Best Piston Engined Fighter Ever...

[Lucky13]

What is the turning radius and roll rate for theTA 152H-1 at different altitudes? Compared to the P-51D and P-47M which I'm sureinto each other at a few occasions?

[Soren]

Davparlr,

You got it wrong about the AoA needed for a low AR wing - a lower AR wing produces less lift pr. AoA than a high AR wing. Another effect of a higher AR is that the critical AoA is lowered - the increase in lift pr. AoA being much greater than that of a lower AR wing.

[davparlr]

Originally Posted by Soren

Davparlr,

You got it wrong about the AoA needed for a low AR wing - a lower AR wing produces less lift pr. AoA than a high AR wing. Another effect of a higher AR is that the critical AoA is lowered - the increase in lift pr. AoA being much greater than that of a lower AR wing.

....processing....

[davparlr]

Originally Posted by Soren

Davparlr,

You got it wrong about the AoA needed for a low AR wing - a lower AR wing produces less lift pr. AoA than a high AR wing. Another effect of a higher AR is that the critical AoA is lowered - the increase in lift pr. AoA being much greater than that of a lower AR wing.

Since I do not have any specific knowledge of how Cl curves react to various variables, I will defer to your expertise. However, I have been doing much (too much) research and calculating. Probably enough to get me into trouble. It is apparent from the formula and calculation that the lower induced drag advantage of the high aspect ratio wing quickly deteriorates with airspeed, specifically, the square of the equivalent airspeed. For instance, at sea level, with equivalent weights, empty weight plus equal loads (1854 lbs added, 10494 lbs total for the Ta-152, 10836 total for the F4U-1, and 11437 total for the F4U-5), I arrived at the following values for drag:

Notes:
1) Hp calculations may be inaccurate. Hp-to-thrust computations are arcane to me and give me a headache and available data provide several ways that had different philosophies and I picked one.
2) Elliptical wing efficiency factor (Oswald efficiency factor) is ignored. I could not get a good reference or calculation formula to generate a good number. This will favor the Ta-152H but I do not believe it will change much scope.
3) Calculation errors are possible, maybe probable.
4) I cannot do a parasite drag comparison since all the data on the F4U-1,4, and 5, is at greater gross weight. I suspect the Ta-152H is cleaner since the SL speed is higher than the F4U-1 with less horsepower, but the F4U-1 is carrying more than equivalent weight. I could probably make some calculations, but I don’t want to right now.
5) Airspeeds are TAS. Calculations were done with EAS.

Case 1. Wings level, non-accelerating flight (W=L, T=D). All values are induced drag numbers. Parasitic drag is not included.

At 150 mph, induced drag on the Ta-152 = 269 lbs, on the F4U-1 = 383, or 114 lbs different. Not much different. However, at 300 mph, the drag of the Ta – 152 = 68 lbs, the F4U-1 = 96, or 28 lbs different. The advantage of the high aspect wing is significantly reduced.

Case 2. Now let’s go into a 60 degree level turn, pulling 2gs.

At 150 mph, Ta-152 = 1080 lbs, F4U-1 = 1532 lbs, or 452 lbs difference, a more significant difference due to the advantage of the high aspect wing (and greater gross weight of the F4U-1). Again, at 300 mph, Ta-152 = 271 lbs, F4U-1 = 385, or 114 lbs difference. Induced drag difference is again significantly reduced.

As can be seen here, far less additional thrust is required by the F4U-1 to maintain performance at high speed than at low speed. In addition, because of increased form area, parasitic drag could increase with a high aspect ratio wing. This would increase with the square of the airspeed.

Case 3. Now if we change the situation where the Ta-152H meets an F4U-5 at 25,000 ft and 449 mph TAS, which is a bit unfair since this is the top speed of the Ta-152H at this altitude, but it will show what is intended.
At wings level, level flight, induced drag of the Ta-152H = 68 lbs, the F4U-5 = 107lbs, or 39 lbs. difference (its about 1000 lbs heavier).

At 60 degree bank (2gs), the induce drag of the Ta-152H = 271 lbs, the F4U-5 = 430 lbs. (the increased F4U-5 weight adds significant induced drag (by the square)), or 159 lbs difference.

However, since the Ta-152H is at max speed, it will quickly lose airspeed and/or altitude. The F4U-5, with 950 shaft hp extra, which would be about 760 thrust hp with a .8 prop/gear efficiency, can easily make up the additional drag load. 760 thrust hp generates about 630 lbs thrust at 449 TAS. This excess energy advantage exists from sea level up to 30k ft. The F4U-5 can pull more gs while maintaining level flight and speed. Energy management is important in air combat and the having an aircraft with inherent energy advantage makes that management much easier. As with fighting the Zero, the F4U-5 pilots would be briefed to not engage the Ta-152H in a very low speed turning fight but to keep your speed up and engage below 30k ft. In those conditions the F4U-5 would be faster, have more energy surplus, pull higher gs without losing airspeed, and dive faster. The Ta-152H pilot would be briefed to try to engage a low airspeed and high altitude; otherwise it would be at a disadvantage across the board.

The high aspect ratio wing is great for high altitude flight or at low indicated airspeeds where low equivalent airspeed operation is common, but at low altitudes and high airspeed (high q), its advantage of induced drag to lift is reduced by the square, and any parasite drag is increased by the square, probably more than offsetting any advantage from the airfoil. I believe that this is why you never see a high aspect wing on a fighter, except, of course, a high altitude fighter like the Ta-152H.

[Lucky13]

Don't you guys get a headache of all this yibbi yabba??

[davparlr]

Originally Posted by Lucky13

Don't you guys get a headache of all this yibbi yabba??

Yep! But it does help understand how these warbirds operate and what forces affect their performance. I have learned alot from my discussions with Soren and others. It inspires me to learn more, and that is what education is all about. I know how you feel when I read all of the ballistic info, but it is still informative and important.

[Lucky13]

Sure is my good man. Just getting back in modelling again and I have to say that doing the research is JUST as FUN as the building itself will be....
This is a great forum to learn at and great people to learn from.

[davparlr]

Originally Posted by Lucky13

Sure is my good man. Just getting back in modelling again and I have to say that doing the research is JUST as FUN as the building itself will be....
This is a great forum to learn at and great people to learn from.

I continue to be amazed at the information these people have about everything associated with warbirds.

[Lucky13]

Or anything else.... I them.....

[Soren]

Davparlr,

Calculating lift and drag is done properly like this:

Lift
Coefficient of lift (CL): Cl = L / (A * .5 * r * V^2) or simply Cl = L / (q * A)
Total Lift (L): L = Cl * A * .5 * r * V^2

Drag
Coefficient of induced drag (Cdi): Cdi = (Cl^2) / (pi * AR * e)
Coefficient of Drag (Cd): Cd = D / (A * .5 * r * V^2) or simply Cd = Cdo + Cdi
Total Drag (D): D = Cd * A * .5 * r * V^2

Using the above equations I can't see how you ever came up with your conclusions Davparlr.

Anyway I'll get back to you on this in detail later, so I'll just address a few things for now..

No, the reason high AR wings are not used on todays fighters isn't for the reasons you claim - its pretty much purely for structural integrity reasons.

As to weight, well weight has an absolute minimal effect on speed, so your explanation that the Ta-152H is faster because its lighter is absurd at best.

[Lucky13]

And the headache gets worse......

[davparlr]

Originally Posted by Soren

Davparlr,

Calculating lift and drag is done properly like this:

Lift
Coefficient of lift (CL): Cl = L / (A * .5 * r * V^2) or simply Cl = L / (q * A)
Total Lift (L): L = Cl * A * .5 * r * V^2

At level flight, L=W, so, using your equation, Cl=W/(A*.5*r* V^2)

Putting this into your induced drag equation and doing a bit of algebra, we get

Cdi = W^2/.25*r^2*V^4*A^2*pi*AR*e, so,

Di=W^2/(.5*R*V^2*A*pi*AR*e) since AR=b ^2/A

Which further gives us Di=(2*W^2)/(R*e*pi*V^2*b*b)

All done properly, see also,

Calculating and Plotting Induced Drag


Now let’s step through a couple of samples:

Say Ta 152 at 150 mph (220 ft/sec), level flight, weight of 10494 lbs, wing span of 47.3 ft., R=.0023769,

Di=(2*10494*10494)/(.0023769*3.1417*220*220*47.3*47.3) gives 272 lbs (I showed 269 since I used 221 for airspeed.

Now if we go the 300 mph (440 ft/sec) we get 68 lbs, which is what I got before.

For a 60 bank level turn, just double the weight.

I think that if you look at the rest of my data, it will be correct. Well, maybe not horsepower, I used Thrust HP = (Thrust (lbs) x TAS(kts))/325.658

Drag
Coefficient of induced drag (Cdi): Cdi = (Cl^2) / (pi * AR * e)
Coefficient of Drag (Cd): Cd = D / (A * .5 * r * V^2) or simply Cd = Cdo + Cdi
Total Drag (D): D = Cd * A * .5 * r * V^2

Using the above equations I can't see how you ever came up with your conclusions Davparlr.

Remember we were talking about how good a high aspect wing is and since we are talking about wing efficiencies, Induced drag is the player.

Just looking at the formula for induced drag, one can see that it will decrease with velocity. All wings behave this way such that, at high speeds the delta in drag between any two wings is reduced significantly. See attached plot.

Anyway I'll get back to you on this in detail later, so I'll just address a few things for now..

No, the reason high AR wings are not used on todays fighters isn't for the reasons you claim - its pretty much purely for structural integrity reasons.

They didn’t use it in WWII either, except on bombers and other high altitude aircraft.

As to weight, well weight has an absolute minimal effect on speed, so your explanation that the Ta-152H is faster because its lighter is absurd at best.

All I said was “I cannot do a parasite drag comparison since all the data on the F4U-1,4, and 5, is at greater gross weight. I suspect the Ta-152H is cleaner since the SL speed is higher than the F4U-1 with less horsepower, but the F4U-1 is carrying more than equivalent weight. I could probably make some calculations, but I don’t want to right now.”

I was saying was that I didn’t know.

I agree that at the speeds we are talking about, induced drag is insignificant compared to parasite drag and the impact on top speed is negligible also. However, at very high altitudes with low equivalent airspeeds, where induced drag is making up a large part of drag, airspeed impact can be significant.

My main point is that, at high airspeeds, the advantages of the high aspect wing is significantly reduced because of the reduced differences in induced drag, therefore L/Di curves of any two airfoils gets closer. Do you disagree with this statement?

[Soren]

Lets do the calculations:´

Everything else being equal we'll assume a CLmax of around 1.4 for both a/c, however CLmax is going to increase slightly with AR. As to 'e' (Oswald efficiency factor, well the Ta-152H's should be higher for obvious reasons, the F4U featuring a gull wing.

The basic figures

Ta-152H CLmax: 1.45 (This is a low educated guess, it might be higher)
Ta-152H 'e': 0.8
F4U-4 CLmax: 1.40
F4U-4 'e': 0.77

Cd0 is unknown for both a/c so therefore total drag will be represented by the Cdi.

Height: Sea Level
Temperature: 15 C
Pressure: 101325 Pascals
Atmosphere: 1.164 Kg/m^3
Speed of sound: 349 m/s

Speed is going to be a high 600 km/h, the absolute for both a/c at SL.

Ta-152H-1 Aerodynamics at SL

Lift:

L = 1.45*23.3*.5*1.164*600^2 = 7078633.2

Drag:

Cdi = (1.45^2)/(pi*8.94*.80) = 0.0935747393
Cd0 = -Unknown-

D = 0.0935747393*23.3*.5*1.164*600^2 = 456 814.66

L/D ratio = 15.49

F4U-4 Aerodynamics at SL:

Lift

L = 1.4*29.17*.5*1.164*600^2 = 8556377.76

Drag:

Cdi = (1.4^2)/(pi*5.35*0.77) = 0.151447355
Cd0 = -Unknown-

D = 0.151447355*29.17*.5*1.164*600^2 = 925600.557

L/D ratio: 9.24

___________________________________________

Ta-152H L/D = 15.49
F4U-4 L/D = 9.24

Thats an extra 59% of lift for the Ta-152H for every unit of drag.

As to why high AR wings hasn't been used much on fighters in time, again its almost purely for structural integrity reasons.

[Lucky13]

....and he's off to the pharmacy to get a few boxes of painkillers....
How was the Mustangs wing load and power/mass compared to the Ta 152H-1?

Ta 152H-1
Wing loading: 202 kg/m² (41.4 lb/ft²)
Power/mass: 0.276 kW/kg (0.167 hp/lb)

Found it....P-51D Mustang
Wing loading: 192 kg/m² (39 lb/ft²)
Power/mass: 300 W/kg (0.18 hp/lb)

How do you figure out the turning and the roll rate of these machines?

[Soren]

The weight of the P-51D is 4585 kg, and the wing area is 21.64 m^2 = a wing-loading of 211.8 kg/m^2. The P-51 also uses a laminar flow wing which is characterized by its low drag and low lift, and it also causes sudden, early and violent stalls in turns.