Maybe the 190s arent wrong....or how to be really unpopular in 1 step

[KAntti]

Originally posted by Angus
And todays fliers give anecdotes.

Some of the old ones still live though. But not many of them flew US, UK and LW planes. The selected who did that would almost exclusively at combat be Finns, then you have the others testing the enemy's stuff.
Oh, got one of them on my sig....goodness

Please tell me the performance data given to any AC is actually been confirmed at least by some flight tests done by a real human, please!!

I gather none of the interviewed flew US UK and LW planes, no where near. And if u still did not get it, my point is not to compare aircraft to each other, but to compare one AC to corresponding AH FM. The whole consept seems to be out of reach to you all who have been beating each other up so long that you dont even understand what I am saying.

This then might be the first time (not really but seems new to you) when 109's performance is not compared to other AC, but only to the FM of AH.

[Bruno]

Anecdotes..? Then here's the game for you:




No math involved!

FM base on pilot's stories!

comes with .50 cal tiger killers...

(image originally posted on SimHQ by 15/JG52_Genie)

[KAntti]

HAHAHAHAHA:rofl

[Angus]

Kantti:
I have some Anecdotes (from autobiographies mostly, - which I have) of pilots that tested RAF&LW, and vice versa with a topping of US.
It gets sour for many in this forum to swallow, that all I have matches nicely, be it a German pilot or an Englishman.
Anyway, if you want to compare the 109's performance to the AH, it's another thread, this one is about the 190's.

[Angus]

Oh, that picture above:
I knew a pilot who got into a scruffle with a 262. Him in a P51 together with a squadron.
262 got thirst for blood. He jumped the P51's and got one alone for himself, but the P51 actually outturned the 262, - by the time the 262 pilot aborted the turnfight it was too late, - the P51 shot him down. A 4 gun P51 BTW....Bruno, are you unemployed?

[Bruno]

Bruno, are you unemployed?

Hardly...

[Kev367th]

Originally posted by KAntti
Holy helll we have a spitmaister here!

Even if it was equally accepted by any one as truth, regardless the side the plane was used Kev? U would actually trust the FM in hands of "trusted and absolutely non partial, non eye beholding" creator?  Do you actually question anything that works in your favor Kev?

As I said before this is a game, AND if u dont believe anecdotes, there still are the real warbirds flying around.

Real warbirds, yup.
But in 'mock' combats theres not the desperation of knowing you could get shot down 'for real'.

Yup regardless of which side.
They are great for reading, but hardly a source of hard quantifiable data.
Not a question of believing or disbelieving anecdotes, you can't build an FM from them, they're not exact.

edit - Apart from that, there's enough disagreement when hard data is posted, imagine the chaos/accusations if FM's started to be built using anecdotes.

Don't fly IL2 so I can't comment, but from speaking to Skuzzy they don't think much of any of the FM's in it.
Apart from that, IL2 has zero to do with this game anyway.

[Crumpp]

No one says anything about Cl.  It's about theory.

Crumpp says:

Anyone can work the following problem...

Gripen says:

Required Cl for both wings are exactly same, regardless shape of the wing and the equation

Now we are making progress.  That is because you have not applied shape to the wing yet, only area.

Before we get a turned around lets clarify some things.

The lift equation is a very powerful formula and can be used to analyze data to a point.  It does return very good agreement but has limitations.  It is not used by engineers for detailed analysis.

In effect it is a 3D CL as it measured actual lift force on a reference area but it is not 3D theory.

Getting back to our sample problem:

You get the same CL for both wings no matter what formula you use, correct?

In our above reference calculation our CL is finshed under the lift equation method and becomes the CL of the wings.  Even though we both know an elliptical wing with no twist will have a higher lift than a rectangular one just due to the influence of induced drag.

Engineers use 3D theory.  3D theory uses a 2D CL and seperates the effects of AR and efficiency.  Once these factors are applied our CL's will be different under the same conditions reflecting the shape of the wings.

http://www.grc.nasa.gov/WWW/K-12/airplane/kitedown.html

Which inspite of your claims is used on much more than rectangular wings.  It was developed for that but has been expanded to cover other wing designs.

All the best,

Crumpp

[gripen]

Originally posted by Crumpp
No one says anything about Cl.  It's about theory.

Nonsense. You have been arguing all the time that the Cl equation:

Cl = L / (A * .5 * r * V^2)

is not 3D theory. It is 3D theory, it gives exact and accurate required 3D lift coefficient for any given airframe at any loading in 3D enviroment. 2D theory would result 2D lift coefficient ie Clo.

Originally posted by Crumpp

Now we are making progress.  That is because you have not applied shape to the wing yet, only area.

Nonsense. We are talking about required lift coefficient, not about induced drag.

Originally posted by Crumpp

Before we get a turned around lets clarify some things.

The lift equation is a very powerful formula and can be used to analyze data to a point.  It does return very good agreement but has limitations.  It is not used by engineers for detailed analysis.

Complete nonsense. The Cl equation has no limitations and it can be used and is used by engineers for exact and detailed analysis.

Originally posted by Crumpp

In effect it is a 3D CL as it measured actual lift force on a reference area but it is not 3D theory.

Complete nonsense. You are trying make your own false definition of 3D theory.

3D theory simply handles finite wings and 3D lift coefficient (ie Cl) for any finite wing at given conditions can be calculated with the formula:

Cl = L / (A * .5 * r * V^2)

2D theory handles infinite wings and 2D lift coefficient is Clo which is usually determined in the wind tunnel.

Originally posted by Crumpp

You get the same CL for both wings no matter what formula you use, correct?

There is only one formula needed to calculate required Cl (3D) and the Cl will be the same for both wings.

Originally posted by Crumpp

In our above reference calculation our CL is finshed under the lift equation method and becomes the CL of the wings.  Even though we both know an elliptical wing with no twist will have a higher lift than a rectangular one just due to the influence of induced drag.

We are not talking about induced drag.

Originally posted by Crumpp

Engineers use 3D theory.  3D theory uses a 2D CL and seperates the effects of AR and efficiency.

Complete nonsense.

3D theory uses 3D lift coefficient ie Cl.

2D theory uses 2D lift coefficient ie Clo.  

Originally posted by Crumpp
Once these factors are applied our CL's will be different under the same conditions reflecting the shape of the wings.

Complete nonsense. The Cl (3D) at given loading is same in all cases despite what ever is the shape of the wings. Required 3D lift coefficient depends just on wing area, density, speed and lift regardless the shape of the wing.

Originally posted by Crumpp

http://www.grc.nasa.gov/WWW/K-12/airplane/kitedown.html

Which inspite of your claims is used on much more than rectangular wings.  It was developed for that but has been expanded to cover other wing designs.

The link shows well the difference between 2D (Clo) and 3D (Cl) lift coefficient, you just can't undersstand how these are used.

Besides, I have not made any claims regarding that page in this thread.

gripen

[BlauK]

Waffle,
Krusty pretty much summed it up.

Also the smaller refraction angle would mean more shifting and therefore more viewing angle (not less). If the refraction angle gets smaller it allows a wider original (from eye) viewing angle and still manages to get by the frame at the outer surface of the glass.

In other words, the more dense material refracts more and makes a bigger shift and provides a wider view.

Thanks anyways for the argument. It made me refresh the old high school stuff on how the refraction is calculated. Maybe I will now try to put in the real numbers for 109 pilot's position etc. and figure out the previously mentioned 1-sided polygon approach in more detail for fixing the problem

[Crumpp]

It is 3D theory, it gives exact and accurate required 3D lift coefficient for any given airframe at any loading in 3D enviroment. 2D theory would result 2D lift coefficient ie Clo.

No Gripen,

YOU have been arguing about CL, I have been arguing that 3D theory is more accurate.

3D theory simply handles finite wings and 3D lift coefficient (ie Cl) for any finite wing at given conditions can be calculated with the formula:

Cl = L / (A * .5 * r * V^2)

Find the CL for an elliptical shaped wing with no twist and an AR of 5:

L = 3000lbs

Area = 150sq ft

V = 400fps

r = 0.0023[slug/ft3]

Now find the CL for a rectangular shaped wing with no twist and an Aspect ratio of 7:

L = 3000lbs

Area = 150 sq ft

V = 400 fps

r = 0.0023[slug/ft3]

What is the CL going to be using the standard forumla??

The SAME!!!  WE are done, right??

Yet we know the lift properties are going to be different for each of these wings so the CL should NOT be the same simply due to AR and induced drag production.

Well not if we want to include the shape of the wing. We start with a 2D CL and factor in the influence of Aspect Ratio and efficiency.

You can see the effect of aspect ratio on the lift produced by a wing quite clearly in the following graph.

 

http://www.aerospaceweb.org/question/aerodynamics/q0167.shtml

You can better understand the effects of induced drag and stall by studying the following graph.

 

http://www.aerospaceweb.org/question/aerodynamics/q0136.shtml

Our answer might be close to the original CL calculated using:

Cl = L / (A * .5 * r * V^2)

But it will not be the same.

Using 3D theory we will have two diffent CL for two different wings.

All the best,

Crumpp

[Crumpp]

The general conclusion I made from this thread is that the frames in the 109 (and 190?) should not be as thick, because in reality when you sit inside the cockpit the refraction makes them look thinner.

That is correct!

All the best,

Crumpp

[Angus]

The 190 frames look too thick too me just look at the pictues above.
The "anecdotes" usually refer to good view.

[gripen]

Originally posted by Crumpp

YOU have been arguing about CL, I have been arguing that 3D theory is more accurate.

Nonsense. You have claimed several times that the formula I linked is not 3D theory and something which you can't specified is more accurate. These claims are completely false. The formula gives directly the exact and accurate required 3D lift coefficient (Cl not Clo) for any given loading at any given flight condition.

Originally posted by Crumpp

The SAME!!!  WE are done, right??

Yep, we are done. We know (if you can actually do the calculation) what is the required Cl for this amount of lift at this flight condition. Required Cl (as well as lift) will be the same despite what ever is the shape of the wing.

Originally posted by Crumpp

Yet we know the lift properties are going to be different for each of these wings so the CL should NOT be the same simply due to AR and induced drag production.

Here we can see from where your confusion starts; AR nor induced drag have no any kind of effect to required Cl nor to required lift at this given flight condition.

Originally posted by Crumpp

Well not if we want to include the shape of the wing. We start with a 2D CL and factor in the influence of Aspect Ratio and efficiency.

The 2D lift coefficient (Clo) has absolute nothing to do with this. The formula gives directly the 3D lift coefficient (Cl) required for given lift.

Here is the 747 calculation:



Please prove with calculation if the calculated lift coefficient is 3D (Cl as marked) or 2D (Clo) as you are trying to argue.

 

Originally posted by Crumpp

Our answer might be close to the original CL calculated using:

Cl = L / (A * .5 * r * V^2)

But it will not be the same.

Nonsense. The required Cl will be exactly same at given loading despite what ever is the geometry of the wing. There is no need to know the angle of attack.

If you want to prove otherwise, please do the math.

Originally posted by Crumpp

Using 3D theory we will have two diffent CL for two different wings.

Nonsense again, the Cl will be exactly same if the area, loading, density and speed are the same.

gripen

[Angus]

Two different wings means two different airfoils right?
So, CL should be different, or could at least.
Well, in reality. Not sure of that theory.