Why still no dive acceleration difference?

[Crumpp]

So, the aircraft weight plays a important role in a 45 degree dive.

Yes it does as does the lower drag of the FW-190.

If you use Take Off Weights, the equilibrium speed of the FW-190 is higher than the P47.

It does not matter though as neither aircraft can achieve equilibrium velocity as they are dynamic pressure and mach limited.

The P47 always has higher limits.

Blackberry, the reason i posted the charts i did, is because the weights you are using appear to be off.

Yes his weights are off and it makes a big difference. Thanks for catching that.

but your not going to be diving much at sea level me thinks!

No but don't think of it as sea level. It is EAS when we use sea level performance. Adjust it for density effects and it becomes actual performance.

As such it is good approximation of Indicated Airspeeds and delivers a very good prediction of the performance trends you should see.

[Crumpp]

HP as 1700.

At SL @1.42ata@2700U/min with allowance for dynamic pressure gains at equivilent airspeeds.

The 2300 hp for the R-2800 is at War Emergency Power and the 1980 hp for the BMW801D2 is at Erhohte Notleistung 1.58ata@2700U/min.

[Crumpp]

If we use a 7-ton P47D, again, P47 has more equilibrium speed.

You will not be comparing fighters....

You guys need a weight and balance chart for the P47 series?

[fruitbat]

Originally Posted by Crumpp

You guys need a weight and balance chart for the P47 series?

I'd be very interested in seeing one thanks

for the FW to clarify, is it 1,980 PS or 1,980hp?

this has turned into a very interesting thread imo.

[Crumpp]

1,980 PS

That is Horsepower.



BTW at the velocity we are talking about, the conversion form PS to HP is really irrelevant.

[fruitbat]

Thanks for posting the weight and balance chart

[BlackBerry]

Originally Posted by Crumpp

That is Horsepower.



BTW at the velocity we are talking about, the conversion form PS to HP is really irrelevant.

Fw190 A8 578km/h@SL 1953HP ,and what 's the weight you want to use? weight and balance chart for fw190A8?

p47D(R2800-59) 345mph=555km/@SL ,2300HP, and what 's the weight you want to use?

[BlackBerry]

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.

The P-47D-15 was produced in response to requests by combat units for increased range. The internal fuel capacity was increased to 375 U.S. gal (1,421 l) and the bomb racks under the wings were made "wet" (equipped with fuel plumbing) to allow a jettisonable drop tank pressurized by vented exhaust air to be carried under each wing, in addition to the belly tank. Five different auxiliary tanks were fitted to the Thunderbolt during its career: