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blairgowrie
02-15-2010, 07:00 AM
Continued from Part 1 :

http://forums.ubi.com/eve/foru...881026238#3881026238 (http://forums.ubi.com/eve/forums/a/tpc/f/23110283/m/2221055328?r=3881026238#3881026238)

M_Gunz
02-15-2010, 08:02 AM
Old school might be the physics you learn when not actually going to aero-engineering school?

I think I have an answer on how to deal with the Gaston Force. You make all your other calculations and then
walk around babbling and waving but essentially doing nothing else, then build the airplane as designed.

IMO drinking lots of booze during the Gaston Phase of Design (TM) would probably help tremendously. Remember
to make noise but otherwise DO NOTHING.

Kettenhunde
02-15-2010, 08:09 AM
Old school might be the physics you learn when not actually going to aero-engineering school?

Ahh, like learning to parrot some formulas from your buddies doing radio controlled models?

K_Freddie
02-15-2010, 10:33 AM
Hopefully there are no missing G's in the 'aerodynamic bible' http://forums.ubi.com/images/smilies/16x16_smiley-wink.gif

M_Gunz
02-15-2010, 10:36 AM
Noooo. Like doing the actual experiments, gaining the actual results, working out the actual formulae and
writing it all up. Sometimes you even do something that flies and learn an aspect of flight in the doing.

Let's see, F=MA... do I merely parrot that or did I actually confirm and understand that through experiment?
I remember doing the experiments but perhaps I was hypnotized into THINKING I did. Man, that's a rough call!

As to the lift formula, I seem to remember learning about the lift curve and some other things, air density
from back in school, was it Boyle's Law and good old Daniel Bernoulli's work that we were wrongly told is
how airplane wings get their lift. I know it's not the whole story by far. The trouble is that I did all that
over 30 years ago and only used parts of it in work. While I did tighten my math up in 88-89, the school I
went to on my GI Bill was deficient in the sciences, the "Physics Prof" didn't know mass from weight.

The most important thing I and many others got from school besides learning basic reading, spelling, math
and science.. and history, literature, sports, theater, a load of other things like getting along -- the
really big thing was you either learned how to learn on your own or you joined the slacker side of the curve.
And slackers did not take the courses that I aced so don't even ask if that's where I fit.

I'm really not so happy with a system that lets me go on about maneuvers at 20k ft that are not possible.
When it comes to EAS, polly wants more than crackers, you dig? How many pages did the debate go into WTF mode
without a clean explanation from the expert? How many pages have passed since that business started of where
did the 1G go before it's understood that the EAS relative turn demonstration was not completely explained
from the start and that actual bank angles, speeds and maneuvers actually possible change with altitude?

We had a clue. EAS does not yield specific performance but more should have been told from the start.
That's no reason to say we only repeat formulae we don't understand. If I was that way then I'd just jump
on the EAS wagon and recite the new gospel without a care.

M_Gunz
02-15-2010, 10:37 AM
Originally posted by K_Freddie:
Hopefully there are no missing G's in the 'aerodynamic bible' http://forums.ubi.com/images/smilies/16x16_smiley-wink.gif

I bet you've followed and understood every post in this thread, parts 1 and 2.

Erkki_M
02-15-2010, 10:49 AM
Kettenhunde, I think you are mistaking me with someone else; I had taken part to this topic before the previous post only twice, perhaps 3 times.

With old school I did not mean "simple", as I say in my previous post. It is simple only seemingly.

You're burning the wrong witch lol

Holtzauge
02-15-2010, 10:51 AM
Originally posted by Holtzauge:
OK, and work on the aircraft can be anything from pratical work to office work I suppose? I assume you are a small outfit so everyone has to help out. When restoring the aircraft to flyable condition I assume you try to refurbish as much as possible? But what about things that need to be replaced? I assume you use drawings and makers specs as far as possible but what about the cases that are not clear cut and obvious how to reverse engineer? Do you have aeronautical structural engineers employed? Apart from board of directors and membership work, do you yourself work on the structural engineering?

Originally posted by Kettenhunde:

I just wanted to know how you solved the issue of structural engineering in the WW2 fighter aircraft foundation and if you did structural engineering work yourself on White 1 that's all.

K_Freddie
02-15-2010, 11:37 AM
Do you realise Blair .. that you just spoilt the potentially longest running topic on the forum..
http://forums.ubi.com/images/smilies/bigtears.gif
http://forums.ubi.com/groupee_common/emoticons/icon_biggrin.gif

Gaston444
02-15-2010, 12:01 PM
Quote, Erkki_M: "I'm not going to translate the interview sentence-to-sentence or word-to-word to you only to give you another topic to argue about(the translation). If you want the interviewer's opinion, go ask Grendel-B @ #Warbirds in ircnet. If you want to stick with the translation, well, Karhila's words should be clear enough for anyone to understand your claims about what his opinions were, let alone tactics used in real life, let alone again tactics that actually worked, to achieve whatever favourable thing in real life aerial combat, are completely wrong, being misunderstood, taken out of context, exaggerated and interviews only partially quoted(and even modified) to support ridiculous claims"

-Well, since he DOES mention the speed of 250 km/h in the context of turns (in the broader context of "I downthrottled and most didn't"!!!!), unless you can prove to me he meant that 250 km/h is the WORST possible speed to sustain turns in, I would say things DO NOT look good for your argument, or, as you say, the "non-ridiculous" side...

Your unwillingness to support your point concretely speaks volumes... I don't remember modifying anything: If I misremembered, can you provide examples?

Hey, Gunz_M, explain to me in the context of talking about the prop thrust CENTER, what is the relevance of your statement:

"Be sure that the prop blades make thrust ALL THE WAY AROUND. Be sure that the rate you'd have to flip the prop to cause zero thrust on the lower part would have to exceed the speed of the plane greatly."

Are you implying the prop thrust has NO center? Are you implying that this thrust center cannot move around into one prop disc half or the other, and STAY there as long as the elevators are deflected?

Can anyone explain to me how elevator deflection will NOT move the prop disc's center of thrust around?

And if the prop thrust center DOES move around, can anyone explain to me how it will NOT match the direction of deflection of the elevator?

Can we stick to LONGITUDINAL forces? I don't think I ever brought up anything else...

Gaston

blairgowrie
02-15-2010, 12:03 PM
Originally posted by K_Freddie:
Do you realise Blair .. that you just spoilt the potentially longest running topic on the forum..
http://forums.ubi.com/images/smilies/bigtears.gif
http://forums.ubi.com/groupee_common/emoticons/icon_biggrin.gif

I know Freddie but I was starting to fall asleep reading it every day. http://forums.ubi.com/images/smilies/16x16_smiley-happy.gif

AndyJWest
02-15-2010, 12:18 PM
Can anyone explain to me how elevator deflection will NOT move the prop disc's center of thrust around?

And if the prop thrust center DOES move around, can anyone explain to me how it will NOT match the direction of deflection of the elevator?

Can we stick to LONGITUDINAL forces? I don't think I ever brought up anything else...

AS has already been pointed out, the P-effect moves the prop's centre of thrust around as the direction of incoming airflow varies. There are also gyroscopic forces. Neither 'match the direction of deflection of the elevator'. As for what you 'brought up', it is up to you to 'support your point concretely': give us the maths, and maybe there will be something to debate.

Kettenhunde
02-15-2010, 02:12 PM
Kettenhunde, I think you are mistaking me with someone else; I had taken part to this topic before the previous post only twice, perhaps 3 times.

With old school I did not mean "simple", as I say in my previous post. It is simple only seemingly.

I was referring to Wurkeri's post unless of course you are him. http://forums.ubi.com/groupee_common/emoticons/icon_smile.gif

The speed and climb rate of the in game P-47D-22 and LFIXC are very close the real life test results at 20k, so I don't see a reason why turning performance should be badly wrong. Besides, an old school calculation supports game results...

http://forums.ubi.com/eve/foru...131006238#9131006238 (http://forums.ubi.com/eve/forums/a/tpc/f/23110283/m/2221055328?r=9131006238#9131006238)

If you use the same methods and assumptions, you will get the same results. I say assumptions because you probably already know that some of these values are set by convention but in fact change depending on the conditions of flight.

M_Gunz,

Almost everything changes with density. Because parasitic drag changes with density our wing efficiency changes as well. Parasitic drag lowers with atmospheric density in a direct relationship and we should experience ~17% increase in wing efficiency. If you factor the correct relationships into TAS calculations, the performance is the same.

If you hold efficiency steady over altitude, you are not going to see the correct relationship if you use TAS.

If you have the proper relationships, formulas, and are consistent in your units, there is no difference between EAS and TAS at this level.

EAS simply levels the environmental effects of density for all aircraft.

K_Freddie
02-15-2010, 02:54 PM
http://forums.ubi.com/groupee_common/emoticons/icon_biggrin.gif

K_Freddie
02-15-2010, 05:25 PM
As a thought ..
A step-by-step analysis of this clip would be appreciated http://forums.ubi.com/groupee_common/emoticons/icon_wink.gif
FW190 move-around (50MB) (http://www.vanjast.com/IL2Movies/Scissors.avi)

K_Freddie
02-15-2010, 05:28 PM
Blair... please don't cut us short http://forums.ubi.com/images/smilies/10.gif

M_Gunz
02-15-2010, 07:15 PM
Originally posted by Gaston444:
Hey, Gunz_M, explain to me in the context of talking about the prop thrust CENTER, what is the relevance of your statement:

"Be sure that the prop blades make thrust ALL THE WAY AROUND. Be sure that the rate you'd have to flip the prop to cause zero thrust on the lower part would have to exceed the speed of the plane greatly."

Are you implying the prop thrust has NO center? Are you implying that this thrust center cannot move around into one prop disc half or the other, and STAY there as long as the elevators are deflected?

Thrust center is just an AVERAGE of all thrust forces across the prop disk. What do you think it would take to move
that center more than a slight amount? Oh yeah, you THINK that a hard turn would do it! You don't bother with finding
out because it's only in defense of your own BS that's gotten you run off more than one forum already.

Can anyone explain to me how elevator deflection will NOT move the prop disc's center of thrust around?

Can YOU explain how you come up with 2000 lb-force difference between top and bottom half of the prop disk?
Can YOU back your numb-A claim to that? YOU have been asked again and again and yet FAIL to do so.

And if the prop thrust center DOES move around, can anyone explain to me how it will NOT match the direction of deflection of the elevator?

AMOUNT PLEASE! That's one "All on the Top and None on the Bottom" to go.

It's really this simple, the average is not the whole. Average moves up but the whole disk makes thrust.

Kettenhunde
02-15-2010, 07:29 PM
Crumpp says:
Almost everything changes with density.

Crumpp told FC:
You used TAS which means you have to use formulas that account for density

http://forums.ubi.com/eve/foru...941004138#3941004138 (http://forums.ubi.com/eve/forums/a/tpc/f/23110283/m/2221055328?r=3941004138#3941004138)

Lastly, for all the "experts"...

If we want to hold our assumptions the same then we need to vary thrust with density.

2300 hp * .85 = 1955thp

1955thp*325 / 180KEAS = 3529.861111lbs of drag

sigma = .53281 @ 20,000ft

SQRT(.53281)* 3529.861111lbs of drag = 2576.58lbs of drag

2300 hp * .85 = 1955thp

1955thp*325 / 246.6KTAS = 2576.5lbs of drag

Holtzauge, Wurkeri, or JtD seem to have passed up the opportunity to demonstrate their knowledge of aerodynamics!

http://forums.ubi.com/groupee_common/emoticons/icon_smile.gif

Wurkeri
02-16-2010, 12:16 AM
Originally posted by na85:
I always try to solve a problem backwards to check my work. Let's put the EAS/TAS issue to bed.

Using
S=27.73m^2
W=13583lbf=60420N
Ra=aspect ratio=5.554
Cdo = 0.032215
density = 0.64938 kg/m^3

for a 2.57G turn at 247 knots true

Power required = q*v*S*[Cdo + Cl^2/(pi*e*Ra)]

where
q = 5.25 kPa
Cl = lift/q*S = 1.066
e = 0.65 (oswald's)

I get Pr = 1786 kW which is ~2395 hp.

This one was not replied in the old thread.

Using the same numbers I get:

247 KTAS = 127.0678 m/s

Power required = q*v*S*[Cdo + Cl^2/(pi*e*Ra)]

=> Pr = 5250Pa * 127.0678m/s * 27.73m^2 * [0,032215 + (1.066^2/(pi*5,554*0,65)] = 2449kW = 3285hp

So there appears to be an error in your calc despite the formula is almost right (misses the propeller efficiency).

With the same propeller efficiency 0.85 as earlier, the result would be 3864hp. You also used lower e (0.65 vs 0.85) than in the earlier calcs hence the much higher Pr than calculated by me, FatCat, Holtzauge, JtD and apparently also Oleg because the Spit LFIXC in the game easily maintains higher g load at 20k than the P-47D-22.

Edit: Edited the typos out, thanks Gunz.

M_Gunz
02-16-2010, 04:45 AM
Originally posted by Wurkeri:
<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">Originally posted by na85:
Power required = q*v*S*[Cdo + Cl^2/(pi*e*Ra)]

Pr = 5250Pa * 127.0678m/s * 27.73m^2 * [0,032215 * (1.066^2/(pi*5,554*0,65) = 2447kW = 3281hp </div></BLOCKQUOTE>

Nate adds the drag coefficients where you, errrr, multiply them? Never seen it done that way before....

Wurkeri
02-16-2010, 05:50 AM
Originally posted by M_Gunz:
Nate adds the drag coefficients where you, errrr, multiply them? Never seen it done that way before....

Ah... thanks! Edited. That was a typo and now the calculation should be correct. You can see from the result that it was originally calculated correctly in the form Cd + Cl^2/(pi*AR*e), I just made a typo while writing the post.

Kettenhunde
02-16-2010, 06:17 AM
Wurkeri,

I am going to try to help you out. It is really a simple fix if you will listen. You will learn something too.

By definition EAS and TAS are the same aerodynamic forces. Think of this as a law. You cannot violate it.

That means you are not correctly handling the density change in your procedure. I am not a big SI user so I am not 100% familiar with everything but the principles still very much apply.

First, let's keep our power separate from our force. It will be easier to see what is going on for you.

Power in a supercharged engien does not vary with density below FTH. However, the force of THRUST still varies as a propeller is nothing but a series of wings moving in a circle.

In EAS the corresponding dynamic pressure for TAS are the exact same.

@ 20,000ft:

Dynamic pressure at 180KEAS q = 180^2/295 = 109.93psf

Dynamic pressure at 246.6KTAS q = (.53281*246.6^2)/295 = 109.83psf

That .53281 is our density ratio and how we are accounting for the change in the atmospheric environment for the Power Required our airplane is operating in.

Drag in lbs = Coefficient of Drag * Dynamic pressure* reference area

If the Coefficients are the same, the dynamic pressure is the same, and our reference area is the same....what is going to vary our results?

Power, when we convert the Power Available we must factor in a density change. Dynamic pressure is not a density change, it is dynamic pressure and by definition equal to our EAS value.

Your formula is not designed to examine power production over altitude. It is designed to look at one altitude and one dynamic pressure. Density changes are not accounted for in your power available.

Power Available = Power Required

Two sides of the equal sign and basic math tells us that what is done to one side, must be done to the other in order to remain equal.

So just calculate your drag in Newtons for now. We will do power conversion and account for density changes when we all agree on the force of drag involved.

blairgowrie
02-16-2010, 07:46 AM

Holtzauge
02-16-2010, 11:22 AM
Originally posted by Kettenhunde:
<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content"> Crumpp says:
Almost everything changes with density.

Crumpp told FC:
You used TAS which means you have to use formulas that account for density

http://forums.ubi.com/eve/foru...941004138#3941004138 (http://forums.ubi.com/eve/forums/a/tpc/f/23110283/m/2221055328?r=3941004138#3941004138)

Lastly, for all the "experts"...

If we want to hold our assumptions the same then we need to vary thrust with density.

2300 hp * .85 = 1955thp

1955thp*325 / 180KEAS = 3529.861111lbs of drag

sigma = .53281 @ 20,000ft

SQRT(.53281)* 3529.861111lbs of drag = 2576.58lbs of drag

2300 hp * .85 = 1955thp

1955thp*325 / 246.6KTAS = 2576.5lbs of drag

Holtzauge, Wurkeri, or JtD seem to have passed up the opportunity to demonstrate their knowledge of aerodynamics!

http://forums.ubi.com/groupee_common/emoticons/icon_smile.gif </div></BLOCKQUOTE>

You yourself point out that neither me, FC, JtD or Wurkeri come close to your results and we all think your calculations are way off. This does not cause you to pause and reflect?

It's obvious from your resumed lecturing and increasing swagger factor that you have not learned a thing from what has passed and need to be reminded.

Do you seriously think everyone has just forgotten and suddenly bought into your pathetic explanation of how loadfactor 3.75 becomes 2.57 which “is not that far from 2.2” and that now everything is hunky-dory?

You behave like an ostrich with your head in the sand. Are you just bluffing or are you really so blissfully unaware how ruffled your feathers are Polly?

I think your continual reference to limitations in understanding and “parroting” formulas without understanding how they should be used is very close to the mark. I just don’t understand why you would shoot yourself in the foot like that....

PS: Must say Wurkeri did a commendable job above requiring just one short equation to show how wrong you are. Maybe this time you will learn to close that big beak of yours.

It's like I said before, Kettenhunde, your limitations in the understanding of aerodynamics is only surpassed by your baseless arrogance.

Holtzauge
02-16-2010, 12:38 PM
Originally posted by Kettenhunde:
<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content"> Crumpp says:
Almost everything changes with density.

Crumpp told FC:
You used TAS which means you have to use formulas that account for density

http://forums.ubi.com/eve/foru...941004138#3941004138 (http://forums.ubi.com/eve/forums/a/tpc/f/23110283/m/2221055328?r=3941004138#3941004138)

Lastly, for all the "experts"...

If we want to hold our assumptions the same then we need to vary thrust with density.

2300 hp * .85 = 1955thp

1955thp*325 / 180KEAS = 3529.861111lbs of drag

sigma = .53281 @ 20,000ft

SQRT(.53281)* 3529.861111lbs of drag = 2576.58lbs of drag

2300 hp * .85 = 1955thp

1955thp*325 / 246.6KTAS = 2576.5lbs of drag

Holtzauge, Wurkeri, or JtD seem to have passed up the opportunity to demonstrate their knowledge of aerodynamics!

http://forums.ubi.com/groupee_common/emoticons/icon_smile.gif </div></BLOCKQUOTE>

Just had a closer look at your calculation and it does seem you are in dire need of an “expert” to “demonstrate” aerodynamics to you :

You get confused by your own equations.

By definition, EAS means that you have the same dynamic pressure q at both altitudes. You don’t seem to have grasped this since you suddenly show 2576.5 lbs drag at 20000 ft and 3529.86111 lb at SL.

You see the drag will be the same at both altitudes, i.e 3529.86111 lb. Otherwise it’s not EAS.

So entering the correct drag into the thrust equation yields:

Ps hp * .85 = Pt thp

Pt thp*325 / 246.6KTAS = 3529.861111lbs of drag

So Pt=2678.35

Which gives Ps=2678.35/0.85=3151 hp

I guess you shot a really big hole in that foot now Polly. All the while you have been parroting on how important it is to use EAS and you can’t seem to be able to use it properly yourself…

M_Gunz
02-16-2010, 03:35 PM
See, I don't know just how Crumpp got drag at 20,000ft (sigma? I don't know any sigma!) but I am
guessing that it has something to do with specific performance and altitude and stepping out from
the fits-all-alts equation.

OTOH I don't understand why Holtz is plugging TAS speed into an EAS equation... is that where he
left the one to do the other to find the specific performance.

They can't both be right since the results differ and I sense a number of posts coming with noise
to signal ratio becoming very high followed by a LOCK as usual. If I knew more than I could maybe
tell the one wrong just how it is so but that's an awfully thin to none maybe it would get looked
at and accepted.

Kettenhunde
02-16-2010, 06:19 PM
You yourself point out that neither me, FC, JtD or Wurkeri come close to your results and we all think your calculations are way off. This does not cause you to pause and reflect?

LOL and I it though it was very amusing and quite telling that none of you understood the density relationships.

FC surprised me....you and Wurkeri not so much and JtD is just kind on the coattails of you two.

I even threw a bone out on several post's hinting at the answer.

You see the drag will be the same at both altitudes, i.e 3529.86111 lb. Otherwise it’s not EAS.

It is the same, we are just describing the relationship mathematically.

If you work in TAS and you hold your assumptions constant, then you must account for density effects.

When we account for density effects, they are the same amount of drag Holtzauge.

If we want to hold our assumptions the same then we need to vary thrust with density.

2300 hp * .85 = 1955thp

1955thp*325 / 180KEAS = 3529.861111lbs of drag

sigma = .53281 @ 20,000ft

SQRT(.53281)* 3529.861111lbs of drag = 2576.58lbs of drag

2300 hp * .85 = 1955thp

1955thp*325 / 246.6KTAS = 2576.5lbs of drag

2576.5lbs@180KEAS = 2576.5lbs@246.6TAS at FL20

BillSwagger
02-17-2010, 12:15 AM
Whats a better term for energy retention?

You have two planes with similar top speeds, and similar drag characteristics. One weighs 7500lbs and the other weighs 4000lbs.

They both dive at the same angle and level off, the heavier one holds its speed longer, right?

Whats that called?

Would it be better to say one has more inertia than the other?

Bill

FatCat_99
02-17-2010, 03:04 AM
Originally posted by Kettenhunde:
If we want to hold our assumptions the same then we need to vary thrust with density.

2300 hp * .85 = 1955thp

1955thp*325 / 180KEAS = 3529.861111lbs of drag

sigma = .53281 @ 20,000ft

SQRT(.53281)* 3529.861111lbs of drag = 2576.58lbs of drag

2300 hp * .85 = 1955thp

1955thp*325 / 246.6KTAS = 2576.5lbs of drag

2576.5lbs@180KEAS = 2576.5lbs@246.6TAS at FL20
And when you put T=2576lb and V=246KTAS into turn equation at 20000ft how big is the load factor?

Kettenhunde
02-17-2010, 04:45 AM
And FC what is your Wing efficiency at 20,000feet?

Still the same as at sea level?

If you use TAS you have to account for <span class="ev_code_YELLOW">ALL of the effects of density</span> to get the correct results.

You do understand this?

Wurkeri
02-17-2010, 05:46 AM
Originally posted by M_Gunz:
OTOH I don't understand why Holtz is plugging TAS speed into an EAS equation... is that where he left the one to do the other to find the specific performance.

This is rather simple issue, see pages 18-20 of the old thread, the posts where I hint na85 that EAS can't be used without conversion and after that na85 finaly realized what was the problem.

Shortly by definition power is force multiplied with speed:

W = F * v

and the speed here is of course true speed. So actually TAS is the correct way to use this formula. In other words Holtzauge is using the formula correctly and the other side has realized this now, over a week after I originally pointed out the error and FatCat did the same couple days later.

M_Gunz
02-17-2010, 05:52 AM
Originally posted by BillSwagger:
Whats a better term for energy retention?

You have two planes with similar top speeds, and similar drag characteristics. One weighs 7500lbs and the other weighs 4000lbs.

They both dive at the same angle and level off, the heavier one holds its speed longer, right?

Whats that called?

The words "Not always true" come to mind. Also where do you find a 4000 lb plane A and a 7500 lb plane B
with similar top speeds and drag characteristics? Even the same plane with and without heavy loading won't
be so close in speed and drag.

M_Gunz
02-17-2010, 06:17 AM
Then walk with me please and tell me what I miss or have wrong, I am not 'getting' it all anyway.

Originally posted by Kettenhunde:

If we want to hold our assumptions the same then we need to vary thrust with density.

2300 hp * .85 = 1955thp

1955thp*325 / 180KEAS = 3529.861111lbs of drag

sigma = .53281 @ 20,000ft

SQRT(.53281)* 3529.861111lbs of drag = 2576.58lbs of drag

2300 hp * .85 = 1955thp

1955thp*325 / 246.6KTAS = 2576.5lbs of drag

It seems to me that he is finding the specific drag both ways here and that using EAS
requires an adjustment due to .. sigma-something-to-do-with-density.
However I can't say if that was done in the proper step of the 180KEAS should have been 247KEAS
before then. I just know what the equation reads like, I don't know what the proper 'grammar' is.

................

By definition, EAS means that you have the same dynamic pressure q at both altitudes. You don’t seem to have grasped this since you suddenly show 2576.5 lbs drag at 20000 ft and 3529.86111 lb at SL.

You see the drag will be the same at both altitudes, i.e 3529.86111 lb. Otherwise it’s not EAS.

I would say that this is the difference to be addressed if I got the above correct.
I am pretty sure that if it's going to be EAS then 180KEAS is the speed to use and not 247KTAS because if
you use 247KTAS it's not EAS either!

Since I don't know about the sigma part, I can't say though. EAS is supposed to be the same dynamic pressure
and I can't see how the drag -at- the EAS speed should be different or need adjustment due to changing density
since AFAIK using EAS speed should already *do* that!

Something about units or systems perhaps? I dunno, waiting for something to be revealed that turns a light on
and a lot of I read comes across as noise.

Kettenhunde
02-17-2010, 06:47 AM
This is rather simple issue, see pages 18-20 of the old thread, the posts where I hint na85 that EAS can't be used without conversion and after that na85 finaly realized what was the problem.

Let's not re-write things. Your declared you cannot use EAS.

Using relative units like EAS can give hilarious results as can be seen above.

http://forums.ubi.com/eve/foru...991020138#4991020138 (http://forums.ubi.com/eve/forums/a/tpc/f/23110283/m/2221055328?r=4991020138#4991020138)

Wurkeri says:
Note that the speed must be true air speed because power is defined as force multiplied by speed.

http://forums.ubi.com/eve/foru...991020138#4991020138 (http://forums.ubi.com/eve/forums/a/tpc/f/23110283/m/2221055328?r=4991020138#4991020138)

http://img694.imageshack.us/img694/6959/wurkerinotagain.jpg (http://img694.imageshack.us/i/wurkerinotagain.jpg/)

You said nothing about conversion or with any particular formulation. You then proceeded to argue with na85 and when he showed you the forces were equal, that is when you decided to incorporate the power into the formula without converting it.

Wurkeri, what you did to Na85 is called a “bamboozle” where I come from...

I don't know if it is intentional or you just don't know the relationships. It might just be a mistake as I always see you using sea level density of 1.225kg/m^3 in your calculations.

Maybe that is why you get results like Cdo increasing with density?

Wurkeri says:

Pr = 5250Pa * 127.0678m/s * 27.73m^2 * [0,032215 + (1.066^2/(pi*5,554*0,65)] = 2449kW = 3285hp

Density changes are not accounted for in your power available.

You are using sea level dynamic pressure with True Airspeed velocity. No wonder you are getting goofy results.

I pointed out in the other thread your formula does not use a density ratio. It uses density and there for Dynamic pressure will change over altitude. It is not kept equalized through the formulation and must be worked by hand. It is a rule and requirement of the formulation that you use True velocity to get the correct density relationships. That limitation has nothing to do with the validity or accuracy of using EAS in other methods.

Dynamic pressure at 20,000 feet @ 127m/s velocity

q = .5 * 0.408427kg/m^3 * (127.0678 m/s)^2 = 3297 Pa

Dynamic pressure at sea level using

q = .5 * 1.22500 kg/m^3* (92.6^2)^2 = 5252Pa

If we use the correct dynamic pressure values for the SI units and formulation we will get the same answers:

Now let’s correct align our velocity with the proper density relationship for Power Available…

EAS Power relationship = 5250Pa * 92.6 m/s * 27.73m^2 * [0,032215 + (1.066^2/(pi*5,61*0,85)] = 1457400.361N/m-s = 1954thp 1954/.85 = 2298bhp

If you combine the power available and power required into one formula to derive power, you must be consistent in your units for velocity or you will not have the correct relationship.

<span class="ev_code_YELLOW">Remember in basic math, what is done to one side of the equal sign must be balanced to the other.

Power available EAS = Power required EAS

(Power available EAS Plus correction for density = Power Available TAS) = (Power required EAS + correction for density = Power available TAS)

Even though all the values have changed, the basic relationship REMAINS TRUE:

Power available EAS = Power required EAS</span>

You have convoluted the issue by not keeping your relationships separate. It is harder for people to understand what is going on.

Maybe that is intentional on your part?

Wurkeri
02-17-2010, 08:35 AM
Originally posted by M_Gunz:
Then walk with me please and tell me what I miss or have wrong, I am not 'getting' it all anyway.

Gunz,
Let's walk through this first from power side. If power is:

W = F * v

Then the force is:

F = W / v

And if we add prop efficiency, n, we got:

F = W * n / v

This works with SI units without any conversions, example is same values as

W =2300hp = 1715110 W
v = 246.6 KTAS = 126,862 m/s
n = 0.85

F = W*n/v = 11491,57 N

Note here that we don't need to know anything about density nor altitude; this is correct thrust at this given true speed and given prop efficiency. This is the way me, Holtzauge, FatCat, JtD and the most of the world do this because it's simple, easy and logical. Same works from thrust (drag) side at reverse order as was shown earlier.

Now, EAS is same as true speed only at sea level so we can't calculate other altitudes without conversion. But this is what na85 and the other side did earlier before I pointed out error on above mentioned pages, over a week ago. Funny thing here is that above, in this new thread, the other side blames us (me, FatCat and others) for this error despite we did not need to do the conversion due to fact that our calcs were based on true speeds. And it was actually na85 and the other side who made this error; na85 at least did admit the error but as can be seen above, the other side behaves different way... not even mentioning the original 3.57g claim nor peeling the 1g off...

As concluding remarks, check Holtzauge's results from page 15 of the old thread and you can also use JtD's spreadsheet and do your own analysis.

M_Gunz
02-17-2010, 08:55 AM
Originally posted by Kettenhunde:
Dynamic pressure at 20,000 feet @ 127m/s velocity

q = .5 * 0.408427kg/m^3 * (127.0678 m/s)^2 = 3297 Pa

Dynamic pressure at sea level using

q = .5 * 1.22500 kg/m^3* (92.6^2)^2 = 5252Pa

I look at these and I don't see the same thing at two different alts.

(127.0678 m/s)^2 ......... speed squared

(92.6^2)^2 ............... something to the 4th power

Quite often these things turn into something only someone in the know can follow
as values appear that I can't what they are or where they came from, and they lack
labels so even trying to look them up is useless. And that goes for all yins.

Is there a typo? Is it right? Where's the girl with the clues cart?

M_Gunz
02-17-2010, 09:22 AM
Originally posted by Wurkeri:
<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">Originally posted by M_Gunz:
Then walk with me please and tell me what I miss or have wrong, I am not 'getting' it all anyway.

Gunz,
Let's walk through this first from power side. If power is:

W = F * v

Then the force is:

F = W / v

And if we add prop efficiency, n, we got:

F = W * n / v

This works with SI units without any conversions, example is same values as

W =2300hp = 1715110 W
v = 246.6 KTAS = 126,862 m/s
n = 0.85

F = W*n/v = 11491,57 N

Note here that we don't need to know anything about density nor altitude; this is correct thrust at this given true speed and given prop efficiency. This is the way me, Holtzauge, FatCat, JtD and the most of the world do this because it's simple, easy and logical. Same works from thrust (drag) side at reverse order as was shown earlier.

Now, EAS is same as true speed only at sea level so we can't calculate other altitudes without conversion. But this is what na85 and the other side did earlier before I pointed out error on above mentioned pages, over a week ago. Funny thing here is that above, in this new thread, the other side blames us (me, FatCat and others) for this error despite we did not need to do the conversion due to fact that our calcs were based on true speeds. And it was actually na85 and the other side who made this error; na85 at least did admit the error but as can be seen above, the other side behaves different way... not even mentioning the original 3.57g claim nor peeling the 1g off...

As concluding remarks, check Holtzauge's results from page 15 of the old thread and you can also use JtD's spreadsheet and do your own analysis. </div></BLOCKQUOTE>

Note here that we don't need to know anything about density nor altitude

So this is the force of thrust generated at a certain TAS from a certain engine power regardless of altitude?
And that is the thrust at that speed whether the plane is accelerating or not? Thrust at TAS only?

For a while there I thought it was how much force required to go a certain TAS but hey that would vary with
air density wouldn't it? Sometimes this is not clear to us dummies, not to know if it is power made or needed
kind of distinction is easy to get turned around and lost. I pay attention to all the words and sometimes all
I get is an idea of caution, what I think I am told and what I am told are not the same. Sometimes one simple
word is all the difference. I even spend some time wondering if y'all are talking about the same things.
Of course this is because engineers NEVER disagree!

Nexion_835
02-17-2010, 09:41 AM
If i understand this correctly, then:

at a certain speed and at a certain altitude

thrust = thrust required ( to overcome drag) for this speed

Power may be roughly constant up to FTH.

n also constant

F= n*W /v

so at higher altitudes, because of the lower density, the plane flies faster. When v rises Thrust (F) drops (because of the lower drag at higher alts with lower density). So no need for drag changes with altitude or density. It will work at all altitudes as long as you know the TAS at this altitude.
Am I right on this?

regards

nexion

M_Gunz
02-17-2010, 09:55 AM
If I run the formula with power as constant for 300kph I get one value for thrust and at 400kph another.
What makes either of these power required for either speed? That is why I think it is thrust produced only.

Kettenhunde
02-17-2010, 10:08 AM
Is there a typo?

Yep

Nexion_835
02-17-2010, 10:12 AM
Originally posted by M_Gunz:
If I run the formula with power as constant for 300kph I get one value for thrust and at 400kph another.
What makes either of these power required for either speed? That is why I think it is thrust produced only.

I think to get required power, the plane must fly at a constant speed so that thrust equals drag.
If a plane is flying at 300kph at known power, you can calculate the required thrust. Or if a plane is flying at 300kph and you know the drag, you can calculate the required power.

regards

nexion

Kettenhunde
02-17-2010, 11:16 AM
Note here that we don't need to know anything about density nor altitude; this is correct thrust at this given true speed and given prop efficiency.

For TAS but that has nothing to do with EAS or working in it.

Which is why:

If we want to hold our assumptions the same then we need to vary thrust with density.

2300 hp * .85 = 1955thp

1955thp*325 / 180KEAS = 3529.861111lbs of drag

sigma = .53281 @ 20,000ft

SQRT(.53281)* 3529.861111lbs of drag = 2576.58lbs of drag

2300 hp * .85 = 1955thp

1955thp*325 / 246.6KTAS = 2576.5lbs of drag

Think of it like this since you don't know fluid mechanics...

Armed with an ice cream scoop and I get ONE swipe with it in a bucket.

Will I get more in my scoop with a bucket of soft ice cream or more with a bucket of water?

Now think of the scoop as a propeller blade. The soft ice cream is the air at low altitude and the water high altitude.

http://forums.ubi.com/images/smilies/16x16_smiley-wink.gif

M_Gunz
02-17-2010, 11:27 AM
Originally posted by Nexion_835:
<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">Originally posted by M_Gunz:
If I run the formula with power as constant for 300kph I get one value for thrust and at 400kph another.
What makes either of these power required for either speed? That is why I think it is thrust produced only.

I think to get required power, the plane must fly at a constant speed so that thrust equals drag.
If a plane is flying at 300kph at known power, you can calculate the required thrust. Or if a plane is flying at 300kph and you know the drag, you can calculate the required power.

regards

nexion </div></BLOCKQUOTE>

I'm just trying make sure what the formula is giving me first, application comes after.

Holtzauge
02-17-2010, 11:29 AM
Exhibit 1:

Originally posted by Kettenhunde:

1955thp*325 / 180KEAS = 3529.861111lbs of drag

Exhibit 2:

Originally posted by Kettenhunde:

In EAS the corresponding dynamic pressure for TAS are the exact same.

@ 20,000ft:

Dynamic pressure at 180KEAS q = 180^2/295 = 109.93psf

Dynamic pressure at 246.6KTAS q = (.53281*246.6^2)/295 = 109.83psf

Exhibit 3:

Originally posted by Kettenhunde:

Drag in lbs = Coefficient of Drag * Dynamic pressure* reference area

If the Coefficients are the same, the dynamic pressure is the same, and our reference area is the same....what is going to vary our results?

I could not agree more…..

So we agree that the dynamic pressure q is the same 109.83 psf for both 180 KEAS and 246.6 KTAS (Exhibit 2)

We also agree that the drag D=3529.86 lbs at 180 KEAS (Exhibit 1)

Coefficient of Drag is the same (Exhibit 3)

So let’s now plug in the numbers for 246.6 KTAS:

The dynamic presuure q is the same……..

The Coefficient of Drag is the same……..

Hopefully you agree that the wing area is still the same……..

D=q*Cd*S

So D=3529.86 lbs also for 246.6 KTAS since neither q, Cd or S are changed……..

Exhibit 4:

Originally posted by Kettenhunde:

2300 hp * .85 = 1955thp

1955thp*325 / 246.6KTAS = 2576.5lbs of drag

So, using your own calculation, the thrust we have is 2576.5 lbs (Exhibit 4)……..

But we just concluded that the drag is 3529.86 lbs also for 246.6 KTAS……..

Doing the sums we find that you lack 953.4 lbs in thrust…..

You know why?

That’s because you are trying to prove that you can sustain a loadfactor of 2.57 with only Ps=2300 when what you really need Ps=3151 hp…………

Q.E.D

M_Gunz
02-17-2010, 12:03 PM
I can already tell you what he will show to be different with TAS:

This is TAS:

You know the Cdo for the P47 goes up with altitude.

It should have a direct relationship so that Cdo2 = Cdo1(sigma2/sigma1).

As density goes down, parasitic drag should go down.

This is EAS?

Drag in lbs = Coefficient of Drag * Dynamic pressure* reference area

If the Coefficients are the same, the dynamic pressure is the same, and our reference area is the same....what is going to vary our results?

Here he does both and the TAS has the sigma (=.53281... just what *is* this sigma? density? a ratio?):

@ 20,000ft:

Dynamic pressure at 180KEAS q = 180^2/295 = 109.93psf

Dynamic pressure at 246.6KTAS q = (.53281*246.6^2)/295 = 109.83psf

But here he doesn't:

Exhibit 4:

quote:
Originally posted by Kettenhunde:

2300 hp * .85 = 1955thp

1955thp*325 / 246.6KTAS = 2576.5lbs of drag

And there's already been two typos in this thread picked up already, I think that they qualify
as honest mistakes cause if we had to shoot everyone who made a slip the human population of
the planet would be few or none.

Could we please just get this straight without handing out life sentences even?
Yeah right... the UBI Zoo, home of the Spanish Inquisition and threads that go nowhere in flames.

Art-J
02-17-2010, 01:54 PM
Originally posted by M_Gunz:
just what *is* this sigma? density? a ratio?

Option b), It's a ratio of density at given altitude to the density at sea level. Invented to make flight mechanics calculations less confusing, but in Ubizoo, it seems to be working exactly the opposite way http://forums.ubi.com/groupee_common/emoticons/icon_biggrin.gif

Cheers - Art

M_Gunz
02-17-2010, 02:17 PM
Thank you, Art. Someone may come by to flame you later but since you know how it is.. http://forums.ubi.com/groupee_common/emoticons/icon_biggrin.gif no harm!

Kettenhunde
02-17-2010, 04:13 PM
I can already tell you what he will show to be different with TAS:

That is not supposed to be different if you use TAS, M_Gunz. That is just plain wrong.

M_Gunz
02-17-2010, 04:33 PM
What difference are you talking about? The one I am or something else?
Did you read the post I responded to? Did you read what I posted?
How about a clue? This is a bit like asking do you want coffee or tea and getting told "yes".

ADD: of course you may be meaning that Holtzauge is completely right in his last post above.

Frequent_Flyer
02-17-2010, 05:49 PM
M_Gunz you are not likely to learn much if anything from KettenHunde. However, following the various threads he/she posts in, you can certainly learn quite a bit about him/her.

M_Gunz
02-17-2010, 06:45 PM
You mean learn anything more? Since you don't know how much I already have learned or
who-all else I have shared posts with? Or how many book scans I've been given?

I think that right now the main 'technical issue' has been cornered but just how it
works out, or doesn't, is not in my ability to predict and I'm not gonna let my
personal politics turn me ignorant. I'm not leaning to either side, just pushing
forward. It still appears to me as if there is more than one 'right' because there
seems to me to be more than one line of reason going on. Perhaps this is because I
have seen different versions of this kind of dance to different bands for most of
my life where disagreements happen and have some hope that people can stop and
consider what the others are really saying and not just how it fits their own views.
That's how it is with intangibles like ideas.

OTOH this could all end pretty soon except for the flaming which then goes on forever.

Kettenhunde
02-17-2010, 08:10 PM
What difference are you talking about?

I am sorry if it was not clear. Parasitic drag varies with density. This is a fixed relationship and if the math is right, we should see some basic properties.

Take Jtd's turn performance sheet for the P47D10.

AT 20,000 feet the CDo = 0.0240
At 30,000 feet the CD0 = 0.0289

Parasitic drag increases with velocity. The P47D10 is pretty speed stable however.

We should see a change in drag from velocity:

Velocity @ FL30 / Velocity @ FL20)^2

(424/406.5)^2 = 1.09 or ~9% increase in drag.

I just picked a speed off the chart, say 360kph:

Parasitic drag at FL 20 @ 360kph = 4095N

Parasitic drag at FL 300 @ 360kph = 5548N

However, density should reduce our Cdo as parasitic drag has a direct relationship with density:

( sigma@FL30 / Sigma@FL20)

(.37413/.53281)= .7 or a 30% reduction in drag...

Parasitic drag should be decreasing, not increasing...

Might as well have water running up-hill modeled in your game too.

Changing subjects back to the EAS/TAS calculation debate...

If you have all the correct data which is usually measured data from a wind tunnel, then EAS and TAS calculations will pretty much align. Everything has a margin of error and it not a perfect world but they will be very very close. I can get the load factor to 2.41G just by varying wing efficiency in a realistic manner.

That is without correct propeller data too.

Many computer FM's use EAS methods. It completely removes the effects as density is an environmental effect and effects all the airplanes equally.

If you hold these assumptions fixed and use TAS, you get 2.2G's at FL20 and 246.6KTAS for the P47.

If you vary wing efficiency in a realistic manner, you will get ~2.41G's under the same conditions.

If you use EAS and just eliminate the density effects, you get 2.57G's.

All of them give good agreement and if your methodolgy is correct should return the exact same relative performance.

The Spitfire Merlin 66 is unable to match the sustained turn performance of the P47D22 at ~FL27 - FL30. Since the Spitfire has a higher ceiling though, regains the superiority.

Frequent_Flyer
02-17-2010, 08:16 PM
M_Gunz you appear to be on the 'defensive'. My observation was regarding the posture and tone of contempt in KettenHunde posts, delivered under the auspice of 'teaching'. Your better off going to:http://www.aerospaceweb.org/ web-site for this type of information

AndyJWest
02-17-2010, 08:32 PM
If we can't build a plane to settle this argument, can we at least hire an aerodynamicist?

Kettenhunde
02-17-2010, 08:40 PM
tone of contempt in KettenHunde posts,

I don't think you know anything about me or my feelings.

I can assure there is none of that in my post's towards any person on these boards.

That is purely your own projections.

I do not like and won't tolerate certain behaviors. The folks who cannot maintain a mature discussion will get ignored. It is effective.

I would like to see people who are interested in this stuff learn about it. That is all.

The other side to that is it not my place to correct every misconception on the internet.

Kettenhunde
02-17-2010, 09:06 PM
Could we please just get this straight without handing out life sentences even?

If you use TAS in BGS, you stick to TAS and convert the EAS at the end if needed.

I was taught be consistant in your units. The SI formulas Wurkeri and Holtzauge include actual density not density ratio or SMOE.

The rule in BGS is if you use density in a formula, it is the density at the altitude you are working at for the prediction.

If you use actual density, the value of q will change and EAS and TAS will not be the same force. Still the same relationship however corrected for density effects.

AndyJWest
02-17-2010, 09:11 PM
...not my place to correct every misconception on the internet.
I'm sorry, Kettenhunde, but I think that this is why you tend to elicit the response you do. Nobody is in any position to correct everything, because we all lack knowledge in some areas, an have a little extra in others. Given the anonymity of the net, the only way we can judge the value of technical information provided is by how well it is presented, and by how understandable it is. There have clearly been a series of misunderstandings in this topic, mostly I suspect due to inexact definition of the terms used, and invalid assumptions about what others mean. This isn't 'misconception', but miscommunication. As such it is a two-way process, best dealt with in an amicable way, rather than with the somewhat arrogant tone adopted by many of the participants. In commenting on your response, I'm not attempting to single you out, but rather to encourage all the participants to be a little less condescending, if for no reason other than to let the rest of us understand what the heck you are all arguing about...

For what it's worth, I think I've gained a little insight into the complexities of aerodynamics in this thread, and think it has the potential to teach me more, but only if 'egodynamics' remains in the background.

JtD
02-17-2010, 11:32 PM
Originally posted by AndyJWest:
If we can't build a plane to settle this argument, can we at least hire an aerodynamicist?

You mean we put another nick into the debate that posts the correct things and gets attacked on a personal level for doing so?

It's not going to help.

You can use my spreadsheet and see for yourself. (http://forums.ubi.com/eve/forums/a/tpc/f/23110283/m/2881049238)

M_Gunz
02-17-2010, 11:44 PM
Originally posted by Frequent_Flyer:
M_Gunz you appear to be on the 'defensive'.

No, I was just trying to explain my position in clear terms to someone who may be trying to instigate trouble.
And for you I will stick to full and clear explanations so as to leave the fewest opportunities for creative
guesswork on your part. Note that I haven't used the T-word and don't throughout this post. http://forums.ubi.com/images/smilies/halo.gif

My observation was regarding the posture and tone of contempt in KettenHunde posts, delivered under the auspice of 'teaching'.

I guess I have enough "life-experience" to look past that and still be able to communicate and learn.

FWIW he, Holtz, Wurkeri, JtD, Viper and others like them are all heads and shoulders above me in this area.
What I learned over 40 years ago is that you find someone like that who is willing to show you things then
you have a usually unique opportunity to learn/communicate and I put aside my petty concerns to do so!
That has gotten me into many interesting and life-enriching experiences even if it didn't make me a huge
load of cash. I'd rather be poor than dull. Perhaps my willingness to deal with and learn from my betters
in so many different areas has expanded my "life-experience" enough to get through where you don't bother.

And I have learned a LOT from ALL those guys! The web sites are a plus but they rarely answer so many
questions as these guys do. I've worked with engineers, even held engineer job titles though I have no
actual degree and one thing I know is engineers disagreeing on approach happens a lot! Gee, humans

Oh well, even if these guys get polite there's always the ones waiting to stir the animosity back up.
NOT that there's any sign of dropping the flames between old rivals/enemies anyway, nice to know there's
a backup crew to make sure that getting along ain't gonna happen here.

M_Gunz
02-17-2010, 11:49 PM
Originally posted by Kettenhunde:
If you use actual density, the value of q will change and EAS and TAS will not be the same force. Still the same relationship however corrected for density effects.

This is one of the things I intuited way back in the first-part thread but didn't have the words to explain.
( http://forums.ubi.com/images/smilies/compsmash.gif damn language skills!) I tried to get at it, honest!

M_Gunz
02-17-2010, 11:51 PM
Originally posted by AndyJWest:
<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">...not my place to correct every misconception on the internet.
I'm sorry, Kettenhunde, but I think that this is why you tend to elicit the response you do. Nobody is in any position to correct everything, because we all lack knowledge in some areas, an have a little extra in others. Given the anonymity of the net, the only way we can judge the value of technical information provided is by how well it is presented, and by how understandable it is. There have clearly been a series of misunderstandings in this topic, mostly I suspect due to inexact definition of the terms used, and invalid assumptions about what others mean. This isn't 'misconception', but miscommunication. As such it is a two-way process, best dealt with in an amicable way, rather than with the somewhat arrogant tone adopted by many of the participants. In commenting on your response, I'm not attempting to single you out, but rather to encourage all the participants to be a little less condescending, if for no reason other than to let the rest of us understand what the heck you are all arguing about...

For what it's worth, I think I've gained a little insight into the complexities of aerodynamics in this thread, and think it has the potential to teach me more, but only if 'egodynamics' remains in the background. </div></BLOCKQUOTE>

+100%! (ie, +1!)

You've had to deal with these kind of things before, haven't you? http://forums.ubi.com/images/smilies/partyhat.gif

Kettenhunde
02-18-2010, 04:25 AM
JtD says:

Parasitic drag should be decreasing, not increasing...

http://img52.imageshack.us/img52/4563/parasiticdrag.jpg (http://img52.imageshack.us/i/parasiticdrag.jpg/)

Might as well have water running up-hill modeled in your game too.

http://forums.ubi.com/eve/foru...421003338#1421003338 (http://forums.ubi.com/eve/forums/a/tpc/f/23110283/m/2891026238?r=1421003338#1421003338)

Kettenhunde
02-18-2010, 04:55 AM
I'm sorry, Kettenhunde, but I think that this is why you tend to elicit the response you do.

I don't think it is reasonable to expect me to continue to put up with the behavior in this thread.

It is not my issue or place to convince people who refuse to listen and argue that violating basic principles is "the correct method".

Xiolablu3
02-18-2010, 05:17 AM
None of the game creators or programmers are present on this forum as far as I know.

M_Gunz
02-18-2010, 05:19 AM
Originally posted by Kettenhunde:
<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">What difference are you talking about?

I am sorry if it was not clear. Parasitic drag varies with density. This is a fixed relationship and if the math is right, we should see some basic properties. </div></BLOCKQUOTE>

Sorry, missed this post last time around.

We should see a change in drag from velocity:

Velocity @ FL30 / Velocity @ FL20)^2

(424/406.5)^2 = 1.09 or ~9% increase in drag.

Okay, seen this before. Drag increase with square of speed so square of speed ratio gives drag due to speed ratio.

However, density should reduce our Cdo as parasitic drag has a direct relationship with density:

( sigma@FL30 / Sigma@FL20)

(.37413/.53281)= .7 or a 30% reduction in drag...

DOH! Yes, straightforward. So technique is break down to factors and multiply those 1.09 * 0.7 = 0.763 * drag at FL20?
When it was CD0 and sigma I wasn't sure what-all else might have been there but I should have at least remembered
that CD0 =is= parasitic drag. Once again; DOH!

Parasitic drag should be decreasing, not increasing...

Might as well have water running up-hill modeled in your game too.

And when I understand the why and how of the formula I can guess what you left unsaid pretty certainly.
Parasitic drag decreases with increase in alt -- checks against all I've seen and heard for more than 20 years.

Changing subjects back to the EAS/TAS calculation debate...

If you have all the correct data which is usually measured data from a wind tunnel, then EAS and TAS calculations will pretty much align. Everything has a margin of error and it not a perfect world but they will be very very close. I can get the load factor to 2.41G just by varying wing efficiency in a realistic manner.

Huh? 2.41G's from the 2.2G's below with assumptions from what post above or in JtD's thread? Not sure here but
I could take a chance, find something and it might even be the right one!

That is without correct propeller data too.

So the tolerance on "correct SWAG" widens.

Many computer FM's use EAS methods. It completely removes the effects as density is an environmental effect and affects http://forums.ubi.com/images/smilies/blush.gif all the airplanes equally.

If you hold these assumptions fixed and use TAS, you get 2.2G's at FL20 and 246.6KTAS for the P47.

If you vary wing efficiency in a realistic manner, you will get ~2.41G's under the same conditions.

If you use EAS and just eliminate the density effects, you get 2.57G's.

All of them give good agreement and if your methodolgy is correct should return the exact same relative performance.

If I use EAS with the values that give 2.2G's or 2.41G's? Difference is 0.37G's (not much until you lose, LOL!)
and 0.16G's. A little too much or too little stick can cover that and more.

The Spitfire Merlin 66 is unable to match the sustained turn performance of the P47D22 at ~FL27 - FL30. Since the Spitfire has a higher ceiling though, regains the superiority.

So now there's examples of all the parts though some don't get worked out the same as others, I will assume typos
for now or hey I just didn't understand and parts out of context and not throw a screaming fit. Of course when I
try and put it all together then if I get it wrong I won't be happy getting the razz either.

Wurkeri
02-18-2010, 06:12 AM
Originally posted by M_Gunz:
So this is the force of thrust generated at a certain TAS from a certain engine power regardless of altitude?
And that is the thrust at that speed whether the plane is accelerating or not? Thrust at TAS only?

For a while there I thought it was how much force required to go a certain TAS but hey that would vary with
air density wouldn't it? Sometimes this is not clear to us dummies, not to know if it is power made or needed
kind of distinction is easy to get turned around and lost. I pay attention to all the words and sometimes all
I get is an idea of caution, what I think I am told and what I am told are not the same. Sometimes one simple
word is all the difference. I even spend some time wondering if y'all are talking about the same things.
Of course this is because engineers NEVER disagree!

I see that you and Nexion are in right track to understand how this works. I'll try to explain whole analysis below so you can see where the errors were and how JtD's spreadsheet works.

I think we all now agree that in the power and thrust side, the relation between power and thrust is TAS and propeller efficiency dependant only so rewriting the power formula gives (Force is replaced with Thrust):

T = W*n/v

As Nexion noted above, at steady flight drag equals thrust so:

D = T

or

D = W*n/v

where

D = drag
T = thrust
W = power
v = true speed
n = propeller efficiency

Now let's take a look to the formula na85 used for to calculate required power in the original thread:

Power required (W) = q*v*S*[Cd0 + Cl^2/(pi*AR*e)]

where
q = dynamic pressure
S = wing area
Cd0 = zero lift drag coefficient
Cl = lift coefficient
e = efficiency factor
AR = aspect ratio

Dynamic pressure q can be rewritten as:

q = 0.5*p*v^2

where
p = density

And drag coefficient part can be rewritten:

Cd0 + Cl^2/(pi*AR*e) = Cd

So showing the density and simplifying the drag coefficient the formula becomes:

W = 0.5*p*v^2*v*S*Cd

As noted earlier na85 forgot the propeller efficiency so rewrite gives:

W = 0.5*p*v^2*v*S*Cd/n

And to see the relation to the thrust and drag balance, it can be rewritten:

0.5*p*V^2*S*Cd = W*n/v

Now the drag side is the basic drag formula and thrust side is the basic propeller thrust formula. Notable thing here is that the density affects only in the drag side when the true speed is used. And this where na85 and the other side made the error; they used EAS instead TAS at the thrust side without conversion. With proper conversion they would have got exactly same results as the others.

So how JtD's spreadsheet works then? I have not checked but I quess it's exactly like this:

The basic assumption is again that at top speed and at given altitude the thrust equals drag ie:

D = T

And we assume following things known:

W = engine power
S = wing area
Sp = wing span
AR = Aspect ratio, calculated from span and area
p = density at this given altitude
M = mass of the plane
n = propeller efficiency
e = efficiency factor
Vs = stall speed
g = load factor, 1 in level flight, limited to certain value.

We can now expand the thrust and drag balance further:

0.5*p*V^2*S*Cd = W*n/v

can be written:

0.5*p*v^2*S*[Cd0+Cl^2/(pi*AR*e)] = W*n/v

And as lift coefficient is:

Cl = L /(0.5*S*v^2*p)

Where:
L = lift force = M*g*9.81

we can finaly write whole balance formula:

0.5*p*v^2*S*[Cd0+([M*g*9.81]/[0.5*p*v^2*S])^2/(pi*AR*e)] = W*n/v

And as an example I picked a P-47B at 15k from WW2aircraftperformance site:

V = 386 mph = 172,52 m/s
W = 2000 hp = 1491400 W
M = 12560 lbs = 5697 kg
Sp = 41 ft = 12.5 m
S = 300 sqft = 27.87 m^2
AR = 5.6
p = 0.7708 kg/m^3
n = 0.85 estimated
e = 0.8 estimated
Vs = 105 mph = 46,93 m/s
g = 6g limit

The only missing value in the balance formula is now Cd0 and using the other values it is found to be 0.0208. Now we can check what kind of load factor can be sustained as an example at 400km/h at this same altitude and putting values to the formula gives a bit over 2.25g. Note that there might be typos http://forums.ubi.com/groupee_common/emoticons/icon_smile.gif

The balance formula might look complicated but after all it's just combination of four basic formulas:

Thrust, T = W*n/v
Drag, D = 0.5*p*V^2*S*Cd
Lift coefficient, Cl = L / (0.5*S*v^2*p)
Drag coefficient, Cd = Cd0 + Cl^2/(pi*AR*e)

Kettenhunde
02-18-2010, 07:01 AM
DOH! Yes, straightforward. So technique is break down to factors and multiply those 1.09 * 0.7 = 0.763 * drag at FL20?
When it was CD0 and sigma I wasn't sure what-all else might have been there but I should have at least remembered
that CD0 =is= parasitic drag. Once again; DOH!

As a method to ensure our relationships are correct and our performance prediction makes sense.

If these relationships do not hold true, then our prediction does not fit how the world works.

Wurkeri says:
it's just combination of four basic formulas:

Kettenhunde says:

You have convoluted the issue by not keeping your relationships separate. It is harder for people to understand what is going on.

http://forums.ubi.com/eve/foru...591021338#9591021338 (http://forums.ubi.com/eve/forums/a/tpc/f/23110283/m/2891026238?r=9591021338#9591021338)

M_Gunz
02-18-2010, 08:03 AM
Originally posted by Wurkeri:

0.5*p*v^2*S*[Cd0+([M*g*9.81]/[0.5*p*v^2*S])^2/(pi*AR*e)] = W*n/v

And as an example I picked a P-47B at 15k from WW2aircraftperformance site:

V = 386 mph = 172,52 m/s
W = 2000 hp = 1491400 W
M = 12560 lbs = 5697 kg
Sp = 41 ft = 12.5 m
S = 300 sqft = 27.87 m^2
AR = 5.6
p = 0.7708 kg/m^3
n = 0.85 estimated
e = 0.8 estimated
Vs = 105 mph = 46,93 m/s
g = 6g limit

The only missing value in the balance formula is now Cd0 and using the other values it is found to be 0.0208. Now we can check what kind of load factor can be sustained as an example at 400km/h at this same altitude and putting values to the formula gives a bit over 2.25g. Note that there might be typos http://forums.ubi.com/groupee_common/emoticons/icon_smile.gif

The balance formula might look complicated but after all it's just combination of four basic formulas:

Thrust, T = W*n/v
Drag, D = 0.5*p*V^2*S*Cd
Lift coefficient, Cl = L / (0.5*S*v^2*p)
Drag coefficient, Cd = Cd0 + Cl^2/(pi*AR*e)

So the 1G top speed is used to find Cd0 which allows other speed (or load?) values to be used to find the load
(or speed?) values that match?

JtD
02-18-2010, 08:32 AM
Yes. http://forums.ubi.com/groupee_common/emoticons/icon_smile.gif

It is the best estimate one can do with the limited data usually around.

Kettenhunde
02-18-2010, 10:30 AM
p = 0.7708 kg/m^3

Varying density with altitude is a step in the right direction over the sea level 1.225kg/m^3 found in the original sheet.

M_Gunz
02-18-2010, 10:37 AM
Hehe and ouch! RW density calculations. (http://wahiduddin.net/calc/density_altitude.htm)

Down towards the bottom there is a simplified method that doesn't need dew point, etc.

Otherwise perhaps a table is used along with interpolation?

Kettenhunde
02-18-2010, 11:13 AM
That is it. You can look up values on a Standard Atmospheric table.

BGS

http://www.pdas.com/e2.htm

SI

http://www.pdas.com/m1.htm

Kettenhunde
02-19-2010, 03:31 AM
Just to make sure nobody is confused on how our Power Required to Power Available relationship is broken down.

sigma = density ratio which is a mathematical representation of <span class="ev_code_RED">the environment</span> the airplane is flying in.

Lets keep is simple so it is easy to follow and nobody gets bamboozled.

To use True Airspeed....

True Airspeed = Velocity in Knots KEAS <span class="ev_code_YELLOW">/sqrt sigma</span>

<span class="ev_code_YELLOW">Power Available</span>

(Thrust Horsepower * Conversion Factor for Knots) / (Velocity in Knots KEAS <span class="ev_code_YELLOW">/sqrt sigma</span> )

<span class="ev_code_YELLOW">EQUALS</span>

<span class="ev_code_YELLOW">Power Required</span>

Drag = {[sigma * (Velocity in KEAS <span class="ev_code_YELLOW">/sqrt sigma</span> )^2]/ conversion factor for Knots}* Coefficient of Drag * Reference Area

Remove sigma and you have EAS.

As we rise in altitude and our aircraft goes faster because of the less dense environment, our parasitic drag decreases giving us the correct relationship with density.

If we want to hold our assumptions the same then we need to vary thrust with density.

2300 hp * .85 = 1955thp

1955thp*325 / 180KEAS = 3529.861111lbs of drag

sigma = .53281 @ 20,000ft

SQRT(.53281)* 3529.861111lbs of drag = 2576.58lbs of drag

2300 hp * .85 = 1955thp

1955thp*325 / 246.6KTAS = 2576.5lbs of drag

sigma is the same value for all airplanes at the same altitude. It is impossible for the same value equally applied to alter the relative performance line up of two aircraft in any way.

M_Gunz
02-19-2010, 07:52 AM
Is that the whole concern for going from EAS to TAS or back?

na85
02-19-2010, 02:35 PM
I forget where, but Kettenhunde said something along the lines of "If you use EAS to calculate X and desire specific performance it is necessary to convert back to TAS"

My understanding of this is that using EAS to calculate say, Pr gives a value that is "incorrect" i.e. not the same value to which a test of that aircraft would agree within some acceptable margin of error.

Is this the case?

Kettenhunde
02-19-2010, 03:38 PM
Is this the case?

You can only use the EAS numbers as relative performance. That relative line up between two aircraft will not change however when you convert to TAS.

If one aircraft is a superior turner in EAS, it will remain a superior turner in its corresponding TAS speed.

It is impossible for the same value equally applied to alter the relative performance line up of two aircraft in any way.

EAS is also very close to IAS and more useful for pilotage.

When converted to TAS, it gives an equally accurate prediction as using TAS with fixed assumptions.

To get the most accurate specific performance prediction within the 1% tolerance expected from operational type aircraft, you have the correct measured values for the aircraft for such things as wing efficiency.

As Cdo goes down due to density, wing efficiency gets better as a result.

It is normal for the aircraft aspect ratios of WWII fighters to see ~.7 at sea level all the way up to ~.97 or so at altitude.

If you had measured propeller data with the correct power and advance ratio, our propeller efficiency will also vary accordingly but mach limits tend to make it so that is not as dramatic.

So the most accurate prediction lies in between an EAS prediction and a TAS prediction that does not vary its assumption for "e" IAW density.

edited for clarity

Kettenhunde
02-19-2010, 03:39 PM
Is that the whole concern for going from EAS to TAS or back?

That is it, M_Gunz. EAS is the same performance as TAS for relative standing between two aircraft.

It is a great way to accurately and quickly predict if the line up of aircraft performance is correct.

So when the TAS "predictions" presented using TAS speeds and sea level density of 1.225Kg/M^3 showed a different line up, it is pretty easy to see that something is not correct.

Now Wurkeri has not admitted to an error in his use of density. I don't expect him too. I could be mistaken and it might be a rule with SI units that density always means sea level values. That is not how BGS system works for when the formula calls for density, it means density at the altitude the aircraft is flying.

So using sea level density for all values does not align with the principles I was taught in college or use today.

If you know the correct principles and can apply some proportional thinking the errors jump out.

M_Gunz
02-19-2010, 08:05 PM
Where did Wurkeri do that?

And are you sure that JtD's spreadsheet that you showed SL density as a constant uses TAS?

Kettenhunde
02-19-2010, 09:12 PM
I thought Wurkeri did the sheet for JtD.

I did not catch the EAS in m/s.

It is easy to see what he is doing now! Thanks.

He has EAS aligned with TAS power. Wow! No wonder his Cdo is going up with altitude.

http://img4.imageshack.us/img4/2604/soops.jpg (http://img4.imageshack.us/i/soops.jpg/)

In TAS our power is....

1716.225*1000*.85/41.09 = 35494.30695N

In EAS our power is....

1716.225*1000*.85/30 = 48,626.375N

His performance estimate robs the P47 of usable thrust as he is not consistent with his units.

I don't think JtD understands his TAS calculation is really:

1716.225*1000*.85/(30m/s divided by sqrt sigma)

Crumpp says:

If we want to hold our assumptions the same then we need to vary thrust with density.

2300 hp * .85 = 1955thp

1955thp*325 / 180KEAS = 3529.861111lbs of drag

sigma = .53281 @ 20,000ft

SQRT(.53281)* 3529.861111lbs of drag = 2576.58lbs of drag

2300 hp * .85 = 1955thp

1955thp*325 / 246.6KTAS = 2576.5lbs of drag

Crumpp says:

To use True Airspeed....

True Airspeed = Velocity in Knots KEAS /sqrt sigma

Power Available

(Thrust Horsepower * Conversion Factor for Knots) / (Velocity in Knots KEAS /sqrt sigma )

EQUALS

Power Required

Drag = {[sigma * (Velocity in KEAS /sqrt sigma )^2]/ conversion factor for Knots}* Coefficient of Drag * Reference Area

Remove sigma and you have EAS.

Power available = Power required

Or some folks have seen it written as:

Thrust = Drag

Grade school math teaches us that what you do to one side of the equal sign has to be done to the other. If you remove sigma from one side, you have to remove sigma from the other or they are no longer equal.

Xiolablu3
02-19-2010, 11:48 PM
I think most people have given up responding...

So maybe you could produce something which is designed to help the community rather than just argue?

Its easy to pick apart someones work when they lay it all on the line like that spreadsheet.

I dont know if its right or wrong, but I would think whoever made it, put a lot of work into it. If you are going to pick it apart maybe you could make a spreadsheet with 'your' calculations?

Can something good finally come out of this, or is it just an ego contest?

JtD
02-20-2010, 12:15 AM
Ok, I've been turning the other cheek, then the other, then the other and so on. I'm out of cheeks now. So I feel I have to make a few points now.

For instance, this

Drag = {[<span class="ev_code_RED">sigma</span> * (Velocity in KEAS /sqrt sigma )^2]/ conversion factor for Knots}* Coefficient of Drag * Reference Area

Highlight is mine. Why is it highlighted in red? Because it is wrong. It's not sigma, it's density. And in this case, actual density.

Remove sigma and you have EAS.

There is no way to do that without falsifying the equation. One side contains a sqrt(sigma)^2, the other a sqrt(sigma). So the best you can do is have sqrt(sigma) on one side, and none on the other.

It is impossible for the same value equally applied to alter the relative performance line up of two aircraft in any way.

This is just plain wrong. To illustrate,

5 - x*3 = 7 - x*5 (which is pretty much the structure we have)

x = 1:
2 = 2 (correct! (and a relation of 1:1)
x = 2:
-1 = -3 (wrong!) (and a relation of 1:3)

He has EAS aligned with <span class="ev_code_red">TAS power</span>.

There is no such thing as "TAS power".

Wow! No wonder his Cdo is going up with altitude.

Cdo is being calculated depending on the input. So, if your cdo goes up, you're having an input problem, and that's nothing you can blame on the tool.

In TAS our <span class="ev_code_red">power</span> is....

1716.225*1000*.85/41.09 = 35494.30695N

In EAS our <span class="ev_code_red">power</span> is....

1716.225*1000*.85/30 = 48,626.375N

You are calculating <span class="ev_code_red">forces</span>.

Also:

Power Available

(Thrust Horsepower * Conversion Factor for Knots) / (Velocity in Knots KEAS /sqrt sigma )

That's what YOU said. Where is sqrt(sigma) in your latter equation?

1716.225*1000*.85/(30m/s divided by sqrt sigma)

Yeah, exactly, try it!

JtD
02-20-2010, 12:18 AM
Originally posted by Xiolablu3:

Can something good finally come out of this, or is it just an ego contest?

The spreadsheet is something good that came out of this.

I don't think an ego contest would actually be much of a contest. If your ego is big enough to declare that 2.12 = 3.75 and can make you believe so, you're really a class on your own. http://forums.ubi.com/groupee_common/emoticons/icon_wink.gif

M_Gunz
02-20-2010, 01:59 AM
Originally posted by Xiolablu3:
I think most people have given up responding...

So maybe you could produce something which is designed to help the community rather than just argue?

Its easy to pick apart someones work when they lay it all on the line like that spreadsheet.

I dont know if its right or wrong, but I would think whoever made it, put a lot of work into it. If you are going to pick it apart maybe you could make a spreadsheet with 'your' calculations?

Can something good finally come out of this, or is it just an ego contest?

Only way to find out if it is right is by going through line by line, same as you would with math or science.
Reason why we like chalkboards is they are easy to read from a distance and the erase and rewrite quickly.
Nowadays, last 20 years or so at least, the whiteboards and markers are the preferred but work the same.

If this business doesn't take you back then you weren't there very long. No biggie at all, it's a niche like
any other. But that means you don't know what to expect. Two or more have different ideas what is right and in
hard math or science it is possible to find out which is and is not. It's not about ego except when someone is
wrong, actually knows it and still pushes the view while clouding all attempts at clarity.

You are right about one thing. If this can go through verification we end up with a spreadsheet. If it goes
through the right way we even end up with one we can trust.

Wurkeri
02-20-2010, 04:03 AM
Originally posted by M_Gunz:
You are right about one thing. If this can go through verification we end up with a spreadsheet. If it goes
through the right way we even end up with one we can trust.

It's very difficult to go through and check a large spreadsheet: there is endless ways to do something correctly and even more ways to do same wrong. However, it's easy to do your own calculation with the same parameters and check if the outcome is correct.

M_Gunz
02-20-2010, 05:07 AM
That is why I am glad that you provided that well annotated post a page or two ago.
You think spreadsheet is hard, try 30,000+ lines of code!

Kettenhunde
02-20-2010, 05:15 AM
JtD says:

Highlight is mine. Why is it highlighted in red? Because it is wrong. It's not sigma, it's density. And in this case, actual density.

It is density ratio JtD, we are working in BGS and Knots.

Equation 2.10 is highlighted for you.

Drag = {[sigma * (Velocity in KEAS /sqrt sigma )^2]/ conversion factor for Knots}* Coefficient of Drag * Reference Area

http://img168.imageshack.us/img168/3604/swrongagain.jpg (http://img168.imageshack.us/i/swrongagain.jpg/)

JtD's spreadsheet is wrong. It undervalues the amount of thrust the aircraft are producing.

I don't think JtD understands his TAS calculation is really:

1716.225*1000*.85/(30m/s divided by sqrt sigma)

This overly favors light wing loaded aircraft like the Spitfire in predicting sustained turning performance.

If this can go through verification we end up with a spreadsheet. If it goes
through the right way we even end up with one we can trust.

M_Gunz
02-20-2010, 05:25 AM
I would think that if JtD has density value plugged in there he would know.

ADD: I think that would give drag = {[density * TAS^2]/conversion for knots} * Cd * Ref Area.

Kettenhunde
02-20-2010, 05:27 AM
However, it's easy to do your own calculation with the same parameters and check if the outcome is correct.

In this case, your math is correct but the principles are wrong. Do you know understand that formulation has rules of use? In this case, density is factored into the equation for True Airspeed. In order to keep our equal sign correct, we need to reflect that in our power.

If we want to hold our assumptions the same then we need to vary thrust with density.

2300 hp * .85 = 1955thp

1955thp*325 / 180KEAS = 3529.861111lbs of drag

sigma = .53281 @ 20,000ft

SQRT(.53281)* 3529.861111lbs of drag = 2576.58lbs of drag

2300 hp * .85 = 1955thp

1955thp*325 / 246.6KTAS = 2576.5lbs of drag

POWER:

http://img705.imageshack.us/img705/2604/soops.jpg (http://img705.imageshack.us/i/soops.jpg/)

In TAS our power is....

1716.225*1000*.85/41.09 = 35494.30695N

In EAS our power is....

1716.225*1000*.85/30 = 48,626.375N

However his parasitic drag calculation is in EAS:

http://img21.imageshack.us/img21/22/sooops.jpg (http://img21.imageshack.us/i/sooops.jpg/)

His performance estimate robs the P47 of usable thrust as he is not consistent with his units.

Using TAS, our calculation is really:

1716.225*1000*.85/(30m/s divided by sqrt sigma)

Power available = Power required

Or some folks have seen it written as:

Thrust = Drag

Grade school math teaches us that what you do to one side of the equal sign has to be done to the other. If you remove sigma from one side, you have to remove sigma from the other or they are no longer equal.

I have told you from the beginning you must work either in TAS or EAS. Then your relative performance picture will be the same no matter which method you choose.

Crumpp says:

So the most accurate prediction lies in between an EAS prediction and a TAS prediction that does not vary its assumption for "e" IAW density.

M_Gunz
02-20-2010, 05:36 AM
Originally posted by Kettenhunde:
In TAS our power is....

1716.225*1000*.85/41.09 = 35494.30695N

In EAS our power is....

1716.225*1000*.85/30 = 48,626.375N

His performance estimate robs the P47 of usable thrust as he is not consistent with his units.

The TAS calculation is really:

1716.225*1000*.85/(30m/s divided by sqrt sigma)

I wish I knew just what these refer to.

Kettenhunde
02-20-2010, 05:53 AM
They refer to the highlighted blocks on the picture.

Choose any block in the "thrust available" Column I on the "turning performance" page.

The basic formula used is SI units:

Power in KW * 1000 * Propeller Efficiency / Column E which is our True Airspeed values.

Now click on Column J Parasitic Drag.

The basic formula in SI units is:

Parasitic Drag = Cdo * wing area * (Column A velocity in m/s EAS) ^2 * .5 * density at sea level.

We are calculating drag in EAS and power in True Airspeed.

You can see from the math below that is not correct at all. What you do to one side of the equal sign, has to be done to the other. If it is not then our equation is not longer valid.

Here is the formula the spreadsheet is using for power worked in both TAS and EAS:

In TAS our power is....

1716.225*1000*.85/41.09 = 35494.30695N

In EAS our power is....

1716.225*1000*.85/30 = 48,626.375N

His performance estimate robs the P47 of usable thrust as he is not consistent with his units.

The TAS calculation is really:

1716.225*1000*.85/(30m/s <span class="ev_code_YELLOW">divided by sqrt sigma</span> )

Sorry about adding that other stuff. It was added in the hopes of making things clearer for everyone.

Kettenhunde
02-20-2010, 06:00 AM
I would think that if JtD has density value plugged in there he would know.

It is my formula from BGS JtD is referring too M_Gunz.

JtD
02-20-2010, 06:42 AM
Originally posted by Kettenhunde:

It is density ratio JtD, we are working in BGS and Knots.

Ok, then the "conversion factor for knots" contains the density information, in this case the standard density. So it isn't a conversion factor for knots, but a factor that accounts for all the constants in the equation. Maybe you should label it so.

You have claimed that many times by now, but it still is correct. If you mean by "wrong" that it does not produce the fairy tales you share around here, then, yeah, it's wrong.

I don't think JtD understands his TAS calculation is really:

1716.225*1000*.85/(30m/s divided by sqrt sigma)

You've said that already, and I've replied. Only that it is now a "TAS calculation". Why don't you try something new?

I might add, that as much as you enjoy attacking a persons credibility instead of arguing the technical side, you really have no clue about what I understand and what not.

Maybe you can manage to illustrate your point by estimating the power required for the following performance:

A plane with 9200 lbs of weight, a wing area of 230 square feet and a wing span of 37 feet reaches a speed of 370 mph TAS at sea level, needing 1600 hp for that.
How much power do you predict will it need to reach 445 mph TAS at 30000 feet, while the weight has been reduced to 8900 lbs?
Personally, I'd expect the prop efficiency to drop while going up, maybe from 90% to 80% due to the higher mach numbers reached by the tips of the blades. Maybe you can post three results, one for a constant prop efficiency of 85%, one for a prop efficiency dropping down from 90% to 80% and one for prop efficiencies as you see fit.
If you think it necessary, please state the margin of error you think your results may have.

Thanks for taking the time.

Holtzauge
02-20-2010, 07:05 AM
JtD: Like Xio concluded, I think we had all given up responding to Kettenhundes nonsense and concluded it would be an exercise in futility to continue.

However, seeing he is jumping all over the turn performance spreadsheet that you were nice enough to share with the forum, I don’t want to stand idly by and see this happen.

First of all let me say that I agree with all the points you made in your reply.

Secondly, I think it looks like he’s introduced yet another Kettenhunde factor K into the equation:

I’ll run it by you and see if you agree:

Originally posted by Kettenhunde:
Drag = {[sigma * (Velocity in KEAS /sqrt sigma )^2]/ conversion factor for Knots}* Coefficient of Drag * Reference Area

As you pointed out, the first sigma is wrong and should be air density instead.

With that correction made, then the first part becomes the dynamic pressure q.

Entering 180 KEAS or 246.6 KTAS gives us a q=110 psf

With Cd=0.1071306 and S=300 sqft

D=110*0.1071306*300= circa 3535 lb

So far we agree with Kettenhunde what is needed for a loadfactor of 2.57 right?

A while back I used this to show that in order to balance this amount of drag then around Ps=5131 Hp was needed.

I assume you agree with this as well….

Now, concerning thrust Kettenhunde posted this:

Originally posted by Kettenhunde:

2300 hp * .85 = 1955thp

1955thp*325 / 246.6KTAS = 2576.5lbs

So the thrust we can get with Ps=2300 hp using a prop efficiency of 0.85 is 2576.5 lb which we also agree with I suppose.

This leaves about 950 lb lacking in thrust that Kettenhunde needs to go away in order to balance his equation.

This is when he throws this into the debate:

Originally posted by Kettenhunde:

sigma = .53281 @ 20,000ft

SQRT(.53281)* 3529.861111lbs of drag = 2576.58lbs of drag

Abra cadabra! Poof! The annoying 953 lb are now gone!

But wait a minute….. Where did the SQRT(.53281) come from?

If one reverse engineers the drag formula:

D=q*Cd*S

Now above we concluded that dynamic pressure q and wing area S are the same right?

So this means that the only thing that could have changed is the Cd.

OK, so now Kettenhunde has reduced Cd by a factor of SQRT(.53281):

Cd=Cd* SQRT(.53281)

I know Kettenhunde posted a hare-brained idea that Cdo is somehow connected to density based on a book excerpt that was talking about drag. I think most of us understood that the change in drag was due to the change in dynamic pressure that follows a change in density, not Cdo. But that, as they say, is another story………..

Anyway, it now seems that Kettenhunde in addition to applying a correction to Cdo is doing this to Cdi as well since:

Cd=Cdo + Cdi

In addition, the hare-brained Cdo equation he posted before had a linear relationship to SIGMA, not a square root relationship like here.

To me this just looks like another Kettenhunde factor pulled out of the drawer.

Looks to me like the new Kettenhunde factor is K= SQRT(.53281) and that Kettenhunde has “made an honest mistake” yet again (remember the one Loadfactor=Lodafactor-K where K=1?) and forgotten to tell us that the factor K of course needs to be applied to the induced drag Cdi as well.

Only problem I see is that massaging the Cd in this way is not something supported by aerodynamic literature AFAIK….

What do you think? Maybe I just misunderstood something…….

Holtzauge
02-20-2010, 08:04 AM
Originally posted by Kettenhunde:
<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">JtD says:

Highlight is mine. Why is it highlighted in red? Because it is wrong. It's not sigma, it's density. And in this case, actual density.

It is density ratio JtD, we are working in BGS and Knots.

Equation 2.10 is highlighted for you.

Drag = {[sigma * (Velocity in KEAS /sqrt sigma )^2]/ conversion factor for Knots}* Coefficient of Drag * Reference Area

http://img168.imageshack.us/img168/3604/swrongagain.jpg (http://img168.imageshack.us/i/swrongagain.jpg/)

JtD's spreadsheet is wrong. It undervalues the amount of thrust the aircraft are producing.

I don't think JtD understands his TAS calculation is really:

1716.225*1000*.85/(30m/s divided by sqrt sigma)

This overly favors light wing loaded aircraft like the Spitfire in predicting sustained turning performance.

If this can go through verification we end up with a spreadsheet. If it goes
through the right way we even end up with one we can trust.

Either way your community wins..... </div></BLOCKQUOTE>

It seems to me that the one who is misunderstanding his equations is you Kettenhunde, not JtD.

You don't seem to be able to calculate dynamic pressure.....

But maybe I got it wrong since I'm an SI guy and struggle with the slugs per cubic foot thingy. Maybe you can explain where I went wrong below?

Originally posted by Kettenhunde:
Drag = {[<span class="ev_code_YELLOW">sigma</span>* (Velocity in KEAS /sqrt sigma )^2]/ conversion factor for Knots}* Coefficient of Drag * Reference Area

Using the by now well known drag formula : D=q*Cd*s

Removing Cd and S leaves us to conclude that:

q={[sigma * (Velocity in KEAS /sqrt sigma )^2]/ conversion factor for Knots

So no, it’s NOT <span class="ev_code_YELLOW">sigma</span> . It’s DENSITY just like JtD says.

Let’s plug in some numbers:

ra=0.0012673 slugs/ft**3 at 20000 ft

SIGMA=0.5332 at 20000 ft

V=180 KEAS or 246.6 KTAS

Beginning with EAS:

V=180 KEAS= 304.6 ft/s EAS

q=0.5*0.0012673*(304.6/sqrt(0.5332))**2=110 psf

Using TAS:

V=246.6 KTAS= 417.3 ft/s TAS

q=0.5*0.0012673*417.3**2=110 psf

This tabs pretty well with what you posted before:

Originally posted by Kettenhunde:

In EAS the corresponding dynamic pressure for TAS are the exact same.

@ 20,000ft:

Dynamic pressure at 180KEAS q = 180^2/295 = 109.93psf

Dynamic pressure at 246.6KTAS q = (.53281*246.6^2)/295 = 109.83psf

Putting SIGMA in there instead of density yields:

q=46390.6 psf

Such a dynamic pressure would surely break up even the sturdy P-47 no?

So show me where I went wrong Kettenhunde. Surely you would not post an equation where the dynamic pressure was 46000 psf?……

Edit: The SIGMA in question highlighted in yellow to avoid any misunderstanding....

Kettenhunde
02-20-2010, 08:54 AM
You have claimed that many times by now, but it still is correct. If you mean by "wrong" that it does not produce the fairy tales you share around here, then, yeah, it's wrong.

It is wrong JtD.

You are using EAS to figure your drag and TAS to figure your power.

The effect of this is the airplanes do not have the correct values of thrust.

Your basic equation of thrust = drag is not correct as you do not account for density effects in your parasitic drag production.

Your thrust equation is using TAS:

1716.225*1000*.85/(30m/s divided by sqrt <0.532811526&gthttp://forums.ubi.com/groupee_common/emoticons/icon_wink.gif

But your drag equation only uses EAS.

Look at block H1 on the turning performance page, the Cdo calculation.

The Yellow portion is the power part of the spreadsheets Cdo calculations. You can look on the spreadsheet at Plane data page E11 that the velocity of 181.6969444 m/s is in TAS.

= <span class="ev_code_YELLOW">((1000*plane_data!C17*plane_data!B32/plane_data!E11)</span> -(plane_data!C8*9.81)^2/(plane_data!B35*0.5*PI()*K10^2*plane_data!C26^2* <span class="ev_code_RED">1.225</span> ))/(K10^2*plane_data!C23* <span class="ev_code_RED">1.225</span> *0.5)

All of the drag portion is done at sea level pressure because the speeds are EAS. You can see the density in red.

Go through the sheet, all of the drag data is done in EAS and all of the power in TAS.

To use True Airspeed....

True Airspeed = Velocity in Knots KEAS /sqrt sigma

Power Available

(Thrust Horsepower * Conversion Factor for Knots) / (Velocity in Knots KEAS /sqrt sigma )

EQUALS

Power Required

Drag = {[sigma * (Velocity in KEAS /sqrt sigma )^2]/ conversion factor for Knots}* Coefficient of Drag * Reference Area

Remove sigma and you have EAS.

Kettenhunde
02-20-2010, 09:08 AM
It is pretty cut and dry.

It is a fact that JtD's spreadsheet values power under TAS but parasitic drag under EAS.

It is not dynamic pressure where the mistake lies. The mistake lies in the fact our equations are not equal.

Crumpp says:
His performance estimate robs the P47 of usable thrust as he is not consistent with his units or application of density.

The TAS calculation is really:

1716.225*1000*.85/(30m/s divided by sqrt sigma )

Thrust will not equal drag as our TAS velocity includes a density correction while our value for drag does not.

This basic relationship of power available to power required sets the point in which we derive all of our performance. If you set that point in TAS, you need to stick to TAS.

If you set the thrust equation to be divided by A16 or the EAS velocity you will fix this. I don't how that aligns with the scales of the charts.

Column I "turning_performance!" formula should read:

=plane_data!\$C\$17*1000*plane_data!\$B\$32/A16

If you swap out the existing value of "E16" of TAS for "A16" EAS, then at least our relationships are correct.

The sheet is a mishmash of concepts.

JtD
02-20-2010, 09:23 AM
Stop making claims, show proof. Use your methods to solve the below task. We can take it from there.

Maybe you can manage to illustrate your point by estimating the power required for the following performance:

A plane with 9200 lbs of weight, a wing area of 230 square feet and a wing span of 37 feet reaches a speed of 370 mph TAS at sea level, needing 1600 hp for that.
How much power do you predict will it need to reach 445 mph TAS at 30000 feet, while the weight has been reduced to 8900 lbs?
Personally, I'd expect the prop efficiency to drop while going up, maybe from 90% to 80% due to the higher mach numbers reached by the tips of the blades. Maybe you can post three results, one for a constant prop efficiency of 85%, one for a prop efficiency dropping down from 90% to 80% and one for prop efficiencies as you see fit.
If you think it necessary, please state the margin of error you think your results may have.

Thanks for taking the time.

JtD
02-20-2010, 09:25 AM
@Holtzauge, thanks for the support, but to be honest I only skip read your post. I wouldn't want invest a load of time. But I'd agree on that it is goofy.

I'm just waiting for Kettenhundes numbers.

Kettenhunde
02-20-2010, 09:55 AM
Stop making claims, show proof.

I can do performance problems for you all day.

JtD
02-20-2010, 10:10 AM
Please do not dodge, just do the calculation. Thanks.

na85
02-20-2010, 10:16 AM
Originally posted by JtD:
Stop making claims, show proof. Use your methods to solve the below task. We can take it from there.

<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">Maybe you can manage to illustrate your point by estimating the power required for the following performance:

A plane with 9200 lbs of weight, a wing area of 230 square feet and a wing span of 37 feet reaches a speed of 370 mph TAS at sea level, needing 1600 hp for that.
How much power do you predict will it need to reach 445 mph TAS at 30000 feet, while the weight has been reduced to 8900 lbs?
Personally, I'd expect the prop efficiency to drop while going up, maybe from 90% to 80% due to the higher mach numbers reached by the tips of the blades. Maybe you can post three results, one for a constant prop efficiency of 85%, one for a prop efficiency dropping down from 90% to 80% and one for prop efficiencies as you see fit.
If you think it necessary, please state the margin of error you think your results may have.

Thanks for taking the time. </div></BLOCKQUOTE>

Ok here is where I think the crux of the issue lies:

JtD's question necessitates the use of TAS to find the correct answer since he does not ask for a relative comparison of 2 types.

On the other hand, Kettenhunde contends that if we are not interested in the actual number, and only want to know which of 2 or more aircraft performs better relative to the other(s), using EAS is acceptable.

2 different but related issues.

JtD
02-20-2010, 10:55 AM
No nate, this is about Kettenhunde claiming my Excel sheet is wrong, without backing it up. This is not how things are supposed to work. So he either gets started with the calculations to provide proof, or he can shove his claims and all his pretty formulas up his backside. Simple as that.

It is still down to the same issue P = F * v which got us mad at each other.

na85
02-20-2010, 11:10 AM
Originally posted by JtD:

It is still down to the same issue P = F * v which got us mad at each other.

What do you mean?

Edit: FYI the site where your spreadsheet is hosted is reported as an attack site.

K_Freddie
02-20-2010, 11:21 AM
Originally posted by JtD:
or he can shove his claims and all his pretty formulas up his backside. Simple as that.

I detect a bit of gayness here... and only after 26 pages.
Formulae, Formulae !!... for correct sequencing of the letters.
http://forums.ubi.com/images/smilies/16x16_smiley-very-happy.gif

JtD
02-20-2010, 11:59 AM
Originally posted by na85:

What do you mean?

I recall we had a heated exchange of words a couple of days ago, over P = F * v, same thing Kettenhunde is now getting lost about.

Edit: FYI the site where your spreadsheet is hosted is reported as an attack site.

Yes, I realized it when I tested the download. This stupid warning is another unnecessary software feature, which sadly cannot be switched off.

@Freddie: Ok, formulas is a nice word that doesn't exist. Didn't even know I was being creative. Thanks for pointing that out. http://forums.ubi.com/groupee_common/emoticons/icon_biggrin.gif

Holtzauge
02-20-2010, 12:36 PM
Originally posted by na85:
On the other hand, Kettenhunde contends that if we are not interested in the actual number, and only want to know which of 2 or more aircraft performs better relative to the other(s), using EAS is acceptable.

na85: Since you seem to have an interest in the relative performance aspect as well, how about giving us your opinion on the Spitfire Mk9 versus P47 turn capabilities Kettenhunde posted here?

Spitfire Mk9 v. P-47 D22 loadfactor at 20kft (http://forums.ubi.com/eve/forums/a/tpc/f/23110283/m/2221055328/p/10)

Kettenhunde has the Spitfire Mk9 Merlin 66 at +18 boost at a maximum loadfactor of about 3.49 and the P-47D22 at 56" Hg at 3.75 at 20000 ft.

Gauging "relative" performance this means that the P47 D22 can sustain a maximum loadfactor that is about 7.4% higher than the Spitfire Mk9 at 20000 ft according to Kettenhunde.

Do you think this sounds reasonable?

Originally posted by Kettenhunde:
The Spitfire is woefully inadequate when matched in a sustained turning contest with the P-47 under these conditions.

Do you agree?

M_Gunz
02-20-2010, 12:52 PM
Originally posted by Holtzauge:
<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">Originally posted by Kettenhunde:

2300 hp * .85 = 1955thp

1955thp*325 / 246.6KTAS = 2576.5lbs

So the thrust we can get with Ps=2300 hp using a prop efficiency of 0.85 is 2576.5 lb which we also agree with I suppose.

This leaves about 950 lb lacking in thrust that Kettenhunde needs to go away in order to balance his equation.

This is when he throws this into the debate:

Originally posted by Kettenhunde:

sigma = .53281 @ 20,000ft

SQRT(.53281)* 3529.861111lbs of drag = 2576.58lbs of drag

Abra cadabra! Poof! The annoying 953 lb are now gone!

But wait a minute….. Where did the SQRT(.53281) come from? </div></BLOCKQUOTE>

It's from turning EAS into TAS which if you quote the WHOLE block of what he posted is very clear.
He has showed that many times and explained it so WHY turn it into something else?

Here is from Crumpp.
The <span class="ev_code_RED">RED</span> parts are working it out using EAS.
The <span class="ev_code_BLUE">BLUE</span> parts are working it out using TAS.

Notice how when using EAS (sea level speed) you have to correct for density at higher altitude.
That is NOT a question, H! There IS no mysterious K-factor, there is only textbook (we've been shown the scans)
correction of EAS value to higher altitude where the air is thinner and drag is less. Nothing else.

If we want to hold our assumptions the same then we need to vary thrust with density.

<span class="ev_code_RED">2300 hp * .85 = 1955thp

1955thp*325 / 180KEAS = 3529.861111lbs of drag

sigma = .53281 @ 20,000ft

SQRT(.53281)* 3529.861111lbs of drag = 2576.58lbs of drag</span>

<span class="ev_code_BLUE">2300 hp * .85 = 1955thp

1955thp*325 / 246.6KTAS = 2576.5lbs of drag</span>

Your little romp after that reminds of someone else and stinks the same way.
How is Miss-Nancy-Representation getting along? All worried something clear might happen here?

M_Gunz
02-20-2010, 01:23 PM
Originally posted by Holtzauge:

Let’s plug in some numbers:

ra=0.0012673 slugs/ft**3 at 20000 ft

SIGMA=0.5332 at 20000 ft

V=180 KEAS or 246.6 KTAS

Beginning with EAS:

V=180 KEAS= 304.6 ft/s EAS

q=0.5*0.0012673*(304.6/sqrt(0.5332))**2=110 psf

Using TAS:

V=246.6 KTAS= 417.3 ft/s TAS

q=0.5*0.0012673*417.3**2=110 psf

This tabs pretty well with what you posted before:

It really should. (304.6/sqrt(0.5332) is ft/s EAS converted to TAS and equals, big surprise, 417.3 ft/s TAS!

And here below we see TAS being converted to handle the same way as EAS and big surprise again = SAME ANSWER!

Originally posted by Kettenhunde:

In EAS the corresponding dynamic pressure for TAS are the exact same.

@ 20,000ft:

Dynamic pressure at 180KEAS q = 180^2/295 = 109.93psf

Dynamic pressure at 246.6KTAS q = (.53281*246.6^2)/295 = 109.83psf

So he is using each EAS and TAS with proper conversion to work the formula as you would with TAS and EAS respectively.

Putting SIGMA in there instead of density yields:

q=46390.6 psf

Such a dynamic pressure would surely break up even the sturdy P-47 no?

Nice job of rodeo clowning. How about some banners, a page or two of long quotes with disagree 100% as the only
remark and a humorous picture or five? Distract from and bury the "enemy posts" under a pile of misdirection?

If *I* can follow the math better than that it's a good thing JtD hasn't been reading your posts.

M_Gunz
02-20-2010, 01:39 PM
Originally posted by JtD:
@Holtzauge, thanks for the support, but to be honest I only skip read your post. I wouldn't want invest a load of time. But I'd agree on that it is goofy.

Well I spent the time and I agree that what H did in both posts is real goofy.

I'm just waiting for Kettenhundes numbers.

I don't understand what you're trying to do there. Do the planes CHANGE WEIGHT in the spreadsheet?
Why not stick to one example at a time and work it out or at least not add elements?

I'm smelling smoke and seeing distorted reflections in another yet another technical discussion and whoopie,
there's fingers pointing in different directions. Everybody with an ego problem, stand up and BS, it's getting
near truth time! If this kind of stuff had gone on where I went to school we'd have learned very little.

Let’s plug in some numbers:

ra=0.0012673 slugs/ft**3 at 20000 ft

SIGMA=0.5332 at 20000 ft

V=180 KEAS or 246.6 KTAS

Beginning with EAS:

V=180 KEAS= 304.6 ft/s EAS

q=0.5*0.0012673*(304.6/sqrt(0.5332))**2=110 psf

Using TAS:

V=246.6 KTAS= 417.3 ft/s TAS

q=0.5*0.0012673*417.3**2=110 psf

This tabs pretty well with what you posted before:

One step forward....

Kettenhunde
02-20-2010, 02:13 PM
I don't understand what you're trying to do there. Do the planes CHANGE WEIGHT in the spreadsheet? Why not stick to one example at a time and work it out or at least not add elements?

Valid series of questions...we need to answer this before moving on.

I completed JtD's analysis but will wait until the relationship of thrust available and density is worked out to everyone's satisfaction so we don't muddy the waters.

Xiolablu3
02-20-2010, 02:20 PM
Originally posted by JtD:
<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">Originally posted by Xiolablu3:

Can something good finally come out of this, or is it just an ego contest?

The spreadsheet is something good that came out of this.

I don't think an ego contest would actually be much of a contest. If your ego is big enough to declare that 2.12 = 3.75 and can make you believe so, you're really a class on your own. http://forums.ubi.com/groupee_common/emoticons/icon_wink.gif </div></BLOCKQUOTE>

I know mate, but at the time Kettenhunde was slating your spreadsheet, but giving nothing back. As the rest of you guys had stopped responding, all we had was Kettenhunde saying 'this is wrong' and 'this is wrong' and 'I know everything', 'you are all wrong', but with no wish to help the community with something such as that spreadsheet.

It was basically 'OK so you've said everyone else is wrong, you are right, what now? Was it just for your ego or are you going to produce something?'

I dont wish to see anyone who creates something for the community in good faith to be put off because 'xxxxxx will just say its wrong and xxxx all over it'

Some of my best gaming moments were playing your maps online in IL2, JTD. Please don't stop creating.

Kettenhunde
02-20-2010, 02:33 PM
'I know everything',

but with no wish to help the community with something such as that spreadsheet.

Nobody has asked me to help with that spreadsheet. If it was correct, I would have not said a thing.

If you go read the other thread, I do readily admit my mistakes despite the fact I never seem to get credit for it with certain BBS members.

How about let's not discuss me but stay on subject by sticking to the facts and the math?

Holtzauge
02-20-2010, 02:58 PM
MGunz: Since you seem to understand Kettenhundes math I’ll cut out the step by step business which seems to bother you and get right to the point:

I understand where he got the 3529.86 lbs drag from for 180 KEAS:

D=0.5*ra*v**2*Cd*S

D=0.5*0.0023796*304.6**2*0.1071306*300=3548 lbs (close enough)

However, for 246.5 KTAS at 20000 ft Kettenhunde gets 2576.5 lbs and this is what I get:

D=0.5*ra*v**2*Cd*S

D=0.5*0.0012673*417.3**2*0.1071306*300=3546 lbs

So I get around 3546 lbs and Kettenhunde get’s 2576.5 lbs.

Clearly, one of us is wrong.

Where, in your opinion, did I go wrong in my calculation?

K_Freddie
02-20-2010, 03:03 PM
Originally posted by Kettenhunde:
I do readily admit my mistakes despite the fact I never seem to get credit for it with certain BBS members.
Sure, but it took a few pages worth of disbelief, for this to happen.
You guys must not get upset if you put yourselves 'up on the pedestal', only to find others knocking it down - this is the way of science and discovery. http://forums.ubi.com/groupee_common/emoticons/icon_wink.gif

After all, for how many 100s (1000s) of years did the 'community' believe the planets orbited around the earth, while the few said otherwise at the risk of persecution (sounds familiar). http://forums.ubi.com/images/smilies/16x16_smiley-very-happy.gif

Holtzauge
02-20-2010, 03:58 PM
Originally posted by M_Gunz:

Well I spent the time and I agree that what H did in both posts is real goofy.

<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content"> Originally posted by Holtzauge
Let’s plug in some numbers:

ra=0.0012673 slugs/ft**3 at 20000 ft

SIGMA=0.5332 at 20000 ft

V=180 KEAS or 246.6 KTAS

Beginning with EAS:

V=180 KEAS= 304.6 ft/s EAS

q=0.5*0.0012673*(304.6/sqrt(0.5332))**2=110 psf

Using TAS:

V=246.6 KTAS= 417.3 ft/s TAS

q=0.5*0.0012673*417.3**2=110 psf

This tabs pretty well with what you posted before:

One step forward.... </div></BLOCKQUOTE>

So the one step forward is that we agree that q=110 psf. Good.

Since you think that what I posted was goofy here is your chance to correct me:

D=q*Cd*S

Now above we concluded that dynamic pressure q=110 psf

S=300 sqft

Cd=0.1071306

This gives me the following drag:

D=110*0.1071306*300=3535 lbs which is darn close to what I got in my TAS calculation above as well: 3546 lbs.

But according to Kettenhunde the drag is:

SQRT(.53281)* 3529.861111lbs of drag = 2576.58lbs

So if we agree on q and S then the only thing left in the equation that can change in D=q*Cd*S is Cd, no?

So the only way I can see the equation working out is if Kettenhunde has changed the Cd:

Cd=0.1071306*SQRT(.53281)

What is the goofy part of my reasoning?

Kettenhunde
02-20-2010, 04:01 PM
You are correct but you are only looking at Thrust required.

Thrust available = Thrust required

That means what is done to one side of the equal sign must be done to the other or our relationship is no longer equal.

You cannot figure power available in TAS and drag in EAS, it will NOT work out.

If you want to change altitudes when working with TAS, you MUST go from TAS to EAS and then back to TAS at the new density before proceeding to figure anything else at the new altitude. You must stay in TAS or your power available picture will be skewed.

If you work in EAS, you must stay in EAS until the analysis is complete, then convert the final answer to TAS.

You must be consistent in your methods.

Holtzauge
02-20-2010, 04:30 PM
Originally posted by Kettenhunde:
You are correct but you are only looking at Thrust required.

Thrust available = Thrust required

That means what is done to one side of the equal sign must be done to the other or our relationship is no longer equal.

You cannot figure power available in TAS and drag in EAS, it will NOT work out.

If you want to change altitudes when working with TAS, you MUST go from TAS to EAS and then back to TAS at the new density before proceeding to figure anything else at the new altitude. You must stay in TAS or your power available picture will be skewed.

If you work in EAS, you must stay in EAS until the analysis is complete, then convert the final answer to TAS.

You must be consistent in your methods.

You yourself said it does not matter if one uses EAS or TAS as long as one is consistent.

Let’s be consistent:

Using TAS:

Cd=0.1071306

D=q*Cd*S

D=0.5*ra*v**2*Cd*S

D=0.5*0.0012673*417.3**2*0.1071306*300=3546 lbs

Using EAS:

You somehow get 2576.5 lbs drag by using EAS and throwing in SQRT(.53281)

D=q*Cd*S

But we have previously concluded that q=110 psf and S=300 sqft.

2576.5 =110*Cd*300

So Cd=0.078076 to get the result you want.

Or perhaps you want to back away from q=110 psf now?

Kettenhunde
02-20-2010, 05:31 PM
You yourself said it does not matter if one uses EAS or TAS as long as one is consistent.

Exactly...consistent...that means you work in either one and convert at the end, Holtzauge.

You say I am going to use one or the other.

If you work in TAS, then you need to vary your density appropriately and use EAS to change altitudes.

Kettenhunde
02-20-2010, 05:34 PM
You somehow get 2576.5 lbs drag by using EAS and throwing in SQRT(.53281)

NO...I got Power available and I illustrated the relationship with density over TAS.

Kurfurst__
02-20-2010, 06:19 PM
Can someone summerize me the goal of this discussion in one sentence please...? http://forums.ubi.com/groupee_common/emoticons/icon_biggrin.gif

Kettenhunde
02-20-2010, 06:29 PM
A bunch of ya-hoo's who think they know better and don't seem to realize .97/.72 = 34% variation in results.

Crumpp says:
However gamers seem to expect an impossible standard. I expect this is another side effect of internet education.

Even today with all of our technology, we cannot put aircraft performance margin of error free from flight testing or calculations. We have teams of engineers that study one particular problem or facet of a designs performance. The rudimentary performance formulas here are very broad in scope.

In fact, nothing is real life is margin of error free It does not matter if you are making socks for Walmart or airplanes for Boeing.

For example, the 2.2G's found in one prediciton vs 2.57G in another is absolutely nothing. It is in fact very good agreement for turn performance calculations. As long as acceptable procedures, formulation, and consistent units is applied both are valid.

The lift generated between those two number is ~16% difference. Wing efficiency changes over flight conditions. It is not a static thing. The difference in wing efficiency between those two conditions is the difference between an efficiency of .85 and .72 which is well within normal.

I use EAS because it eliminates such environmental effects to give what I feel is a closer relative analysis. Those effects can completely change any prediction. The design team has made a mistake when that happens. It something human being do.

The only thing unrealistic is the expectation of anything being free from a margin of error.

There is a reason why you don't see much turn performance calculations in the real world of aircraft design. More than any other parameter of aircraft performance, it has the largest variation of acceptable results.

Kettenhunde
02-20-2010, 07:11 PM
Here is what is confusing people.

Lets break down what is happening on the Thrust required <drag> side of the equation.

in TAS:

Dynamic pressure = sigma *(KTAS)^2 / 295

Dynamic pressure = sigma *(KEAS/SQRT sigma)^2 / 295

Dynamic pressure = sigma *(KEAS/SQRT sigma)*(KEAS/SQRT sigma) / 295

Dynamic pressure = [sigma*(KEAS*KEAS/SQRT sigma*SQRT sigma)]/ 295

Dynamic pressure = [sigma*(KEAS*KEAS/SQRT sigma^2)]/ 295

Of course, the square root of sigma squared is sigma.

Dynamic pressure = [sigma*(KEAS*KEAS/sigma]/ 295

Sigma cancels and we are left with:

Dynamic pressure = (KEAS*KEAS]/ 295 or the expression for EAS dynamic pressure.

Just because it cancels does not mean the correct thing to do is add it in to the other side of the equal sign on the Power Available side when we are using EAS.

You still have to follow the basic rules of math.

If you work in EAS, sigma is not factored into the equation and the results are converted at the end.

If you work in TAS, you must add in sigma on both sides of the equation. Your parasitic drag will vary IAW density effects and subsequently vary the value of e.

The end result, done correctly will be very close in either method.

JtD
02-20-2010, 10:57 PM
I don't understand what you're trying to do there. Do the planes CHANGE WEIGHT in the spreadsheet?
Why not stick to one example at a time and work it out or at least not add elements?

I have a real life reference for the requested parameters which I can check the results against.

If you want to use the spreadsheet for the analysis, then you'll have to run two sets of data anyway.

JtD
02-20-2010, 11:00 PM
Originally posted by Kettenhunde:

Valid series of questions...we need to answer this before moving on.

I completed JtD's analysis but will wait until the relationship of thrust available and density is worked out to everyone's satisfaction so we don't muddy the waters.

No, you can post the results right away. No need to find excuses. Just two numbers, maybe a third, can't be so hard.

In the meantime you've repeated your claims several times again, yet have not managed to provide the little bit of proof you've been asked for.

JtD
02-20-2010, 11:02 PM
Originally posted by Xiolablu3:

Some of my best gaming moments were playing your maps online in IL2, JTD. Please don't stop creating.

Oh, thanks. Now I feel obliged to get back to map making...and I'm already short of il-2 time. http://forums.ubi.com/images/smilies/16x16_smiley-very-happy.gif

AndyJWest
02-20-2010, 11:07 PM
Originally posted by Kurfurst__:
Can someone summerize me the goal of this discussion in one sentence please...? http://forums.ubi.com/groupee_common/emoticons/icon_biggrin.gif

Probably not, but I doubt it has much to do with "Soviet FW-190A combat evaluation and the He-100 issue". I also suspect it has little to do with the mathematical analysis of aircraft turn performance at altitude. http://forums.ubi.com/images/smilies/16x16_smiley-wink.gif

M_Gunz
02-21-2010, 12:18 AM
Originally posted by Holtzauge:
MGunz: Since you seem to understand Kettenhundes math I’ll cut out the step by step business which seems to bother you and get right to the point:

I understand where he got the 3529.86 lbs drag from for 180 KEAS:

D=0.5*ra*v**2*Cd*S

D=0.5*0.0023796*304.6**2*0.1071306*300=3548 lbs (close enough)

Assuming that ra you use is density that Wurkeri labels p? And this is the TAS formula that EAS must
be corrected for altitude to give TAS in order to get a correct answer. [EAS/sqrt(sigma)]=TAS to plug in.

AT sea level, no? Notice ra=0.0023796 and not "ra=0.0012673 slugs/ft**3 at 20000 ft"

When you do it that way, "it tabs pretty well" as I quote from you below.

Originally posted by Holtzauge:

Let’s plug in some numbers:

ra=0.0012673 slugs/ft**3 at 20000 ft

SIGMA=0.5332 at 20000 ft

V=180 KEAS or 246.6 KTAS

Beginning with EAS:

V=180 KEAS= 304.6 ft/s EAS

q=0.5*0.0012673*(304.6/sqrt(0.5332))**2=110 psf

Using TAS:

V=246.6 KTAS= 417.3 ft/s TAS

q=0.5*0.0012673*417.3**2=110 psf

This tabs pretty well with what you posted before:

Same q AT 20,000 ft.

However, for 246.5 KTAS at 20000 ft Kettenhunde gets 2576.5 lbs and this is what I get:

D=0.5*ra*v**2*Cd*S

D=0.5*0.0012673*417.3**2*0.1071306*300=3546 lbs

With the same Cd at 20,000 ft as at sea level where thrust calculates to appx 3546 lbs.

Wurkeri: 0.5*p*v^2*S*[Cd0+([M*g*9.81]/[0.5*p*v^2*S])^2/(pi*AR*e)] = W*n/v

And in this example at 20,000ft W*n/v = 2576.5, how many g's can you turn at 417.3 f/s at that altitude?

What AR, e and mass are you using? Wurkeri's example is not the same starting with 2000hp instead of 2300hp.
There are at least 3 different "examples" floating around and if you don't want to answer this then there
will be at least 4. Is this a discussion or a shell game? Quick! Shuffle-shuffle-shuffle. Where's the pea?
Screw that for a fool's chase, I'm not playing your silly game.

I am pretty sure that Crumpp has stated that Cd decreases with altitude but you experts know better than I.

So I get around 3546 lbs and Kettenhunde get’s 2576.5 lbs.

Clearly, one of us is wrong.

Where, in your opinion, did I go wrong in my calculation?

IMO? I showed where you diddled with the procedures, even I could see and show that. Instead of answering that
you pose a different question. Deja-vu, I've seen this many times before.

Why not fix what you were doing and follow through with that? You took foul shots and got caught out, admit it.
Trying to snowplow me doesn't change what you did, it's just an attempt to change the subject. Standard tactic.

M_Gunz
02-21-2010, 12:32 AM
Originally posted by Holtzauge:
What is the goofy part of my reasoning?

I showed that plainly. You go attributing what is done right as some mysterious K-factor and I showed that
where you did that was wrong.

Now you choose to ignore that and press on to justify what you did. Your numbers differ so there must be
something wrong and it must be the K-factor.

I told you that WHERE you decided to call it is bogus. I didn't offer anything else. HOW you prove your
point is up to you, all I showed was that you did not find mistakes by K where you said you did.

IE your disproof is not. Can't you see that? Can't you admit it? Deja-F-ing-VU!

Cd=0.1071306*SQRT(.53281) --- must have something to do with ALTITUDE.
Seems I remember him posting the same business in more than one way all along now.

If you can show that Cd does not change with altitude then I guess you'll have him there!
Why bother with the shell game? Just go for it!

M_Gunz
02-21-2010, 12:41 AM
Originally posted by Kurfurst__:
Can someone summerize me the goal of this discussion in one sentence please...? http://forums.ubi.com/groupee_common/emoticons/icon_biggrin.gif

You can ask the OP but somehow I doubt you want to do that as that subject DIED as ways back.

Right now it's trying to verify a turn-estimate spreadsheet in order to answer a previous issue, etc, etc, etc.
If solved it might even circle back onto the original issue, close the circle and shoot that down yet again!

We don't so much debate as dogfight subjects around here. Lose sight, lose the fight might just explain all
the diversions, chaff and *actual honest maneuvering to get at the point* you see constantly in any technical

And yeah, as if you didn't already know!

Salute!

JtD
02-21-2010, 12:53 AM
A plane going the same EAS in level flight at different altitudes produces the same drag and thus has the same drag coefficient, independent of altitude.

If the EAS varies, or the flight condition changes, it will produce different drag and thus have a different drag coefficient, independent of altitude.

M_Gunz
02-21-2010, 12:53 AM
Originally posted by JtD:
<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">I don't understand what you're trying to do there. Do the planes CHANGE WEIGHT in the spreadsheet?
Why not stick to one example at a time and work it out or at least not add elements?

I have a real life reference for the requested parameters which I can check the results against.

If you want to use the spreadsheet for the analysis, then you'll have to run two sets of data anyway. </div></BLOCKQUOTE>

Oh. Yes, that has a good sense about it but how close will be close enough using these equations with assumed
efficiencies? I am guessing that the spreadsheet gets sufficiently close? But I only guess you have checked,
no meaning attached.

M_Gunz
02-21-2010, 12:59 AM
Originally posted by JtD:
A plane going the same EAS in level flight at different altitudes produces the same drag and thus has the same drag coefficient, independent of altitude.

Can I take that Cd and plug it in to TAS formula?

JtD
02-21-2010, 01:40 AM
I'm not willing to present my results before Kettenhunde presents his or until he has found enough excuses to not post them. If I posted before he did, he'd just go on and claim how they are wrong, throwing fancy formula around. Sorry.

You can plug the cd into TAS formula, if you use the current density, not the standard one. Which, as we know by now, is just the same as using standard density and EAS or the same as using standard density, TAS and sigma for density correction.

The symbol of density is the lower case greek letter rho, which looks like this: ? or similar to p, which is what Wurkeri uses, or in Holtzauges use it is ra for "rho actual". (Or current density as I say sometimes.)

FatCat_99
02-21-2010, 02:22 AM
Originally posted by Kurfurst__:
Can someone summerize me the goal of this discussion in one sentence please...? http://forums.ubi.com/groupee_common/emoticons/icon_biggrin.gif
Not a sentence but might be appropriate.

To those interested in aircraft turn performance I can ,at this point, only say that they should use Google, library or call local university. Turning performance is simple, it takes just few pages of University textbook, math involved is elementary school level and it requires minimal knowledge about aerodynamics.

I would say that person of average intelligence can learn and comprehend this subject in less than 1 hour. But this thread is living proof that we are not dealing with average persons at UBI http://forums.ubi.com/images/smilies/88.gif Couple of loonies, few trolls and even few people who knows how to use books.

Rock on UBI http://forums.ubi.com/images/smilies/metal.gif

FC

Holtzauge
02-21-2010, 02:52 AM
Originally posted by M_Gunz:
<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">Originally posted by Holtzauge:
MGunz: Since you seem to understand Kettenhundes math I’ll cut out the step by step business which seems to bother you and get right to the point:

I understand where he got the 3529.86 lbs drag from for 180 KEAS:

D=0.5*ra*v**2*Cd*S

D=0.5*0.0023796*304.6**2*0.1071306*300=3548 lbs (close enough)

Assuming that ra you use is density that Wurkeri labels p? And this is the TAS formula that EAS must
be corrected for altitude to give TAS in order to get a correct answer. [EAS/sqrt(sigma)]=TAS to plug in.

AT sea level, no? Notice ra=0.0023796 and not "ra=0.0012673 slugs/ft**3 at 20000 ft"

When you do it that way, "it tabs pretty well" as I quote from you below.

Originally posted by Holtzauge:

Let’s plug in some numbers:

ra=0.0012673 slugs/ft**3 at 20000 ft

SIGMA=0.5332 at 20000 ft

V=180 KEAS or 246.6 KTAS

Beginning with EAS:

V=180 KEAS= 304.6 ft/s EAS

q=0.5*0.0012673*(304.6/sqrt(0.5332))**2=110 psf

Using TAS:

V=246.6 KTAS= 417.3 ft/s TAS

q=0.5*0.0012673*417.3**2=110 psf

This tabs pretty well with what you posted before:

Same q AT 20,000 ft.

However, for 246.5 KTAS at 20000 ft Kettenhunde gets 2576.5 lbs and this is what I get:

D=0.5*ra*v**2*Cd*S

D=0.5*0.0012673*417.3**2*0.1071306*300=3546 lbs

With the same Cd at 20,000 ft as at sea level where thrust calculates to appx 3546 lbs.

Wurkeri: 0.5*p*v^2*S*[Cd0+([M*g*9.81]/[0.5*p*v^2*S])^2/(pi*AR*e)] = W*n/v

And in this example at 20,000ft W*n/v = 2576.5, how many g's can you turn at 417.3 f/s at that altitude?

What AR, e and mass are you using? Wurkeri's example is not the same starting with 2000hp instead of 2300hp.
There are at least 3 different "examples" floating around and if you don't want to answer this then there
will be at least 4. Is this a discussion or a shell game? Quick! Shuffle-shuffle-shuffle. Where's the pea?
Screw that for a fool's chase, I'm not playing your silly game.

I am pretty sure that Crumpp has stated that Cd decreases with altitude but you experts know better than I.

So I get around 3546 lbs and Kettenhunde get’s 2576.5 lbs.

Clearly, one of us is wrong.

Where, in your opinion, did I go wrong in my calculation?

IMO? I showed where you diddled with the procedures, even I could see and show that. Instead of answering that
you pose a different question. Deja-vu, I've seen this many times before.

Why not fix what you were doing and follow through with that? You took foul shots and got caught out, admit it.
Trying to snowplow me doesn't change what you did, it's just an attempt to change the subject. Standard tactic. </div></BLOCKQUOTE>

Sorry, I did not get that. There are so many attacks on my person above so I'm not sure where you stand on the drag issue:

I get 3546 lbs drag and Kettenhunde gets 2576.5 lbs.

I assume you think 2576.5 lbs is right since you slash my calculations and said that you could follow Kettenhundes?

Yes or no?

Holtzauge
02-21-2010, 03:15 AM
Originally posted by Kettenhunde:
<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">You somehow get 2576.5 lbs drag by using EAS and throwing in SQRT(.53281)

NO...I got Power available and I illustrated the relationship with density over TAS. </div></BLOCKQUOTE>

You are avoiding the question: The question I put was what DRAG will we have at 180 KEAS at 20000 ft:

So the thrust we have is 2576.5 lbs right?

Hopefully you agree the T=D?

If you agree then this means that the drag is also is 2576.5 lbs.

D=q*Cd*s

2576.5=110*Cd*S

Cd=2576.5/(110*300)

=0.078076

So using q=110 psf and wing area 300 sqft the Cd needs to be 0.078076 no?

Kettenhunde
02-21-2010, 05:16 AM
Right now it's trying to verify a turn-estimate spreadsheet in order to answer a previous issue, etc, etc, etc.
If solved it might even circle back onto the original issue, close the circle and shoot that down yet again!

Cut and past the formula from column I turning_performance to change it from this:

=plane_data!\$C\$17*1000*plane_data!\$B\$32/E16

to this:

=plane_data!\$C\$17*1000*plane_data!\$B\$32/A16

That should fix it.

All of your units are now consistent.

When you vary altitude, input the aircraft's power and Vmax at that altitude.

Kettenhunde
02-21-2010, 05:21 AM
To those interested in aircraft turn performance I can ,at this point, only say that they should use Google, library or call local university. Turning performance is simple, it takes just few pages of University textbook, math involved is elementary school level and it requires minimal knowledge about aerodynamics.

I would say that person of average intelligence can learn and comprehend this subject in less than 1 hour. But this thread is living proof that we are not dealing with average persons at UBI Too Happy Couple of loonies, few trolls and even few people who knows how to use books.

Problem is there are quite a few internet educated parrots who regurgitate a formula but do know basic principles like parasitic drag reduces with denisty using TAS or efficiency is linked to Cdo......

This causes them to make stupid statements like there is only ONE valid method or their simple TAS calculations are more accurate.

They don't understand that without the right data the margin of error is huge on turning performance.

JtD
02-21-2010, 06:20 AM
Originally posted by Kettenhunde:

That should fix it.

Yes, we all know that this is your claim.

However, despite of repeated requests you have no yet provided anything to prove that claim. So please post the numbers I asked you for, and we can move on.

Until you do, your suggestions have no value of any kind.

Kettenhunde
02-21-2010, 07:11 AM
Yes, we all know that this is your claim.

However, despite of repeated requests you have no yet provided anything to prove that claim. So please post the numbers I asked you for, and we can move on.

Until you do, your suggestions have no value of any kind.

Oh Baloney JtD.

The readers in this thread have asked not to muddy the waters further for the time being.

Quit making immature hay out of nothing. Let the people reading that are trying to follow get the concepts.

The point is not to confuse and bamboozle folks.

JtD
02-21-2010, 07:19 AM
Yet another excuse. I'm certain people following this topic are very interested in your numbers, and no one has asked you to not post them. It will help to clarify the issue.

If you still think this topic is not the right place, you can open a separate topic and post there.

Please do not post any more excuses, just post the results. Thank you.

M_Gunz
02-21-2010, 07:41 AM
Originally posted by Holtzauge:

I get 3546 lbs drag and Kettenhunde gets 2576.5 lbs.

I assume you think 2576.5 lbs is right since you slash my calculations and said that you could follow Kettenhundes?

Yes or no?

You assume. I pointed out where you were wrong in your attack on Crumpp, not who has the final answer.
Instead of answering that, you change the subject to pose another puzzle. I can tell you when you say
that 2 + 2 does not equal 5 that you are wrong there, call it slash what you did was wrong anyway.

I see typos and pointed out, those were corrected without asking me anything. Only the naughty kid with

Crumpp may be wrong, that does not mean he is wrong where you said. Simple enough. Chase your own shadow.

M_Gunz
02-21-2010, 07:59 AM
There are more than one example floating around, how about narrow it back down to only one?

This is what Wurkeri had used in his post and I'd rather use his method.
Before walking through, who objects to using his values below?

Apple or Orange, not both.

And as an example I picked a P-47B at 15k from WW2aircraftperformance site:

V = 386 mph = 172,52 m/s
W = 2000 hp = 1491400 W
M = 12560 lbs = 5697 kg
Sp = 41 ft = 12.5 m
S = 300 sqft = 27.87 m^2
AR = 5.6
p = 0.7708 kg/m^3
n = 0.85 estimated
e = 0.8 estimated
Vs = 105 mph = 46,93 m/s
g = 6g limit

I am sure the values for 2300HP P-47 are somewhere in the extended thread but I'm not going
to spend the hours re-reading posts to find them then look for more posts objecting to same.

I will make an assumption. Those demanding to see it worked out already have the initial data
so if the same people are going to want different than the above, speak up now or shut up later.

Kettenhunde
02-21-2010, 09:18 AM
and no one has asked you to not post them.

Pfft...

M_Gunz says:
I don't understand what you're trying to do there. Do the planes CHANGE WEIGHT in the spreadsheet?
Why not stick to one example at a time and work it out or at least not add elements?

http://forums.ubi.com/eve/foru...241031438#3241031438 (http://forums.ubi.com/eve/forums/a/tpc/f/23110283/m/2891026238?r=3241031438#3241031438)

The idea JtD is not to bamboozle folks or impress them with knowledge to stroke your own ego. We are discussing some complicated ideas which despite the claims of some, most people cannot just go down to the local library, open a book and do this stuff correctly.

In fact, that is probably how we got here in long a conversation about applying basic math and principles.

JtD
02-21-2010, 09:29 AM
So please open a new topic and post your numbers. It doesn't make a difference to me. Thanks.

If you fail to do so, and continue with just posting excuses, I take it you're not qualified to accurately assess whether or not my Excel sheet is correct.

Holtzauge
02-21-2010, 10:01 AM
OK, I can see that Kettenhunde is not planning to answer my question above about drag and Cd so I will post a calculation showing my view of what you can actually get with a thrust of 2576.5 lb and be done with it.

Cdo=0.032215 (Note that this is a bit high but was what Kettenhunde posted)
e=0.85
A=5.554
S=300 sqft
W=13583 lb
ra=0.0012673 slugs/ft**3 at 20000 ft altitude
v=246.6 KTAS (417.3 ft/s)
Ps=2300 hp
np=0.85 (prop efficiency)

T=Ps*np*325/v

T=2300*0.85*325/246.6=2576.5 lbs thrust as previously posted by Kettenhunde.

Assuming T=D:

D=0.5*ra*v**2*Cd*S

Since I am only comfortable using TAS measured in ft/s in US units I will use that:

2576.5=0.5*0.0012673*417.3**2*Cd*300

So Cd=0.077834

But Cd=Cdo+Cdi

Cdi=0.077834-0.032215=0.045619

Cdi=Cl**2/(pi*A*e)

0.045619=Cl**2/(pi*5.554*0.85)

Gives Cl=0.82255

L=0.5*0.0012673*417.3**2*0.82255*300
=27228.8 lb

n=27228.8/13583

n=2.004

So nowhere near the 2.57 Kettenhunde claims using Ps=2300 hp and calculating with EAS.

To forestall any questions why this is even lower than the n=2.108 I posted 20 pages back for my C++ simulation I can inform that the difference is due to me using a slightly lower Cdo since I think the value of 0.032215 is overly pessimistic.

OK, so that was the TAS calculation step by step and right now I do not think it will be fruitful to continue any more so I will leave M_Gunz and Kettenhunde to work over my calculation and post their own view on the matter.

Finally, as someone wisely said before: You have to judge the evidence put forward and how it is presented and draw your own conclusions.

I’ve had my say here with this TAS calculation and I’ll return to monitor mode for now.

Holtzauge
02-21-2010, 10:02 AM
And M_Gunz: You can take all your personal insults and trollish misinterpretations of my previous posts and shove them up the same orifice where you keep your understanding of Kettenhundes maths.

Over and out.

blairgowrie
02-21-2010, 10:35 AM
Oh my goodness! I am going to give one warning and one only. NO MORE PERSONAL INSULTS PLEASE.

Kettenhunde
02-21-2010, 11:53 AM
0.045619=Cl**2/(pi*5.554*0.85)

Gaston444
02-21-2010, 12:58 PM
Originally posted by FatCat_99:
<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">Originally posted by Kurfurst__:
Can someone summerize me the goal of this discussion in one sentence please...? http://forums.ubi.com/groupee_common/emoticons/icon_biggrin.gif
Not a sentence but might be appropriate.

To those interested in aircraft turn performance I can ,at this point, only say that they should use Google, library or call local university. Turning performance is simple, it takes just few pages of University textbook, math involved is elementary school level and it requires minimal knowledge about aerodynamics.

I would say that person of average intelligence can learn and comprehend this subject in less than 1 hour. But this thread is living proof that we are not dealing with average persons at UBI http://forums.ubi.com/images/smilies/88.gif Couple of loonies, few trolls and even few people who knows how to use books.

Rock on UBI http://forums.ubi.com/images/smilies/metal.gif

FC </div></BLOCKQUOTE>

-Oh yeah? If it was so simple, how come the actual 1989 flight test done by the "Association of Test Pilots" found that the P-47D, FG-1, P-51D and F6F ALL had "The Corner Speed very close to the maximum level speed"?

Does that sound like the "calculated" 270 MPH P-51D Corner Speed with flaps?

This was confirmed to me as plausible by an aircraft designer I contacted.

Gaston

AndyJWest
02-21-2010, 01:28 PM
Ok, Gaston. Can you explain what you think 'Corner Speed' means? When you have done this, we might see how relevant it is to the debate.

TheGrunch
02-21-2010, 01:53 PM
Originally posted by Gaston444:
Oh yeah? If it was so simple, how come the actual 1989 flight test done by the "Association of Test Pilots" found that the P-47D, FG-1, P-51D and F6F ALL had "The Corner Speed very close to the maximum level speed"?
Does that sound like the "calculated" 270 MPH P-51D Corner Speed with flaps?
As I said to you last time you posted this, corner speed is the speed at which an aircraft can achieve maximum INSTANTANEOUS turn performance. It's the lowest speed at which the aircraft can reach its limiting load factor before exceeding the wing's critical angle of attack. That's why it's close to level max speed.
Your aircraft designer calls that plausible because it is. That doesn't mean it's the speed at which the aircraft achieves its best sustained turn performance.

The problem is that a lot of information provided to air combat enthusiasts confuses best sustained turn speed with corner velocity. I know because I made the same mistake until I was corrected.

na85
02-21-2010, 02:27 PM
Originally posted by Holtzauge:

Kettenhunde has the Spitfire Mk9 Merlin 66 at +18 boost at a maximum loadfactor of about 3.49 and the P-47D22 at 56" Hg at 3.75 at 20000 ft.

Gauging "relative" performance this means that the P47 D22 can sustain a maximum loadfactor that is about 7.4% higher than the Spitfire Mk9 at 20000 ft according to Kettenhunde.

Do you think this sounds reasonable?

Certainly it sounds reasonable. A 7% difference is quite small; well inside the 10% margin that most agree denotes similar aircraft. I'd have more to say if I could actually see/read the way those charts were formulated in excel.

<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">Originally posted by Kettenhunde:
The Spitfire is woefully inadequate when matched in a sustained turning contest with the P-47 under these conditions.

Do you agree? </div></BLOCKQUOTE>

I wouldn't choose the words "woefully inadequate" perhaps. Do I feel the Spitfire holds a turn advantage over all speeds/altitudes? Certainly not.

Kettenhunde
02-21-2010, 03:29 PM

Yeah I couldn't help but poke a little fun at Hop.

http://forums.ubi.com/images/smilies/16x16_smiley-tongue.gif

M_Gunz
02-21-2010, 04:55 PM
Originally posted by TheGrunch:
<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">Originally posted by Gaston444:
Oh yeah? If it was so simple, how come the actual 1989 flight test done by the "Association of Test Pilots" found that the P-47D, FG-1, P-51D and F6F ALL had "The Corner Speed very close to the maximum level speed"?
Does that sound like the "calculated" 270 MPH P-51D Corner Speed with flaps?
As I said to you last time you posted this, corner speed is the speed at which an aircraft can achieve maximum INSTANTANEOUS turn performance. It's the lowest speed at which the aircraft can reach its limiting load factor before exceeding the wing's critical angle of attack. That's why it's close to level max speed.
Your aircraft designer calls that plausible because it is. That doesn't mean it's the speed at which the aircraft achieves its best sustained turn performance.

The problem is that a lot of information provided to air combat enthusiasts confuses best sustained turn speed with corner velocity. I know because I made the same mistake until I was corrected. </div></BLOCKQUOTE>

It's close to the maximum IAS those guys would push over 40 year old warbirds through maneuvers.
Did the test pilots association say they used flaps in such turns?
Who in their right mind pops flaps in a 270mph IAS turn when you easily pull 6G's at around 250?

Gaston444
02-21-2010, 08:35 PM
Originally posted by TheGrunch:
<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">Originally posted by Gaston444:
Oh yeah? If it was so simple, how come the actual 1989 flight test done by the "Association of Test Pilots" found that the P-47D, FG-1, P-51D and F6F ALL had "The Corner Speed very close to the maximum level speed"?
Does that sound like the "calculated" 270 MPH P-51D Corner Speed with flaps? <pre class="ip-ubbcode-code-pre"> </pre>
As I said to you last time you posted this, corner speed is the speed at which an aircraft can achieve maximum INSTANTANEOUS turn performance. It's the lowest speed at which the aircraft can reach its limiting load factor before exceeding the wing's critical angle of attack. That's why it's close to level max speed.
Your aircraft designer calls that plausible because it is. That doesn't mean it's the speed at which the aircraft achieves its best sustained turn performance.

The problem is that a lot of information provided to air combat enthusiasts confuses best sustained turn speed with corner velocity. I know because I made the same mistake until I was corrected. </div></BLOCKQUOTE>

-Your definition is correct: Now can anyone tell me what is the "Corner Speed" for the P-51D at the WWII standard limit of 7G in Il-2? In "Aces High", with one notch of flaps, it is at a low 270 MPH, though this may or may not include the 8G of lightweight post-war standard. It is about 290 MPH "clean".

Note the 1989 test was limited to 6Gs, owing to airframe age, and STILL the minimum speed was above 350 MPH at least...

As the 1989 test pilots concluded: "This high value entails a rapid loss of energy whenever turning at near the structural limit (more like the angle of attack limit as you said). They say this because the engine has little reserve power at these speeds to maintain the turn, so the instantaneous max G turn always burns a LOT of speed.

In theory, this means ANY downthrottling reduces the sustained turn rate, since speed in sustained turns should be kept as near the 350 MPH+ instantaneous peak as the engine power can allow.

Any mention of SUSTAINED downthrottling that INCREASES turn rate over several 360°s, followed by FURTHER downthrottling, not acceleration, with resulting FURTHER increases in the sustained turn rate in an at-level-speed P-51D, means that the notion that getting away (slower) from this correct 350 MPH+ Corner Speed being a BAD thing for sustained turns is false.

And in any case, a 350-400 MPH "Corner Speed" in a P-51D no longer looks likes an doghouse curve at all: It is an assymetrical peak way off to the right side...

So WHAT is Il-2's P-51D's Corner Speed at 6-7, or 8Gs?

Since we appear to agree on the 350 MPH+ reality at last...

Gaston

AndyJWest
02-21-2010, 08:53 PM
Does anyone understand this?

Consider this gem:

FURTHER downthrottling, not acceleration

At corner speed, as defined, you can only move the throttle in the direction necessary to maintain such speed (if possible) or you are no longer at said speed. Full stop. If you accelerate, or decelearate, you aren't at corner speed. I may not know as much aerodynamics as other posters in this thread, but I think I understand basic logic.

Gaston444
02-21-2010, 09:19 PM
Downthrottling in SUSTAINED turns.

And the accepted relationship between the peak SUSTAINED turn rate and the "Corner Speed", which I don't accept, is that the first should be as close as possible to the second for optimal performance.

If that's too complicated, please tell us...

Gaston

P.S. And it would be really nice to know what Il-2's F6F, P-51D, P-47D and F4U "Corner Speeds" are like, as I have slight doubts that they are as high as the 1989 tests, especially at a modest 6Gs... (Though admittedly "very close to maximum level speed" is not extremely precise, I doubt anything under 350 MPH makes the cut)

G.

G.

TheGrunch
02-21-2010, 09:22 PM
Originally posted by Gaston444:
So WHAT is Il-2's P-51D's Corner Speed at 6-7, or 8Gs?

Since we appear to agree on the 350 MPH+ reality at last...

Gaston
It's actually higher than that for every aircraft you mention in Il-2, because the game is a bit easy-mode about structural failure. This is projected to be fixed in one of the future patches by Daidalos Team, along with omission of compressibility effects for most aircraft. In the game at the moment any aircraft can pull up to 13.5 Gs I believe. The P-51 is one of the few examples that is easier to break, but I believe you can take it up to about 400-450mph before losing wings.
As I said Gaston, the speed at which you can pull maximum Gs is NOT a good speed for maximum *sustained* turning rate because you can't actually sustain this speed while pulling enough Gs to achieve best turn rate. It's a good speed to be at at the beginning of a turning contest, since you can gain quite a bit of advantage to begin with before you must reduce elevator deflection to maintain your best sustained turn speed.
I've already told you why I don't think your TWO cherry-picked pilot accounts are solid enough evidence to make a GENERAL conclusion about aircraft performance.
As for doghouse charts...you can't just look at the critical AoA and limiting load factor. There's a power limit for sustained turns as well. Please read this clear and concise explanation (http://www.simhq.com/_air/air_011a.html) of energy-manoeuverability diagrams, because your understanding is incomplete and I'm certain it will clear up some of your misconceptions.

na85
02-21-2010, 10:02 PM
Originally posted by Gaston444:

So WHAT is Il-2's P-51D's Corner Speed at 6-7, or 8Gs?

Corner speed doesn't occur "at a G level." It occurs at the intersection of the structural and lift limits of the airframe.

It is dependent on a combination of the aircraft's weight, power, altitude, and configuration. But over the typical range of speeds at which WW2 air combat occurs, any change in Vc is negligible.

To ask for "corner speed at 6 G" doesn't make any sense. That's like asking for "best climb speed at 6G"

TheGrunch
02-21-2010, 10:08 PM
Originally posted by na85:
Corner speed doesn't occur "at a G level." It occurs at the intersection of the structural and lift limits of the airframe.
He's talking about the fact that different tests use a different limiting load factor, I think. It's still a strange way to put it, mind.

M_Gunz
02-22-2010, 02:58 AM
The pilot is the limiting factor and for WWII that was considered 6 G's to many testing establishments.

The thing is that you can turn at 6 G's at very high speed and hold it while your speed bleeds down.
But the Corner Speed will be at the *slowest* you can hold 6 G's. That gives you the least radius and
with these planes in level flight lasts all of an instant, LOL! That occurs very close to 2 1/2 times
stall IAS. Play with altitudes and convert to TAS and you have enough bits to misinterpret as you will.

Then there is the "Sustained Corner Speed" that so many Warbirds and Aces High players make a huge point
of knowing and generally mislabel as above, it is rather the maximum sustained turn IAS and a very useful
speed to maintain during turning virtual combat in sims. You get closer to 3 G's there.

M_Gunz
02-22-2010, 03:03 AM
Originally posted by M_Gunz:
There are more than one example floating around, how about narrow it back down to only one?

This is what Wurkeri had used in his post and I'd rather use his method.
Before walking through, who objects to using his values below?

Apple or Orange, not both.

<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">And as an example I picked a P-47B at 15k from WW2aircraftperformance site:

V = 386 mph = 172,52 m/s
W = 2000 hp = 1491400 W
M = 12560 lbs = 5697 kg
Sp = 41 ft = 12.5 m
S = 300 sqft = 27.87 m^2
AR = 5.6
p = 0.7708 kg/m^3
n = 0.85 estimated
e = 0.8 estimated
Vs = 105 mph = 46,93 m/s
g = 6g limit

I am sure the values for 2300HP P-47 are somewhere in the extended thread but I'm not going
to spend the hours re-reading posts to find them then look for more posts objecting to same.

I will make an assumption. Those demanding to see it worked out already have the initial data
so if the same people are going to want different than the above, speak up now or shut up later. </div></BLOCKQUOTE>

I find maybe something better at the same site. For P-47B, top speed data at 3 different altitudes and
many different props. This will be fun! Thank you for the tip Wurkeri! Salute!

JtD
02-22-2010, 09:25 AM
I would like to conclude my intervention, and since Kettenhunde has only managed to post excuses, I'll provide the numbers I asked him for.

As I said, I had a real life reference, which can be found here (http://www.wwiiaircraftperformance.org/mustang/p-51b-6883.html).

The task was to estimate the engine power at an altitude of 300000 feet, which I would now like to lower to the 29400 stated in the report, for easier reference. Not much of an issue, I was using a wing efficiency of 0.9 here.

Power estimate for 440.5 mph at 29400 feet with the plane from the test and a constant 85% prop efficiency:
- Method Kettenhunde: 705 hp
- Method Excel Sheet: 1140 hp
- Flight test protocol: 1292 hp
Relative error:
- Method Kettenhunde: 83%
- Method Excel Sheet: 13%

Power estimate for 440.5 mph at 29400 feet with the plane from the test and a prop efficiency decreasing from 90% to 80% with altitude and speed:
- Method Kettenhunde: 784 hp
- Method Excel Sheet: 1270 hp
- Flight Test Protocol: 1292 hp
Relative Error:
- Method Kettenhunde: 65%
- Method Excel Sheet: 2%

Now obviously the Excel Sheet is not perfect, which is why it is recommend for relative performance estimates only, but it certainly is not wrong. An about 10% error is a good result for an as simple tool as that one.
If we now look at the proposition of Kettenhunde, we can see that his method is far off, an error of 80% can hardly be described as acceptable.
We can therefore conclude that the Excel sheet is correct and Kettenhundes claim is wrong.

I would be surprised if Kettenhunde did not come back to again declare how wrong I am, attack my person, but I will not be responding to that. He had his chance to prove his claims, he chose to let it pass. Case closed.

If any of you has further questions, you can send me a PM.

Kettenhunde
02-22-2010, 10:27 AM
I would be surprised if Kettenhunde did not come back to again declare how wrong I am, attack my person, but I will not be responding to that. He had his chance to prove his claims, he chose to let it pass. Case closed.

Pitiful....You attack my credibility and berate me if I respond.

Be a man and learn to accept that people can disagree with you, it is not a crime against nature.

Aeronautical Science happens to be an area I got my degree in and if I see things that are not correct, I will disagree with them.

The basic math relationships are laid out in the thread.

It will work on the power if you input the correct parameters and work in the same units. Since the report is dealing in TAS, then of course it will work out....it is set by the speed and power available. Does that surprise you?

I have told you from the beginning you must work either in TAS or EAS. Then your relative performance picture will be the same no matter which method you choose.

Where do you get this 784hp in TAS units?? I bet you are confusing EAS and TAS predictions again. That seems to have been a very common theme with both you and Holtzauge.

Crumpp says:

Here is what is confusing people.

Lets break down what is happening on the Thrust required <drag> side of the equation.

in TAS:

Dynamic pressure = sigma *(KTAS)^2 / 295

Dynamic pressure = sigma *(KEAS/SQRT sigma)^2 / 295

Dynamic pressure = sigma *(KEAS/SQRT sigma)*(KEAS/SQRT sigma) / 295

Dynamic pressure = [sigma*(KEAS*KEAS/SQRT sigma*SQRT sigma)]/ 295

Dynamic pressure = [sigma*(KEAS*KEAS/SQRT sigma^2)]/ 295

Of course, the square root of sigma squared is sigma.

Dynamic pressure = [sigma*(KEAS*KEAS/sigma]/ 295

Sigma cancels and we are left with:

Dynamic pressure = (KEAS*KEAS]/ 295 or the expression for EAS dynamic pressure.

Just because it cancels does not mean the correct thing to do is add it in to the other side of the equal sign on the Power Available side when we are using EAS.

You still have to follow the basic rules of math.

If you work in EAS, sigma is not factored into the equation and the results are converted at the end.

If you work in TAS, you must add in sigma on both sides of the equation. Your parasitic drag will vary IAW density effects and subsequently vary the value of e.

The end result, done correctly will be very close in either method.

It is a fact that is like claiming water runs uphill.....

http://forums.ubi.com/eve/foru...211083338#6211083338 (http://forums.ubi.com/eve/forums/a/tpc/f/23110283/m/2891026238?r=6211083338#6211083338)

http://forums.ubi.com/eve/foru...291088338#5291088338 (http://forums.ubi.com/eve/forums/a/tpc/f/23110283/m/2891026238?r=5291088338#5291088338)

Your spreadsheet confuses units as well and I pointed out how to correct that.

Why don't you stick to one issue at a time instead of constantly changing parameters. It does not make it any clearer to anyone when you do this.

M_Gunz
02-22-2010, 10:33 AM
ADD: @#\$%! -- half the time I type a post, someone else posts one in between! This is reply to JtD above!

I guess 440.5 is close enough to 442.5 and don't look now but you made a meaningless typo, 300000 ft?

You don't show how you worked the "Kettenhunde Method" so why can't he make an honest objection if he
gets different values for his way than you did?

Looking at some of these examples (not saying yours) there are some differences in AR popped up associated
with much pointing and yelling. Sleight of math should at least be prefaced with "nothing up my sleeves"
or "at no time do my fingers leave my hands" sort of declaration.

I thought that power is simply thrust x TAS. How hard is that? I also thought that the examples I've seen
here for the P-47 use either 2300HP or 2000HP. What kind of work backward from that gives different value?
I MUST be confused as to what you call power there.

na85
02-22-2010, 11:29 AM
Originally posted by JtD:

Power estimate for 440.5 mph at 29400 feet with the plane from the test and a constant 85% prop efficiency:
- Method Kettenhunde: 705 hp
- Method Excel Sheet: 1140 hp
- Flight test protocol: 1292 hp
Relative error:
- Method Kettenhunde: 83%
- Method Excel Sheet: 13%

Actually the Kettenhunde method that he PMed to me on the 21st at 08:41 returns 1150 hp, using constant ep = 0.85

Could you please show the work behind Method Kettenhunde?

Gaston444
02-22-2010, 11:55 AM
Originally posted by TheGrunch:
<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">Originally posted by Gaston444:
So WHAT is Il-2's P-51D's Corner Speed at 6-7, or 8Gs?

Since we appear to agree on the 350 MPH+ reality at last...

Gaston
It's actually higher than that for every aircraft you mention in Il-2, because the game is a bit easy-mode about structural failure. This is projected to be fixed in one of the future patches by Daidalos Team, along with omission of compressibility effects for most aircraft. In the game at the moment any aircraft can pull up to 13.5 Gs I believe. The P-51 is one of the few examples that is easier to break, but I believe you can take it up to about 400-450mph before losing wings.
<pre class="ip-ubbcode-code-pre"> </pre>As I said Gaston, the speed at which you can pull maximum Gs is NOT a good speed for maximum *sustained* turning rate because you can't actually sustain this speed while pulling enough Gs to achieve best turn rate. It's a good speed to be at at the beginning of a turning contest, since you can gain quite a bit of advantage to begin with before you must reduce elevator deflection to maintain your best sustained turn speed.
I've already told you why I don't think your TWO cherry-picked pilot accounts are solid enough evidence to make a GENERAL conclusion about aircraft performance.
As for doghouse charts...you can't just look at the critical AoA and limiting load factor. There's a power limit for sustained turns as well. Please read this clear and concise explanation (http://www.simhq.com/_air/air_011a.html) of energy-manoeuverability diagrams, because your understanding is incomplete and I'm certain it will clear up some of your misconceptions. </div></BLOCKQUOTE><pre class="ip-ubbcode-code-pre"> </pre>

- 13.5 Gs is the correct structural SAFETY MARGIN (15 Gs on the Me-109G), it has nothing to do with how much Gs you can actually pull...

WWII standard, with special G-suit maybe, was for the P-51D set at 7G: WHAT IS THE LOWEST SPEED ON IL-2 THAT THIS CAN BE ACHIEVED?

Gaston

P.S. People on Aces High know damn well the difference between 3-4 Gs sustained turn rates and Corner Speed, and they have the P-51D Corner Speed at 270 MPH at 7 Gs with flaps, and at about 280-290 MPH without.

Do people on Il-2 know as much about their own game?

G.

Holtzauge
02-23-2010, 02:10 AM
Originally posted by Kettenhunde:
Aeronautical Science happens to be an area I got my degree in

Is that an Bsc, Msc or a Phd? In what discipline? At what University? What year did you graduate?

You have implied a connection to Embry-Riddle before so what kind of a degree do you have from them?

Why don't you come clean on this here and now.

It would certainly strengthen your position immensely in this debate if are an aeronautical engineer.

So let's go for it: You post a resume of your academic credentials and work experience in the aeronautical business and I will post mine.

Kettenhunde
02-23-2010, 03:05 AM
Could you please show the work behind Method Kettenhunde?

I bet he is getting it off the spreadsheet where I told him to input the correct formula into his sheet.

Without any formal classes he does not realize that @ 30,000ft:

705hp / sqrt <.37413> = 1152hp

705hp EAS = the same exact performance as 1152hp TAS......

Kettenhunde
02-23-2010, 05:22 AM
You are not anything in any field of aerosciences Holtzauge and it is a waste of time to discuss it.

M_Gunz
02-23-2010, 05:35 AM
Demanding personal info...
tsk tsk tsk what's next? Address, phone #, proof-proof-proof! family photos, ID card, SSN, credit check.... http://forums.ubi.com/images/smilies/51.gif
And the penalty for not meeting those demands is?

Funny but I've never seen anyone DUMB ENOUGH to provide a map to their front door here and hope I never do.

That's why I stick to WHAT is posted rather than WHO posts it.

Holtzauge
02-23-2010, 05:39 AM
And there we have it plain for all to see: The troll and the fraud in perfect harmony!

Holtzauge
02-23-2010, 06:36 AM
Originally posted by Kettenhunde:
You are not anything in any field of aerosciences Holtzauge and it is a waste of time to discuss it.

I have an Msc in aerodynamics and structural engineering from the Royal Institute of Technology, Stockholm, Sweden. Graduated 1986. I have worked in the aerospace business from 1987 to 2000. I have worked on the JA37 Viggen and JAS39 Gripen fighters systems. My Msc thesis accessed and calculated the performance impact of external stores on figther aircraft.

M_Gunz
02-23-2010, 06:48 AM
Originally posted by Holtzauge:
And there we have it plain for all to see: The troll and the fraud in perfect harmony!

Oooh-ooh-ooh! PERSONAL INSULTS! PERSONAL INSULTS! MOOOOOOOOMMMMMY!!!!

Oh and I this time I'm not writing the H-word. It should be obvious enough anyway.

yuuppers
02-23-2010, 06:49 AM
Holtzauge, can you have one of your aeronautic engineer friends to post in this thread?

blairgowrie
02-23-2010, 07:00 AM
You can take a vacation Holtzauge. You were warned not to continue personal insults.

FatCat_99
02-23-2010, 07:30 AM
Originally posted by yuuppers:
Holtzauge, can you have one of your aeronautic engineer friends to post in this thread?
What for,do you really think that would help? http://forums.ubi.com/images/smilies/16x16_smiley-very-happy.gif

AndyJWest
02-23-2010, 09:38 AM
Things I've learned from this thread:

(a) Some aerodynamic principles I didn't understand before.

(b) That I need better maths to understand aerodynamics.

(c) That I'm unlikely to get a complete understanding of the subject here.

(d) That it doesn't need 'politics' for a disagreement to go downhill rapidly.

A shame, really. It looked at one point as if a consensus could have emerged.

M_Gunz
02-23-2010, 10:19 AM
Don't give up yet, I've got a dilly coming up working with that nice clear post from Wurkeri.
Still checking numbers. I have Cd0 at one alt of multiple alts using data from the same plane.

Does Cd0 = 0.2866 sound ballpark for a P-47B at 5,000 ft... anyone?

na85
02-23-2010, 10:22 AM
I think that's too high, Gunz. Probably more in the area of 0.02

TheGrunch
02-23-2010, 10:32 AM
Originally posted by Gaston444:
WHAT IS THE LOWEST SPEED ON IL-2 THAT THIS CAN BE ACHIEVED?
It's certainly higher than 270mph. I don't know exactly, I play the damn game, I don't sit with Infomod on all the time looking at numbers from DeviceLink, jeez. If you want to know that much try it yourself, the game's only 10 quid and Infomod and LesniHU's autopilot are free. Can't you EVER answer any of your own questions about this game with research? Most of your objections to this game so far have been unfounded. What has Oleg Maddox done to personally offend you so much?

M_Gunz
02-23-2010, 10:42 AM
Originally posted by TheGrunch:
<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">Originally posted by Gaston444:
WHAT IS THE LOWEST SPEED ON IL-2 THAT THIS CAN BE ACHIEVED?
It's certainly higher than 270mph. I don't know exactly, I play the damn game, I don't sit with Infomod on all the time looking at numbers from DeviceLink, jeez. If you want to know that much try it yourself, the game's only 10 quid and Infomod and LesniHU's autopilot are free. Can't you EVER answer any of your own questions about this game with research? Most of your objections to this game so far have been unfounded. What has Oleg Maddox done to personally offend you so much? </div></BLOCKQUOTE>

It should be about 2.5x stall speed, sqrt of 6 is just a bit less.
Clean 1G stall for a P-51 with combat flaps is over 100mph??

M_Gunz
02-23-2010, 10:46 AM
Originally posted by na85:
I think that's too high, Gunz. Probably more in the area of 0.02

Here's what I have so far. The method is from Wurkeri but the mistakes are mine!
This is not finished but I could use checking anyway. The answers are important, not the ego.

================================================== ================================================== =====

Turn performance estimation -- working out details using data from the same plane at different altitudes.

WAR DEPARTMENT
AIR CORP, MATERIAL DIVISION
Wright Field, Dayton, Ohio
June 18, 1942

P-47B Airplane, A.C. No. 41-5902
Acceptance Performance Tests

High speed in level flight with oil cooler flaps and intercooler flaps flush and throttle wide open with turbo on to give military rated power or 18,250 limiting turbo r.p.m.

Altitude Feet____________________ 5,000_____15,000_____25,000_____27,800
True Speed m.p.h._________________352________386_______420___ _____429

At these altitudes the B.H.P. is 2000HP, engine P.R.M. is 2700 and MP is 51" to 52" Hg.

I will use some of the values from Wurkeri's post (Thanks W!) since this is more about method than hair-counting.

================================================== =
W = 2000 hp = 1491400 W -------- it's from the same data, same plane AFAIK
M = 12560 lbs = 5697 kg ----------- actually 12565 lbs but I'm sure less before actual takeoff. ;^)
Sp = 41 ft = 12.5 m------------------- wingspan
S = 300 sqft = 27.87 m^2------------ wing area
AR = 5.6 ------------------------------- aspect ratio being wingspan^2 / wing area
n = 0.85 estimated
e = 0.8 estimated
Vs = 105 mph = 46,93 m/s---------- clean 1G stall IAS
================================================== =

1 foot = 0.3048 meters
1 mile = 1.609344 kilometers
air density taken from "A Sample Atmosphere Table (SI units)" at http://www.pdas.com/m1.htm

Altitude Feet____________________ 5,000______15,000_____25,000_____27,800
True Speed m.p.h._________________352________386_______420___ _____429
Altitude Meters___________________1,524______4,572______7,6 20______8,473
True Speed m/s___________________157.4______172.5______187.8___ ___191.8
Air density kg/m^3_________________1.059______0.774______0.551___ ___0.499 -- values interpolated from table below

--------------------------------------------------------------------------------------------------
air density taken from "A Sample Atmosphere Table (SI units)" at http://www.pdas.com/m1.htm
km alt____ kg/cu.meter
0________1.225
2________1.007
4________0.8193
6________0.6601
8________0.5258
10_______0.4135
---------------------------------------------------------------------------------------------------

Symbols used -- thanks Wurkeri!
===========================
W = engine power
S = wing area
Sp = wing span
AR = Aspect ratio, calculated from span and area
p = density at this given altitude
M = mass of the plane
n = propeller efficiency
e = efficiency factor
Vs = stall speed
g = load factor, 1 in level flight, limited to certain value.
===========================

Also from Wurkeri:
===========================
whole balance formula ------- without checking units and cancels, where's my cracker?

0.5*p*v^2*S*[Cd0+([M*g*9.81]/[0.5*p*v^2*S])^2/(pi*AR*e)] = W*n/v
===========================

I look at thrust alone first = W * n / v
W = 1491400 Watts
n = 0.85 estimated

Altitude Meters___________________1,524______4,572______7,6 20______8,473
True Speed m/s___________________157.4______172.5______187.8___ ___191.8
Thrust_kg_______________________8,054______7,349__ ____6,750______6,609

e = 0.8 estimated, Ar = 5.6; (pi * AR * e) = 14.07
S = 27.87 m^2 -------- wing area
M = 5697 kg ---------- mass of the plane
g = 1 ------------------- data values are from level flight

.5 * p * v^2 * 27.87 * [Cd0 + ( [ 5697 * 9.81 ] / [ .5 * p * v^2 * 27.87 ] )^2 / 14.07 ] = Thrust

Altitude 1524m given all our assumtions from above:
.5 * 1.059 * 24775 * 27.87 * [Cd0 + ( 55888 / [ .5 * 1.059 * 24775 * 27.87 ] )^2 / 14.07 ] = 8054

And with help from and thanks to na85:

<span class="ev_code_RED">365609 * [Cd0 + ( 55888 / 365609 )^2 / 14.07 ] = 8054</span>
Cd0 + ( 55888 / 365609 )^2 / 14.07 = 8054/365609
Cd0 = 8054/365609 - (55888/365609 )^2 / 14.07
Cd0 = 0.022 - 0.153^2 / 14.07
Cd0 = 0.022 - 0.0233/14.07
Cd0 = 0.022 - 0.0017
Cd0 = 0.021

na85
02-23-2010, 11:07 AM
Originally posted by M_Gunz:

Altitude 1524m given all our assumtions from above:
.5 * 1.059 * 24775 * 27.87 * [Cd0 + ( 55888 / [ .5 * 1.059 * 24775 * 27.87 ] )^2 / 14.07 ] = 8054
<span class="ev_code_RED">365609 * [Cd0 + ( 55888 / 365609 )^2 / 14.07 ] = 8054</span>
Cd0 + ( 55888 / 365609 )^2 = 14.07 * 8054 / 365609 = 0.30995
Cd0 + 0.023367 = 0.30995
Cd0 = 0.2866

Writing equations in forum posts is irritating isn't it? http://forums.ubi.com/images/smilies/16x16_smiley-very-happy.gif I wish we had some kind of equation writer.

If you look at the line I highlighted in red, the next line you multiply the 14.07 and the 365609 to the other side, but you can't actually do that because there are TWO terms on the left side of the eqn. I make that mistake all the time. You can only move the 365609 over. If you wanted to multiply the 14.07 over you would also have to multiply Cd0 by 14.07. That is a poor explanation on my part; I hope that makes sense.

It should go like this:

<span class="ev_code_RED">365609 * [Cd0 + ( 55888 / 365609 )^2 / 14.07 ] = 8054</span>
Cd0 + ( 55888 / 365609 )^2 / 14.07 = 8054/365609
Cd0 = 8054/365609 - (55888/365609 )^2 / 14.07
Cd0 = 0.022 - 0.153^2 / 14.07
Cd0 = 0.022 - 0.0233/14.07
Cd0 = 0.022 - 0.0017
Cd0 = 0.021

TS_Sancho
02-23-2010, 11:17 AM
Originally posted by Gaston444:

P.S. People on Aces High know damn well the difference between 3-4 Gs sustained turn rates and Corner Speed, and they have the P-51D Corner Speed at 270 MPH at 7 Gs with flaps, and at about 280-290 MPH without.

Do people on Il-2 know as much about their own game?

G.

Gaston could you please provide a historical source listing VFE for any P51 as 270 mph?

Somebody mentioned earlier that Oleg Maddox has a degree in aeronautical engineering, He holds multiple flight certificates as well.

Dale Addink is a software designer and amateur flight enthusiast.

Who do you feel is the more qualified to develop an accurate computer model predicting aircraft performance?

TheGrunch
02-23-2010, 11:27 AM
Originally posted by TS_Sancho:
Who do you feel is the more qualified to develop an accurate computer model predicting aircraft performance?
Oh, Gaston doesn't care about that, he thinks that the science of calculating and modelling aircraft performance is wrong and prefers to rely upon ambiguous pilot accounts.

na85
02-23-2010, 11:36 AM
Sancho at risk of taking an unwanted vacation from these boards I will tell you that Gaston seemingly has no interest in providing proof to back up his claims. He has been asked to do so multiple times by multiple members and either ignores the request or retorts with something along the lines of "you need mathematics to prove what amounts to a kindergarten discussion?" i.e. thinly veiled insults.

M_Gunz
02-23-2010, 11:55 AM
Originally posted by na85:
Sancho at risk of taking an unwanted vacation from these boards I will tell you that Gaston seemingly has no interest in providing proof to back up his claims. He has been asked to do so multiple times by multiple members and either ignores the request or retorts with something along the lines of "you need mathematics to prove what amounts to a kindergarten discussion?" i.e. thinly veiled insults.

Same risk. Kindergarten discussion fits the rhetoric he poses.

M_Gunz
02-23-2010, 11:58 AM
Originally posted by na85:
<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">Originally posted by M_Gunz:

Altitude 1524m given all our assumtions from above:
.5 * 1.059 * 24775 * 27.87 * [Cd0 + ( 55888 / [ .5 * 1.059 * 24775 * 27.87 ] )^2 / 14.07 ] = 8054
<span class="ev_code_RED">365609 * [Cd0 + ( 55888 / 365609 )^2 / 14.07 ] = 8054</span>
Cd0 + ( 55888 / 365609 )^2 = 14.07 * 8054 / 365609 = 0.30995
Cd0 + 0.023367 = 0.30995
Cd0 = 0.2866

Writing equations in forum posts is irritating isn't it? http://forums.ubi.com/images/smilies/16x16_smiley-very-happy.gif I wish we had some kind of equation writer.

If you look at the line I highlighted in red, the next line you multiply the 14.07 and the 365609 to the other side, but you can't actually do that because there are TWO terms on the left side of the eqn. I make that mistake all the time. You can only move the 365609 over. If you wanted to multiply the 14.07 over you would also have to multiply Cd0 by 14.07. That is a poor explanation on my part; I hope that makes sense.

It should go like this:

<span class="ev_code_RED">365609 * [Cd0 + ( 55888 / 365609 )^2 / 14.07 ] = 8054</span>
Cd0 + ( 55888 / 365609 )^2 / 14.07 = 8054/365609
Cd0 = 8054/365609 - (55888/365609 )^2 / 14.07
Cd0 = 0.022 - 0.153^2 / 14.07
Cd0 = 0.022 - 0.0233/14.07
Cd0 = 0.022 - 0.0017
Cd0 = 0.021 </div></BLOCKQUOTE>

Ah yes, you are right! Thank you. Now going to edit the post above (with a note) and the draft.
Oh geez! What a stupid mistake!

na85
02-23-2010, 12:07 PM
No big deal. That's one of those silly mistakes everyone makes from time to time. http://forums.ubi.com/groupee_common/emoticons/icon_cool.gif

I've got a dilly coming up working with that nice clear post from Wurkeri.

What's your goal for this? By that I mean what are you attempting to show/what problem are you solving?

M_Gunz
02-23-2010, 12:09 PM
Originally posted by na85:
No problem.

<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">I've got a dilly coming up working with that nice clear post from Wurkeri.

What's your goal for this? </div></BLOCKQUOTE>

Since I don't know how it's going to turn out let's just say I hope to put a finish to the farting around.
And that to me should constitute a dilly of a post indeed!

na85
02-23-2010, 12:14 PM
PM me if you would like any help.

Wurkeri
02-23-2010, 12:57 PM
Originally posted by na85:
What's your goal for this? By that I mean what are you attempting to show/what problem are you solving?

My idea was to post a clear presentation how to calculate this properly so anyone can check, as an example, if JtD's spreadsheet works as supposed or pretty much everything else claimed in this thread (yes, even Gaston's stuff).

And I see that Gunz is going to right direction here. You see the point; if you know, you don't need to believe.

M_Gunz
02-23-2010, 01:41 PM
Exactly. Some of my biggest days back in school were when I could verify myself what I had taken
on faith for years. Weighing an electron and getting within 2% of the accepted value without rigging
that lab. Seeing why the volume of a sphere is 4/3 pi r^3... the 4/3 comes from the limits of the
integration. And when we learned finite series, to make our own trig tables were all big days and
all 35+ years ago for me. I just wish everyone could have such days.

Yes Nate, my old head needs help checking the math but still I have to do it myself.
And Fatcat, maybe some day you will understand that not everyone does even this every day or week.
I can't run 20 miles with full gear any more either, doesn't mean I never did. How many you know
who have done both, and more? ;^P Hope you have such a good run yourself! Take care how you judge!

na85
02-23-2010, 01:51 PM
Originally posted by M_Gunz:

Yes Nate, my old head needs help checking the math but still I have to do it myself.

Yeah I realized that might be the case after I had posted. I've been doing similar throughout the course of this thread.

M_Gunz
02-23-2010, 03:19 PM
Turn performance estimation -- working out details using data from the same plane at different altitudes.
With many thanks to community members for help in not only making this but in correcting errors on the way.

WAR DEPARTMENT
AIR CORP, MATERIAL DIVISION
Wright Field, Dayton, Ohio
June 18, 1942

P-47B Airplane, A.C. No. 41-5902
Acceptance Performance Tests

High speed in level flight with oil cooler flaps and intercooler flaps flush and throttle wide open with turbo on to give military rated power or 18,250 limiting turbo r.p.m.

Altitude Feet____________________ 5,000_____15,000_____25,000_____27,800
True Speed m.p.h._________________352________386_______420___ _____429

At these altitudes the B.H.P. is 2000HP, engine P.R.M. is 2700 and MP is 51" to 52" Hg.

I will use some of the values from Wurkeri's post (Thanks W!) since this is more about method than hair-counting.

================================================== =
W = 2000 hp = 1491400 W -------- it's from the same data, same plane AFAIK
M = 12560 lbs = 5697 kg ----------- actually 12565 lbs but I'm sure less before actual takeoff. ;^)
Sp = 41 ft = 12.5 m------------------- wingspan
S = 300 sqft = 27.87 m^2------------ wing area
AR = 5.6 ------------------------------- aspect ratio being wingspan^2 / wing area
n = 0.85 estimated
e = 0.8 estimated
Vs = 105 mph = 46,93 m/s---------- clean 1G stall IAS
================================================== =

1 foot = 0.3048 meters
1 mile = 1.609344 kilometers
air density taken from "A Sample Atmosphere Table (SI units)" at http://www.pdas.com/m1.htm

Altitude Feet____________________ 5,000______15,000_____25,000_____27,800
True Speed m.p.h._________________352________386_______420___ _____429
Altitude Meters___________________1,524______4,572______7,6 20______8,473
True Speed m/s___________________157.4______172.5______187.8___ ___191.8
Air density kg/m^3_________________1.059______0.774______0.551___ ___0.499 -- values interpolated from table below

--------------------------------------------------------------------------------------------------
air density taken from "A Sample Atmosphere Table (SI units)" at http://www.pdas.com/m1.htm
km alt____ kg/cu.meter
0________1.225
2________1.007
4________0.8193
6________0.6601
8________0.5258
10_______0.4135
---------------------------------------------------------------------------------------------------

Symbols used -- thanks Wurkeri!
===========================
W = engine power
S = wing area
Sp = wing span
AR = Aspect ratio, calculated from span and area
p = density at this given altitude
M = mass of the plane
n = propeller efficiency
e = efficiency factor
Vs = stall speed
g = load factor, 1 in level flight, limited to certain value.
===========================

Also from Wurkeri:
===========================
whole balance formula ------- without checking units and cancels, where's my cracker?

0.5*p*v^2*S*[Cd0+([M*g*9.81]/[0.5*p*v^2*S])^2/(pi*AR*e)] = W*n/v
===========================

I look at thrust alone first = W * n / v
W = 1491400 Watts
n = 0.85 estimated

Altitude Meters___________________1,524______4,572______7,6 20______8,473
True Speed m/s___________________157.4______172.5______187.8___ ___191.8
Thrust_kg_______________________8,054______7,349__ ____6,750______6,609

e = 0.8 estimated, Ar = 5.6; (pi * AR * e) = 14.07
S = 27.87 m^2 -------- wing area
M = 5697 kg ---------- mass of the plane
g = 1 ------------------- data values are from level flight

.5 * p * v^2 * 27.87 * [Cd0 + ( [ 5697 * 9.81 ] / [ .5 * p * v^2 * 27.87 ] )^2 / 14.07 ] = Thrust

Altitude 1524m given all our assumtions from above:
.5 * 1.059 * 24775 * 27.87 * [Cd0 + ( 55888 / [ .5 * 1.059 * 24775 * 27.87 ] )^2 / 14.07 ] = 8054
365609 * [Cd0 + ( 55888 / 365609 )^2 / 14.07 ] = 8054
Cd0 + ( 55888 / 365609 )^2 / 14.07 = 8054 / 365609
Cd0 = 8054 / 365609 - ( 55888 / 365609 )^2 / 14.07
Cd0 = 0.022 - 0.153^2 / 14.07
Cd0 = 0.022 - 0.0233 / 14.07
Cd0 = 0.022 - 0.0017
Cd0 = 0.021 -------------------- at 5,000 ft with these assumptions and this method

Altitude 4572m given all our assumtions from above:
.5 * 0.774 * 24775 * 27.87 * [Cd0 + ( 55888 / [ .5 * 0.774 * 24775 * 27.87 ] )^2 / 14.07 ] = 7349
267215 * [Cd0 + ( 55888 / 267215 )^2 / 14.07 ] = 7349
Cd0 + ( 55888 / 267215 )^2 / 14.07 = 7349 / 267215
Cd0 = 7349 / 267215 - ( 55888 / 267215 )^2 / 14.07
Cd0 = 0.0275 - 0.209^2 / 14.07
Cd0 = 0.0275 - 0.0437 / 14.07
Cd0 = 0.0275 - 0.0031
Cd0 = 0.024 -------------------- at 15,000 ft with these assumptions and this method

Altitude 7620m given all our assumtions from above:
.5 * 0.551 * 24775 * 27.87 * [Cd0 + ( 55888 / [ .5 * 0.551 * 24775 * 27.87 ] )^2 / 14.07 ] = 6750
190227 * [Cd0 + ( 55888 / 190227 )^2 / 14.07 ] = 6750
Cd0 + ( 55888 / 190227 )^2 / 14.07 = 6750 / 190227
Cd0 = 6750 / 190227 - ( 55888 / 190227 )^2 / 14.07
Cd0 = 0.0355 - 0.294^2 / 14.07
Cd0 = 0.0355 - 0.0863 / 14.07
Cd0 = 0.0355 - 0.0061
Cd0 = 0.029 -------------------- at 25,000 ft with these assumptions and this method

Altitude 8473m given all our assumtions from above:
.5 * 0.499 * 24775 * 27.87 * [Cd0 + ( 55888 / [ .5 * 0.499 * 24775 * 27.87 ] )^2 / 14.07 ] = 6609
172275 * [Cd0 + ( 55888 / 172275 )^2 / 14.07 ] = 6609
Cd0 + ( 55888 / 172275 )^2 / 14.07 = 6609 / 172275
Cd0 = 6609 / 172275 - ( 55888 / 172275 )^2 / 14.07
Cd0 = 0.0384 - 0.324^2 / 14.07
Cd0 = 0.0384 - 0.105 / 14.07
Cd0 = 0.0384 - 0.0075
Cd0 = 0.031 -------------------- at 27,800 ft with these assumptions and this method

At this point I think there has already been debate about Cd0 *actually* increasing and whether or not our assumptions
concerning fixed efficiencies of wing and prop are correct or not. The equation is never the reality and sometimes
things stand out that you wonder if they are artifacts of the math and what the difference is.

Still not done, more later, please anyone check and if you find a problem say where and what.

na85
02-23-2010, 04:08 PM
I notice you kept the same true airspeed (157.4m/s) through all altitudes in your computations:

Altitude 1524m given all our assumtions from above:
.5 * 1.059 * <span class="ev_code_red">24775</span> * 27.87 * [Cd0 + ( 55888 / [ .5 * 1.059 * <span class="ev_code_red">24775</span> * 27.87 ] )^2 / 14.07 ] = 8054
365609 * [Cd0 + ( 55888 / 365609 )^2 / 14.07 ] = 8054
Cd0 + ( 55888 / 365609 )^2 / 14.07 = 8054 / 365609
Cd0 = 8054 / 365609 - ( 55888 / 365609 )^2 / 14.07
Cd0 = 0.022 - 0.153^2 / 14.07
Cd0 = 0.022 - 0.0233 / 14.07
Cd0 = 0.022 - 0.0017
Cd0 = 0.021 -------------------- at 5,000 ft with these assumptions and this method

Altitude 4572m given all our assumtions from above:
.5 * 0.774 * <span class="ev_code_red">24775</span> * 27.87 * [Cd0 + ( 55888 / [ .5 * 0.774 * <span class="ev_code_red">24775</span> * 27.87 ] )^2 / 14.07 ] = 7349
267215 * [Cd0 + ( 55888 / 267215 )^2 / 14.07 ] = 7349
Cd0 + ( 55888 / 267215 )^2 / 14.07 = 7349 / 267215
Cd0 = 7349 / 267215 - ( 55888 / 267215 )^2 / 14.07
Cd0 = 0.0275 - 0.209^2 / 14.07
Cd0 = 0.0275 - 0.0437 / 14.07
Cd0 = 0.0275 - 0.0031
Cd0 = 0.024 -------------------- at 15,000 ft with these assumptions and this method

Altitude 7620m given all our assumtions from above:
.5 * 0.551 * 24775 * 27.87 * [Cd0 + ( 55888 / [ .5 * 0.551 * 24775 * 27.87 ] )^2 / 14.07 ] = 6750
190227 * [Cd0 + ( 55888 / 190227 )^2 / 14.07 ] = 6750
Cd0 + ( 55888 / 190227 )^2 / 14.07 = 6750 / 190227
Cd0 = 6750 / 190227 - ( 55888 / 190227 )^2 / 14.07
Cd0 = 0.0355 - 0.294^2 / 14.07
Cd0 = 0.0355 - 0.0863 / 14.07
Cd0 = 0.0355 - 0.0061
Cd0 = 0.029 -------------------- at 25,000 ft with these assumptions and this method

Altitude 8473m given all our assumtions from above:
.5 * 0.499 * <span class="ev_code_red">24775</span> * 27.87 * [Cd0 + ( 55888 / [ .5 * 0.499 * <span class="ev_code_red">24775</span> * 27.87 ] )^2 / 14.07 ] = 6609
172275 * [Cd0 + ( 55888 / 172275 )^2 / 14.07 ] = 6609
Cd0 + ( 55888 / 172275 )^2 / 14.07 = 6609 / 172275
Cd0 = 6609 / 172275 - ( 55888 / 172275 )^2 / 14.07
Cd0 = 0.0384 - 0.324^2 / 14.07
Cd0 = 0.0384 - 0.105 / 14.07
Cd0 = 0.0384 - 0.0075
Cd0 = 0.031 -------------------- at 27,800 ft with these assumptions and this method

I'm wondering if it might be more instructive to compute it using the same equivalent speeds for each altitude, so as to more clearly show the relationship between alt and Cd,0.

Then again maybe TAS is what is needed, not EAS. *goes to do some math*

M_Gunz
02-23-2010, 04:43 PM
Too tired to mess with it now and EAS is a "not yet". First TAS.
I probably should have stopped at 5,000 ft until I was fresh again. I'll fix it and re-post later.
I wonder which way Cd0 will go? Maybe be the same like it should!

na85
02-23-2010, 07:52 PM
Well I wasn't suggesting using EAS yet if you haven't had time to look at it. I was thinking more along the lines of choosing speeds TAS that correspond to the same dynamic pressure Q. That's just speculation on my part as to whether it will give a nicer result or not.

Gaston444
02-23-2010, 11:33 PM
Originally posted by TS_Sancho:
<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">Originally posted by Gaston444:

P.S. People on Aces High know damn well the difference between 3-4 Gs sustained turn rates and Corner Speed, and they have the P-51D Corner Speed at 270 MPH at 7 Gs with flaps, and at about 280-290 MPH without.

Do people on Il-2 know as much about their own game?

G.

Gaston could you please provide a historical source listing VFE for any P51 as 270 mph?

Somebody mentioned earlier that Oleg Maddox has a degree in aeronautical engineering, He holds multiple flight certificates as well.

Dale Addink is a software designer and amateur flight enthusiast.

Who do you feel is the more qualified to develop an accurate computer model predicting aircraft performance? </div></BLOCKQUOTE>

-In philosophy circles that is called a "demmand for priviledge", by the way. HighTech, if that's him, has actually flown the P-51D in mock air battles, but apparently has learned little of value from them: When cornered with THIS:

http://www.spitfireperformance...hanseman-24may44.jpg (http://www.spitfireperformance.com/mustang/combat-reports/339-hanseman-24may44.jpg)

His reaction was: "It doesn't make sense." His accolyte Widewing even went as far as to say Hanseman "didn't know what he was doing, and lived"... I kid you not, and even though SUSTAINED downthrottling is pasted all over the 700+ P-51 accounts found in the same place (as are the P-51-only, but including Ds, serial G-load gun jams, BTW)...

To answer the CS issue: On direct comparisons reviews that can be found online, both "Aces High" and "Il-2" pretty much agree with each other on most things, and why shouldn't they?: They are derived from the same assumptions...

A significant difference was noted for the FW-190A's turn performance, which was better on "IL-2": An influence I guess from all these Russian combat SUMMARIES, NOT individual pilot opinions but collective SUMMARIES of experience, meaning: "Guys, what have we observed on the front, and what can we all AGREE on based on actual experience?"

And the conclusions?: FW-190A out-turns BF-109F or G, very much likes to offer low speed turning combat, doesn't like vertical maneuvers and sinks 220 M(!!!!) AFTER lifting the nose up to level from a modest dive:

http://www.ww2f.com/eastern-eu...iences-fw-190-a.html (http://www.ww2f.com/eastern-europe/21828-russian-combat-experiences-fw-190-a.html)

In agreement with Johnny Johnson's own POST-WAR assesment, AND this:

http://img105.imageshack.us/img105/3950/pag20pl.jpg

AND with ANY actual German ace that ever held the FW-190A's stick in hand, including one who answered questions directly on the "Aces High" forum through a relative, AND Rechlin itself that, according to Rall, said the same thing but that his 900 lbs lighter 109F could "in my hands", defeat it (Likely by downthrottling all the way down to 160 MPH, as explained HERE: http://www.virtualpilots.fi/fe...icles/109myths/#g6r6 (http://www.virtualpilots.fi/feature/articles/109myths/#g6r6) )

But against all that, the LOGICAL choice is to go with computer game designers, math theories, and, best of all, gamer's opinion who don't even know their own GAME well enough to answer a basic question: What is the lowest speed you can pull 7Gs on a P-51D in Il-2?...

Gaston

P.S. And since I have already established with an aircraft designer that THIS is plausible:

http://bbs.hitechcreations.com.../topic,261798.0.html (http://bbs.hitechcreations.com/smf/index.php/topic,261798.0.html)

2.5 times stall does indeed sound exactly like 263 MPH, or exactly the calculated values all games agree on at around 270 MPH, and THAT certainly doesn't sound like "very close to the maximum level speed" now does it?

Maybe that is the cause of this sudden collective "memory lapse"?

Yes it's true; asking for Corner Speed is asking for a lot...

And none of this is "evidence" anyway, right? It's all very "ambiguous". Yup.

G.

mhuxt
02-24-2010, 12:01 AM
There is no collective memory lapse.

You are delusional.

That is all.

M_Gunz
02-24-2010, 02:23 AM
Originally posted by Gaston444:
2.5 times stall does indeed sound exactly like 263 MPH, or exactly the calculated values all games agree on at around 270 MPH, and THAT certainly doesn't sound like "very close to the maximum level speed" now does it?

IAS at not terribly high alt... how fast you think the P-51 is and how close to top speed is "close"?

What it doesn't sound like is max TAS at optimum alt. Easy to mistake the two when you have an agenda in mind.

TheGrunch
02-24-2010, 02:39 AM
Restored WWII warbirds are also flown well below their maximum rated boost pressures.

Kettenhunde
02-24-2010, 04:05 AM
"It doesn't make sense."

It makes perfect sense. If the airplane is above Va or its best turn velocity, then you want to slow down.

Cutting the throttle and dropping flaps is necessary to do so especially in a slick airplane like the P51 series.

Kettenhunde
02-24-2010, 04:12 AM
Maybe be the same like it should!

It definitely goes down with altitude in the P-47 series, it is relatively speed stable.

Xiolablu3
02-24-2010, 06:30 AM
Originally posted by Kettenhunde:

Yeah I couldn't help but poke a little fun at Hop.

http://forums.ubi.com/images/smilies/16x16_smiley-tongue.gif </div></BLOCKQUOTE>

hmmm, a bit late in the day to say you were 'joking', me thinks.

In which other parts were you 'joking'? Just so we can be sure.

AndyJWest
02-24-2010, 08:14 AM
-In philosophy circles that is called a "demmand for priviledge"

This isn't a philosophical debate, it is a historical/technical one, and as such should be based on verifiable evidence.

M_Gunz
02-24-2010, 09:33 AM
Originally posted by Kettenhunde:
<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">"It doesn't make sense."

It makes perfect sense. If the airplane is above Va or its best turn velocity, then you want to slow down.

Cutting the throttle and dropping flaps is necessary to do so especially in a slick airplane like the P51 series. </div></BLOCKQUOTE>

A couple seconds of beyond corner speed pulling 6 G's and you'll have slowed down easily as fast.
The idea is enter the fight at that speed at the highest in the first place.

Kettenhunde
02-24-2010, 10:14 AM
hmmm, a bit late in the day to say you were 'joking', me thinks.

In which other parts were you 'joking'? Just so we can be sure.

Yeah I was joking with hop. It does not change the performance analysis, Xio.

What is your issue with me?

How about if you don't like what I say, you ignore it or at least be an adult about it if you are going to place yourself above the average board member and moderate.

The discussion is far from finished on the aerodynamics.

M_Gunz
02-24-2010, 04:58 PM
Further adventures in correcting stupid mistakes later....

Turn performance estimation -- working out details using data from the same plane at different altitudes.
With many thanks to community members for help in not only making this but in correcting errors on the way.

WAR DEPARTMENT
AIR CORP, MATERIAL DIVISION
Wright Field, Dayton, Ohio
June 18, 1942

P-47B Airplane, A.C. No. 41-5902
Acceptance Performance Tests

High speed in level flight with oil cooler flaps and intercooler flaps flush and throttle wide open with turbo on to give military rated power or 18,250 limiting turbo r.p.m.

Altitude Feet___________________ 5,000_____15,000_____25,000_____27,800
True Speed m.p.h._________________352________386_______420___ _____429

At these altitudes the B.H.P. is 2000HP, engine P.R.M. is 2700 and MP is 51" to 52" Hg.

I will use some of the values from Wurkeri's post (Thanks W!) since this is more about method than hair-counting.

================================================== =
W = 2000 hp = 1491400 W -------- it's from the same data, same plane AFAIK
M = 12560 lbs = 5697 kg ----------- actually 12565 lbs but I'm sure less before actual takeoff. ;^)
Sp = 41 ft = 12.5 m------------------- wingspan
S = 300 sqft = 27.87 m^2------------ wing area
AR = 5.6 ------------------------------- aspect ratio being wingspan^2 / wing area
n = 0.85 estimated
e = 0.8 estimated
Vs = 105 mph = 46,93 m/s---------- clean 1G stall IAS
================================================== =

1 foot = 0.3048 meters
1 mile = 1.609344 kilometers

Altitude Feet____________________ 5,000______15,000_____25,000_____27,800
True Speed m.p.h._________________352________386________420__ ______429
Altitude Meters___________________1,524______4,572______7,6 20______8,473
True Speed m/s____________________157.4______172.5______187.8__ ____191.8
Air density kg/m^3________________1.059______0.774______0.551____ __0.499 -- values interpolated from table below

--------------------------------------------------------------------------------------------------
air density taken from "A Sample Atmosphere Table (SI units)" at http://www.pdas.com/m1.htm
km alt____ kg/cu.meter
0________1.225
2________1.007
4________0.8193
6________0.6601
8________0.5258
10_______0.4135
---------------------------------------------------------------------------------------------------

Symbols used -- thanks Wurkeri!
===========================
W = engine power
S = wing area
Sp = wing span
AR = Aspect ratio, calculated from span and area
p = density at this given altitude
M = mass of the plane
n = propeller efficiency
e = efficiency factor
Vs = stall speed
g = load factor, 1 in level flight, limited to certain value.
===========================

Also from Wurkeri:
===========================
whole balance formula ------- without checking units and cancels, where's my cracker?

0.5*p*v^2*S*[Cd0+([M*g*9.81]/[0.5*p*v^2*S])^2/(pi*AR*e)] = W*n/v
===========================

I look at thrust alone first = W * n / v
W = 1491400 Watts
n = 0.85 estimated

Altitude Meters___________________1,524______4,572______7,6 20______8,473
True Speed m/s____________________157.4______172.5______187.8__ ____191.8
Thrust_N__________________________8,054______7,349 ______6,750______6,609

e = 0.8 estimated, Ar = 5.6; (pi * AR * e) = 14.07
S = 27.87 m^2 -------- wing area
M = 5697 kg ---------- mass of the plane
g = 1 ------------------- data values are from level flight

.5 * p * v^2 * 27.87 * [Cd0 + ( [ 5697 * 9.81 ] / [ .5 * p * v^2 * 27.87 ] )^2 / 14.07 ] = Thrust

Altitude 1524m given all our assumtions from above:
.5 * 1.059 * 24775 * 27.87 * [Cd0 + ( 55888 / [ .5 * 1.059 * 24775 * 27.87 ] )^2 / 14.07 ] = 8054
365609 * [Cd0 + ( 55888 / 365609 )^2 / 14.07 ] = 8054
Cd0 + ( 55888 / 365609 )^2 / 14.07 = 8054 / 365609
Cd0 = 8054 / 365609 - ( 55888 / 365609 )^2 / 14.07
Cd0 = 0.022 - 0.153^2 / 14.07
Cd0 = 0.022 - 0.0233 / 14.07
Cd0 = 0.022 - 0.0017
Cd0 = 0.021 -------------------- at 5,000 ft with these assumptions and this method

Altitude 4572m given all our assumtions from above:
.5 * 0.774 * 29756 * 27.87 * [Cd0 + ( 55888 / [ .5 * 0.774 * 29756 * 27.87 ] )^2 / 14.07 ] = 7349
320939 * [Cd0 + ( 55888 / 320939 )^2 / 14.07 ] = 7349
Cd0 + ( 55888 / 320939 )^2 / 14.07 = 7349 / 320939
Cd0 = 7349 / 320939 - ( 55888 / 320939 )^2 / 14.07
Cd0 = 0.023 - 0.174^2 / 14.07
Cd0 = 0.023 - 0.0303 / 14.07
Cd0 = 0.023 - 0.0022
Cd0 = 0.021 -------------------- at 15,000 ft with these assumptions and this method

Altitude 7620m given all our assumtions from above:
.5 * 0.551 * 35269 * 27.87 * [Cd0 + ( 55888 / [ .5 * 0.551 * 35269 * 27.87 ] )^2 / 14.07 ] = 6750
270802 * [Cd0 + ( 55888 / 270802 )^2 / 14.07 ] = 6750
Cd0 + ( 55888 / 270802 )^2 / 14.07 = 6750 / 270802
Cd0 = 6750 / 270802 - ( 55888 / 270802 )^2 / 14.07
Cd0 = 0.025- 0.206^2 / 14.07 ------ the 0.206 is rounded, when I square the unrounded I get .0426
Cd0 = 0.025- 0.0426 / 14.07 ------- when I square the rounded I get .0424
Cd0 = 0.025- 0.003 ------------------ but either one results in about .00301 anyway
Cd0 = 0.022 -------------------- at 25,000 ft with these assumptions and this method

Altitude 8473m given all our assumtions from above:
.5 * 0.499 * 36787 * 27.87 * [Cd0 + ( 55888 / [ .5 * 0.499 * 36787 * 27.87 ] )^2 / 14.07 ] = 6609
255801 * [Cd0 + ( 55888 / 255801 )^2 / 14.07 ] = 6609
Cd0 + ( 55888 / 255801 )^2 / 14.07 = 6609 / 255801
Cd0 = 6609 / 255801 - ( 55888 / 255801 )^2 / 14.07
Cd0 = 0.026 - 0.2185^2 / 14.07
Cd0 = 0.026 - 0.0477 / 14.07
Cd0 = 0.026 - 0.0034
Cd0 = 0.0226 -------------------- at 27,800 ft with these assumptions and this method

At this point I think there has already been debate about Cd0 *actually* increasing and whether or not our assumptions
concerning fixed efficiencies of wing and prop are correct or not. With this method and these assumptions Cd0 is not
increasing enough to bother me at all.

The equation is never the reality. I only investigate how the equation works with some fixed assumed (and reasonable)
values along with historic data.

Still not done, more later, please anyone check and if you find a problem say where and what.

Erkki_M
02-25-2010, 08:08 AM
Gaston:

http://images.uncyc.org/commons/thumb/d/db/Mtg_hitler.png/200px-Mtg_hitler.png

http://forums.ubi.com/images/smilies/16x16_smiley-very-happy.gif

M_Gunz
02-25-2010, 03:56 PM
Turn performance estimation -- working out details using data from the same plane at different altitudes.
With many thanks to community members for help in not only making this but in correcting errors on the way.

WAR DEPARTMENT
AIR CORP, MATERIAL DIVISION
Wright Field, Dayton, Ohio
June 18, 1942

P-47B Airplane, A.C. No. 41-5902
Acceptance Performance Tests

High speed in level flight with oil cooler flaps and intercooler flaps flush and throttle wide open with turbo on to give military rated power or 18,250 limiting turbo r.p.m.

Altitude Feet___________________ 5,000_____15,000_____25,000_____27,800
True Speed m.p.h._________________352________386_______420___ _____429

At these altitudes the B.H.P. is 2000HP, engine P.R.M. is 2700 and MP is 51" to 52" Hg.

I will use some of the values from Wurkeri's post (Thanks W!) since this is more about method than hair-counting.

================================================== =
W = 2000 hp = 1491400 W -------- it's from the same data, same plane AFAIK
M = 12560 lbs = 5697 kg ----------- actually 12565 lbs but I'm sure less before actual takeoff. ;^)
Sp = 41 ft = 12.5 m------------------- wingspan
S = 300 sqft = 27.87 m^2------------ wing area
AR = 5.6 ------------------------------- aspect ratio being wingspan^2 / wing area
n = 0.85 estimated
e = 0.8 estimated
Vs = 105 mph = 46,93 m/s---------- clean 1G stall IAS
================================================== =

1 foot = 0.3048 meters
1 mile = 1.609344 kilometers

Altitude Feet____________________ 5,000______15,000_____25,000_____27,800
True Speed m.p.h._________________352________386________420__ ______429
Altitude Meters___________________1,524______4,572______7,6 20______8,473
True Speed m/s____________________157.4______172.5______187.8__ ____191.8
Air density kg/m^3________________1.059______0.774______0.551____ __0.499 -- values interpolated from table below

--------------------------------------------------------------------------------------------------
air density taken from "A Sample Atmosphere Table (SI units)" at http://www.pdas.com/m1.htm
km alt____ kg/cu.meter
0________1.225
2________1.007
4________0.8193
6________0.6601
8________0.5258
10_______0.4135
---------------------------------------------------------------------------------------------------

Symbols used -- thanks Wurkeri!
===========================
W = engine power
S = wing area
Sp = wing span
AR = Aspect ratio, calculated from span and area
p = density at this given altitude
M = mass of the plane
n = propeller efficiency
e = efficiency factor
Vs = stall speed
g = load factor, 1 in level flight, limited to certain value.
===========================

Also from Wurkeri:
===========================
whole balance formula ------- without checking units and cancels, where's my cracker?

0.5*p*v^2*S*[Cd0+([M*g*9.81]/[0.5*p*v^2*S])^2/(pi*AR*e)] = W*n/v
===========================

I look at thrust alone first = W * n / v
W = 1491400 Watts
n = 0.85 estimated

Altitude Meters___________________1,524______4,572______7,6 20______8,473
True Speed m/s____________________157.4______172.5______187.8__ ____191.8
Thrust_N__________________________8,054______7,349 ______6,750______6,609

e = 0.8 estimated, Ar = 5.6; (pi * AR * e) = 14.07
S = 27.87 m^2 -------- wing area
M = 5697 kg ---------- mass of the plane
g = 1 ------------------- data values are from level flight

.5 * p * v^2 * 27.87 * [Cd0 + ( [ 5697 * 9.81 ] / [ .5 * p * v^2 * 27.87 ] )^2 / 14.07 ] = Thrust

Altitude 1524m given all our assumtions from above:
.5 * 1.059 * 24775 * 27.87 * [Cd0 + ( 55888 / [ .5 * 1.059 * 24775 * 27.87 ] )^2 / 14.07 ] = 8054
365609 * [Cd0 + ( 55888 / 365609 )^2 / 14.07 ] = 8054
Cd0 + ( 55888 / 365609 )^2 / 14.07 = 8054 / 365609
Cd0 = 8054 / 365609 - ( 55888 / 365609 )^2 / 14.07
Cd0 = 0.022 - 0.153^2 / 14.07
Cd0 = 0.022 - 0.0233 / 14.07
Cd0 = 0.022 - 0.0017
Cd0 = 0.021 -------------------- at 5,000 ft with these assumptions and this method

Altitude 4572m given all our assumtions from above:
.5 * 0.774 * 29756 * 27.87 * [Cd0 + ( 55888 / [ .5 * 0.774 * 29756 * 27.87 ] )^2 / 14.07 ] = 7349
320939 * [Cd0 + ( 55888 / 320939 )^2 / 14.07 ] = 7349
Cd0 + ( 55888 / 320939 )^2 / 14.07 = 7349 / 320939
Cd0 = 7349 / 320939 - ( 55888 / 320939 )^2 / 14.07
Cd0 = 0.023 - 0.174^2 / 14.07
Cd0 = 0.023 - 0.0303 / 14.07
Cd0 = 0.023 - 0.0022
Cd0 = 0.021 -------------------- at 15,000 ft with these assumptions and this method

Altitude 7620m given all our assumtions from above:
.5 * 0.551 * 35269 * 27.87 * [Cd0 + ( 55888 / [ .5 * 0.551 * 35269 * 27.87 ] )^2 / 14.07 ] = 6750
270802 * [Cd0 + ( 55888 / 270802 )^2 / 14.07 ] = 6750
Cd0 + ( 55888 / 270802 )^2 / 14.07 = 6750 / 270802
Cd0 = 6750 / 270802 - ( 55888 / 270802 )^2 / 14.07
Cd0 = 0.025- 0.206^2 / 14.07 ------ the 0.206 is rounded, when I square the unrounded I get .0426
Cd0 = 0.025- 0.0426 / 14.07 ------- when I square the rounded I get .0424
Cd0 = 0.025- 0.003 ------------------ but either one results in about .00301 anyway
Cd0 = 0.022 -------------------- at 25,000 ft with these assumptions and this method

Altitude 8473m given all our assumtions from above:
.5 * 0.499 * 36787 * 27.87 * [Cd0 + ( 55888 / [ .5 * 0.499 * 36787 * 27.87 ] )^2 / 14.07 ] = 6609
255801 * [Cd0 + ( 55888 / 255801 )^2 / 14.07 ] = 6609
Cd0 + ( 55888 / 255801 )^2 / 14.07 = 6609 / 255801
Cd0 = 6609 / 255801 - ( 55888 / 255801 )^2 / 14.07
Cd0 = 0.026 - 0.2185^2 / 14.07
Cd0 = 0.026 - 0.0477 / 14.07
Cd0 = 0.026 - 0.0034
Cd0 = 0.0226 -------------------- at 27,800 ft with these assumptions and this method

At this point I think there has already been debate about Cd0 *actually* increasing and whether or not our assumptions
concerning fixed efficiencies of wing and prop are correct or not. With this method and these assumptions Cd0 is not
increasing enough to bother me at all.

The equation is never the reality. I only investigate how the equation works with some fixed assumed (and reasonable)
values along with historic data.

Now I turn to working this out for turns at speed without resorting to anything but TAS, probably get this wrong the first
few times as well:

Given:
0.5*p*v^2*S*[Cd0+([M*g*9.81]/[0.5*p*v^2*S])^2/(pi*AR*e)] = W*n/v

Given all our assumtions from above and new speeds, the TAS of the IAS at altitude for about a 3G turn.
Vs = 105 mph = 46.93 m/s---------- x 1.732 = 81.29 m/s IAS

And really it's not IAS but CAS except on Tuesdays between 7AM and 9AM when they're cleaning this side of the street.
I'm not going through OAT and all that since AFAIK the historic data was corrected for Standard Atmosphere in 1942
which leaves room for correction right there that I'm not going into anyway. I'm using a table and saying close enough.

From that IAS/TAS table I get factors at altitudes to interpolate:
Altitude Meters ____________1,500____2,000____4,000____4,500____7, 500____8,000____8,500
IAS x value = TAS__________1.099____1.131____1.263____1.295____1 .493____1.525____1.558

Altitude Meters___________________1,524______4,572______7,6 20______8,473
TAS/IAS factor___________________1.101______1.268______1.5 01______1.556
Turn TAS m/s 81.29 x factor above___89.50_____103.08_____122.02______126.49
Thrust_N________________________14164_____12212___ ___10389______10022

Thrust as W * n / v = 149400 *.85 / v
These are only the TAS that is sqrt(3) x Vs, it is not saying a 3G turn must result. It's just nice to have something to
compare with later.

.5 * 1.059 * 8010 * 27.87 * [ 0.021 + ( [ 55888 * g-load ] / [ .5 * 1.059 * 8010 * 27.87 ] )^2 / 14.07 ] = 14164
118205 * [ 0.021 + ( [ 55888 * g-load ] / 118205 )^2 / 14.07 ] = 14164
0.021 + ( [ 55888 * g-load ] / 118205 )^2 / 14.07 = 14164 / 118205 = 0.1198
( [ 55888 * g-load ] / 118205 )^2 = ( 0.1198 - 0.021 ) * 14.07 = 1.39
[ 55888 * g-load ] / 118205 = sqrt( 1.39 ) = 1.179
g-load = 1.179 * 118205 / 55888 = 2.49 G's -------------------- at 5,000 ft **with these assumptions and this method**.

Okay, let's see what I did wrong this time as far as the method.

Still not done, more later, please anyone check and if you find a problem say where and what.

Kettenhunde
02-25-2010, 08:46 PM
With this method and these assumptions Cd0 is not increasing enough to bother me at all.

Rounding errors...If so it is not an issue.

Gaston444
02-26-2010, 12:46 AM
Originally posted by Kettenhunde:
<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">"It doesn't make sense."

It makes perfect sense. If the airplane is above Va or its best turn velocity, then you want to slow down.

Cutting the throttle and dropping flaps is necessary to do so especially in a slick airplane like the P51 series. </div></BLOCKQUOTE>

-That's what I was, desperately, clinging to when I first read these "types" of account (fortunately, my game design is cheaper to change than Oleg"s!)... Notice I say these "types" because I have encountered these downthrottling-type of WWII accounts for decades on end, also angrily ignoring them, before this one finally opened my eyes...

Please examine your hypothesis and this text, Kettenhunde, and realize how every single aspect of the text contradicts any notion of any kind of significant above-Corner Speed velocity:

http://www.spitfireperformance...hanseman-24may44.jpg (http://www.spitfireperformance.com/mustang/combat-reports/339-hanseman-24may44.jpg)

The very first part of the action is shooting down a Me-109 that was coming down for LANDING: He closes, knowing he was not seen, firing from 300 to 50 yards at 0 degrees at 150 ft. of altitude, AND, a key detail, has time to see it crash BEFORE he pulls up...

I would say a speed of 400 MPH to gain and shoot at a landing aircraft, which has not seen you, is unlikely...

He pulls up 350 ft. and a turning dogfight immediately develops at 500 ft. ("WAS maneuvering to get on my tail"). I would say it is highly unlikely the speed is above 400 MPH then... (To put it mildly)

The Me-109G-6 (May 1944) starts off well by gaining in the turn. "At FIRST, he began to turn inside me". I hope anyone here agrees that the Merlin P-51 would NOT have trouble at least turning with a Me-109G-6 above 350 MPH, or even above 300 MPH...

Then the P-51 pilot makes a change: "THEN he stopped cutting me off as I cut throttle, dropped 20 degrees of flaps and increased prop pitch."

So now, even by being overly generous, speed is here down to 300 MPH IAS. Note this is already hopelessly below the "very close to maximum level speed" Corner Speed oulined by professional test pilots in 1989:

http://bbs.hitechcreations.com.../topic,261798.0.html (http://bbs.hitechcreations.com/smf/index.php/topic,261798.0.html)

But it is far from over yet, as AFTER all this is the fun part: "EVERY TIME I got close to the edge of the airdrome, they opened fire with light AA guns"

EVERY TIME: Meaning: Every time a 360 degree circle was made, they would take the shot: If that means at least two 360 degrees turns (no slackening of the circle because of the tailing 109), AND BOTH these tight 360s WERE MADE WITH POWER DOWN, FLAPS DOWN, and with no significant diving...

So what follows this now "at or less than 270 MPH Corner Speed" turning? : "GRADUALLY I "WORKED" the 109 away from the field"

Meaning he did NOT break out of the turn and lead it away, which would have been instantaneous: He instead laboriously "worked" the circles further and further away from the airdrome that kept shooting at him...

So now we are definitely at 250 MPH or less, as the number of tense 360 degree dogfight circles is certainly more than two for it to be "work"... Likely 5 or 6 in fact...

So what follows this now at best 200-250 MPH speed?: " and I COMMENCED to turn inside of him as I decreased throttle setting."

So downthrottling ONCE... AFTER shooting an unaware landing aircraft> AFTER climbing 350 ft.> and WHILE doing at least two 360s at 500 ft.> AND "working him away" with a TAILING fighter was not enough: He realized downthrottling ONCE was not efficient enough. His power and speed had to go down further, AWAY from the 350+ MPH Corner Speed determined in 1989... And this because of the off-center airframe drag center leverage to prop thrust center leverage the elevator depressing the tail creates.

While I argued previously for a RISE in the prop thrust center, perhaps saying that the elevator depressing the tail LOWERS the airframe drag center in relation to the prop thrust center will be more convincing? I would say the total leverage against turning could be a combination of both...

It amounts to the same thing anyway...

His first downthrottling had achieved this: The tailing, and GAINING, enemy "stopped cutting me off".

Then, after several 360s, obligingly punctuated by AA fire, further downthrottling obtained "I COMMENCED to turn inside him"

The full sentence: "Gradually I worked the Me-109 away from the field AND COMMENCED turning inside him as I decreased throttle setting" makes it pretty clear the causality link between decreasing the throttle, after several 360 on the deck, and the gaining on a previously TAILING enemy...

And if he used the pathetic 500 ft. altitude for spiraling down, then his apparent fixation on downthrottling seems counter-productive. He mentions no diving at all...

If you see any evidence here of someone desperately trying to slow down to the actual 350 MPH+ Corner Speed, or the even the fictional 270 MPH "Corner Speed", enlighten us...

Gaston

P.S. What HighTech specifically said did not make sense was the combination of downthrottling with the coarsening of the prop pitch. It makes sense to me, IF you are churning REAL air, but I am sure math formulas will prove the athmosphere wrong...

G.

M_Gunz
02-26-2010, 04:34 AM
Not the same thing as best sustained turn. Especially with the added fictions.

Kettenhunde
02-26-2010, 03:25 PM
Are we done yet? M_Gunz are you comfortable with has transpired?

M_Gunz
02-26-2010, 09:23 PM
Been waiting on discussion and error reports. Hard to believe there's no math errors there.
I should finish and go on to a spreadsheet. If we keep power through the next few days there
should be time during the son of mighty blizzard of 2010 supposed to happen now.

Wurkeri
02-26-2010, 11:00 PM
Gunz,
I suggest a bit different approach, just calculate the Clmax from 1g stall speed at sea level and use it to calculate the g load up to the speeds where the thrust (or g) limit is reached. You can improve the stall speed estimate if position error and other corrections are available. Above the thrust limit speed, use the original formula. This way you can use TAS for the entire calculation, the speed conversions just confuse things as seen earlier.

Kettenhunde
02-27-2010, 05:15 AM
I haven't checked your math but the results are looking better.

Cdo should stay the same in the technique used. If we have measured results for our assumptions then you will see it be reduced as the derived relationships from dynamic pressure show us.

There is a couple of ways to do this stuff and contrary to the baloney and blustering, you cannot learn this in an hour.

You might can learn one technique in an hour with help depending on your level. You might be sold a bill of goods in an hour that the technique has the absolute definitive answers to the questions asked.

Facts are there are multiple techniques returning different results that are acceptable as long as you follow basic math rules, don't violate basic aerodynamic principles, and the adhere to any instructions specific to the formula.

The relative performance lineup should be the same although it is possible to shift errors in the systems to either axis such as thrust or speed.

If you use EAS your relative performance lineup is exactly the same as using TAS in the BGS system I am using. If you use TAS correctly with all of the measured data; it will return the exact same results.

Kettenhunde
02-27-2010, 07:20 AM
I suggest a bit different approach, just calculate the Clmax from 1g stall speed at sea level and use it to calculate the g load up to the speeds where the thrust (or g) limit is reached.

That won't tell you much about thrust limited turn performance.

M_Gunz
02-27-2010, 09:45 AM
If it's not a waste of time I should finish the other three turns at the IAS(s) I planned just to
show any trend. Meantime I goofed with some values and purely by math I could come to some wrong
conclusions, I have the thrust to make a really hard turn at Vs for example if I forget how the
plane actually turns.

If the whole thing isn't dead (waste of time) then I agree with others that a spreadsheet is an
obvious way to go. C code doesn't share as widely. I'd just adapt with JtD's but he mixes EAS in
there, or maybe that's the same as where I used IAS to get same-dynamic-pressure turn speeds?
I'd have to change his sheet to do some of what I want as well, and learn all that's in it and
how it works.

It also seems like plotting curves and finding intersections might give a whole range of answers
much more quickly and compactly as per real examples. Lift limit would require finding Cl as well
I suppose.

I picked up a nasty head cold earlier this week and feeling pretty miserable anyway, slower than
usual.

Gaston444
02-27-2010, 12:24 PM
Originally posted by M_Gunz:
Not the same thing as best sustained turn. Especially with the added fictions.

Isn't it part of YOUR fiction that the best sustained turn rate speed should be kept as close as possible to the "Corner Speed", hence the fictional need to keep power HIGH when your speed drops below it? (Which without diving would be pretty fast if it was at 270 MPH, and is in fact near instantaneous for the 350+ reality)

Also I have asked here about ten times now what is the Corner Speed at 7G of Il-2's P-51D, and no one here can even give me a straight answer?

Are you guys for real?

Gaston

na85
02-27-2010, 01:16 PM
Originally posted by Gaston444:

Also I have asked here about ten times now what is the Corner Speed at 7G of Il-2's P-51D, and no one here can even give me a straight answer?

Are you guys for real?

You've been asked about 10 times by JtD to provide power on stall speeds for the Me-109G and FW190A. You've also been asked more than 10 times to provide any sort of mathematical proof of your assertions. Yet you can't give anyone a straight answer?

Are you for real?

M_Gunz
02-27-2010, 01:16 PM
Find me a WWII fighter that could sustain corner speed before asking a question like that.

Lift at any AOA goes by speed squared. It is lift that turns the plane. Lift makes drag.
Without thrust to sustain the speed the lift decreases, the turn becomes less.

Unless you have a weird idea that with the same plane a 2G turn will be quicker than a 3G
turn by some chance? Don't bother responding with some apples to oranges anecdotes, partial
in a fight... no you don't have the full details and you're full of it filling them in as
you do.

Kettenhunde
02-27-2010, 01:59 PM
If it's not a waste of time I should finish the other three turns at the IAS(s) I planned just to show any trend.

It is not a waste of time at all.

If the whole thing isn't dead (waste of time) then I agree with others that a spreadsheet is an obvious way to go. C code doesn't share as widely. I'd just adapt with JtD's but he mixes EAS in there, or maybe that's the same as where I used IAS to get same-dynamic-pressure turn speeds?
I'd have to change his sheet to do some of what I want as well, and learn all that's in it and
how it works.

A spreadsheet is the way to go. I built a couple for analyzing performance.

I haven't played with JtD's spreadsheet enough but using consistent units is a step in the right direction. That way you are correctly accounting for density effects.

Gaston444
02-27-2010, 02:50 PM
Originally posted by M_Gunz:
Find me a WWII fighter that could sustain corner speed before asking a question like that.

Lift at any AOA goes by speed squared. It is lift that turns the plane. Lift makes drag.
Without thrust to sustain the speed the lift decreases, the turn becomes less.

Unless you have a weird idea that with the same plane a 2G turn will be quicker than a 3G
turn by some chance? Don't bother responding with some apples to oranges anecdotes, partial
in a fight... no you don't have the full details and you're full of it filling them in as
you do.

-Corner Speed BY DEFINITION IS NOT SUSTAINED.

What is your argument against my question about the P-51D's Corner Speed in Il-2?

Gaston

M_Gunz
02-27-2010, 07:28 PM
Originally posted by Gaston444:
<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">Originally posted by M_Gunz:
Find me a WWII fighter that could sustain corner speed before asking a question like that.

Lift at any AOA goes by speed squared. It is lift that turns the plane. Lift makes drag.
Without thrust to sustain the speed the lift decreases, the turn becomes less.

Unless you have a weird idea that with the same plane a 2G turn will be quicker than a 3G
turn by some chance? Don't bother responding with some apples to oranges anecdotes, partial
in a fight... no you don't have the full details and you're full of it filling them in as
you do.

-Corner Speed BY DEFINITION IS NOT SUSTAINED.

What is your argument against my question about the P-51D's Corner Speed in Il-2?

Gaston </div></BLOCKQUOTE>

Gaston444

Posted Sat February 27 2010 14:24 Hide Post

quote:
Originally posted by M_Gunz:
Not the same thing as best sustained turn. Especially with the added fictions.

Isn't it part of YOUR fiction that the best sustained turn rate speed should be kept as close as possible to the "Corner Speed", hence the fictional need to keep power HIGH when your speed drops below it? (Which without diving would be pretty fast if it was at 270 MPH, and is in fact near instantaneous for the 350+ reality)

Also I have asked here about ten times now what is the Corner Speed at 7G of Il-2's P-51D, and no one here can even give me a straight answer?

Are you guys for real?

Gaston

Isn't it part of YOUR fiction that the best sustained turn rate speed should be kept as close as possible to the "Corner Speed",

I never made such a statement.

YOU seem to have a problem knowing the difference between POWER and SPEED. Or the meaning of much anything that gets
in your way. And yet you have no trouble making up BS and putting other people's names on it, again and again.

I post this for others to read as exhibition of your BS. I have no hope that you will change at all.
Don't bother trolling me any more.

M_Gunz
02-27-2010, 08:29 PM
Turn performance estimation -- working out details using data from the same plane at different altitudes.
With many thanks to community members for help in not only making this but in correcting errors on the way.

WAR DEPARTMENT
AIR CORP, MATERIAL DIVISION
Wright Field, Dayton, Ohio
June 18, 1942

P-47B Airplane, A.C. No. 41-5902
Acceptance Performance Tests

High speed in level flight with oil cooler flaps and intercooler flaps flush and throttle wide open with turbo on to give military rated power or 18,250 limiting turbo r.p.m.

Altitude Feet___________________ 5,000_____15,000_____25,000_____27,800
True Speed m.p.h._________________352________386_______420___ _____429

At these altitudes the B.H.P. is 2000HP, engine P.R.M. is 2700 and MP is 51" to 52" Hg.

I will use some of the values from Wurkeri's post (Thanks W!) since this is more about method than hair-counting.

================================================== =
W = 2000 hp = 1491400 W -------- it's from the same data, same plane AFAIK
M = 12560 lbs = 5697 kg ----------- actually 12565 lbs but I'm sure less before actual takeoff. ;^)
Sp = 41 ft = 12.5 m------------------- wingspan
S = 300 sqft = 27.87 m^2------------ wing area
AR = 5.6 ------------------------------- aspect ratio being wingspan^2 / wing area
n = 0.85 estimated
e = 0.8 estimated
Vs = 105 mph = 46,93 m/s---------- clean 1G stall IAS
================================================== =

1 foot = 0.3048 meters
1 mile = 1.609344 kilometers

Altitude Feet____________________ 5,000______15,000_____25,000_____27,800
True Speed m.p.h._________________352________386________420__ ______429
Altitude Meters___________________1,524______4,572______7,6 20______8,473
True Speed m/s____________________157.4______172.5______187.8__ ____191.8
Air density kg/m^3________________1.059______0.774______0.551____ __0.499 -- values interpolated from table below

--------------------------------------------------------------------------------------------------
air density taken from "A Sample Atmosphere Table (SI units)" at http://www.pdas.com/m1.htm
km alt____ kg/cu.meter
0________1.225
2________1.007
4________0.8193
6________0.6601
8________0.5258
10_______0.4135
---------------------------------------------------------------------------------------------------

Symbols used -- thanks Wurkeri!
===========================
W = engine power
S = wing area
Sp = wing span
AR = Aspect ratio, calculated from span and area
p = density at this given altitude
M = mass of the plane
n = propeller efficiency
e = efficiency factor
Vs = stall speed
g = load factor, 1 in level flight, limited to certain value.
===========================

Also from Wurkeri:
===========================
whole balance formula ------- without checking units and cancels, where's my cracker?

0.5*p*v^2*S*[Cd0+([M*g*9.81]/[0.5*p*v^2*S])^2/(pi*AR*e)] = W*n/v
===========================

I look at thrust alone first = W * n / v
W = 1491400 Watts
n = 0.85 estimated

Altitude Meters___________________1,524______4,572______7,6 20______8,473
True Speed m/s____________________157.4______172.5______187.8__ ____191.8
Thrust_N__________________________8,054______7,349 ______6,750______6,609

e = 0.8 estimated, Ar = 5.6; (pi * AR * e) = 14.07
S = 27.87 m^2 -------- wing area
M = 5697 kg ---------- mass of the plane
g = 1 ------------------- data values are from level flight

.5 * p * v^2 * 27.87 * [Cd0 + ( [ 5697 * 9.81 ] / [ .5 * p * v^2 * 27.87 ] )^2 / 14.07 ] = Thrust

Altitude 1524m given all our assumtions from above:
.5 * 1.059 * 24775 * 27.87 * [Cd0 + ( 55888 / [ .5 * 1.059 * 24775 * 27.87 ] )^2 / 14.07 ] = 8054
365609 * [Cd0 + ( 55888 / 365609 )^2 / 14.07 ] = 8054
Cd0 + ( 55888 / 365609 )^2 / 14.07 = 8054 / 365609
Cd0 = 8054 / 365609 - ( 55888 / 365609 )^2 / 14.07
Cd0 = 0.022 - 0.153^2 / 14.07
Cd0 = 0.022 - 0.0233 / 14.07
Cd0 = 0.022 - 0.0017
Cd0 = 0.021 -------------------- at 5,000 ft with these assumptions and this method

Altitude 4572m given all our assumtions from above:
.5 * 0.774 * 29756 * 27.87 * [Cd0 + ( 55888 / [ .5 * 0.774 * 29756 * 27.87 ] )^2 / 14.07 ] = 7349
320939 * [Cd0 + ( 55888 / 320939 )^2 / 14.07 ] = 7349
Cd0 + ( 55888 / 320939 )^2 / 14.07 = 7349 / 320939
Cd0 = 7349 / 320939 - ( 55888 / 320939 )^2 / 14.07
Cd0 = 0.023 - 0.174^2 / 14.07
Cd0 = 0.023 - 0.0303 / 14.07
Cd0 = 0.023 - 0.0022
Cd0 = 0.021 -------------------- at 15,000 ft with these assumptions and this method

Altitude 7620m given all our assumtions from above:
.5 * 0.551 * 35269 * 27.87 * [Cd0 + ( 55888 / [ .5 * 0.551 * 35269 * 27.87 ] )^2 / 14.07 ] = 6750
270802 * [Cd0 + ( 55888 / 270802 )^2 / 14.07 ] = 6750
Cd0 + ( 55888 / 270802 )^2 / 14.07 = 6750 / 270802
Cd0 = 6750 / 270802 - ( 55888 / 270802 )^2 / 14.07
Cd0 = 0.025- 0.206^2 / 14.07 ------ the 0.206 is rounded, when I square the unrounded I get .0426
Cd0 = 0.025- 0.0426 / 14.07 ------- when I square the rounded I get .0424
Cd0 = 0.025- 0.003 ------------------ but either one results in about .00301 anyway
Cd0 = 0.022 -------------------- at 25,000 ft with these assumptions and this method

Altitude 8473m given all our assumtions from above:
.5 * 0.499 * 36787 * 27.87 * [Cd0 + ( 55888 / [ .5 * 0.499 * 36787 * 27.87 ] )^2 / 14.07 ] = 6609
255801 * [Cd0 + ( 55888 / 255801 )^2 / 14.07 ] = 6609
Cd0 + ( 55888 / 255801 )^2 / 14.07 = 6609 / 255801
Cd0 = 6609 / 255801 - ( 55888 / 255801 )^2 / 14.07
Cd0 = 0.026 - 0.2185^2 / 14.07
Cd0 = 0.026 - 0.0477 / 14.07
Cd0 = 0.026 - 0.0034
Cd0 = 0.0226 -------------------- at 27,800 ft with these assumptions and this method

At this point I think there has already been debate about Cd0 *actually* increasing and whether or not our assumptions
concerning fixed efficiencies of wing and prop are correct or not. With this method and these assumptions Cd0 is not
increasing enough to bother me at all.

The equation is never the reality. I only investigate how the equation works with some fixed assumed (and reasonable)
values along with historic data.

Now I turn to working this out for turns at speed without resorting to anything but TAS, probably get this wrong the first
few times as well:

Given:
0.5*p*v^2*S*[Cd0+([M*g*9.81]/[0.5*p*v^2*S])^2/(pi*AR*e)] = W*n/v

Given all our assumtions from above and new speeds, the TAS of the IAS at altitude for about a 3G turn.
Vs = 105 mph = 46.93 m/s---------- x 1.732 = 81.29 m/s IAS

And really it's not IAS but CAS except on Tuesdays between 7AM and 9AM when they're cleaning this side of the street.
I'm not going through OAT and all that since AFAIK the historic data was corrected for Standard Atmosphere in 1942
which leaves room for correction right there that I'm not going into anyway. I'm using a table and saying close enough.

From that IAS/TAS table I get factors at altitudes to interpolate:
Altitude Meters ____________1,500____2,000____4,000____4,500____7, 500____8,000____8,500
IAS x value = TAS__________1.099____1.131____1.263____1.295____1 .493____1.525____1.558

Altitude Meters___________________1,524______4,572______7,6 20______8,473
TAS/IAS factor___________________1.101______1.268______1.5 01______1.556
Turn TAS m/s 81.29 x factor above___89.50_____103.08_____122.02______126.49
Thrust_N________________________14164_____12212___ ___10389______10022

Thrust as W * n / v = 149400 *.85 / v
These are only the TAS that is sqrt(3) x Vs, it is not saying a 3G turn must result. It's just nice to have something to
compare with later.

.5 * 1.059 * 8010 * 27.87 * [ 0.021 + ( [ 55888 * g-load ] / [ .5 * 1.059 * 8010 * 27.87 ] )^2 / 14.07 ] = 14164
118205 * [ 0.021 + ( [ 55888 * g-load ] / 118205 )^2 / 14.07 ] = 14164
0.021 + ( [ 55888 * g-load ] / 118205 )^2 / 14.07 = 14164 / 118205 = 0.1198
( [ 55888 * g-load ] / 118205 )^2 = ( 0.1198 - 0.021 ) * 14.07 = 1.39
[ 55888 * g-load ] / 118205 = sqrt( 1.39 ) = 1.179
g-load = 1.179 * 118205 / 55888 = 2.49 G's -------------------- at 5,000 ft **with these assumptions and this method**.

.5 * 0.774 * 103.8^2 * 27.87 * [ 0.021 + ( [ 55888 * g-load ] / [ .5 * 0.774 * 103.8^2 * 27.87 ] )^2 / 14.07 ] = 12212
116210 * [ 0.021 + ( [ 55888 * g-load ] / 116210 )^2 / 14.07 ] = 12212
0.021 + ( [ 55888 * g-load ] / 116210 )^2 / 14.07 = 12212 / 116210 = 0.1051
( [ 55888 * g-load ] / 116210 )^2 = ( 0.1051 - 0.021 ) * 14.07 = 1.183
[ 55888 * g-load ] / 116210 = sqrt( 1.183 ) = 1.088
g-load = 1.088 * 116210 / 55888 = 2.26 G's -------------------- at 15,000 ft **with these assumptions and this method**.

.5 * 0.551 * 122.02^2 * 27.87 * [ 0.022 + ( [ 55888 * g-load ] / [ .5 * 0.551 * 122.02^2 * 27.87 ] )^2 / 14.07 ] = 10389
114320 * [ 0.022 + ( [ 55888 * g-load ] / 114320 )^2 / 14.07 ] = 10389
0.022 + ( [ 55888 * g-load ] / 114320 )^2 / 14.07 = 10389 / 114320 = 0.0909
( [ 55888 * g-load ] / 114320 )^2 = ( 0.0909 - 0.022 ) * 14.07 = 0.969
[ 55888 * g-load ] / 114320 = sqrt( 0.969 ) = 0.984
g-load = 0.984 * 114320 / 55888 = 2.01 G's -------------------- at 25,000 ft **with these assumptions and this method**.

.5 * 0.499 * 126.49^2 * 27.87 * [ 0.0226 + ( [ 55888 * g-load ] / [ .5 * 0.499 * 126.49^2 * 27.87 ] )^2 / 14.07 ] = 10022
111255 * [ 0.0226 + ( [ 55888 * g-load ] / 111255 )^2 / 14.07 ] = 10022
0.022 + ( [ 55888 * g-load ] / 111255 )^2 / 14.07 = 10022 / 111255 = 0.0901
( [ 55888 * g-load ] / 111255 )^2 = ( 0.0901 - 0.0226 ) * 14.07 = 0.949
[ 55888 * g-load ] / 111255 = sqrt( 0.949 ) = 0.974
g-load = 0.974 * 111255 / 55888 = 1.94 G's -------------------- at 27,800 ft **with these assumptions and this method**.

================================================== =========================
================================================== =========================

Altitude Feet_____________________5,000_____15,000_____25,0 00_____27,800
Altitude Meters___________________1,524______4,572______7,6 20______8,473
Turn TAS m/s 81.29 x factor above___89.50_____103.08_____122.02______126.49 --- for IAS/CAS = 182 mph at sea level.
Max sustainable g-load at speed______2.49______2.26_______2.01_______1.94 ----- using this method and assumptions

Okay, let's see what I did wrong this time as far as the method and the math.

At some lower IAS or higher power or higher thrust (1942 P-47B) it can probably sustain higher G's.
With less power... nope, do the math or STFU.

For the next steps a spreadsheet seems the best way to go, one with all parts well commented which takes time.

Kettenhunde
02-28-2010, 05:27 AM
Okay, let's see what I did wrong this time as far as the method and the math.

Nothing that I can see. Except of course for things that are beyond your control.

Since you have no measured data for any of the assumptions, you are pushing all of the error in our relative performance onto the velocity our performance occurs when you factor density effects.

TheGrunch
02-28-2010, 06:45 AM
Originally posted by Gaston444:
Are you guys for real?
Test it yourself or stop your idiotic trolling. It'll cost you a maximum of \$10 + shipping, which if you want to know so much should not be too much. This is a forum for Il-2 players, not science-skeptics trying to find fault in computer games with ambiguous anecdotes. If you can't be bothered to work up the effort to perform tests to validate your own conclusions about a game, then why the HELL should anyone bother to answer your questions?
I'll even show you how to set up Infomod and/or LesniHU's Autopilot, which I have never yet attempted myself, but at least I would be willing to display some initiative unlike yourself.

Kettenhunde
02-28-2010, 10:17 AM
For the next steps a spreadsheet seems the best way to go, one with all parts well commented which takes time.

Now compare the Spitfire Mk IX Merlin 66 +18 with the P47 at 28,000 feet at 246.6KTAS...

M_Gunz
02-28-2010, 12:24 PM
Originally posted by Kettenhunde:
<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">Okay, let's see what I did wrong this time as far as the method and the math.

Nothing that I can see. Except of course for things that are beyond your control.

Since you have no measured data for any of the assumptions, you are pushing all of the error in our relative performance onto the velocity our performance occurs when you factor density effects. </div></BLOCKQUOTE>

Same power and mass turning at increasing radius and speed, I kind of expect less G's.

http://www.aerospaceweb.org/qu...formance/q0146.shtml (http://www.aerospaceweb.org/question/performance/q0146.shtml)

radius = v^2 / [ g * sqrt( g-load^2 - 1 ) ] ------ since I have g-load already, I don't need to calculate bank angle
turn rate radians/sec = [ g * sqrt( g-load - 1 ) ] / v

per the turn estimate post, all in metric:
Alt Feet _____Alt Meters _______ TAS m/s _____ g-load _____ Radius Meters _____ turn-rate deg/sec
5,000 ________ 1,524 __________ 89.5 ________ 2.49 __________ 358 ______________ 9.36
15,000 _______ 4,572 _________ 103.08 _______ 2.26 __________ 534 ______________ 6.12
25,000 _______ 7,620 _________ 122.02 _______ 2.01 __________ 870 ______________ 4.63
27,800 _______ 8,473 _________ 126.49 _______ 1.94 __________ 981 ______________ 4.31

I am not up to playing with angular momentum, how the same mass at higher speed and radius works out
but I get this idea that the same engine power won't sustain the same g-load as those increase.

Kettenhunde
02-28-2010, 01:39 PM
M_Gunz:

Same power and mass turning at increasing radius and speed, I kind of expect less G's.

That is what happens but remember.......

Crumpp says:

Since you have no measured data for any of the assumptions, you are pushing all of the error in our relative performance onto the velocity our performance occurs when you factor density effects.

Same aerodynamic forces to the airplane at a given EAS, M_Gunz.

http://forums.ubi.com/images/smilies/16x16_smiley-wink.gif

M_Gunz
02-28-2010, 04:58 PM
But not the same kinetic nor the same momentum! AFAICT those increase.

And yeah, I expect there is error due to fixed assumptions, but how much?

Kettenhunde
02-28-2010, 06:21 PM
But not the same kinetic nor the same momentum!

No, the forces are the exact same....the relationship is equal.

That energy is used to create the dynamic pressure in the less dense atmosphere that equalizes the forces.

And yeah, I expect there is error due to fixed assumptions, but how much?

You can expect wing efficiency to vary between ~.72 and ~.97 or so depending on the altitude

M_Gunz
02-28-2010, 07:19 PM
Originally posted by Kettenhunde:
<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">But not the same kinetic nor the same momentum!

No, the forces are the exact same....the relationship is equal.

That energy is used to create the dynamic pressure in the less dense atmosphere that equalizes the forces. </div></BLOCKQUOTE>

We must be talking about two different things because what I've trying to express has nothing to do with
drag or dynamic pressure at all. I'll have to find a way to show the angular momentum. Same as spinning
with your arms in as opposed to with your arms out, the latter slows you down and if you bring them back
in you speed up. I'd like to see how 2.74x the radius at 46% the turn rate is the same.

That is not aerodynamic forces. It's momentum.

<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content"> And yeah, I expect there is error due to fixed assumptions, but how much?

Well you can expect a measured Cdo to vary about 56% from sea level to 30,000 feet.

Of course our wing efficiency will reflect this relationship as well. </div></BLOCKQUOTE>

It might be time to introduce some textbook scans on that. Or perhaps the NASA site covers it?
Something that won't turn into 10+ pages of "Is." and "Isn't", if that's even remotely possible.

Kettenhunde
02-28-2010, 07:30 PM
That is not aerodynamic forces. It's momentum.

The airplane does not care, it only feels the aerodynamic forces.

I edited that comment out as I did not want to get into another 10 pager either.

You can ballpark the change required by:

Vmax velocity in fps = Cubed root(power/flatplate area)

Density effects drops our power from 2300hp to ~1424hp.

You can use this relationship in EAS to get exactly the same results as TAS with much fewer steps.

Cdo is generally a measured component when we start using TAS calculations.

Kettenhunde
02-28-2010, 08:25 PM
You mean the quick search that turns up all the different techniques with different results for calculating turn performance.

The ones that are obviously not as accurate as the formulas you learn in an hour at the public library??

http://img697.imageshack.us/img697/2066/turnperformancesearch.jpg (http://img697.imageshack.us/i/turnperformancesearch.jpg/)

http://img132.imageshack.us/img132/6775/andyetanotherone.jpg (http://img132.imageshack.us/i/andyetanotherone.jpg/)

http://img412.imageshack.us/img412/6186/anotheroneg.jpg (http://img412.imageshack.us/i/anotheroneg.jpg/)

http://img517.imageshack.us/img517/7504/bootstrap.jpg (http://img517.imageshack.us/i/bootstrap.jpg/)

http://img695.imageshack.us/img695/2486/engineernotes.jpg (http://img695.imageshack.us/i/engineernotes.jpg/)

http://img62.imageshack.us/img62/279/yetanotherone.jpg (http://img62.imageshack.us/i/yetanotherone.jpg/)

M_Gunz
02-28-2010, 09:00 PM
Hey now, we are looking at steady-state horizontal turns in planes with variable-pitch props.
Nothing about how you start the turn, no trick turns and no fixed prop considerations please!

T = D and constant bank angle just looking at performance.

In virtual combat, this virtual pilot does not DO horizontal 360s, 180s or even 90 degree turns.
Well maybe if I'm in the better stall turner with an enemy coming onto my tail and I think I can
sucker him into following but otherwise I can't think of a situation where I'd fly like that.
The information has its own uses though.

Kettenhunde
03-01-2010, 04:04 AM
Hey now, we are looking at steady-state horizontal turns in planes with variable-pitch props.

Those are how airplanes turn, M_Gunz. Those reports are looking at steady state turns too.

Same exact questions you guys are asking.

this virtual pilot does not DO horizontal 360s, 180s or even 90 degree turns.

Come on now...

Rate x Time = Distance....

90 degress /14 degrees a second turn rate = 6.43 seconds

The only difference in the reports is the technique used to answer the question.

M_Gunz
03-01-2010, 07:08 AM
So there is evidence that the efficiency should change.

And BTW:

quote:
this virtual pilot does not DO horizontal 360s, 180s or even 90 degree turns.

Come on now...

why did you leave the "In virtual combat," part out? Kind of changes the meaning doesn't it?

Kettenhunde
03-01-2010, 04:40 PM
So there is evidence that the efficiency should change.

Of course....I have a report that shows this very thing with measured data. I will dig it up.

why did you leave the "In virtual combat," part out? Kind of changes the meaning doesn't it?

I did not mean to change any meaning of what you wrote. The "virtual combat" does not change what you wrote for me in regards to determining aircraft performance.

Kettenhunde
03-01-2010, 05:45 PM
Here is the data, efficiency changes not only with altitude but velocity as well.

Conditions of flight...

http://img39.imageshack.us/img39/3708/flighttest.jpg (http://img39.imageshack.us/i/flighttest.jpg/)

Data:

http://img697.imageshack.us/img697/9666/cdoande.jpg (http://img697.imageshack.us/i/cdoande.jpg/)

M_Gunz
03-02-2010, 10:31 AM
My data such as it is already contains the results of such efficiencies so I look to where the error due
to assumed values should be pushed and what I come up with that I can deal with is the calculated Cd0.
So I look at the differences in calculated Cd0 and I see .021, .021, .022, .0226 with the increased
values at higher altitude and speed.

Maybe what I should do is recalculate the 25,000 ft and 27,800 using Cd0 of .021 and call it close enough
given the method.

Kettenhunde
03-02-2010, 03:01 PM
Maybe what I should do is recalculate the 25,000 ft and 27,800 using Cd0 of .021 and call it close enough given the method.

In fact the proper way to do TAS in subsonic incompressible flow is to use EAS speeds at sea level and go from there keeping Cdo the same.

Your power required and available will be the same for a given design over altitude.

If it takes 1424hp to reach a given TAS at sea level, it will take 1424hp to reach the same velocity at altitude. So we use EAS to change altitudes which fixes our Cdo.

Once again, there is not difference in the relative performance between an EAS or a TAS prediction. Provided of course, the prediction is done correctly.

M_Gunz
03-02-2010, 04:36 PM
Lowest alt data I have for that same plane (and loading, etc) is 5,000 ft.
It gives the same Cd0 as at 15,000 ft though, to 3 places and nowhere much
after or I would have used 4 as I did for 27,800 ft. The change of .001 at
25,000 was a surprise and a further .0006 at only 2800 ft more was bigger
per foot (as opposed to 15,000 ft to 25,000 ft giving +.001).
I'm thinking it's not a case of rounding which if I tried for 4 significant
digits would have started to be the case from what I was looking at.

Kettenhunde
03-02-2010, 04:39 PM
Here is one with sea level data...

http://www.wwiiaircraftperform...d-75035-11oct43.html (http://www.wwiiaircraftperformance.org/p-47/p-47d-75035-11oct43.html)

M_Gunz
03-02-2010, 04:52 PM
Ahhh, very good! So with the method about as good as it will probably get I feel pretty clear about
laying out a spreadsheet to suit the task as I see it. I dislike the "messy code" that comes from
jumping straight in and using the top-to-bottom-and-evolve-to-fit approach.

Kettenhunde
03-05-2010, 01:32 PM
Keep in mind M_Gunz that Wurkeri is using a power required formula to define power available.

That is not the correct application of the principles.

The engine produces a finite amount of power available. That is why we go from EAS or we use:

Lastly, for all the "experts"...

If we want to hold our assumptions the same then we need to vary thrust with density.

2300 hp * .85 = 1955thp

1955thp*325 / 180KEAS = 3529.861111lbs of drag

sigma = .53281 @ 20,000ft

SQRT(.53281)* 3529.861111lbs of drag = 2576.58lbs of drag

2300 hp * .85 = 1955thp

1955thp*325 / 246.6KTAS = 2576.5lbs of drag

Holtzauge, Wurkeri, or JtD seem to have passed up the opportunity to demonstrate their knowledge of aerodynamics!

If you have an error in speed measurement, it will manifest itself as a change in Cdo using the technique he showed you.