No it doesn't Hop, what we were looking for was the difference between both a/c and as you can see the difference is the same wether you use km/h or m/s.
You could use cubits per sar, the proportion would still be the same, but you will not get even a roughly correct idea of rate of turn.
Look, Soren, you claimed I had my figures wrong. I didn't. You did so in an extremely patronising way, which is why I pointed out you had your units wrong. Accept it, move on.
Listen if Clmax is 1.36 then you're not really turning very well at 0.5 now are you ??
Depends on speed, doesn't it? Just doing the rough calc:
Spitfire - 37418 = 0.5*22.48*.5*1.164*125^2 = 2.7 G. So at 125 m/s, the Spitfire is pulling not far off 3 Gs at a CL of 0.5.
If we use the 600 km/h you used the first time:
Spitfire - 37418 = 0.5*22.48*.5*1.164*167^2 = 4.9 G.
What we want to be looking at is CLmax, as this is the region both a/c are most likely going to reach when they turn fight each other.
It really does depend on speed. You try reaching too high a Cl at too high a speed, you are going to rip the wings off.
For example, at the 600 km/h you originally used:
Spitfire - 37418 = 1.4*22.48*.5*1.164*167^2 = 13.7 G. Think you are going to be pulling that in a turn fight?
More to the point, CL max is usually unsustainable (think you can sustain 4.9 G in a WW2 fighter?)
Yes I do understand physics, however you're having big trouble indeed Hop. Ever wondered why a/c need to adjust pitch settings as speed goes up ?? Or are you under the impression that as speed increases lift stayes the same ??
No. What you wrote was:
Oh and btw generating a 2:1 lift to weight ratio is very normal in straight flight - so yes I do believe that.
If you are generating lift equal to twice you weight, you will not be flying in a straight line, it's a physical impossibility. If you increase speed, you decrease the angle of attack so that lift continues to equal weight. At least, if you want to continue flying straight you do.
For Christ's sake what is it you don't understand man ?!!
I don't understand what you believe. I only understand the physics. When someone approaches the discussion with a wrong concept, it can be very hard to understand what their concept is.
Yes like I said a higher L/W ratio gives you a higher INITIAL turn rate,
You gave calculations showing the Spitfire with a substantially higher lift/weight ratio. That means the Spitfire, in your calculations, was turning tighter.
It will, of course, generate considerably more drag than the Ta 152 if it is turning much tighter than the Ta152.
however since the Spitfire has got loads more drag pr. lift its going to loose out quickly in a sustained turn fight.
It doesn't have "loads" more, it has approx 3% more, a difference that is more than made up by the greater power (and much greater power to weight, incidentally) of the Spitfire.
Having written that, I have got a grasp on your wrong concept.
Induced drag increases with the square of the Cl. That means the tighter a plane turns, the less efficient its lift production becomes.
First, the Ta 152 as a baseline
please check all these calculations, I did them the first time using a CL of 1.42 for the Ta 152, and as I had intended to do them at the 1.45 CL you wanted, I redid them all, so please check if any mistakes have crept in)
Lift = 1.45 * 23.3 * .5 * 1.225 * 125^2
Lift = 323 333 N. That's 6.95 times weight.
Cdi = (1.45^2)/(3.142*8.94*.80) = 0.0936
Drag = 0.0936*23.3*.5*1.225*125^2 = 20,872
Lift/drag = 15.5/1
Now, you worked out the Spitfire at a Cl of 1.35:
Lift = 1.35 * 22.48 * .5 * 1.225 * 125^2
Lift = 290,440 N. That's 7.76 times weight.
Cdi = (1.35^2)/(3.142 * 5.61 * .83) = 0.1246
Drag = 0.1246*22.48*.5*1.225*125^2 = 26 807
Lift/drag = 10.8/1
From this you come up with the Spitfire having 43% more drag/lift. Fair enough.
However, the error is that the Spitfire is turning tighter here. It's generating more lift in relation to its weight, in other words it's pulling more G. When you reduce the turn to the same G as the Ta 152, you get a very different figure:
Lift = 1.21 * 22.48 * .5 * 1.225 * 125^2
Lift = 260 320 N. That's 6.95 times weight, the same turn as the Ta 152.
Cdi = (1.21^2)/(3.142 * 5.61 * .83) = 0.1
Drag = 0.1*22.48*.5*1.225*125^2 = 21 514
Lift/drag = 12.1/1
See how when it's only turning as tightly as the Ta 152, the lift/drag ratio for the Spitfire improves. But wait, there's more.
however since the Spitfire has got loads more drag pr. lift its going to loose out quickly in a sustained turn fight
Ok. The Ta 152 now has approx 30% lift/drag advantage over the Spitfire. So the Ta 152 still has a huge advantage, right?
The problem is, the Spitfire is LIGHTER. It doesn't have to generate the same amount of lift as the Ta 152 to match its turn. In fact, if you look at both planes making the same turn above, the drag figures are:
Ta 152 - 20,872
Spitfire - 21,514
The Spitfire has 3% more induced drag when making the same turn as the Ta 152.
Now, those figures are nowhere near exact, of course, and as we don't have exact figures for Oswald efficiency, and don't have parasitic drag figures at all, there's no point in trying to make them more exact. But the induced drag figures for the 2 aircraft are very, very close, and the Spitfire still has that power advantage.
Rough figures ?? No, exact figures Hop. But yes there is a little guesswork involved as we don't have the 'e' figure for both aircraft,
Exact figures, plus or minus a guess?
however knowing the difference in 'e' of a completely elliptical with no twist and an AR of 6, and also knowing what approximate effect leading edge guns have on 'e' I can make a reasonable and educated guess.
So a guess, as I said.
There can be many reasons why the international std. is different, but again wether we use the international std. figure or the other more regional one doesn't matter, the difference stayes the same.
No, it doesn't, which is why I didn't say you had the "wrong" density figure, just not standard atmosphere.