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It is time to rethink FM policy... :) |
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Anyway, you are more than welcome to do the research, dig up a few tests and take it from there. |
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I have a question, does bf109g6as outdive spitfire IX or fw190A8 during the initial phase in 4.11m? |
Is this data correct in 4.11m?
Bf-109-G2
[Mass] kg Empty 2320.0 TakeOff 2830.0 [Squares] m^2 Wing 16.16 Aileron 1.02 Flap 2.00 Stabilizer 1.90 Elevator 1.20 Keel 0.70 Rudder 1.10 [Polares] lineCyCoeff 0.094 Cy0_0 0.15.............................................. . AOACritH_0 21.0............................................ AOACritL_0 -16.0........................................... CyCritH_0 1.48.............................................. . CyCritL_0 -1.0230048..................................... CxMin_0 0.027............................................. .. parabCxCoeff_0 6.7E-4.................................... 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 Bf-109G-2 0.43632/2830 = 1.5417667844522968197879858657244e-4 Spitfire.LF.IXC 0.4408/3300 = 1.3357575757575757575757575757576e-4 P-47D-27 0.662272/6583 = 1.0060337232264924806319307306699e-4 |
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Excess thrust......not the same! :rolleyes: |
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And P-47 falls like the brick in comparison with most other aircrafts. Quote:
FM data you posted for several planes looks like 4.11 data. Quote:
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Take the force triangle for a dive. A component of weight contributes to thrust based on the angle of dive. The difference between the force on the axis of motion in the dive and the force on the axis of motion for level flight is your initial excess force that will move the aircraft to its new equilibrium point velocity. The derivative between that and equilibrium is your average excess force along that vector.... Then apply the same formula... Force = Mass x Acceleration Rearrange it to solve for Acceleration: The acceleration of gravity is considered constant but acceleration is not constant. Acceleration = Force/Mass You then have the aircrafts acceleration rate to the equilibrium point. Now I am not a computer programmer but I am sure there is a way to look at the code to see if it following those principles. |
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The answer was specifically given to the question asked. |
The amount of excess thrust determines an aircraft dive acceleration.
The acceleration of gravity is constant but that excess thrust is not constant. It is a characteristic of the design and each aircraft will have a different acceleration in a dive. |
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