![]() |
#11
|
|||
|
|||
![]() Quote:
Fw190A8, 3.5 tons weight, 2000HP, prop efficiceny 80%, Vmax at sea level=580km/h=161m/s P47D,7 tons, 2000HP, prop efficiceny 80%, Vmax at sea level=550km/h=153m/s Vmax is a equilirium point. Formular is below, on the left side of formular is the thrust part, including engine thrust and gravity vector, on the right side,is air braking force which is proportional square of speed. g*sin(0)+enginethrust/mass == dragcoefficent*(580^2) Enginethrust=engineoutput*efficiecny/speed For fw190a8 at Vmax, level flight =dive angle with 0 Total thrust=enginethrust=2000*735*80%/161=7300N=21.3% of weight that is to say, with total thrust/ratio of 21.3%, fw190 could achieve 161m/s. When dive at 45 degree and assume double speed=322m/s. Add 70.7% gravity thrust/weight ratio, minus 10.6% engineThrust/weight ratio, finally we get 81.35% thrust/weight ratio, thus 81.35/21.3=3.82 times thrusts/weight as before, and 1.95 times of airspeed is needed to counteract. 1.95 is roughly equal to 2, so the new equilirium speed is around 580*2=1160km/h. Let's see P47, when Vmax level flight. Total thrust=enginethrust=2000*735*80%/153=7686N=11.2% of weight When dive at 45 degree ,assume the speed is 322m/s=2.1 times Speed before. Add 70.7% gravity thrust/weight ratio, minus 5.87% engineThrust/weight ratio, finally we get 76% thrust/weight ratio, thus 76/11.2=6.79 times thrusts/weight as before, and 2.6 times of airspeed is needed to counteract 2.6>2.1 so 322m/s air is not enough to counteract the increased thrust/weight ratio, so 322m/s is below P47 new equilirium speed. To sum up, when dive at 45 degree, P47 get more total thrust/weight ratio increase (6.79 times vs 3.82 times) than fw190a8, thus higher equilibrium speed. Last edited by BlackBerry; 06-05-2012 at 04:38 PM. |
|
|