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What Goes Up

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Articles

What Goes Up - Pt.1

What Goes Up - Pt.1

Videos to accompany the Dec issue article - 17/11/10

What Goes Up - Pt.2

What Goes Up - Pt.2

Videos to accompany the Jan 2011 issue article - 20/12/10

What goes up - the stall!

What goes up - the stall!

new videos to accompany the May issue article - 10/5/11

David Burton 220/12/2010 20:32:23
22 forum posts
2 photos
Hi,
 
thanks everyone for the kind comments - much appreciated. I'm glad you are finding the articles interesting and useful.
 
Simon - I very much suspect you are right that the video of the A380 is a test of some sort - the fact that the aircraft is in airbus livery and was fitted with a tail protector is, as you say. a bit of a give away. I'm not sure what type of test though, it could a high alpha take off test or (the one I suspect) a max take-off load plus test. Either way its an exciting demo of the effects of high angle of attack!
 
Ross, thanks for your comments. Glad you are enjoying it. Different folks are at different stages in their understanding of this stuff. Sometimes those who are a lot further along the track can forget that they too had to start learning somewhere! They can also get a bit impatient to get at the more detailed stuff. But we should all be able to move along together given a bit of understanding!
 
Myron, gas turbine aerofoils are, as you will know, a highly specialised field of aerodynamics in their own right. The various tricks employed - especially in engines designed to operate at supersonic flying speeds are amazing! As we work through we may well steadily introduce the odd design equation. If you think about it all of part two is in fact about one equation - the so-called lift equation - but I quite cunningly avoid ever writting it out in symbols! Its there in words, and all the terms are there - just not the symbols!
 
Oliver, your comments on the "big blow" are very interesting. People often don't realise the method by which the air can damage their property. Your Dad was quite right - its like the video I showed in the part one set of how to blow a coin into a glass - it by blowing over the top of it not at it that the trick is done. Same with your roof tiles but on a bigger scale!
 
Anyway - thanks again for the feedback everyone. I'm just putting the finishing touches to part three at the moment - meaty stuff all about vortex flow and the wing!
 
If there is any topic within aerodynamics you would particularly like to see covered just mention it here - or PM me, or drop an email to the address given at the start of this thread
Mark Pritchard31/12/2010 00:35:11
12 forum posts
David,
Nice videos, but please read up on "V" speeds. The A380 was actually undergoing Vmu (Velocity Minimum Unstick) tests which is to confirm the minimum speed at which the aircraft can take off, hence the temporary tail skid (painted red) to avoid damaging the tail.
Rgds,
Mark 
David Burton 231/12/2010 01:15:35
22 forum posts
2 photos
Hi Mark,
 
thanks for the comment. I am familar with the Vu test. That the 380 was undergoing a test I agree is fairly obvious - from both the Airbus paint scheme and the presence of the tailskid. But what makes you so sure it was minimum unstick speed test? I mean it might have been  - I'm not saying it isn't - but there are any number of tests it could have been! Including the one I speculated on - i.e. max takeoff load test. Who knows, unless they were actually there? If you were, and know this specific incident, I freely accept that obviously.
 
The point of the video really, as I'm sure you appreciate, is simply to demonstrate the effect of increasing alpha - but equally the need to rapidly reduce it on climb out in this particular situation!
 
Glad you enjoyed the videos.
 
David
Phil W05/01/2011 20:37:20
1 forum posts
Hi, yes nice videos indeed. I can confirm that the 380 video is indeed one of the Vmu tests which took place at Istres sometime in July 2005. It is a certification requirement that the calculated Vmu must be demonstrated to be within +1/- 0 KIAS of the verified Vmu over a range of weights before the aircraft is allowed into service.
 
Anyhow, keep up the good work, looking forward to the 3rd instalment.....
 
Cheers
 
Phil
David Burton 205/01/2011 21:23:04
22 forum posts
2 photos
Hi Phil,
 
thanks for that clarification much appeciated. As I say I was just interested in a demonstration of the effect of high alpha - and its certainly that! With little or no information on the You Tube thread one is left guessing at the nature of the test. So thanks again!
 
David
 
PS I don't think Graham will be too cross if I let out of the bag that in part 3 we are going to be looking at the "bound vortex"!
John Cole13/01/2011 12:15:03
615 forum posts
24 photos
"Wings develop a good 10% more lift in winter than they do in summer".
 
Well I guess it must be colder in Liverpool than I thought.  Nearest location the BBC publishes average conditions for is Aberystwyth:
January average day max 7 degrees C (280 Kelvin)
August average day max 18 degrees C (291 Kelvin)
 
Suggests nearer 4% to me.
David Burton 213/01/2011 18:42:43
22 forum posts
2 photos
Hi John,
 
Thanks for your comment.
 
For the figures you quote I totally agree, but I must admit to taking somthing closer to the max and min, rather than the means! 
 
For a cold winter's day here I took 0 degrees C. Not that untypical at the moment - in fact a week ago I was flying at the patch and it was -4!
 
For a warm summer's day I took 30 degrees C. OK, a little optimistic maybe - but I have known it to reach that here! Not very often - but it does happen and we can all dream!
 
At 0 degrees air density is 1.292Kg/m3
While at 30 degrees air density is 1.1644Kg/m3
 
A difference of 9.88%
 
In the original draft of the article I did in fact cite the temperatures - but in the inevitable editing for length that detail was one of the casulaities! Anway thanks for your point.
 
David
John Cole13/01/2011 19:35:58
615 forum posts
24 photos
I used to live near Chester Zoo, and I remember cheering when summer temperatures got beyond 20!  JSC
 
Oh, and don't forget humidity.  Water vapour's lighter than dry air.
John Bunting25/01/2011 16:05:15
105 forum posts
David:
Having asked about the lift of symmetrical airfoils, after Part 1, I've been looking forward to your comments on it in Part 3.
The nub of the matter seems to lie in your words, "this wing wil 'flip' from significant down-force to significant lift as we pass through the zero singularity that it never manages to actually occupy!".
This leads one to wonder just what the lift curve actually looks like, for these very small angles of attack. It seems to suggest that the slope of the curve may be larger there than it is for larger angles; or else that there may be a small hysteresis loop around zero, so the wing goes very slightly beyond the theoretical zero lift angle, before the up or down force 'flips' from one state to the other. I wonder if any wind-tunnel tests, or even mathematical modelling, show any such behaviour.
Talking of maths in general, I read the 'Aeromodeller' from 1945 onwards, and the amount of maths then was frightening, compared to today. Most of it, unfortunately, based on full-size data, at much higher Reynolds numbers than for model flight. I remember covering sheets of paper with figures, to work out the angle of incidence for maximum L/D or power factor. But the one that really sticks in my mind was a series of articles, stuffed with equations, entitled "Counteracting the effects of engine failure in twin-engined model aircraft". Happy days!
David Burton 225/01/2011 21:57:32
22 forum posts
2 photos
Hi John,
 
Good question. About three years ago I developed an interest in what happens to symmetrical aerofoils at very small angles of attack. I had notice that very few published lift curves actually indicated that the zero alpha position had actually been measured. There were measurements at, say plus minus 1 degree, but at the zero point the line was simply drawn through the origin - as you might expect without an actual experimental point. But I wanted to know what happened at very small angles - less than half a degree.
 
I would need a very accurate symmetrical wing section to do this. So I designed a symmetrical aerofoil using "Solid Works" (a professional 3D design software suite suite) This meant that, at least in the computer, the section was perfectly symmetrical because I simply designed one half and mirrored it.
 
To make the section - I had it machined from a single block of Araldite using a Bridgeport 5 axes CNC Machining Centre. (An advantage of working in a University - we have lots of specialist kit around!) Using this means that the section is going to be about as accurate as I could possibly get it.
 
To check it was accurate I measured it using a optical 3D measurement system we had developed - it showed that the section was accurate to the computer model, at all points, to better than 50microns (i.e. 1/20 of a millimetre). It also showed that such errors as there where were random - there was no structured error leading to unintentional camber.
 
So I now had a "perfect" symmetrical section aerofoil. The next problem was that the force balance and the section rotational stages fitted to our wind tunnel would not be accurate enough for what I wanted to do. So I used a micrometer screw rotational head to control the position of the section and a new fibre optic strain sensor based on a Bragg grating system that had been developed by the research of a close colleague to measure the force. The important thing about this force sensor was that it had a very high sensitivity and responses almost flat to signals from static all the way to 2MHz - very fast.
 
With this I set about obtaining a lift curve. The micrometer screw limited the angles I could use to only approximately plus minus one and half degress - very small but that didn't matter - as I was only interested in small angles.
 
Continued in the next post.......
David Burton 225/01/2011 22:19:19
22 forum posts
2 photos
The first lift curve we got is shown below:
 

Nothing too surprising there. The points even fit a straight line very well, and the line pretty well goes through zero. So far everthing is as expected.
 
Then I tried at smaller and smaller angles. The result of this is probably best summarised by the lift curve below:
 

Between 0.4 degrees and 0.1 the section is behaving more or less as we would expect. But once we got below, somewhere around 0.07-0.05 degrees the Cl value started to become constant - at approximately 0.06-0.07. This was the case on both sides of the zero point. The two data points at angle zero are in reality the mean of a series of readings around that point - plotting them all confuses the graph.
 
So, in this range close to zero the Cl value plateaus - remaining almost constant at a value we might call Clcrit. If you start at say minus 0.05 degrees and steady increase the angle to plus 0.05 what happens is that the Cl value remains around minus 0.06-0.07 and then suddenly switches to plus 0.06-0.07. It prooved impossible - despite many attempts to get any stable reading between these two points. And no amount of careful adjustment would produce a "zero-lift" condition.
 
So I think this is the answer to your question - at small angles, less than about minus 0.05 degrees, the lift curve runs almost horizontal at minus Clcrit, nominally at zero it switches to plus Clcrit and it runs at this value until about plus 0.05 degrees.
 
The cost of producing the wing section meant I was not able to repeat this experiment for other symmetrical sections - but I have no reason to believe this result is untypical in any way.
 
I hope this is helpful.
 
David
 
John Bunting26/01/2011 11:27:48
105 forum posts
Thanks, David: all most interesting. I hardly imagined that my guesswork would turn out to be so closely investigated in your experiments! Can you also tell us the chord and thickness of the test airfoil, and the tunnel flow speed(s)?
It occurs to me that if the facilities are still available, one further case might be examined: a flat plate airfoil, which should be cheap to make.
John Cole26/01/2011 12:05:59
615 forum posts
24 photos
Looking at the graph, I think you are referring to a hysteresis effect between -0.05 and 0.05 degrees, with Cl being at either -0.006 or 0.006, not 0.06 as you say in your text. Do I read your graph correctly? The effect appears to be present at e.g. 0.1 degrees, where you show Cl of about 0.008, where a straight line would give nearer 0.005.
 
I don't think you say what aerofoil section you used, but the results of your first graph show a low slope of less than 0.05 / degree. I would expect end-corrected data to be significantly higher that this for common symmetric sections. Is your data end-corrected? If not, that might explain the low value.
 
Fortunately, these numbers are quite small (order of 1% of Cl at cruise). Although symmetrical wings are not often flown at such low (subsonic) alpha as you are examining, stabilisers may be - though I have to note that the range in which you've observed the effect is small compared with e.g. wing-induced downwash at the horizontal stabiliser, If the effect were larger, or observable at greater angles, then control would be hard to maintain!
John Cole26/01/2011 12:58:15
615 forum posts
24 photos
Yes, wind tunnel data for low and zero alpha is hard to find. But this is from Moran's "An Introduction to Theoretical and Computational Aerodynamics"
ISBN: 0486428796
 

The graph is a bit of a mess as it shows Cm as well as Cl, and lines for both the basic NACA0012 but also this section with a deflected split flap! So ignore the two Cm lines, effectively horizontal at Cm = 0.0 and -0.23, and the upper Cl line which shoots off the page.
 
The Cl line for "straight NACA0012" has points plotted at both +1 and -1 degrees (triangle marker for Re of 6*10**6), and also at zero degrees (hard to say what's plotted there though!).
 
The effect that you have observed is quite small compared to the scale of this diagram: smaller than the thickness of the lines as drawn, I think. To show the scale, I estimate the height of e.g. the individual triangle point-markers to be equivalent to Cl of 0.045, about 7.5 times the size of the effect you have observed.
David Burton 231/01/2011 10:47:53
22 forum posts
2 photos
Hi,
 
sorry I haven't been back for little while, but to respond to some of these points:
 
John B: I'll have to go off memory as I am away from the office at the moment, but if I recall correctly the aerofoil section was about 17% thickness ratio and its chord would have have been of the order of 35-40cm. Air speed, again from memory, about 30m/s or thereabouts. So Re approx 7x10**5.
 
One day I will get round to the flat plate experiment - I need to convince an MSc student it would make a good project! And then dig out the fibre optic lift balance kit again!
 
John C: no the data is not end corrected.
 
It could be a hysteresis effect (let's face it, it could be a lot of things!) but I doubt it. The reason for that is the instability the flow exhibits at this point. Anywhere in that range the flow can continually "flip" between plus and minus about 0.06. So it isn't just that we "only get these values and no zero" its the ocsillation between them at a fixed (albeit very small) alpha which is so distinctive. and unusual.
 
David
 
 
John Cole31/01/2011 14:43:20
615 forum posts
24 photos
You did not comment on my spotting what semed to be a typo (from examining your graph): saying Cl = 0.06 when it looks like 0.006 to me. And you have repeated that in your latest posting (saying 0.06), making me wonder if the graph is incorrect? The values of +/- 0.06 are the greatest-magnitude values on your upper chart, at alpha = +/- 1.5 degrees. Please note: if the effect is the smaller value, that doesn't make it less important or interesting!
 
I suggest that your first description does suggest a hysteresis loop, and that if the alpha were e.g. exactly zero and Cl was continually flipping between +0.006 and -0.006 then your balance would record it as zero unless it had extremely quick response!
David Burton 231/01/2011 14:59:48
22 forum posts
2 photos
Sorry John, you're quite right 0.006 not 0.06! Tired eyes!
 
The balance actually did have a very fast response because of the fibre optic strain gauge. The data logging was at 2MHz sampling rate - which is slower that the frequency response of the fibre optic system. So we could get very good temporal resolution. I did have some graphics of this, Cl vs time - not sure I can find them - I'll take a look.
 
The basic effect was that the aerofoil would not oscillate a constant speed between the two values - oscillate is not really the right word. I know it is the one I used - but in tetrospect it does not describe accurately what we observed. The lift force would sit at one value of Cl, stable for a short period, and then flip to the other, again for a short period before flipping back again. The period for which it was stable would be anything from approx 0.1sec to 1-2sec. Without any decernable pattern. And this behaviour was the same for all alpha values between about -0.07 and +0.07 degress.
 
David
John Cole31/01/2011 15:58:24
615 forum posts
24 photos
Then what you describe is not hysteresis or Cl would be stable at -0.006 with increasing alpha up to the flip angle (of say alpha = +0.07 degrees) when it would flip to +0.006 and remain at that value with alpha decreasing until it reached -0.07 degrees when it would flip back. I believe hysteresis is associated with +ve feedback, which would not seem to be the case here.
 
Maybe the reason others have not spotted this (apart from not looking for it!) is the timescale: maybe they weren't looking for such short-period signals. And I have only found that one graph (above) with points plotted AT alpha = zero. And the relatively-small size of the effect may mean it was too small to register on their equipment.
 
There's another "funny" in your second graph: you might expect the slope of Cl vs. alpha to be constant as alpha moves towards zero, and then reduce as it approaches the zone between +/- 0.07 degrees. But there seems to be something else happening at around -0.3 and +0.3 degrees. Both these points seem to be well off the straight line drawn through zero / zero. To help make the point, here's your graph, superimposed on itself and inverted (rotated through 180 degrees), to show the effect is the same + and -. Comments?
 


Edited By John Cole on 31/01/2011 16:05:45

John Bunting01/02/2011 01:39:08
105 forum posts
If you ever do the flat plate tests, David, I wonder if you'll get similar results to F W Schmitz. I think he got a funny boomerang-shaped lift curve, with a sudden decrease of slope at about α=5°, Cl=0.5.
John Cole01/02/2011 11:24:03
615 forum posts
24 photos
I had difficulty finding flat-plate data. Here's what I found. It's from NACA report 319 by Briggs & Dryden of the Bureau of Standards. The report is dated 1928; they were looking at a range of sections for use in airscrews, and therefore looked at high speed airflow from Mach 0.5 to 1.08. The data is not end-corrected.
 
I have re-plotted their data for the flat plate for Mach 0.5 (c. NTP, I guess), from the tables on page 31. The flat plate has a thickness of 4% and a chord of 1 inch, Re about 5x10**5 I think. They also looked at wedges and curved plates as well as typical (Clark / RAF) sections of that day.
 
It all supports my view: you don't need fancy airfoil sections to fly: just a powerful engine. Fancy sections are all about reduced drag, not increased lift. I wonder if that's what's in Part 4!
 

 
 

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