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/