Author Topic: Maybe the 190s arent wrong....or how to be really unpopular in 1 step (Read 12892 times)

Crumpp · « **Reply #150 on:** February 02, 2006, 08:37:32 AM »

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I offer my apologies to Crumpp in particular,

No problem whatsoever. The stupid assumptions we all make on these boards leads to so many misunderstandings.

I certainly have to take a "hunk of the blame cookie" for it as well and offer my apologies for the attitude.

I meant that you are most welcome to come down to the Museum and make those measurements. I think it would be interesting!

All the best,

Crumpp

gripen · « **Reply #151 on:** February 02, 2006, 09:31:33 AM »

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Originally posted by Crumpp
Wow!

Well, your own post [2-01-2006 10:03 PM]:

"If your not including the effects of induced drag along with the AR, it's not 3D."

Which completely false statement. Lift coefficient (3D) can be easily calculated without knowing the AR or induced drag as pointed out several times above.

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Originally posted by Crumpp

Not just area.
...
So while the wing is "finite" under your definition. It does not fit the engineering defintion for 3D theory.

Yet another logical problem in your thinking; if a wing has a finite area, it has also the tips despite what ever is the AR.

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Originally posted by Crumpp

3D theory would be more accurate in this case for exactly the reason Justin posted.

Well, as you can see from the 747 example which justin_n posted, the equation I linked is 3D. It is accurate and exact.

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Originally posted by Crumpp

No and I never said you did need them. Unless your using 3D theory. If your not using using 3D theory then your correct.

Well, let's quote your statement again:

"If your not including the effects of induced drag along with the AR, it's not 3D."

Which completely false statement. Required lift coefficient (3D) can be easily calculated without knowing the AR or induced drag as pointed out several times above.

Are you saying that the example justin_g posted is not 3D?

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Originally posted by Crumpp

Gripen you can twist things however you want.

No matter what formula used our results remained the same when the same conditions of flight were used for both planes.

Things came out a little different when we did it your way by having the planes at different speeds or altitudes.

I have no idea what you are trying to argue. That thread was about determining e factor and it was pointed out that supposed "generalized formulas" were actually not such but for specific wing shape and for wing only.

It was agreed that the best way to determine e factor is polar analysis.

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Originally posted by Crumpp

No your trying to claim I am saying you cannot calculate CL?????

I don't see such claim in my reply.

Generally you have problems to put up logical argument.

gripen

Waffle · « **Reply #152 on:** February 02, 2006, 09:38:03 AM »

Maybe this will help show what I'm trying to explain is happening.

first pic is an image of what would be hitting the the outside of the armored glass which view is blocked by the frame.

Second pic is that same image refreacted to the pilot side of the glass, say 2" closer. - Same feild of view.

Inside / outside

BlauK · « **Reply #153 on:** February 02, 2006, 09:44:37 AM »

Waffle, your picture would mean that the entrance and exit angles for the glass are not equal. Sorry, but that is not what they teach in schools.
Try drawing your picture from above and you will also realize it.

Waffle · « **Reply #154 on:** February 02, 2006, 09:48:17 AM »

lol - Don't have time to photchop everything to a T, nor would I want to try but you can get a general idea.

Waffle · « **Reply #155 on:** February 02, 2006, 10:00:40 AM »

Here's a rough top view of 109 construction, feel free to run refraction lines on it.

Crumpp · « **Reply #156 on:** February 02, 2006, 10:06:16 AM »

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Yet another logical problem in your thinking; if a wing has a finite area, it has also the tips despite what ever is the AR.

Can you point out the tips of the helicopter area or the 747 used in this example please and how the differences are accounted for in your formula by using raw area??

http://www.aerospaceweb.org/question/aerodynamics/q0015b.shtml

Assume two planes of equal weight, equal wing area, and equal speed flown in the same atmosphere.

One has an elliptical wing with no twist, and the other a square wing with no twist. The CL will calculate out exactly the same using that formula.

The shape of the wing is not considered, only the area. Not to be a Smart A, but you did have Geometry starting in early childhood development that taught you shapes, I am sure. Later in your education you should have had some more advanced geometry that taught you area.

Area is not shape nor does it define shape. It is just area.

For example a square with sides of 2 has the same area as a rectange with a=1 and b=4?

Area is equal but the shape is different, understand now?

http://www.gomath.com/htdocs/ToGoSheet/Geometry/area.html

Because your statement earlier:

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required Cl can be calculated simply knowing the lift, speed, density and wing area without instrumentation or wind tunnel.

Is absolutely true.

It is not 3D theory, however as it does not account for the shape of the wing.

All the best,

Crumpp

Krusty · « **Reply #157 on:** February 02, 2006, 10:13:56 AM »

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Originally posted by bozon
I don't know how good it was but the P47 was not known for poor slow speed handeling or nasty stalls like the F4U or the P51 or the 190. It is extremely hard to tell how the real thing handeled is these extreme conditions, so AH might be spot on or not.

Especially considering that 2 of the 3 planes you listed there are gentle as babies in the stall (p51 and f4u) where in fact they were monsters and both had severe handling problems (not modeled in AH)

BlauK · « **Reply #158 on:** February 02, 2006, 10:17:21 AM »

Sorry Waffle, I wont bite that picture

If the inside "attachment strips" were that wide the frames would not appear thinner in photos. I am beginning to wonder if you are trolling on purpose

Here is a pic from top for you (notice that there is still some more space for the inside strip):

You are assuming the refracted line of sight (LOS) goes along the dotted red line "A" and would offer equal angle with the black LOS without any armoured glass.

Instead the refracted LOS goes along the red line "B", which offers more viewing angle than the black line. The entrance and exit angles on opposite sides of the glass have to be same!

Crumpp · « **Reply #159 on:** February 02, 2006, 10:20:40 AM »

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Maybe this will help show what I'm trying to explain is happening.

What an excellent photo, Waffle!

Are the angles lined up exactly on that 2nd picture animation? If they are please notice the lattice work windows at the top and in the armoured glass view.

When your view minus the armoured glass is on, the pains do not line up. When the armoured glass is in place they do lineup accordingly.

I don't think there is really much to argue about on this subject. The designers of armoured glass were well aware of the optics, need for fighter pilots to be able to see as large a field of view as possible, and the need for that view to be accurate.

IIRC Zeiss manufactuers the armoured glass for most Luftwaffe birds.
I can verify that.

And this is not some Axis vs Allies crap. This is just normal technology of the day and the recessed armoured glass became the WWII standard.

All the best,

Crumpp

Waffle · « **Reply #160 on:** February 02, 2006, 10:22:45 AM »

Blauk -

look at this pic again

Crumpp · « **Reply #161 on:** February 02, 2006, 10:24:48 AM »

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Instead the refracted LOS goes along the red line "B", which offers more viewing angle than the black line. The entrance and exit angles on opposite sides of the glass have to be same!

Exactly, that is why the "inside view" of waffles picture matches the "outside reality".

All the best,

Crumpp

Waffle · « **Reply #162 on:** February 02, 2006, 10:26:00 AM »

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Originally posted by Crumpp
What an excellent photo, Waffle!

Are the angles lined up exactly on that 2nd picture animation? If they are please notice the lattice work windows at the top and in the armoured glass view.

When your view minus the armoured glass is on, the pains do not line up. When the armoured glass is in place they do lineup accordingly.

I don't think there is really much to argue about on this subject. The designers of armoured glass were well aware of the optics, need for fighter pilots to be able to see as large a field of view as possible, and the need for that view to be accurate.

IIRC Zeiss manufactuers the armoured glass for most Luftwaffe birds.
I can verify that.

And this is not some Axis vs Allies crap. This is just normal technology of the day and the recessed armoured glass became the WWII standard.

All the best,

Crumpp

No watermelon they wont line up, because I dont have the "image" of whats hitting the outside glass....lol

All I had was a reduced internal picture after the refraction, and blew that up 10% so what you're seeing is magnification... not reversed refractioN.

edit - should clarify - image was scaled to fit, which would be more / skew / enlargement with no "angular" correction.

Crumpp · « **Reply #163 on:** February 02, 2006, 10:27:11 AM »

That is a nice frame Waffle. Is it yours or did you get the picture off ebay?

It's in pretty good condition for an original. Is it restored or reproduction?

All the best,

Crumpp

Crumpp · « **Reply #164 on:** February 02, 2006, 10:32:11 AM »

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No watermelon they wont line up, because I dont have the "image" of whats hitting the outside glass....lol

How did you make this film btw?