Spit 5

[GScholz]

I simply cannot understand why you don't get that a plane needs the same amount of lift to stay in level flight ... regardless of speed. The weight of the plane still needs to be held up by the same force, or how else can the plane fly?

[GScholz]

And please drop the MiloMoron attitude. We don't need two of you.

[gripen]

Originally posted by GScholz
I simply cannot understand why you don't get that a plane needs the same amount of lift to stay in level flight ... regardless of speed. The weight of the plane still needs to be held up by the same force, or how else can the plane fly?

OK, lets do this from the ground up.

For the Cdi calculation we need Cl values at the speeds we are interested.

In the case of the Spitfire the weight of the plane is 3400kg so the amount of lift it need for level flight at all flyable speeds is:

9,81 m/s2 * 3400 kg = 33354 N

The plane flys at near sea level so the density of the air is 1,229 kg/m3.

The speeds we are studying are:

200 km/h => 55,56 m/s
400 km/h => 111,11 m/s

And the wing area is 22,48 m2

Next we just put values to the Cl formula:

Cl=L/(r * (V^2/2) * A)

200 km/h => 33354/(1,229 * (55,56^2/2) * 22,48 = 0,782
400 km/h => 33354/(1,229 * (111,11^2/2) * 22,48 = 0,196

Now we know that the plane the must have exactly these Cl values to produce lift of 33354 N at unknown AoAs, no wind tunnels or simulations needed. As noted above, this calculation does not need AoA values because we concentrate to the lift only; the entire induced drag is due to lift.

Now you can see that to produce constant amount of lift, the needed Cl value at 400 km/h drops to quarter of the needed Cl value at 200 km/h. In practice this means that the plane flys at lower AoA at 400 km/h than at 200 km/h but again we are not interested.

Now we can simply put Cl values to the Cdi formula, here I use efficiency factor 0,95 and the aspect ratio is 5,61:

Cdi = (Cl^2) / (pi * AR * e)

200 km/h => (0,782^2)/(pi * 5,61 * 0,95) = 0,0366
400 km/h => (0,196^2)/(pi * 5,61 * 0,95) = 0,00229

So here we are, these values are exact if the efficiency factor is right. Here we can also see that the induced drag for constant amount of lift drops to one sixteenth when the speed doubles.

gripen

[Angus]

Bear in mind that lift induced drag is always a factor while heading into airstream, be it vertical or nor.
A Spittie with its creation of lift above most planes will still be creating lift while zooming, in excess of most planes. It may be a liability, for the only thing it does above say 80 degrees is tilt the nose upwards, which needs to be compensated. Also, wings do count as parasite drag partially.
Look at the zoomers of AH. I think it's all rather correct. The zoomers are hevy planes with heavy wingloading.
Just picture this. A wingless projectile weigting 5 tonnes gets launced upwards at 500 kts together with a similar projectile with a 15 yards wingspan. The wingless one flies further, no doubt, before tipping over.
Change this to 45 degrees, and the winged one goes further...
You get the picture, I hope.

[Crumpp]

List your source please for the efficiency factor for both planes please.

Don't you understand you have all the data that the engineers have BEFORE they take it the wind tunnel?

The wind tunnel is used to confirm or deny that data.  The actual numbers for both planes could be completely different based on boundary layer dynamics and a host of other incalculable factors.

How many planes have made it all the way to the runway before it was discovered they did not perform as the "data" said they should.  Come on Gripen, your obviously a smart guy, be smarter than this.

Crumpp

[gripen]

MiloMorai,
When there is no lift needed ie at zero g, then the induced drag is also zero because needed lift coefficient is also zero. In fact you can calculate needed lift coefficient for any wanted speed, altitude and g load combination but you don't know if the plane can reach that high Cl, for that you need tested data.

As and example at 2 g load lift would be:

2 * 33354 N = 66708 N

And at 0 g load:

0 * 33354 N = 0 N

Regarding efficiency factor, we don't know what would be the exact value but for comparison purposes we can use same efficiency factor for both planes, the Fw 190 and Spitfire IX, despite we know that the Spitfire has better, near ideal shape of the wing. Then we can be certain that the calculation does not favor the Spitfire

As an example here are the Cl and Cdi values for both planes at 300 and 600 km/h assuming 3 g load and efficiency factor 0,9:

Spitfire IX:
300 km/h => Cl=1,043 Cdi=0,0686
600 km/h => Cl=0,261 Cdi=0,004287

Fw 190:
300 km/h => Cl=1,451 Cdi=0,1236
600 km/h => Cl=0,363 Cdi=0,007724

So now we can see that even by using numbers which favor Fw 190, the Spitfire IX has clearly lower Cdi (nearly half) than the Fw 190 at any reachable g load except at 0 g load when there is no induced drag and Cdi for both planes is therefore also zero.

gripen

[Charge]

"we know that the Spitfire has better, near ideal shape of the wing"

I'd like to remind you that to counter the stall characteristics of the elliptical wing there is washout in the wings which places the outer portion of the wing into more liftless angle when the mainpart of the wing (the root) is in position where it does the lifting job, so to me it seems that when determining the lift for Spit's wing only part of the wing can be considered effective depending on the speed and AoI of the main part of the wing. So when determining the AoI of the wing the washout and its effects to overall lift capabilities of the wing should not be forgotten.



-C+

[Angus]

Well, when it comes to total lift, the Spitty definately beats the 190. But that's not the same as zoom.
BTW, give me a weight, wing area+loading, and time to 10K and 20K (with starting run) and I'll calculate it to Newtons if you like

[Crumpp]

Gripen

Regarding efficiency factor, we don't know what would be the exact value but for comparison purposes we can use same efficiency factor for both planes, the Fw 190 and Spitfire IX,

Now your numbers don't make ANY sense at all.

The Spitfires wing generates more lift because it has more lifting surface.

Look at the formula.  The larger the Cl the more induced drag.  The lower the Aspect, the higher the induced drag.

Using the same efficiency factor, which is the main advantage of the spitfires wing in induced drag, should clearly yield more favourable results to the Higher aspect ratio and lower lifting wing.  After all, induced drag is a byproduct of lift.

Angus,

I'll look the info up for you or you can give me you email and I will send you the FW-190 wt charts.  There are in this thread too somewhere.  Don't have time now, headin to work.

Crumpp

[Nashwan]

Look at the formula. The larger the Cl the more induced drag.

The higher the wingloading, the higher the CL you need to maintain.

Lower wingloading means you need a lower CL.

Under comparable conditions, the Spitfire will need a lower CL than the 190.

Using the same efficiency factor, which is the main advantage of the spitfires wing in induced drag, should clearly yield more favourable results to the Higher aspect ratio and lower lifting wing. After all, induced drag is a byproduct of lift.

The heavier the plane, the more lift you need. The higher the wing loading, the higher the CL you need.

Low wingloading = lower induced drag.

[bozon]

Originally posted by Crumpp
Don't you understand you have all the data that the engineers have BEFORE they take it the wind tunnel?

The wind tunnel is used to confirm or deny that data.  The actual numbers for both planes could be completely different based on boundary layer dynamics and a host of other incalculable factors.

Crumpp, he's calculating the Cl of a level flying plane at a given speed - this is the same as putting it in the wind tunnel. The engineers want to know this BEFORE they build the actual plane and risk some dumb bellybutton pilot to fly it. But for an actual flying plane the calculation must apply.

the "e" factor debated here, is a "fuzz" factor, determined empirically (though it can be estimated), to the effective area of the wing. This area is reduced by wing-tip turbulance and I belive it is not really constant in AoA. In order raise this value, you can change the shape of the wing-tip (eliptical) or change the aspect ratio of the wing (long thin wings are more efficient).

this wing-tip turbulance and efficiency of the wing will effect high AoA preformance mainly and almost non to max speed. This is why spits bleed less energy when manuvering. The induced drag (optimizing wing efficiency) adds peanuts to the total drag comparing with the parasitic drag at high speed, level cruising. If you are building a speedster, use small wings - you'll have more induced drag, but much less parasitic drag at very high speeds - see the F104 example.


Bozon

[VO101_Isegrim]

Wingloading alone does not defines the amount of lift the wing generates. Let me remind you, the Spit had thin profile wings, which produce less lift proportinally than thick profiled wings on other a/c, such as the FW 190A`s.

The Spitfire`s Max. Lift Coefficient was 1.12. That`s ultra-low, and means the Spit`s huge wing area only produce a relatively tiny amount of lift. It`s not effective for lift generating purposes, also for the reasons Crummp mentioned, re: washout.

In fact, liftloading of the MkIX was already slightly worser than the 109G, despite having a 41% larger wing area a lower 'wingloading'.

So sorry, the "Low wingloading = lower induced drag" theme is just wishful thinking. Just look at how much power the Spit required to haul itself around compared to it`s nemesis.

The FW 190A-5 could reach 542 km/h with 1560 PS (static) at SL (1.3ata).
The Spit IXLF required 1680 HP (static) to reach 539 km/h at SL (+18lbs).

That pretty much summerizes the relative drag differences between the two.

[MiloMorai]

Originally posted by VO101_Isegrim

The FW 190A-5 could reach 542 km/h with 1560 PS (static) at SL (1.3ata).
The Spit IXLF required 1680 HP (static) to reach 539 km/h at SL (+18lbs).

It would be nice to see hp to hp or ps to ps numbers.

Using the power graph for the BMW801D in the A-8 handbook, 1560ps is ~1540hp.

How does one get a hp number greater than the less than 1600hp(ram) as seen in this Merlin 66 graph at SL?





How is that the A-8 could only manage between 325 and 330mph(523 - 531kph) at 1.32ata, SL?

[Crumpp]

Thanks Bozon

Right I understand exactly what you are saying but in viewing the formulas I concluded that there is lots of wiggle room to support both sides of the debate.  The "induced drag" element can be manipulated to seem larger or smaller than it is in the air and still be "factual".

Combining this with the fact AH is a computer program and we don't know how it models induced drag, entering the "number" debate is rather pointless.  Beyond of course establishing whether or not induced drag is a factor in why spits did not enter the vertical fight with 190's.

 The "induced drag" debate, following the formula, pretty much comes down to the 190's better aspect ratio vs. the "e" factor of the Spitfires wingtip. HTC and the "computer environment" in the end determine the variables we would be squabbling over.

Since induced drag is not a factor in zoom climbing.  It's even more pointless.  We can say that in a co-energy state the facts point to Merlin Power Spits never developing the ability to effectively fight 190's in a similar arena.  The fight was always dissimilar, energy vs. angles or as Capt. Eric Brown put it "Blondes and Brunettes".  Only when the Spitfire gained weight was it able to follow a zooming 190 with a chance of catching it without a large energy advantage.  The Spitfire Mk XIV developed the ability to fight the similar fight and it was able to dominate the 190A.

This is confirmed by following the power and weight gains of both the 190 and the spit.  Cross-referenced with performance charts and tactical trials.  

Additionally, the extra weight and Horsepower the FW-190A8 gained improved it's zoom climb and dive acceleration.  All traditional strengths of the 190.  With a 3 lb increase to wing loading over the 190A5, CG adjustment, and a difference of HP-WT ratio in the 100th's, other performance areas suffered little or not at all.  All in a much better armament package than the 190A5.  

Crumpp

[Nashwan]

So sorry, the "Low wingloading = lower induced drag" theme is just wishful thinking.

Isegrim, look at the equations again. The lower the CL, the lower the induced drag.

CDi = CL^2 / pi*AR *e