Explain this and win the prize!

[gripen]

This thread is about determining efficiency factor of the WWII fighters. While the Spiteful did not saw service during war, it was designed during war and it's particularly interesting due to its wing which had no washout. And we are not directly talking about the absolute amount of drag but how the drag changes with the Cl.

This thread is not another X vs Y comparison, start a new thread if you are not interesting about determining efficiency factor.

gripen

[Angus]

From the back of my head, I rememberthe Spiteful being a bit of a disappointment. The wing was just marginally more effective than the Spitfire wing. ?!?!

[gripen]

Yep, there was just marginal improvement in the drag rise due to compressibility if compared to the original Spitfire wing.

gripen

[Crumpp]

This thread is not another X vs Y comparison, start a new thread if you are not interesting about determining efficiency factor.

Actually this thread was about if the general formulas using AR were viable to determine e factor.  It was shown they are a good way to determine it.  Having a drag polar from the actual aircraft is better.

And you cannot compare Wooden Models to actual aircraft of a different design.

Crumpp

[Guppy35]

Originally posted by Charge
"There were modifications withing the Mk IX series, and then there was the superb Mk VIII....."

The VII was stretched a bit and was mounted with 61, 64 or 71 Merlin and was the first to have those two similar radiators under the wings and it had a retractable tail wheel. The VIII was unpressurized variant with Vokes filter with a few different engines.

The tail wheel was however made fixed in IX and the late IX model had again the retractable one where as the other later models again had fixed wheel and again from XII onwards the tailwheel was retractable. Maybe just a feature of parallel development, dunno.

-C+

Basically it came down to the IX being developed off the Spit Vc as more of a lash up of the Merlin 60 series to get it into service faster hence the fixed tail wheel.

The VIII incorperated many of the refinements of the prototype Mk III that never went into production but had a retractable tail wheel, short span ailerons and at one point clipped wings.

The XII had  roughly 50 of the hundred built on Spit Vc airframes so they had fixed tail wheels.  These being the XIIs in the EN serial ranges, and about  50 of them built on Mk VIII airframes in the MB serial ranges which had retractable tail wheels.

And as the IX and VIII in many ways had similar performances, from a production stand point it was smarter to continue to improve on the IX and then XVI airframes without the refinements like retractable tail wheels and short span ailerons.  

While the Spit VIII is considered the best overall of the Merlin Spitfires, the IX clearly was the most significant Merlin variant along with XVI, which was a Spit IX with an American built Packard Merlin 266 in place of the Rolls Royce built Merlin 66 in the IX

Dan/Slack
Late to the party as usual

[gripen]

Originally posted by Crumpp
Actually this thread was about if the general formulas using AR were viable to determine e factor.  It was shown they are a good way to determine it.  

Could you please show us where it was shown out that general formulas are good way to determine it?

The data shows that these gives almost routinely around 10% too high values for WWII fighters.

So far it has been also shown that a generalized value of e around 0,75-0,8 gives much better estimate than generalized formulas.

Originally posted by Crumpp

And you cannot compare Wooden Models to actual aircraft of a different design.

As noted above: If you make an argument you should be able to prove it and the purpose of this thread is not to make comparison.

gripen

[Crumpp]

Could you please show us where it was shown out that general formulas are good way to determine it?

Read Badboys comments on the General Formulas:

 

The number normally represented by the character e in induced drag calculations was originally known as Oswald’s efficiency factor, and his original paper is available for download from the NACA report server. More commonly it has a component of parasite drag lumped in with it and is just called the airplane efficiency factor and can be estimated depending on the aspect ratio, taper ratio, sweep angle and twist. Theoretically an elliptical wing would have an efficiency factor of 1, meaning that it will have a coefficient of induced drag close to the theoretical maximum. Even though it is a function of aspect ratio, sweepback angle, taper ratio and twist, the largest influence on the wing of a WWII fighter with very little sweep or twist comes from aspect ratio and taper ratio and so there are approximate formulae for estimating e that only include aspect and taper ratio, and even more approximate methods that only include aspect ratio.  The important thing you must appreciate is that they are only approximate. But better than just assuming a constant value for every aircraft.

And certainly better than comparing a Wooden Model to an actual aircraft of a different design.

What they do show us is that Lednicer was correct.  The Spitfire does have a slight  advantage.   The general conclusion you can draw off this thread is:

1.  In high G manuvering and in the low speed realm at the induced drag "wall" the Spitfire has the advantage in the overall drag situation.  The advantage in High Speed manuvering is only slight due to the FW-190's greater mass requiring greater braking forces.

2. Angus was correct when he concluded:

 

So, it's not an ALWAYS, but MOST OFTEN subject.

The FW-190 "MOST OFTEN" has less drag than the Spitfire.
It always has less form drag.

Crumpp

[gripen]

Originally posted by Badboy

The important thing you must appreciate is that they are only approximate. But better than just assuming a constant value for every aircraft.

The idea behind the generalized formula is good if we assume tapered wing with near optimal taper ratio. Apparently Wood's and NADC formula are created with somekind of statistics on the measured e factors. The problem is that that the WWII fighters generally had lower e factor which means that the both formulas give too high estimates and therefore also a constant value works better in most cases. As an example here is comparison with constant value of e 0,8:



It can be seen that generalized formulas result about twice larger average error than the constant value. It can be also seen that there is a correlation between the aspect ratio and the e factor, so it is possible to create a better working general formula than Wood's or NADC.

Originally posted by Crumpp
And certainly better than comparing a Wooden Model... blaah blaah

Only you are comparing planes in this thread and please point out what is wrong with model tests. The lift distribution should be similar.

No proof, no argument.

gripen

[Crumpp]

It can be seen that generalized formulas result about twice larger average error than the constant value.

Gripen assuming a constant value for e in all aircraft is very much like trying to compare cars as if they all had the same engine.

It is unworkable from a comparision standpoint.  Garbage in Garbage out.

Margin of error,  that's laughable.  So in order to have a margin we have to know the correct answer to see how far off of it we are in our calculations.  So tell us, Just what exactly is the "e" factor for the REAL Spitfire and the FW-190!!  Not a wooden model to the real plane.....


As you have been pouting about, the whole purpose of this thread to figure out the exact "e" factor.  Now your saying you already have a base to judge the "margin of error"!

But better than just assuming a constant value for every aircraft.

Badboy teaches aeronautics for a living.  I will take his word over yours anyday.  No matter how many "margin of error" charts you throw up.

Crumpp

[gripen]

Originally posted by Crumpp
assuming a constant value for e ... blaah blaah

The reason for the constant value on the above chart is only the show what's the problem with above mentioned generalized formulas; in most cases both give too high value of e.

Originally posted by Crumpp
So tell us, Just what exactly is the "e" factor for the REAL Spitfire and the FW-190!!

My estimates are included above chart. Probably the service planes had even lower values.

gripen

[Crumpp]

My estimates  are included above chart.  Probably the service planes had even lower values.

So what is the exact value that you are using to determine this so called "Margin of Error"?

The reason for the constant value on the above chart is only the show what's the problem with above mentioned generalized formulas; in most cases both give too high value of e.

BS

There is no problem with the generalized formulas.  You should listen to someone who has actually studied aeronautics instead of making up your own theories.  Write NASA and tell them Wood and the NACD are wrong.

It's not the "exact" number that is important but it's place relative to other wings on the same scale.  The "exact" number will change based on the formula and the scale used.  Assuming a constant value creates a false picture of this position by totally ignoring the design of the wing.

Now how is that good?  Except for the fact it artificially inflates the Spitfires standing on the scale.

The more of these design parameters you can account for the better.  This thread started out on the right track.  Too bad it was derailed by your wooden model which you attempted to pass off as the real thing.


Crumpp

[gripen]

Originally posted by Crumpp
So what is the exact blaah blaah...

Those are the numbers I got. I have given the sources as well as the quide lines for calculations. So far I have not seen you posting anything which proves that there is something wrong with the numbers I got.

For others, here is the drag polars for the F2A-3:



The interesting part is that they tested the F2A-3 also with flaps and the values of e are:

Flaps in 0,77
Flaps 22 deg open 0,80
Flaps 66 deg open 0,95

gripen

[Crumpp]

Those are the numbers I got.

Exactly.

So YOUR calculations using a constant value become the center point in which the "Margin of Error" is calculated.

Why...because you don't know the exact value and not having a drag polar off an actual aircraft are unable to even approximate it with any accuracy.

That's pretty funny!

The interesting part is that they tested the F2A-3 also with flaps and the values of e are:

They tested the FW-190 with and without prop, various power settings, landing gear up and down, and flaps in both positions.  It's several pages of polar plots and prop efficiency charts.

Crumpp

[gripen]

Originally posted by Crumpp
Blaah Blaah Blaah...

I don't see anything relevant in your post.

gripen

[JoOwEn]

this is why...