Looking for Spitfire I sustained turn rate charts

[Charge]

4x2745kg=10980kg So that is the weight the wings should be able to lift and the engine to pull through air with no speed loss?

Or at 3,7Gs 10157kg? If that requires some AoA of the wing so there must even be some excess thrust to overcome the drag and still maintain constant speed?

-C+

[Westy]

 Everytime I see Vulcan get into a "discussion" with the sycophants "over there" I start hearing the lyrics to "The Ballad of Brave Sir Robin."

[HoHun]

Hi Angus,

>I did once a Spit I and 109E climb calculations, - i.e. climb converted to NM and NM divided with time.

Sounds interesting! :-) But what is NM?

Regards,

Henning (HoHun)

[Angus]

Umm...what is the definition of a " maximum sustained turn"??

I always thought it was the maximum turn one could sustain without losing altitude.

BTW, does a wide chord give a benefit under banking angles and a turn rather than span?

Regards
Angus

PS...quick edit:
What is NM?

NEWTON METERS

[HoHun]

Hi F4UDOA,

>Hate to be a best but could you show the calculation you preformed to get your results??

I'll mail you my spreadsheet if you like, but it's hell made from numbers. It's very hard to use, and much easier to use than to understand. That's not meant to discourage you, just to prevent disappointment :-)

>How do you know where to find the best sustained turn based on Clmax and engine power at alt?

The best sustained turn is achieved at Clmax. I simply iterate the speed until I find a turn at Clmax where the excess power is zero.

>There should be a simple ratio to come up with a turning "index" like Clmax / wingloading * powerloading.

There's a simple way of comparing fighters with identical wings, like the variants of a specific type. If you're interested, I could outline that method, which is useful for quick estimates.

Regards,

Henning (HoHun)

[HoHun]

Hi Charge,

>Or at 3,7Gs 10157kg? If that requires some AoA of the wing so there must even be some excess thrust to overcome the drag and still maintain constant speed?

Interesting perspective :-)

But yes, a 10-ton-Spitfire could sustain a speed close to that mentioned above, flying at emergency power at the edge of the stall.

I wouldn't like to be the pilot, though ;-)

Regards,

Henning (HoHun)

[HoHun]

Hi Angus,

>Umm...what is the definition of a " maximum sustained turn"??

>I always thought it was the maximum turn one could sustain without losing altitude.

Without losing either altitude or speed.

>BTW, does a wide chord give a benefit under banking angles and a turn rather than span?

A long-span wing is more efficient with regard to induced drag.

>NEWTON METERS

Hm - did you calculate torque or energy?

Regards,

Henning (HoHun)

[Angus]

Torque would be energy divided with time right?

Anyway, I did total energy and energy on a timescale.

[gwshaw]

I do the calculations at rated altitude for the different manifold pressures.

For the P-40E

At 8600 lbs, full fuel load
80% propellor efficiency

42 in Hg mil 1150 hp @ 12,000 ft
2.25 G sustained between 178 mph and 188 mph IAS (about 214 - 226 mph TAS)

56 in Hg WEP 1490 hp @ 4300 ft
2.85 G sustained between 180 mph and 232 mph IAS (about 192 - 247 mph TAS)

60 in Hg WEP 1580 hp @ 2500 ft
3.05 G sustained between 196 mph and 229 mph IAS (about 203 - 238 mph TAS)

I'll have to look into calculating rate and radius as well. Don't remember the formulas offhand.

It looks like HoHun and I are doing things a little differently. He is calculating the speed a max Cl turn can be sustained. I'm calculating the max G that can be sustained at a given power and weight.

Greg Shaw

[HoHun]

Hi Angus,

>Torque would be energy divided with time right?

Torque and energy are the same dimension, actually.

>Anyway, I did total energy and energy on a timescale.

Hm, I hadn't thought of that angle! But what is the gain over a pure altitude-over-time graph? :-)

Regards,

Henning (HoHun)

[HoHun]

Hi Greg,

>For the P-40E

>At 8600 lbs, full fuel load
>80% propellor efficiency

>42 in Hg mil 1150 hp @ 12,000 ft
>2.25 G sustained between 178 mph and 188 mph IAS (about 214 - 226 mph TAS)

I get 16.7 °/s @ 3700 m for a P-40E at 3911 kg/44" Hg.

Your data indicates 13.1 - 12.4 °/s for 42" Hg.

>56 in Hg WEP 1490 hp @ 4300 ft
>2.85 G sustained between 180 mph and 232 mph IAS (about 192 - 247 mph TAS)

I get 20.9 °/s at 4000 ft for a P-40N-1 at 3447 kg/57" Hg.

Your data indicates around 16.2 °/s, but the P-40N I calculated is considerably lighter so it's no suprise your high-powered P-40E doesn't turn as quickly.

>I'll have to look into calculating rate and radius as well. Don't remember the formulas offhand.

T360° = 2 * pi * v / sqrt (a^2 - g)

>It looks like HoHun and I are doing things a little differently. He is calculating the speed a max Cl turn can be sustained. I'm calculating the max G that can be sustained at a given power and weight.

If you're interested, we could swap spreadsheets :-) Your engine model seems really neat, much smarter than the tabulated data I'm using!

Regards,

Henning (HoHun)

[gwshaw]

Originally posted by HoHun

>For the P-40E

>At 8600 lbs, full fuel load
>80% propellor efficiency

>42 in Hg mil 1150 hp @ 12,000 ft
>2.25 G sustained between 178 mph and 188 mph IAS (about 214 - 226 mph TAS)



I get 16.7 °/s @ 3700 m for a P-40E at 3911 kg/44" Hg.

Your data indicates 13.1 - 12.4 °/s for 42" Hg.

That is probably due to my speed range being about 15-20 mph faster than your figure. I'm hitting the Cl limits about the same speed you are, just a different way of looking at things.

>56 in Hg WEP 1490 hp @ 4300 ft
>2.85 G sustained between 180 mph and 232 mph IAS (about 192 - 247 mph TAS)


I get 20.9 °/s at 4000 ft for a P-40N-1 at 3447 kg/57" Hg.

Your data indicates around 16.2 °/s, but the P-40N I calculated is considerably lighter so it's no suprise your high-powered P-40E doesn't turn as quickly.

[/color]
Yah, about 14% heavier, but only about 11-12% more power. Higher induced drag, and lower power/weight, it shouldn't turn as well as the P-40N.

>I'll have to look into calculating rate and radius as well. Don't remember the formulas offhand.


T360° = 2 * pi * v / sqrt (a^2 - g)

[/color]
Cool, what is the a?

>It looks like HoHun and I are doing things a little differently. He is calculating the speed a max Cl turn can be sustained. I'm calculating the max G that can be sustained at a given power and weight.


If you're interested, we could swap spreadsheets :-) Your engine model seems really neat, much smarter than the tabulated data I'm using!

Regards,

Henning (HoHun)

[/color]

Sure, I will shoot you of an email.

Greg Shaw

[gwshaw]

Originally posted by HoHun
If you're interested, we could swap spreadsheets :-) Your engine model seems really neat, much smarter than the tabulated data I'm using!

Regards,

Henning (HoHun) [/B]

My current model is pretty simple, being little more than a Standard Atmosphere chart with delusions of grandeur. But is has proven surprisingly accurate, generally within +-1% of published charts.

Do an Allison F3/4R for a quick example.
1150 hp @ 12,000 ft on 42 in Hg.

We know from the Standard Atmosphere that static pressure at 12,000 ft is 19.03 in Hg and abs temp is 264.56K. That is all the information we need to guesstimate power from SL on up.

Static pressure - 19.03 in Hg
Manifold pressure - 42 in Hg
Blower pressure ratio - 42 / 19.03 = 2.21

From there you can multiply static pressure at any altitude by 2.21 to get the max MAP the blower can provide.

ie SL pressure 29.92 * 2.21 = 66.1 in Hg max MAP
SL temp 288.36K

HP is proportional to MAP * rpm. As long as rpm stays the same that provides the starting point to calculate hp.

66 / 42 * 1150 = 1805 hp

Take temp difference into account:

sqrt (264.56/288.36) = .958

.958 * 1805 = 1730 hp

That is actually about 1.5% below the 1760 hp @ SL on the Allison charts. But that could be just due to slightly different Standard Atmospheres, and is within a decent margin of error anyways.

I'm working on a more detailed calculator, using info from Hooker's "Not Much of an Engineer," but this will do for now.

Greg Shaw

[Badboy]

Just peeked in to see what was happening... Just love this stuff

Badboy

PS
This on the AH Spitfire Mk 1 from another thread, scrole down to my post:

http://www.hitechcreations.com/forums/showthread.php?threadid=103594&referrerid=2314

[Angus]

FromHoHun:

"Hi Angus,

>Torque would be energy divided with time right?

Torque and energy are the same dimension, actually.

>Anyway, I did total energy and energy on a timescale.

Hm, I hadn't thought of that angle! But what is the gain over a pure altitude-over-time graph? :-) "

Ok. I'll explain what I did.
The original concept was to calculate the wing lift efficiency.
The Spitfire MkI (87 oct) and the 109E (87 oct) were the best candidates I could find, since the engine power and weight were very similar. (Actually both rather favouring the 109 from stats)
So, I calculated the mass to Newtons, then onwards with mass to altidude and those divided with the time it took to pull the mass up to 10K and 20K.
For fun, one can play about with this pr. hp. of engine power. Since that is not really my cup of tea, I didn't do that much.
So, looking at the outcome, the 109 was quicker to any altitude in real life, the difference being less marked in Newtons. (Spitfire heavier, so there was more mass lifted to same alt)
UNTIL.......
I tried a Spitfire I with a 3-blade CS airscrew. (The other two were fixed pitch, I rather think one of them was a 2-blade, but not sure)
The Spitfire with 87 oct and an early type Rotol was considerably superior to the 109 regarding climb, be it real time or Newtons, the difference being more marked in Newtons.
I am not good enough at this math to conclude how much the difference could have been had the both aircraft been the same weight. But it really struck me never the less, and for this there can be basically two explanations.

1. The Merlin is quite more powerful than the DB
(109E and Spit I on 87 oct)

2 The Wing of the Spitfire in question generates quite some more lift than the wing of the 109 given roughly the same thrust.

I rather favour number two here as a logical explanation.

Anyway, HoHun, I'll mail this to you if you are interested.

Best regards

Angus