since you obviously don't understand this stuff why don't you just shut up and contemplate a little longer before opening your unhelpful and loutish mouth?

anyhow i've said enough, take it or leave it, it might be useful to some pilots
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Old 8th Oct 2016, 02:11
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and contemplate a little longer before opening your unhelpful and loutish mouth?
You should take up some of your own advice. You work on the principle that bull**** baffles brains, an educator lays out the subject matter in a manner that is appropriate to the audience, in this case, one that doesn't involve maths.

Ascend Charlie has given the simplified detail of what concerns the PIC.
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