Aces High Bulletin Board
General Forums => Aircraft and Vehicles => Topic started by: Vulcan on October 16, 2004, 06:09:32 PM
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Any help appreciated. cheers.
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wwiiol issue? :p
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18 seconds?
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Spit I sustains a 4G turn @ 240mph, 1000ft alt. You think thats right. (No speed or alt loss).
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Hi Vulcan,
>Spit I sustains a 4G turn @ 240mph, 1000ft alt. You think thats right. (No speed or alt loss).
It might be correct, but it's not the quickest turn.
According to my analysis, the quickest turn should be:
Spitfire I (+12 lbs/sqin, 2745 kg) @ 300 m:
- 3.7 G
- 281 km/h TAS
- 26.0 °/s
(Your data gives about 20 °/s, which seems too slow for a Spitfire.)
For comparison:
P-40E (44" Hg, 3911 kg) @ 300 m:
- 2.4 G
- 251 km/h TAS
- 17.6 °/s
Me 109E (1.30 ata, 2610 kg) @ 300 m:
- 2.9 G
- 260 km/h TAS
- 21.4 °/s
Regards,
Henning (HoHun)
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Whooola...nice numbers....very very nice.
Now, two questions for HoHun:
1. Do you also have similar figures for other planetypes?
2. How about a formula to calculate this into seconds for 360 degrees.
BTW, I remember I started a thread like this ages ago, think it was called turn rate and G calculations or something in the direction....Will look.
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Originally posted by Angus
2. How about a formula to calculate this into seconds for 360 degrees.
That's easy, just: 360/angular speed.
For example, for the 109E 360/21.4 °/s = 16.8s :)
@Hohun: Thx for the numbers, those are coming from tests or they are your calculations?
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Hi Angus,
>1. Do you also have similar figures for other planetypes?
Since they require a very detailed analysis, I have them only for these three types with the degree of reliability I consider good enough for posting them :-)
I've run a detailed analysis on the P-40E and the Me 109E, and fortunately, the Spitfire is well documented so it was easy to add.
>2. How about a formula to calculate this into seconds for 360 degrees.
Just divide 360° by the turn rate to get the seconds for a full circle.
Regards,
Henning (HoHun)
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Hi Meyer,
>@Hohun: Thx for the numbers, those are coming from tests or they are your calculations?
These are my calculations, based on tests of the maximum lift coefficients of the three aircraft, a somewhat generic wing model, and the applicable engine powers. I've cross-checked my model pretty thoroughly, and it seems to provide fairly accurate numbers as far as I can tell.
Regards,
Henning (HoHun)
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Hi again,
Here's my Me 109E/Spitfire I comparison.
Note that there are 3 different Emil variants in the picture, and at different power settings.
- DB601A-1 (4.0 km) - early DB601A-1 with low full throttle height
- DB601A-1 (4.5 km) - late DB601A-1 with higher full throttle height, available for the Battle of Britain
- DB601N - 100 octane engine probably introduced while the Battle of Britain was fought. High altitude power looks like it fades too slowly, but there are several sets of engine curves showing that kind of power so I left it that way (for now).
The Spitfire I has two boost settings:
- +6.25 lbs/sqin - maximum using 87 octane fuel, up to mid-1940
- +12 lbs/sqin - cleared mid-1940, avaiable for the Battle of Britain
http://www.x-plane.org/users/hohun/spit_vs_me109_speed.gif
http://www.x-plane.org/users/hohun/me109e_vs_spitfirei_climb.jpg
Regards,
Henning (HoHun)
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Lovely, lovely.
I did once a Spit I and 109E climb calculations, - i.e. climb converted to NM and NM divided with time.
If I haven't already, I'll mail it to you.
Want it?
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Errr Hohun the question isn't the quickest turn... the problem is if you take a Spit I to 1000feet, get it to 240mph, put it into a 4G turn, it does not lose any speed, nor altitude, and can do so til its gas runs out.
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Hi Vulcan,
>Errr Hohun the question isn't the quickest turn... the problem is if you take a Spit I to 1000feet, get it to 240mph, put it into a 4G turn, it does not lose any speed, nor altitude, and can do so til its gas runs out.
Well, here's my answer: Impossible.
Spitfire I (+12 lbs/sqin, 2747 kg) @ 300 m:
- 4 G
- 386 km/h TAS
- 20.3 °/s
- -4.0 m/s
It should lose altitue at 4 m/s (ca. 800 fpm) in that turn.
Regards,
Henning (HoHun)
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HoHun,
Hate to be a best but could you show the calculation you preformed to get your results??
How do you know where to find the best sustained turn based on Clmax and engine power at alt?
There should be a simple ratio to come up with a turning "index" like Clmax / wingloading * powerloading. Of course that makes no sense but you get the idea.
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Vulcan I suppose your inquiry come from this thread (http://discussions.playnet.com/viewtopic.php?t=121115&postdays=0&postorder=asc&start=75)
Please point dochk and warpd to this article :
http://www.simhq.com/_air/air_011a.html
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4x2745kg=10980kg So that is the weight the wings should be able to lift and the engine to pull through air with no speed loss?
Or at 3,7Gs 10157kg? If that requires some AoA of the wing so there must even be some excess thrust to overcome the drag and still maintain constant speed?
-C+
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:D
Everytime I see Vulcan get into a "discussion" with the sycophants "over there" I start hearing the lyrics to "The Ballad of Brave Sir Robin."
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Hi Angus,
>I did once a Spit I and 109E climb calculations, - i.e. climb converted to NM and NM divided with time.
Sounds interesting! :-) But what is NM?
Regards,
Henning (HoHun)
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Umm...what is the definition of a " maximum sustained turn"??
I always thought it was the maximum turn one could sustain without losing altitude.
BTW, does a wide chord give a benefit under banking angles and a turn rather than span?
Regards
Angus
PS...quick edit:
What is NM?
NEWTON METERS
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Hi F4UDOA,
>Hate to be a best but could you show the calculation you preformed to get your results??
I'll mail you my spreadsheet if you like, but it's hell made from numbers. It's very hard to use, and much easier to use than to understand. That's not meant to discourage you, just to prevent disappointment :-)
>How do you know where to find the best sustained turn based on Clmax and engine power at alt?
The best sustained turn is achieved at Clmax. I simply iterate the speed until I find a turn at Clmax where the excess power is zero.
>There should be a simple ratio to come up with a turning "index" like Clmax / wingloading * powerloading.
There's a simple way of comparing fighters with identical wings, like the variants of a specific type. If you're interested, I could outline that method, which is useful for quick estimates.
Regards,
Henning (HoHun)
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Hi Charge,
>Or at 3,7Gs 10157kg? If that requires some AoA of the wing so there must even be some excess thrust to overcome the drag and still maintain constant speed?
Interesting perspective :-)
But yes, a 10-ton-Spitfire could sustain a speed close to that mentioned above, flying at emergency power at the edge of the stall.
I wouldn't like to be the pilot, though ;-)
Regards,
Henning (HoHun)
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Hi Angus,
>Umm...what is the definition of a " maximum sustained turn"??
>I always thought it was the maximum turn one could sustain without losing altitude.
Without losing either altitude or speed.
>BTW, does a wide chord give a benefit under banking angles and a turn rather than span?
A long-span wing is more efficient with regard to induced drag.
>NEWTON METERS
Hm - did you calculate torque or energy?
Regards,
Henning (HoHun)
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Torque would be energy divided with time right?
Anyway, I did total energy and energy on a timescale.
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I do the calculations at rated altitude for the different manifold pressures.
For the P-40E
At 8600 lbs, full fuel load
80% propellor efficiency
42 in Hg mil 1150 hp @ 12,000 ft
2.25 G sustained between 178 mph and 188 mph IAS (about 214 - 226 mph TAS)
56 in Hg WEP 1490 hp @ 4300 ft
2.85 G sustained between 180 mph and 232 mph IAS (about 192 - 247 mph TAS)
60 in Hg WEP 1580 hp @ 2500 ft
3.05 G sustained between 196 mph and 229 mph IAS (about 203 - 238 mph TAS)
I'll have to look into calculating rate and radius as well. Don't remember the formulas offhand.
It looks like HoHun and I are doing things a little differently. He is calculating the speed a max Cl turn can be sustained. I'm calculating the max G that can be sustained at a given power and weight.
Greg Shaw
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Hi Angus,
>Torque would be energy divided with time right?
Torque and energy are the same dimension, actually.
>Anyway, I did total energy and energy on a timescale.
Hm, I hadn't thought of that angle! But what is the gain over a pure altitude-over-time graph? :-)
Regards,
Henning (HoHun)
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Hi Greg,
>For the P-40E
>At 8600 lbs, full fuel load
>80% propellor efficiency
>42 in Hg mil 1150 hp @ 12,000 ft
>2.25 G sustained between 178 mph and 188 mph IAS (about 214 - 226 mph TAS)
I get 16.7 °/s @ 3700 m for a P-40E at 3911 kg/44" Hg.
Your data indicates 13.1 - 12.4 °/s for 42" Hg.
>56 in Hg WEP 1490 hp @ 4300 ft
>2.85 G sustained between 180 mph and 232 mph IAS (about 192 - 247 mph TAS)
I get 20.9 °/s at 4000 ft for a P-40N-1 at 3447 kg/57" Hg.
Your data indicates around 16.2 °/s, but the P-40N I calculated is considerably lighter so it's no suprise your high-powered P-40E doesn't turn as quickly.
>I'll have to look into calculating rate and radius as well. Don't remember the formulas offhand.
T360° = 2 * pi * v / sqrt (a^2 - g)
>It looks like HoHun and I are doing things a little differently. He is calculating the speed a max Cl turn can be sustained. I'm calculating the max G that can be sustained at a given power and weight.
If you're interested, we could swap spreadsheets :-) Your engine model seems really neat, much smarter than the tabulated data I'm using!
Regards,
Henning (HoHun)
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Originally posted by HoHun
>For the P-40E
>At 8600 lbs, full fuel load
>80% propellor efficiency
>42 in Hg mil 1150 hp @ 12,000 ft
>2.25 G sustained between 178 mph and 188 mph IAS (about 214 - 226 mph TAS)
I get 16.7 °/s @ 3700 m for a P-40E at 3911 kg/44" Hg.
Your data indicates 13.1 - 12.4 °/s for 42" Hg.
That is probably due to my speed range being about 15-20 mph faster than your figure. I'm hitting the Cl limits about the same speed you are, just a different way of looking at things.
>56 in Hg WEP 1490 hp @ 4300 ft
>2.85 G sustained between 180 mph and 232 mph IAS (about 192 - 247 mph TAS)
I get 20.9 °/s at 4000 ft for a P-40N-1 at 3447 kg/57" Hg.
Your data indicates around 16.2 °/s, but the P-40N I calculated is considerably lighter so it's no suprise your high-powered P-40E doesn't turn as quickly.
[/color]
Yah, about 14% heavier, but only about 11-12% more power. Higher induced drag, and lower power/weight, it shouldn't turn as well as the P-40N.
>I'll have to look into calculating rate and radius as well. Don't remember the formulas offhand.
T360° = 2 * pi * v / sqrt (a^2 - g)
[/color]
Cool, what is the a?
>It looks like HoHun and I are doing things a little differently. He is calculating the speed a max Cl turn can be sustained. I'm calculating the max G that can be sustained at a given power and weight.
If you're interested, we could swap spreadsheets :-) Your engine model seems really neat, much smarter than the tabulated data I'm using!
Regards,
Henning (HoHun)
[/color]
Sure, I will shoot you of an email.
Greg Shaw
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Originally posted by HoHun
If you're interested, we could swap spreadsheets :-) Your engine model seems really neat, much smarter than the tabulated data I'm using!
Regards,
Henning (HoHun) [/B]
My current model is pretty simple, being little more than a Standard Atmosphere chart with delusions of grandeur. But is has proven surprisingly accurate, generally within +-1% of published charts.
Do an Allison F3/4R for a quick example.
1150 hp @ 12,000 ft on 42 in Hg.
We know from the Standard Atmosphere that static pressure at 12,000 ft is 19.03 in Hg and abs temp is 264.56K. That is all the information we need to guesstimate power from SL on up.
Static pressure - 19.03 in Hg
Manifold pressure - 42 in Hg
Blower pressure ratio - 42 / 19.03 = 2.21
From there you can multiply static pressure at any altitude by 2.21 to get the max MAP the blower can provide.
ie SL pressure 29.92 * 2.21 = 66.1 in Hg max MAP
SL temp 288.36K
HP is proportional to MAP * rpm. As long as rpm stays the same that provides the starting point to calculate hp.
66 / 42 * 1150 = 1805 hp
Take temp difference into account:
sqrt (264.56/288.36) = .958
.958 * 1805 = 1730 hp
That is actually about 1.5% below the 1760 hp @ SL on the Allison charts. But that could be just due to slightly different Standard Atmospheres, and is within a decent margin of error anyways.
I'm working on a more detailed calculator, using info from Hooker's "Not Much of an Engineer," but this will do for now.
Greg Shaw
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Just peeked in to see what was happening... Just love this stuff :)
Badboy
PS
This on the AH Spitfire Mk 1 from another thread, scrole down to my post:
http://www.hitechcreations.com/forums/showthread.php?threadid=103594&referrerid=2314
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FromHoHun:
"Hi Angus,
>Torque would be energy divided with time right?
Torque and energy are the same dimension, actually.
>Anyway, I did total energy and energy on a timescale.
Hm, I hadn't thought of that angle! But what is the gain over a pure altitude-over-time graph? :-) "
Ok. I'll explain what I did.
The original concept was to calculate the wing lift efficiency.
The Spitfire MkI (87 oct) and the 109E (87 oct) were the best candidates I could find, since the engine power and weight were very similar. (Actually both rather favouring the 109 from stats)
So, I calculated the mass to Newtons, then onwards with mass to altidude and those divided with the time it took to pull the mass up to 10K and 20K.
For fun, one can play about with this pr. hp. of engine power. Since that is not really my cup of tea, I didn't do that much.
So, looking at the outcome, the 109 was quicker to any altitude in real life, the difference being less marked in Newtons. (Spitfire heavier, so there was more mass lifted to same alt)
UNTIL.......
I tried a Spitfire I with a 3-blade CS airscrew. (The other two were fixed pitch, I rather think one of them was a 2-blade, but not sure)
The Spitfire with 87 oct and an early type Rotol was considerably superior to the 109 regarding climb, be it real time or Newtons, the difference being more marked in Newtons.
I am not good enough at this math to conclude how much the difference could have been had the both aircraft been the same weight. But it really struck me never the less, and for this there can be basically two explanations.
1. The Merlin is quite more powerful than the DB
(109E and Spit I on 87 oct)
2 The Wing of the Spitfire in question generates quite some more lift than the wing of the 109 given roughly the same thrust.
I rather favour number two here as a logical explanation.
Anyway, HoHun, I'll mail this to you if you are interested.
Best regards
Angus
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1. The Merlin is quite more powerful than the DB
(109E and Spit I on 87 oct)
2 The Wing of the Spitfire in question generates quite some more lift than the wing of the 109 given roughly the same thrust.
Just a few thoughts Angus:
1. But it wasn't, at least according to numbers. The propellors were of different shape and probably rotated at slightly different speed. But what was their efficiency ->generated thrust?
2. Depends on the AoA? I'd imagine the 109 wing generate very good figures for lift in small AoA due to its NACA profile where as Spitfires wing is better when AoA increases because of its wing area alone. I'd also like to see the effects of the washout in Spits lift figures in level flight. The wing tips can't generate much lift as the AoI is different for the root and tip? So their drag figures, too, are quite different in different speeds and angles?
I'd also imagine the angle of climb have a significant effect on results your calculations may give.
-C+
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Came across this and thought of this thread.
(http://www.onpoi.net/ah/pics/users/503_1098657121_spit109turn.jpg)
Hope this helps.
Crumpp
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Lovely, excellent.
Now I just have to read the darned thing. ;)
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Originally posted by Crumpp
Came across this and thought of this thread.
Crumpp
In my last message I posted a link to another thread in which I posted the url to those diagrams at the fourthfightergroup site:
http://www.fourthfightergroup.com/eagles/spit109turn.gif
They are simply a different style of EM diagram sometimes seen in reports dated that far back, but contain almost identical information to the ones I produce, just presented in a different way.
Quite a long time ago now, I decided to reproduce the same type of analysis used in that diagram and piggy back it onto the work I've already done for the Aces High aircraft. Now when I produce an EM analysis for any aircraft, I automatically get both types of diagram. There are advantages and disadvantages to both methods. The diagrams I've been producing were not conceived until the 1960s and they have the advantage that they can be overlaid with one another for a far easier comparison. The disadvantage with the type of EM diagram that originated in the late 30s is that they can't be overlaid. So for example, here is an overlay of the AH Spitfire MkI and the Me109e
(http://www.badz.pwp.blueyonder.co.uk/Files/Images/Spit1c.jpg)
The main advantage to producing the type of EM diagrams seen in those early reports is that they can be compared with the ones that exist for the real aircraft. For example, the Spitfire in the diagram above has this diagram:
(http://www.badz.pwp.blueyonder.co.uk/Files/Images/Spit1b.jpg)
But let's compare that with the diagram for the real Spitfire MkI.
(http://www.badz.pwp.blueyonder.co.uk/Files/Images/Spit1a.jpg)
Here we can see that both the real Spitfire and the AH Spitfire have the same corner velocity at that altitude and configuration, so let's compare a turn. Just for example I've selected a 5g turn at the corner speed of 250mph. I've indicated on the diagram for the real Spitfire that it would need to descend at 16 degrees below the horizon to sustain that turn and it would turn a full circle in about 14.5 seconds with a radius of about 850ft. You can see from the diagram for the AH Spitfire that it would also make the same turn in about 14.5 seconds with a radius of 850ft, and that it would need to descend at an angle of 23 degrees below the horizon, a descending turn only 6 degrees steeper than the real aircraft. But the turn rates and radii for the turn, along with the corner speed are amazingly close. The difference in the angle of descent is probably due to differences in engine power available at that altitude between the real world tests and Aces High, and perhaps some differences in weight.
It is interesting that both diagrams are essentially the same shape, and that they agree quite closely in many respects, indicating that the flight model in Aces High has accounted for all of the aerodynamic factors that would influence the shape of the curves to any significant degree. A worthy achievement indeed. Kudos to HTC.
This is even more significant, because I've made a similar comparison with the Spitfire and 109e from other simulations, and so far Aces High has first place for accuracy. I was thinking of writing up the whole comparison for an article on SimHQ, but other projects have pushed back in the que.
Hope that helps...
Badboy
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Excellent work Badboy. :aok
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Hi Badboy,
My Spitfire I turn calculation agrees with the WW2 chart quite well in turn rate at 12000 ft, but I get a much higher turn speed and accordingly a larger radius. (About 190 mph vs. 160 mph.)
Since I've used the NACA report lift coefficient for the Spitfire which is a bit lower than the British value, that appears normal. Since the NACA value is measured data opposed to the "assumed" data in the WW2 turn rate graph you linked, I feel I'm erring on the side of caution here ;-)
Regards,
Henning (HoHun)
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Originally posted by HoHun
Hi Badboy,
My Spitfire I turn calculation agrees with the WW2 chart quite well in turn rate at 12000 ft, but I get a much higher turn speed and accordingly a larger radius. (About 190 mph vs. 160 mph.)
Since I've used the NACA report lift coefficient for the Spitfire which is a bit lower than the British value, that appears normal. Since the NACA value is measured data opposed to the "assumed" data in the WW2 turn rate graph you linked, I feel I'm erring on the side of caution here ;-)
Regards,
Henning (HoHun)
Hi HoHun
Nice to talk to you again…
I know the graph posted does say that those lift coefficient values are assumed, and that is unfortunate, because I know they are not… At least not in the sense of being guessed at. I read the reports a long time ago, but recal that those values were arrived at after in flight and wind tunnel measurements were used to produce an empirical relationship. I’ve seen the data, the resulting graph, and equation. The in-flight measurements were obtained using swivelling pitot and suspended static heads, and stalls were done at both full throttle and gliding in order to determine the stall boundaries used on the graph. I think they used the word “assumed” to indicate that the intermediate values were not measured directly, but resulted from calculations based on the resulting empirical formulae. I checked the calculations back when I read it, and I think they are likely to yield far better results than using a constant Clmax value, and I use similar methods in my own work.
I’m just mentioning that because I’ve seen the graph used in previous threads, where it was negatively criticised on the basis of that word, which I admit is misleading. I didn’t say anything before, because I’ve discovered there really isn’t any point bringing facts into a discussion when the participants have some other axe to grind.
So it is perhaps unfortunate that the authors of that report used that word, because although they did some very excellent work at the time, they weren’t to know that some 60 years later the appearance of that word on their graph would be used in arguments on a flight sim’ board to invalidate it for reasons of gamesmanship… Not accusing anyone here, just saying. Bottom line... My vote goes to Farnborough not NACA :-)
Badboy
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Hi Badboy :-)
>I know the graph posted does say that those lift coefficient values are assumed, and that is unfortunate, because I know they are not… At least not in the sense of being guessed at.
Roger that, and I actually I assumed they were assumed as quoted for a very good reason :-)
Still, for the NACA data, I had the full report along with the test procedures, and was certain the measured Clmax was indeed achieved under Gs. Do you know whether this was the case for the British tests, too?
>I checked the calculations back when I read it, and I think they are likely to yield far better results than using a constant Clmax value, and I use similar methods in my own work.
Well, I'm using a constant Clmax, but since I picked the Clmax for the acceleration that coincedes with the best turn rate, the error will only catch up with me in some other place ;-)
>So it is perhaps unfortunate that the authors of that report used that word
I agree. Though it's correct terminology, it implies more uncertainty about the Clmax than there actually was.
>Bottom line... My vote goes to Farnborough not NACA :-)
I stick with NACA for now, but I'm ready to admit we're operating at the limits of what is possible in retrospect accuracy anyway :-)
Regards,
Henning (HoHun)
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Here's a little something to keep the allied farmbois panties bunched...
http://mnemeth1.brinkster.net/movies/EAA_Interviews.wmv
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They are simply a different style of EM diagram sometimes seen in reports dated that far back, but contain almost identical information to the ones I produce, just presented in a different way.
I did not follow your link. That diagram was sent to me through another source. Nice work, Badboy.
Crumpp
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Whoa, nice link Wotan.
Do you know what 109G the guy is referring to?
1500 lbs ligter than the Mustang anyway.