Why still no dive acceleration difference?

[BlackBerry]

Originally Posted by Crumpp

This is wrong. You are confusing terminal velocity with aceleration.

At Vmax:

FW190

9418lbs * sin 45 = 6660lbs excess thrust

a = F/m

m = 9418lbs/32.2 = 292 lb-s^2/ft

a = 6660lbs/292lb-s^2/ft

a = 22.8 ft/s^2 for the Focke Wulf

P47D22:

13500lbs * sin 45 = 9546lbs

a = F/m

m = 13500lbs/32.2 = 419 lb-s^2/ft

a= 9546lb/419lb-s^2/ft

a = 22.78 ft/s^2 for the P47D22

That is the best case scenario for the P47D22.

If we dive from say, 260 KEAS, then we see 25.35 ft/s^2 from the FW190 and 24.4ft/s^2 from the P47.

The FW190 has a .95ft/s^2 advantage in aceleration rate.

In your opinion, when dive from Vmax at 45 degree(from SL to a deep valley),

excess thrust=weight vector=weight*sin(45)

for any aircraft with mass=m, there is

a=weight*sin(45)/mass=32.2*sin(45)=22.8ft/s^2

for a 10lb plane, a=22.8
for a 100lb plane,a=22.8
for a 1000lb plane, a=22.8

In my opinion, when dive from Vmax at 90 degree(from SL to a deep valley),

excess thrust=weight vector+ engine thrust - drag force

at Vmax:
For full loaded fw190A8:

a = 9.8+ 7155/4272-0.2778*(160.5 ^2)/4272= 9.8+1.67- 1.67=9.8m/s^2= 32.2ft/s^2

For full loaded P47D:

a = 9.8+ 8767/6152-0.3681*(154.3 ^2)/6152= 9.8+1.425- 1.425=9.8m/s^2= 32.2ft/s^2

when speed building up to 720km/h=200m/s.......

For full loaded fw190A8:

a = 9.8+ 7155*(578/720)/4272-0.2778*(200 ^2)/4272= 9.8+1.35- 2.6=8.55m/s^2


For full loaded P47D:

a = 9.8+ 8767*(555/720)/6152-0.3681*(200 ^2)/6152= 9.8+1.10- 2.39=8.51m/s^2


when speed building up to 800km/h=222m/s......


For full loaded fw190A8:

a = 9.8+ 7155*(578/800)/4272-0.2778*(222 ^2)/4272= 9.8+1.21- 3.2=7.81m/s^2


For full loaded P47D:

a = 9.8+ 8767*(555/800)/6152-0.3681*(222 ^2)/6152= 9.8+0.989- 2.95=7.84m/s^2


when speed building up to 850km/h=236m/s......


For full loaded fw190A8:

a = 9.8+ 7155*(578/850)/4272-0.2778*(236 ^2)/4272= 9.8+1.14- 3.62=7.32m/s^2


For full loaded P47D:

a = 9.8+ 8767*(555/850)/6152-0.3681*(236 ^2)/6152= 9.8+0.93- 3.33=7.4m/s^2


Conclusion, P47D has slightly better dive acceleration when reaching 750-850km/h if both propeller efficiency=80%. I think this is the il2 FM method, if you test both in il2 4.11m, You'll find slightly dive acceleration difference.

However, if fw190A8 propeller efficiency drops from 80% to 50% at 850km/h,

For full loaded fw190A8:

a = 9.8+ (50%/80%)7155*(578/850)/4272-0.2778*(236 ^2)/4272= 9.8+0.71- 3.62=6.89m/s^2 quite smaller than 7.4m/s^2 of P47D.

if fw190A8 propeller efficiency drops from 80% to 50% at 800km/h,

For full loaded fw190A8:

a = 9.8+ 50%/80%*7155*(578/800)/4272-0.2778*(222 ^2)/4272= 9.8+0.76- 3.2=7.36m/s^2 quite smaller than 7.84m/s^2 of P47D.

From fw190A8, wide chord propeller was used in fw190, the new wide propeller outperforms old narrow chord cousin at low -medium speed, but is inferior to old one at high speed(>= vmax).

If we assume that there is 0.5 m/s^2 difference between Fw190A8 and P47D along the 20 seconds dive, in the end of dive, P47D is 10m/s=36km/h faster. In fact, there is quite some air compressibility at 800kmh(500mph) where aircrafts need 200HP+ to overcome the increase of air darg coefficient which is NOT a constant any more. The more air darg coefficent, the more dive acceleration advantage for P47D due to bigger weight.


Conclusion, the il2 FM(without Mach number) is NOT suitable for simulating the high speed dive because of air compressibility which influences the propeller efficiency and air-wing drag coefficient.