Why still no dive acceleration difference?

[BlackBerry]

Originally Posted by MadBlaster

It's good thinking. I'll take a shot.

I think if the props on both planes are nearing 450 TAS and running inefficiently, you must fall back to the drags of the planes themselves. We know that the wing loading on p47-22 is greater than fw 190A. We also know that fw190A out turns a p47-22 based on the fan plots. So, I think we can conclude that the p47 is simple more aerodynamically streamlined for diving (less draggy) and this is why it eventually catches up and surpasses the 190A in a dive. I don't think the p47 prop all of a sudden gets more efficient when it breaks the sound barrier, but I could be wrong about that. Anyway, it is not the weight of the p47, but more so that the 190 wing simply generates more lift and that creates a drag. Yes, there is a weight difference. But if both planes were shaped as same sized spheres and one is twice as heavy as the other, I think you won't get that much separation.

Also ot, don't ever dive after a p51 in 109. Climb, pursue and hope he turns. If he is diving away from his home base, you have him. Simply cut off the angle.

The "no-lift" drag coefficeint for P47D-37 is "0.0256",it's a constant below 0.8 Mach. The test 0f 1943 December between fw190G and P47D was definitely below 0.8 Mach. You can see that P47 has a big wing of 25.87 square meters:

P-47D-27 = 0.0256 * 25.87 = 0.662272

0.662272 is also a constant, if you want to get drag force of the wind, multiple speed^2:

0.662272X(250mph)^2

I have no fw190G's data, fw190G-1 based on A4; G-2 based on A5, they are both 1.42ata( 42").

let's put aside propeller's thrust at first, only gravity and wind drag(there are some induced force but I just calculate roughly).


gXweightXcos(60)-dragcoefficentX(speed)^2= (dive-accelaration)Xweight

that is

dive-accelaration=gXcos(60)-dragcoefficentX(speed)^2/weight

P47's weight is almost twice of fw190A4, so at 250mph speed it's almost impossible for fw190 to outdive P47 in il2 4.11m. But in real world, fw190 pulled away rapidly!

The only factor we didn't include is the detailed airscrew efficeiency curve espicielly when tip reachs 0.8-1.0 Mach and above.
Spitfire.LF.IXC
[Mass]
Empty 2650.0
TakeOff 3300.0

[Squares]
Wing 19.0
Aileron 1.32
Flap 2.125
Stabilizer 1.90
Elevator 1.20
Keel 0.85
Rudder 1.10

[Polares]
lineCyCoeff 0.092
AOAMinCx_Shift 0.0
Cy0_0 0.1
AOACritH_0 16.0
AOACritL_0 -17.0
CyCritH_0 1.4
CyCritL_0 -0.7
CxMin_0 0.0232
parabCxCoeff_0 5.4E-4




P-47D-27
[Mass]
Empty 4630.0
TakeOff 6583.0

[Squares]
Wing 25.87
Aileron 1.45
Flap 2.76
Stabilizer 3.50
Elevator 2.05
Keel 1.30
Rudder 1.10

[Polares]
lineCyCoeff 0.092
AOAMinCx_Shift 0.9
Cy0_0 0.17
AOACritH_0 16.0
AOACritL_0 -15.0
CyCritH_0 1.25
CyCritL_0 -0.8
CxMin_0 0.0256
parabCxCoeff_0 4.8E-4


Bf-109G-2 = 0.027 * 16.16 = 0.43632
Spitfire.LF.IXC = 0.0232 * 19.0 = 0.4408
P-47D-27 = 0.0256 * 25.87 = 0.662272



Someone says

On the R-2800 engines the Reduction Gear ratio was about 50%, the crankshaft would produce about 50% more power strokes per revolution than a direct drive engine. On the dash 21 (P-47 Thunderbolt) engine it was 16:8 ratio. On most other R-2800 engines the ratio is 16:9, 16 revolutions of the engine give 9 revolutions of the prop. The numbers have no common multiple, it's a vibration control function.

The R-1830 of the C-47, and R-2800, and that R-4360 are geared engines, notice how large the nose case is - where the "crankshaft" and the prop meet.

R-2800 engine, 2700rpm, 50% reduction for airscrew=1350rpm, 4m diametre. On the ground when engine at full rpm, the propeller's tip's rotating speed is:

3.14X4X1350/60=282m/s=282/340=0.83 Mach

Wow, it's seems that P47's designer just want to make the tip speed approach sonic as soon as posssible. Why? The supersonic state for airscrew's tip? We all know P47 was intently designed for high altitude escort where the sonic speed is samller than 340m/s on the ground, and P47 often dives at hight speed at high aititude, therefore P47's airscrew tip must often beyond 1 Mach.

airscrew=the twisted and rotating "wing" above 1 Mach, what does this mean in il2?

Again we analysis 1943's test.

(C)

(1) 10000 feet to 3000 feet, starting at 250 m.p.h., diving at angle of 65 degree with constant throttle setting. The FW-190 pulled away rapidly at the beginning but the P-47 passed it at 3000 ft with a much greater speed and had a decidedly better angle of pull out.

When p47 flew on 10000 feet@250 mph IAS,what's the speed of propeller's tip?

At 5,000' TAS = IAS + 9%
At 10,000' TAS = IAS + 16%
At 15,000' TAS = IAS + 25%
At 20,000' TAS = IAS + 36%
At 25,000' TAS = IAS + 49%
At 30,000' TAS = IAS + 64%

250 mph IAS=290mph TAS=130m/s, rotating speed is 282m/s, combination speed is 310m/s, Mach number=310/328=0.945Mach

When slam throttle full forwards and dives 60 degree, P47's airscrew will probably be the first one to suffer from sonic barrier.0.95-1.0 Mach. This is probably the reason why P47 was outdived by fw190G from 250 mph(initial diving stage). As speed building up to 650km/h or so (3000ft altitude), mach number=1.

(Probably)Fw190's airscrew tip entered 0.9-1.0 Mach later than P47, that's why 190 outdove P47 at the begining, but when both of them were all suffering from low airscrew efficiency at high speed, P47 will gain on 190, the formula I'v posted above demonstrates this clearly.


When P47 dives to 7500 altitude @800 km/h TAS, and tip mach number is 1.16. Hamilton standard airscrew is NACA-16 series which is laminar flow airfoil.

<<Static characteristics of Hamilton Standard propellers having Clark Y and NACA 16 series blade sections>>
http://digital.library.unt.edu/ark:/...etadc62146/m1/

[MadBlaster]

hmm, before I read further, I think we need zero lift drag coefficient for 190 to say that weight is the deciding factor. wiki says 27.87 m² for p47 wing area and 18.30 m² for 190A. Agree with you that denominator (i.e., weight) is almost twice as large for p47 verses 190, but numerator? To keep simple math, assume zero drag coeff=1 for both planes, 190 weighs "1" and p47 weighs "2" (weight on relative basis to each other). Then drag coefficient portion of numerator

-dragcoefficentX(speed)^2/weight

where you did this -> P-47D-27 = 0.0256 * 25.87 = 0.662272 (i assume your using 25.87 for wing area)

(i.e., use 1 instead of .0256 and 1 for speed since same for both planes and 2 for weight p47 and 1 for weight of 190)

27.87/2 (p47) or 18.3/1 (190) is bigger??? The latter is bigger, and since it is a subtraction from this gXcos(60), wing area and/or differences in zero drag coeff may be the deciding factor in the calculation of dive acceleration, not the weight. And if this is the case, dive acceleration is less for 190 than p47. Sorry,if this is confusing. It's late here.

[BlackBerry]

Originally Posted by MadBlaster

hmm, before I read further, I think we need zero lift drag coefficient for 190 to say that weight is the deciding factor. wiki says 27.87 m² for p47 wing area and 18.30 m² for 190A. Agree with you that denominator (i.e., weight) is almost twice as large for p47 verses 190, but numerator? To keep simple math, assume zero drag coeff=1 for both planes, 190 weighs "1" and p47 weighs "2" (weight on relative basis to each other). Then drag coefficient portion of numerator

-dragcoefficentX(speed)^2/weight

where you did this -> P-47D-27 = 0.0256 * 25.87 = 0.662272 (i assume your using 25.87 for wing area)

(i.e., use 1 instead of .0256 and 1 for speed since same for both planes and 2 for weight p47 and 1 for weight of 190)

27.87/2 (p47) or 18.3/1 (190) is bigger??? The latter is bigger, and since it is a subtraction from this gXcos(60), wing area and/or differences in zero drag coeff may be the deciding factor in the calculation of dive acceleration, not the weight. And if this is the case, dive acceleration is less for 190 than p47. Sorry,if this is confusing. It's late here.

27.87/2=13.93 much smaller than 18.3, so P47 has smaller subtraction from this g*cos(60).


BTW, this link says VDM 3m diametre has a 0.54 prop reduction gearing. Fw190A8 :VDM 9-12176A10 ft, 11 ¾ in. diameter 390 lbs

BMW-801D 2700rpm

FockeWulf190BMW 801D radialVDM prop mechanism was built into the nose case of the engine itself, along with 1.72 to 1 cooling fan drive, 0.54 prop reduction gearing, magneto, oil pump, and front camshaft drive.

http://www.enginehistory.org/Convent...20Electric.pdf

So in 1943's test, @10000ft ,250mph IAS, fw190A4's propeller tip speed=(254^2+130^2)^0.5=286m/s=0.87Mach, less than 0.95 Mach of P47's.

Some P47's has 16:9 reduction which provides higher Mach number (1.05Mach ).

In conclusion, My opinion is that when p47 and fw190a4 at full engine 2700rpm dive from @10000ft ,250mph IAS, fw190's tip speed is about 0.87mach while p47's is around 0.945-1.05 Mach. Probably at that time P47's propeller's efficiency is quite lower than fw190, so p47 was outdove rapidly at initial diving stage.
111.JPG

[MadBlaster]

Your conclusion seems reasonable to me. I was digging into that hamilton clark link, it says the machs for best efficiencies were in the .7-.9 range. I would assume that plays out at cruising speeds.

[Crumpp]

Wow, it's seems that P47's designer just want to make the tip speed approach sonic as soon as posssible. Why?

4 meter diameter compared to a 3 meter diameter....

4 meters is a big prop and they have to push the tip speeds. Keep in mind, diameter is the most important factor in propeller design.

Good design can compensate, though.

it says the machs for best efficiencies were in the .7-.9 range.

They were not dumb at Republic!

This is probably the reason why P47 was outdived by fw190G from 250 mph(initial diving stage).

I did not check over your math but your final conclusion is absolutely right.

[BlackBerry]

Crumpp, thank your comments，I think I've found the answer！ the NACA report of Hamilton standard tells us everything：1350rpm propeller 13ft diameter CSP just like P47's with the exception of 3-blade vs 4-blade.

http://digital.library.unt.edu/ark:/...etadc62146/m1/

There were two airfoil being tested， Clark Y airfoil was before WWII, not laminar. NACA16 was during WWII, NACA16 airfoil is laminar flow profile, and the test shows that there is no advantage of NACA16 airfoil when propeller's tip speed is 1Mach and when "advance ratio" is above 2.0, but there is no more than 3% efficiency benefit from NACA16 when "advanc ratio" is between 1.2 and 2.0.

advance ratio= TAS/(rpm*diameter)

fw190:3 meter propeller, 1400 rpm
p47d: 4 meter propeller, 1400 rpm

You can see Figure 24, when tip speed is 1 Mach, the more advance ratio, the lower efficiency. So we now come to know why Repulic engineer wanted as big propeller as possible because they wanted smaller advance ratio! When fw190 and P47 dive to same high TAS, the P47 has smaller advance ratio and higher efficiency AS LONG AS BOTH PROPELLER'S TIP SPEED IS AROUND 1 MACH. Republic engineers were right: since high TAS diving(efficiency loss)is inevitable for P47, why not prefer low "advance ratio" while accepting the high mach number of 4 meters big propeller's tip?

The complete formular of diving 65 degree is below:

acceleration=g*cos(65)-dragcoefficent*(TAS)^2/weight+Propellerthrust/weight

A simple math question: if you are P4D7's pilot fighting against a fw190G(both arr 250mph TAS @10000ft ), how can you get higher dive acceleration? On the right of formular there are three parts:

1) g*cos(65)
You have nothing to do with it, every a/c shares same value.

2) dragcoefficent*(TAS)^2/weight

Your huge weight is your advantage, and the bigger TAS, the more important role this part plays. So you should build up speed ASAP.

3) Propellerthrust/weight

Unfortunately the third part is your enemy's advantage. Although P47's efficiency is almost same as fw190's, your huge weight is your shortcoming.The NACA report says when tip speed is above 0.9 mach, the drag coefficient of tip increase rapidly, and when tip speed is 1.0 mach and advance ratio is above 2.0, the efficiency of propeller drops sharply. @250 IAS ,p47's advance ratio is about 1.32, not high, but as speed slightly building up, efficiency drops quicker than 0.5-0.8 Mach curve.

http://digital.library.unt.edu/ark:/...dc62146/m1/43/
http://digital.library.unt.edu/ark:/...dc62146/m1/13/
http://digital.library.unt.edu/ark:/...dc62146/m1/11/

You are now 250mph TAS @10000ft, your tip speed is 1 mach, and you will suffer from compressibility loss while your enemy dose NOT. The fw190 has more thrust than you, his weight is less than you, therefore his thrust/weight is much greater than you so that he can overcome your advantage----the second part of formular.

What should you do ? The answer is very simple and as same as the conclusion we've got from 2nd part of formular:

BUILD UP SPEED ASAP.

Drag him down!!! Yes, your tip speed will be always above 1 mach but now fw190's is also around 1 mach, he is now suffering from compressibility just like you, furthermore, his advance ratio(J) will much bigger than you, so his propeller efficiency drops more sharply than you. Now, the third part of formular is NOT enemy's advantage any more. You've succeded in eliminating his advantage and retaining and enlarging yours.

Congratulations from Republic engineers! You now have energy advantage by diving to high TAS, you are extending your distant now, do what you want to do.

The last thing is that if il2 4.11 models the compressibility loss of propeller efficiency . If not, there are 2 probem with its FM.

1) every piston a/c dives faster than it shoud be @ high TAS.

2)For those a/c like P47, the advantage of high TAS diving acceleration has been ignored, so is it's Low TAS diving accelaration shortcoming.

Now we can perfectly explain the fact of 1943 Dec test between fw190G and P47D, and many other comparation such as spitfire vs bf109 initial and final diving difference.

BTW, it's stupid for P47 to dive in a shallow angle with zeke which demonstrates the 6 ton thunderbolt has only a littile advantage(100yards)to the "kite"----Zeke52. I can image those angry faces of the Republic's engineers.


As for P51d, He has similar high TAS diving acceleration with P47, but the reason is not much depending on huge weight, it is very low coefficient of laminar wing. pls look at 2nd part of formular.

[MadBlaster]

about this part of the equation, Propellerthrust/weight ->

Simply lower the rpms to bring the p47 tips speeds down to optimal range .7-.9? Won't that give you better acceleration? It seems to work that way in game. Starting at 250 mph, full throttle, and 100% pp, nose trim 2 notches down, rads closed. if I nose it down into dive and crank down the prop pitch to 0% quickly and then bring it back up to ~77% and gradually lower throttle to about 77% in a dive from 250 mph, the planes gets to ~ 400 mph ias very quickly.

[BlackBerry]

Originally Posted by MadBlaster

about this part of the equation, Propellerthrust/weight ->

Simply lower the rpms to bring the p47 tips speeds down to optimal range .7-.9? Won't that give you better acceleration? It seems to work that way in game. Starting at 250 mph, full throttle, and 100% pp, nose trim 2 notches down, rads closed. if I nose it down into dive and crank down the prop pitch to 0% quickly and then bring it back up to ~77% and gradually lower throttle to about 77% in a dive from 250 mph, the planes gets to ~ 400 mph ias very quickly.

Yes, that works. But even you can get same propeller efficiency as 190, your weight is too great to overcome, Propellerthrust/weight is still inferior to 190's.

We need to know if 190's airscrew tip compressibility loss is modelled or not at high mach number where 190's huge "advance ratio" making efficiency even worse than P47's.

[MadBlaster]

agree, but better to do it than not if your p47 guy. for 190 guy, you have revealed something that I think goes ignored. sometimes it is good to over ride the vdm auto prop pitch control and go to manual mode for same reason, to keep tip speeds at optimal ranges.

[mayshine]

Thanks to BlackBerry and everyone,

all your threads have done greatly to help me to understand more.

and may be there is a chance to involve the Prop calculation in IL2

to knock down Cliff of Dover and be the most perfect all the way?

Sorry for offensive language to someone l

I am tricky and want this topic to enjoy a long life and drag more men
to research in their full strenth,

So that I deliberately use somewhat offensive language.

sorry again JTD or so