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Let's see if you can find the mystery of the missing 24 000 lbs...
Gaston
I am underwhelmed by this argument. And in fact we dont have to prove anything. The accepted theory is on our side, you are the one promoting a view outside the accepted norms and trying to dispute the universally accepted theory, and that means its up to you to present the supporting facts and arguments, and then even explain it to technophobes like me. Youve done none of that I submit.
I have to find the time to make the graphics, and real busy otherwise...
At 3 Gs sustained, the Spitfire's wings bear about twice the load aero-engineers think it does...
At 3G the Lift = SQRT[(G)>>2 - 1]*W = Sqrt(8)*W=2.83*7500=21,213 pounds At 8G the Lift=sqrt[(8)>>2-1]*W= 7.937*7500=59,529 pounds for turns in which the a/c is in a steep bank. For a dive pullout the applied forces are a direct multiple of the Gross Weight of the airframe times the 'G' load
Aero-engineers assume around 22 500 lbs at 3 Gs (aircraft's weight at 7500 lbs X 3): It's likely in reality around 45-48 000 lbs...
Nope - see above - but at an 8G Corner Velocity (much faster than the 3G sustained turn velocity) the Normal' vector of Lift is 7.937*Weight. see above for formula.
BTW - this is near the limit load for stress for the Spit and the region in which it starts to 'bend'
So where does the extra wing lift, for the extra 24 000 lbs, come from?
From pulling higher G's at CLmax. For a Gross Weight=7500 pounds (heavy for a Spit), pulling 8G's would equate to 60,000 pounds of lift on the wing and very likely be at the limit design stress for the spars... and not sustainable for maintaining a stable altitude
Read what I wrote here (minus a few minor mistakes) and assume wind tunnel tests gave them a perfectly accurate assessment of a wing's lifting abilities: The lift values are confirmed: It was all there right under their noses...
The Maximum Lift = CLmax*Wing Area*1/2*rho*(Vel)>>2
Let's see if you can find why they didn't know about half the lift force effectively borne by the wings at 3 Gs...
Lets see you place lucid and constructive math on your posts and maybe you can figure it out
The bottom line is, even if the Spit does beat the FW-190A in turns (which it does do by a wide margin above 4 Gs unsustained) there is no reason, according to them, why the two should be so close at low sustained speeds...
Speeds near stall (at CLmax) are in the domain of the fearless (and competent) pilot flying in an imperfect fluid with an imperfect aircraft. The a/c with the best tip control (i.e. positive aileron authority) throughout and a slightly better wing loading and enough power available is liable to 'win'
Even Russian TsAGI tests show the FW-190A-5 beating, in right turns, all the Bf-109Gs except one: A Bf-109G-4 that barely matches it at 21 seconds for a sustained 360...
Where the hell is the math that accounts for this match?
Where in the hell is the match that is perfectly flown, with specified weights, perfectly conditioned airframe and engine flown in exactly the same conditions? The math is all over the forum - look for both max sustained turn and rate of turn equations - they are close so long as you have a reasonable CLmax, GW, BHp, and flight test data to derive parasite drag from high speed runs and published, Hp/Altitude and GW
There is plenty of evidence engineer calculations predict nothing when it comes to heavy, powerful nose traction types...
So, trot out your evidence - oh I forgot you don't do math, or engineering, or physics, or wind tunnel tests, or fly..
I think it's about time they noticed how an aircraft flies...
Let's see if you can find the mystery of the missing 24 000 lbs...
Gaston
How does it arrive at such a result? I've yet to see the math demonstrated in any coherent way...
We have yet, here or HiTech or every site you have trolled, to see you coherent in any demonstarble way.
Also at least one very experienced Fin ace quotes the "optimal" sustained turn rate for the Bf-109G-6 as being at 160 MPH.
He is quite correct - that is the speed at which the 109 may be flown (in perfect condition and pilot skill) in a sustained, near 3G turn, with no excess power.
"Accepted math" puts it around 220-240 MPH at least...
Demonstrate your math and prove that it is 'acceptable'
6G unsustained "Corner Speed" on the P-51D Mustang is tested as being at a minimal speed of 320 MPH... Your math? 250-260 MPH...
6.33 G corner speed for a 10,000 pound P-51D is approximately 259mph.
P.S. By the way, for those treating the P-51's Pilot's flight manual figures as gospel, note that the B-29's flight manual was said to be the very first to have, by a very wide margin, serious and complete flight data gathered in a serious scientific way... The P-51, pre-dating the B-29, was obviously not included in that...
G.
and the Troll will be back.
A rather unsustainted assesment.Even Russian TsAGI tests show the FW-190A-5 beating, in right turns, all the Bf-109Gs except one: A Bf-109G-4 that barely matches it at 21 seconds for a sustained 360...
? Either a type mismatch from your sources, or your own error.Where the hell is the math that accounts for this match?
A rather unsustainted assesment.
The TsAGI tests "Samoletostroenie p. 95" are clearly showing that soviet FW-190 A-5 is turning a sustainted 360°, in 22-23 secunds, the 109 F4 in 19.8-20.6s, the 109 G-4 in 21s (that means it's might be a best or a middle value).
In case that you have two different values in soviet tests, one is for the left turn, the other one for the right.
The Spit IX beats them all, due to low wing loading, high power to weight ratio.
Ñïèòôàéð Mk.IX â ÖÀÃÈ.
No mystery in physics
? Either a type mismatch from your sources, or your own error.
Regards