Author Topic: Explain this and win the prize! (Read 23957 times)

Crumpp · « **Reply #75 on:** September 17, 2004, 10:01:07 PM »

Gripen,

You list the Spitfires Taper Ratio as elliptical. It is not. The Spitfires TR should be around .67.

http://www.onemetre.net/Download/Downwash/Downwash.htm

According to this site the TR should be around .4 for elliptical performance:

Quote

A tapered wing concentrates the strength of the wing near to the center. Tapered wings approximate elliptical wings and are thus slightly more efficient than are straight wings. A taper ratio of 0.4 most closely approximates the elliptical. Highly tapered wings can suffer undesirable reynolds numbers causing the tip of the wing to stall earlier than the root at low speeds.

http://ourworld.compuserve.com/homepages/JHopkinson/wing2.html#WTR

Crumpp

Crumpp · « **Reply #76 on:** September 21, 2004, 05:58:52 AM »

Just heard back from Flugwerk.

They are building FW-190A8/N's and have restored several. According to them the "e" factor from FW documentation is:

Quote

I was checking my documents for the FW190 and the tip efficiency factor, or rather the Oswald Ellipse factor is about 1.15 or 0.87 depending on how you look at it. Basically it has about 15% more induced drag than a theoretical elliptical lift distribution.

Crumpp

gripen · « **Reply #77 on:** September 30, 2004, 04:11:40 AM »

Quote

Originally posted by Crumpp

You list the Spitfires Taper Ratio as elliptical. It is not. The Spitfires TR should be around .67.

I don't know if you understand at all what your source says. Basicly there is no exact taper ratio for the elliptical wing, one can calculate theoretical value but it has very little to do with the lift distribution and Oswald's efficiency.

Quote

Originally posted by Crumpp
According to this site the TR should be around .4 for elliptical performance:

Actually your source says that in the case of the tapered wing "A taper ratio 0,4 most closely approximates elliptical ".

That also rises an interesting question, who is right in this case. Your source or Badboy and professor Wood who claim that "taper ratio of 0.57 which yields an almost elliptical lift distribution"?

Who is right; Messerschmitt documentation which claims taper ratio 0,44 for the Bf 109G or the Aeronautical engineering instructor who claims taper ratio 0,52 for the Bf 109?

Quote

I was checking my documents for the FW190 and the tip efficiency factor, or rather the Oswald Ellipse factor is about 1.15 or 0.87 depending on how you look at it. Basically it has about 15% more induced drag than a theoretical elliptical lift distribution.

Who has actually written this?Are these values for the entire airframe (as I have calculated) or for the wing only? What is the Cl range of the measurements?

gripen

Crumpp · « **Reply #78 on:** September 30, 2004, 04:05:03 PM »

Quote

Who is right; Messerschmitt documentation

Quote

Who has actually written this?Are these values for the entire airframe (as I have calculated) or for the wing only? What is the Cl range of the measurements?

Focke-Wulf Gmbh., Bremen wrote it. It's the drag polars and it comes from this report.

Sent Pyro a copy.

Crumpp

gripen · « **Reply #79 on:** September 30, 2004, 05:13:18 PM »

Ah, so apparently you have the drag polars. Now just calculate e factor as described in the Perkins&Hage or the documentation might contain value of the K.

But actually I was asking which of your sources is right on taper ratio because you seem to use quite contradictory sources?

gripen

Crumpp · « **Reply #80 on:** September 30, 2004, 06:05:13 PM »

Quote

Now just calculate e factor as described in the Perkins&Hage or the documentation might contain value of the K.

It does.

The e factor averages out too:

Quote

0.87

You answer this:

Quote

But actually I was asking which of your sources is right on taper ratio because you seem to use quite contradictory sources?

here:

Quote

Basicly there is no exact taper ratio for the elliptical wing, one can calculate theoretical value

One thing is certain. The Spitfire did not have an truely elliptical wet lifting shape on it's wingtips as David Lednicers report concludes. Based on what TR you use will vary your calculations.

What the drag polars show us is that Zigrat does good work. His spreadsheet draws the correct conclusions in this case.

The Spitfire averages .88 and the FW-190 averages .87.
The FW-190 also always has less parasitic drag than the Spitfire.

Pyro and HTC have a copy of the report. I am sure their calculations will confirm this too.

Crumpp

Crumpp · « **Reply #81 on:** September 30, 2004, 06:11:33 PM »

Hey,

Do I get the prize?

Crumpp

joeblogs · « **Reply #82 on:** September 30, 2004, 06:54:31 PM »

My copy came today. Will take some time to digest.

One concern I have about all this debate is the acceptable margin of error. A lot of the engineering calculations substitute constants for other (sometimes unknown) equations. This was apparently good enough for the engineers at the time, but I worry the debate is running on such fine lines that it might affect the conclusions we reach.

-Blogs

Quote

Originally posted by gripen
So above you actually say that the lift coefficient of the swept wing will be lowered at given speed, that means that the angle of attack must be increased to reach given lift at this speed if compared to the straight wing and that means higher induced drag at given speed. Great logic.

If you know the Cd at the given Cl and AR, the e factor is easy to calculate. All what is needed is the drag polar of a given plane. If you can't figure out how, check Perkins&Hage or use solver.

Angus,
Please look other side's arguments a bit more closely.

gripen

Crumpp · « **Reply #83 on:** September 30, 2004, 07:01:02 PM »

Quote

One concern I have about all this debate is the acceptable margin of error. A lot of the engineering calculations substitute constants for other (sometimes unknown) equations. This was apparently good enough for the engineers at the time, but I worry the debate is running on such fine lines that it might affect the conclusions we reach.

I agree with you on this Joe. This debate did run on some fine lines.

Anyway, as far as I am concerned it is over. The question is answered.

Been pooring over my copy of Perkins&Hage, great info.

Crumpp

gripen · « **Reply #84 on:** September 30, 2004, 11:27:08 PM »

Crumpp,
I don't see the documentation here on how did you reach the e factor value 0,87 (or 1,15). According to Lednicer's span loading chart, the lift distribution of the Fw 190 was a bit worse than the lift distribution of the P-51. And the e factor of the P-51 seems to be around 0,75-0,8 so who is right, Lednicer or you?

What is the value of the K in the Fw drag polar?

And still no answer for contradictory taper ratio claims?

gripen

Crumpp · « **Reply #85 on:** October 01, 2004, 07:39:47 AM »

Quote

I don't see the documentation here on how did you reach the e factor value 0,87 (or 1,15).

No you don't. I will share it in the same manner you shared the polar plots you have in your possession.

Quote

And the e factor of the P-51 seems to be around 0,75-0,8 so who is right, Lednicer or you?

Ever consider a third possibility?

Quote

And still no answer for contradictory taper ratio claims?

It is answered, see above.

Crumpp

gripen · « **Reply #86 on:** October 21, 2004, 09:12:03 AM »

Here is the data Crumpp apparently don't want us to see:

It gives directly the value of the K at high speed and in the climb so the value of the e is very easy to calculate. At high speed the Cl is very low and probably not at linear stage of the Cd/Cl^2 curve so the climb speed value (1,24) is the one we are interested:

CwF = Cw0 + 1,24 * Ca^2
=>
Cd = Cd0(f) + 0,0678 * Cl^2
=>
0,0678 = 1 / (pi * AR * e)
=>
e = 0,78

At high speed the e value is 0,83.

Quote

Originally posted by Crumpp
No you don't. I will share it in the same manner you shared the polar plots you have in your possession.

Now you are creating double standards; I have given the source as well as place to get for every dataset I have used in this thread. In addition I have given directly the shape of the curves and the calculations. You have not given your data nor how did you reach your numbers.

Quote

Originally posted by Crumpp
Ever consider a third possibility?

All I see here is that even your own sources don't support your own arguments.

Quote

Originally posted by Crumpp
It is answered, see above.

I don't see your answer above. You have quoted two values, 0,4 and 0,52, which is right?

gripen

Crumpp · « **Reply #87 on:** October 24, 2004, 12:20:42 PM »

Your calculation for the high speed "e" factor of the Spitfire:

Quote

Cl 0,6 0,847682793

Your calculation for the high speed "e" factor of the FW-190

Quote

At high speed the e value is 0,83.

I think you don't have the whole report. If you compared the Spitfires climb "e" factor, I am pretty sure they would exhibit a similar spread.

Crumpp

Angus · « **Reply #88 on:** October 24, 2004, 01:51:39 PM »

Always bear in mind, that if calculations show that an aircraft should do something it doesn't, something could be wrong with the formula.

Edited due to rude wording originally.

gripen · « **Reply #89 on:** October 28, 2004, 03:06:34 AM »

Quote

Originally posted by Crumpp
Your calculation for the high speed "e" factor of the Spitfire:
...
Your calculation for the high speed "e" factor of the FW-190

It would help a lot if you could understand what we are talking about; the K value of the FW 190 at high speed is for the Cl value about 0,17 ie not for linear stage of the Cd/Cl^2 curve while the e value for the Spitfire you quoted is for Cl 0,6.

Basicly you continously quote numbers and link pages without understanding at all what do these mean.

Quote

Originally posted by Crumpp
I think you don't have the whole report.

Here is another example of your logic; using same logic I could say that you have only the headlines of the sheet because that is the part you linked above.

We need just values of the K for this thread and I linked them above, you did not.

gripen

Aces High Bulletin Board