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#291
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p47D(R2800-59) 345mph=555km/@SL ,2300HP, and what 's the weight you want to use? Last edited by BlackBerry; 06-07-2012 at 12:41 AM. |
#292
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Fw190A8
m----mass,4272 kg,max load for a standard A8,100% fuel A----dive angle p----engine output,1953 HP@sea level r----propeller efficeincy=0.8=80% Vmax ----max level speed at SL, 578km/h=160.6m/s d----drag coefficient g----gravity, 9.8 m/s^2 t---- engine thrust, N when level flys at Vmax, fw190a8 in equilirium, zero acceleratiom, all forces are balanced. t=P*r/V=1953*735*0.8/160.5=7155N m*g* sin(0)+t=d*Vmax^2 d=t/Vmax^2=7155/160.5^2=0.2778 P47D m----mass,5675kg, 12500lb, normal combat load(55% fuel) A----dive angle p----engine output,2300 HP r----propeller efficeincy=0.8=80% Vmax ----max level speed at SL 345mph=555km/h=154.3m/s d----drag coefficient g----gravity, 9.8 m/s^2 t---- engine thrust, N when level flys at Vmax, P47D in equilirium, zero acceleratiom, all forces are balanced. t=P*r/V=2300*735*0.8/154.3=8765N m*g* sin(0)+t=d*Vmax^2 d=t/Vmax^2=8765/154.3^2=0.3681 55% fuel P47D vs 100% fuel Fw190A8 in 45 dive degree for fw190A8, let the new equilirium speed as V: new engine thrust should be 7155*(160.5/V) m*g* sin(45)+t=d*V^2 4272*9.8*0.707+7155*(160.5/V)=0.2778*V^2 thus v^3-106548V-4133828=0 we get fw190A8 V=344.5m/s=1240km/h. And BMW801 still produce thrust as 7155*(160.5/344.5V)=3333N for P47D let the new equilirium speed as V: new engine thrust should be 8765*(154.3/V) m*g* sin(45)+t=d*V^2 5675*9.8*0.707+8765*(154.3/V)=0.3681*V^2 v^3-106818V-3674109=0 we get P47D V=342.8m/s=1234km/h=1240km/h of Fw190A8 And R2800-59 still produce thrust as 8765*(154.3/342.8 )=3945N 55% fuel P47D vs 100% fuel Fw190A8 in 65 dive degree for fw190A8, let the new equilirium speed as V: new engine thrust should be 7155*(160.5/V) m*g* sin(65)+t=d*V^2 4272*9.8*0.906+7155*(160.5/V)=0.2778*V^2 thus v^3-136538V-4133828=0 we get fw190A8 V=384m/s=1382km/h. for P47D let the new equilirium speed as V: new engine thrust should be 8765*(154.3/V) m*g* sin(45)+t=d*V^2 5675*9.8*0.906+8765*(154.3/V)=0.3681*V^2 v^3-136884V-3674109=0 we get P47D V=383m/s=1379km/h=1382km/h of Fw190A8 Conclusion: 1) 55% P47D and 100% fuel Fw190A8 share same equilirium speed at 45-65 degree dive. 2) 5675kg P47D=normal combat load, 4272kg Fw190A8=max load A8. There is only 0.454(12500-10700)=817kg difference between empty and normal load P47, but there is (4272-3050)=1222kg difference for fw190A8. In fact, 5675kg P47D is only 200 US galon fuel, not 100% fuel. if internal fuel is full(375 US gal), 6152kg for P47D which has better dive acceleration. Quote:
Last edited by BlackBerry; 06-07-2012 at 06:25 AM. |
#293
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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. |
#294
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anyone know what the Vne limits are for these planes?
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#295
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At low altitude: FW190 - 466 mph IAS P47 - 500 mph IAS |
#296
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thanks for the figures, appreciated. ![]() |
#297
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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. Last edited by BlackBerry; 06-08-2012 at 12:24 PM. |
#298
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Duh...
![]() damn Ti89 and my old brain. Quote:
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Just a thought, a generic modification of efficiency based on some basics of propeller design might enhance things. Problem is we don't have enough information on the propellers to implement any kind of detail. Last edited by Crumpp; 06-08-2012 at 01:02 AM. |
#299
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Efficiency: assume 3-blade naca16= 3-blade clark= 3-blade gotingen @0.7Mach 4-Blade naca of P47=88% at 0.4 Mach 3-Blade naca of P47=82% at 0.4 Mach (See P47 data I'v posted) 3-Blade naca of P47d=63% at 0.7 Mach (out of envelop,so <80%) see data posted 3-Blade naca/gottingen of fw190A5= 50% at 0.7 Mach (3.3 m diameter, bigger advance ratio) 3-Blade naca/gottingen of fw190A8= 40% at 0.7 Mach(wide blade even lose 8% at Vmax) 4-Blade naca of P47=?????% at 0.7 Mach????? It should be bigger than 63% or not? It seems that US has never unclassified the 4-blade NACA16 high Mach number wind tunnel data. ![]() If 70%, there is 30% efficiency advantage over fw190A8, this could explain a lot of 45 degree dive test in history. Last edited by BlackBerry; 06-08-2012 at 01:26 AM. |
#300
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And then, someone could simulate propeller efficiency by Xfoil or Ansys(software). I agree with you this is a hard work, 1C developping team needs a lot of money and time to do so. |
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