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

[gripen]

Originally posted by Crumpp
NO it is not.  Cut and dry it is 2D theory equation.

It uses finite wing so it's 3D. Please follow the link justin_g posted and see how it is used 3D.

Originally posted by Crumpp

You confuse reference area with the influence of induced drag and AR.

I wonder what you might mean, the equation is simply for the Cl.

Originally posted by Crumpp

Goes back to the monsterously silly thread you linked, Gripen.  The one you argued that woods formula was wrong.

Hm... Actually I pointed out in that thread that Wood's formula for rectangular wing was used wrongly. It's not for other wing types nor for entire airframe.

gripen

[Lye-El]

Originally posted by KAntti

I bet the only way one might get close is to sit couple of todays pilots in front of the game and let them tell what they think of it. There are lots of warbirds flying around, ask the pilots of them to visit AH convention or smthin and tell us (and HT) how a plane should or should not handle.

We have pilots playing the game. Including a Active duty F/A 18 pilot, don't remember is callsign though.  As far as pilots flying 60 year old warbirds I would think that they are't racking them around the sky and most assuredly don't routinely ride around with the stall horn blareing. They only get one life.

[BlauK]

Waffle (.. and Simaril)
AFAIK, bending is not same as curving. Something bent can have a sharp angle.





I claim that the current view (black line) offers less view and viewing angle than the refracted view (green line). Th erefraction makes one see "around the frame" and therefore it makes more field of view! It is all about the initial angle of meeting the inside surface of the glass... which is larger than without teh glass. There is no magic or curved lenses needed.

If you still claim otherwise, then explain the area between the black and green lines on the outside of the armoured glass!

[Waffle]

Blauk - thats nice in that drawing, but I'm fairly certain that they had bracing and brackets on the back side of the glass.

[BlauK]

Originally posted by Waffle BAS
Blauk - thats nice in that drawing, but I'm fairly certain that they had bracing and brackets on the back side of the glass.

Surely, and on the outside too! Ar yuo saying that there were more and wider bracings on the inside? I dont think so, because in such case one could not see any such difference in frame thickness as is visible in many photos.

I was simply talking about the phenomenon and the wider field of view it produces

[Waffle]

Put it this way - it has to support the force from any pressure - be it wind / debris or impact. It has to support little force from the pilots side outward...unless the poor soul is trying to kick his way out through armored glass.  Look at the 109  - you'll see 2 bars on the inside with bolts through them. I'm pretty certain they were inset and wider than the front exterior, as to give the glass more support.

[Simaril]

Sigh.

The drawing shows a nice green line, but its wrong.


Use the exact link you (crumpp) posted above. Choose any angle (I used 20 degrees). Set the density to air, and have the bottom density be glass. You'll get roughly 13.1 degrees.

Now set the top to 13.1 degrees, the density to glass, and have the beam pass to air density. You get the same 20 degrees you started with. Net effect is...NOTHING.


You're ignoring the fact that the near edge of the glass block takes up some physical space, and that the front face of the armor block covers up materials deep to it....one of those geometry things I dont feel like calculating, but you have a triangle. The lateral face of the block (the one to the side) forms the base. The line between the near corner of the block and the far side of the framing, extended from the view position till you reach plane of the frame, is the hypotenuse. The far face of the block and the front face of the canopy form the short arm of the right triangle.

Those angles define an invisble blind spot, invisible BECAUSE of the doubel refraction hiding the frame from view. You still have a blind spot, it just isnt a big black line. Ever try looking through that really thick glass at a bank drive through window? Have to move your head around to see all over....becasue of the refraction. In fact, most people find views through thick armor glass distorting and disturbing, not at all a view advantage.

You are right that the thickness of the block will increase view angles in a miniscule way -- precisely as if the canopy frame were farther from the viewer (and thus visbly smaller. BUT the separation between the actual and visible distance IS ONLY THE THICKNESS OF THE ARMOR GLASS. Pretty trivial.

[Waffle]

Eyeball this pic....

Note the right side (as if you were sitting behind it) where the glass would join  the exterior frame. Then look at the brace/bracket top and follow it's left edge. You can see that it's well inside of the line of where the exterior side of the glass meets the frame.

[Crumpp]

Crumpp, I give up. I showed you that turn radius does depend on wing area, if you want, through reducing the "power required", but you refuse to listen.

Find where I disputed you on this bozon??

I haven't and your correct.  It is exactly what I have said from the begining.  Wingloading is a reflection of the power available to power required.

Only thing I claimed is that addition of thrust improves turn performance and a small increase in weight can be overcome with additional thrust.  Designers were well aware of this.

Generally speaking, If you track the weight increases of most designs you will find they are accompanied by a power increase of appropriate magnitude.

The FW-190A8 gains 3.8% weight over the FW-190A5.  It gains 22.7% more horsepower and a 10% more efficient propeller to increase it's effective thrust.

I have also said from the begining exactly what is in Perkins & Hage.  Turning is a fundamental relationship of power available to power required.  That relationship is the basis for turn performance.  Not wingloading.  Looking at wingloading alone is like looking at piece of the pie and not the whole thing.

Everybody keeps looking at this from the POV of which matters more, wieght or thrust.  It's completely irrelevant.  They both influence turn ability.

It uses finite wing so it's 3D.

Is not a finite wing.  It's an area.  Any shape can have an equal area.  Finite defines the shape of the wing.  Not just a random area.

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

Hm... Actually I pointed out in that thread that Wood's formula for rectangular wing was used wrongly. It's not for other wing types nor for entire airframe.

Right.  Fortunately people can read the thread.  If they want to muddle through it.

Even if the canopy frame looks thinner (because it's shifted due to the refraction) you are not going to magically get more feild of view from a flat piece of glass.

Nobody is claiming you are but the fact remains you can place a coin of the nose of the FW190 against the cockpit and see it.  In AH, the bottom of the "frame shape" prevents this from happening.  It's not the same FOV.

I don't think your an expert on the physics of light nor am I.  I suspect those that built and designed the cockpit knew what they were doing inspite of what some gamers in this thread want to claim and that goes for all cockpits too, not just the Luftwaffe ones.

All the best,

Crumpp

[Crumpp]

Note the right side )as if you were sitting behind it) where the glass would join the exterior frame. Then look at the brace/bracket top and follow it's left edge. You can see that it's well inside of the line of where the exterior side of the glass meets the frame.

When you put the glass in it, let me know!  We can finish this discussion.

BTW, that is exactly the reason I have not participated in many of the "view threads".  The view looked very close to the framing without the glass.


All the best,

Crumpp

[Waffle]

You put the glass in and the feild of view is blocked by the exterior frame Crumpp. You cannot see through and around metal....

The only thing that happens is that after the light bends through the glass -you are seeing the image which hits the exterior side -  only it's 50-60mm closer to the pilot, which is what makes the rails look thinner due the refraction. The image is just shifted closer.

[Crumpp]

Th erefraction makes one see "around the frame" and therefore it makes more field of view!

That is exactly the effect, BlauK.

Whatever the theory hunters want to claim.  You nailed exactly the effect you see.

Good diagram.

All the best,

Crumpp

[Simaril]

Crump,

"they knew what they were doing" does not equate to "the AH canopy frame is wrong because it ignores armor block refraction."

Your explanation of refraction is in error. There IS a blind spot, it just isnt black liek framing. Its easier for AH to model a black line for framing, but it isnt unrealistic -- its just darker and more visible.

Forget whether there is glass in the frame Waffle posted or not -- you've been throwing aerodynamic calculations around freely, so you're good with numbers. Look at the angles I talked about, adn the results of the apllet you directed us to. Your explanation isnt correct.

[Simaril]

Originally posted by Crumpp
That is exactly the effect, BlauK.

Whatever the theory hunters want to claim.  You nailed exactly the effect you see.

Good diagram.

All the best,

Crumpp

When you try to prove your point, its science; when we show something you dont like, we're "theory hunters"?


I agree that you see less black due to refraction, but THERE IS A BLIND SPOT that is covered with a "visible" refracted image. Just like your eye's blind spot, you dont know its there unless you test it -- otherwise its invisible.


The "I know what I see, I dont care what your calculations show", just gotta beleive approach is antithetical to your previous approach. I wonder why you've changed ....

[Crumpp]

Look at the angles I talked about, adn the results of the apllet you directed us to. Your explanation isnt correct.

Where do you think the LOS picks up when it reaches the flat exterior surface?

Do you think it redirects to a new angle of vision independant of the LOS of the angle of entry on the interior glass?

Or do you think it keeps the same angle only it has been refracted to the exterior glass surface?

All the best,

Crumpp