Aces High Bulletin Board
General Forums => Aircraft and Vehicles => Topic started by: Kweassa on September 07, 2003, 02:27:20 AM
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While in a debate at UBI soft's IL-2 Forums, apparent problems concerning physics have come up, and it's got me stumped.
Read this thread first please! :) (http://forums.ubi.com/messages/message_view-topic.asp?name=us_il2sturmovik_gd&id=yzppl)
The discussion is about the pitching capabilities of the Fw190A under high speeds. The case presented, is a 800km/h(492mph) near-90 degrees dive, plane probably not properly trimmed, so the pilot probably would have to push the stick to generate no pitch movement at all.
At about 500m(1500ft) the Fw190 pilot, (momentarily notwithstanding the effects of high Gs dumped onto the pilot, or the G forces that might destroy the plane) pulls full stick deflection possible. The plane recovers level flight with about 300m loss of altitude.
The primary question of interest is:
1) How high would be the G forces generated for a typical Fw190A to pull out in such a manner?
2) Would the elevator authority of a Fw190A allow such an input technically?(not withstanding effects given to a pilot, or G loads to the plane)
The point of view I stand in the afore mentioned discussion is this:
I am willing to accept the elevator authority of the Fw190As are overdone to a certain point, but I certainly don't think a pilot 490mph vertical dive would necessarily need a 'vector-thrust' technology to initiate a safe pull-out from 1500 feet. Pulling out within a 300m radius might be overdone, but I think it was possible with a close call.
I've been searching through almost every physics equation I could find about aerodynamics, but being a layman in physics in the first place, applying them to my needs are very tough! :o
The centripetal calcualtion of force as presented in the thread, does not seem to be right, or at least not to be applied as such in simple turms, but I could not find a method of calculating speed, G forces, sustained AoA needed per given time to be used as a correct method.
I'd appreciate some help from what you guys think about the discussion! I've been going through things like Lift Factor, ClMax, air density, cosine this, tangent that.. and my head is practically spinning wild :p
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The centripetal acceleration formula presented by "No.1RAAF_edin" is correct. So the total acceleration from 90 deg. dive would be about 18g. Too much for the pilot and the plane, clearly. Of the elevator authority of RL aircraft I cannot say, but it is quite academic matter in such a situation, anyway.
60 deg dive angle would halve the acceleration, so I guess that would be just possible, for (some) sim and RL aircraft.
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Hi Kweassa,
>1) How high would be the G forces generated for a typical Fw190A to pull out in such a manner?
Centripetal acceleration is
a = v^2 / r
Let's say the speed is true air speed: 800 km/h = 222.2 m/s.
The radius of the turn is 300 m.
So we arrive at
a = 165 m/s^2
Since 1 G = 9.81 m/s^2, this equals 16.8 Gs. A bit much :-)
>2) Would the elevator authority of a Fw190A allow such an input technically?(not withstanding effects given to a pilot, or G loads to the plane)
The desired elevator authority at high speed is just enough to gain full performance, but not enough to break the plane. As far as I know, the Fw 190's elevator authority was considered satisfactory, so it probably wouldn't allow 17 Gs to be pulled :-)
>I certainly don't think a pilot 490mph vertical dive would necessarily need a 'vector-thrust' technology to initiate a safe pull-out from 1500 feet.
Vector thrust wouldn't help much, anyway. Even at 90° deflection, it would add less than 1.5 G to the pull-out (for current jets). With the engine in the fuselage, these Gs wouldn't further stress the wings so they could be added on top of the structural load at least.
>The centripetal calcualtion of force as presented in the thread, does not seem to be right, or at least not to be applied as such in simple turms, but I could not find a method of calculating speed, G forces, sustained AoA needed per given time to be used as a correct method.
Well, the problem with the above formula is that it neglects that in reality, pulling major Gs will decelerate the airframe even in a dive so that the speed decays throughout the manoeuvre. That means the same Gs will give you a smaller radius the further you're into the pull-out. However, considering the high initial speed, I'd think the answer would still be "impossible".
Regards,
Henning (HoHun)
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the big problem is that during the hard pullout, the plane will loose a lot of speed. this is IMO the hardest thing to take into account and might make a big difference (especially if you are the pilot...)
second, when pulling constant G (generating constant lift) a plane pulling out of a vertical dive will not make a circullar pattern, since the acceleration acting on it will not be constant:
G-g*cos(a)
where 'a' is the instantenious dive angle (changes from 90 to 0).
This means that you will actually need more initial alt to be able to pull out then the previous calculation.
a modern plane can handle 9-10G manuvers, so I assume a WWII can handle less. In a 10G manuver, the correction for gravity will be under 10%. HoHun calculated that it will take almost 17G to pull out in the circular approximation. I dont belive any reasonable speed loss will compensate for that.
(I don't suppose anyone have the Cd at maximum AoA for a 190?)
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Also consider the the Pilot place card restriction for dive posted in the cockpit was 466MPH below 10K in the FW190A5.
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Can you elaborate on that, F4UDOA?
Also, from what Hohun posted..
The desired elevator authority at high speed is just enough to gain full performance, but not enough to break the plane. As far as I know, the Fw 190's elevator authority was considered satisfactory, so it probably wouldn't allow 17 Gs to be pulled :-)
Would that mean under certain conditions, the elevator authority of the Fw190 would give around 7~8Gs? I saw the HTC Help pages on Planes citing "-4G/+8G" as the max loading of the Fw190A-8.
Would there be conditions in a desparate, steep pull outs, that would momentarily exceed very high Gs(9~10Gs) in WW2 planes? If in that case, does the airframe succumb immediately at the point it 'crosses the line'?
:confused:
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Hi Kweassa,
>Would that mean under certain conditions, the elevator authority of the Fw190 would give around 7~8Gs?
Well, it's actually a bit more complicated than I pictured it. The design goal is to make the control forces required for destructive manoeuvres so heavy that the pilot doesn't destroy the plane accidentally. Even a well-designed aircraft might allow destructive Gs to be pulled, but the pilot has to make a conscious effort.
>I saw the HTC Help pages on Planes citing "-4G/+8G" as the max loading of the Fw190A-8.
In WW2, the designs typically had safety factors of 1.33 or 1.5 before the structure could be expected to yield. And the G numbers were rated for a specific flying weight, so with a half-empty tank, no external ordnance etc., you might get even more. And if the Gs were applied for a fraction of the second only, even higher loads could be survived!
However, a steady 17 G pull-out would be beyond any WW2 aircraft's possbilities, not to mention the pilots'.
>Would there be conditions in a desparate, steep pull outs, that would momentarily exceed very high Gs(9~10Gs) in WW2 planes? If in that case, does the airframe succumb immediately at the point it 'crosses the line'?
I'd say for every plane that broke were several that came home bent :-) Crinkled skin, popped rivets, increased dihedral - telltale signs of an overstressed structure.
Regards,
Henning (HoHun)
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Hi F4UDOA,
>Also consider the the Pilot place card restriction for dive posted in the cockpit was 466MPH below 10K in the FW190A5.
That's just so you can't sue Focke-Wulf if you get harmed in the plane at higher speeds :-)
The placard restrictions were exceeded for example by the US Navy in their comparative trials.
(They probably didn't care about the manufacturer's warranty anyhow ;-)
The USN found the Fw 190 behaved very well even beyond placard speed.
(The Me 109 had the same placard limit, but not the same nice behaviour beyond it!)
Regards,
Henning (HoHun)
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Hi Bozon,
>the big problem is that during the hard pullout, the plane will loose a lot of speed. this is IMO the hardest thing to take into account and might make a big difference
Good point. After some further thinking, I'd estimate the difference isn't that big, though.
Here's my rough estimate:
Let's say that at full angle-of-attack, the lift-to-drag ratio of the Fw 190 might be as low as 10.
Pulling 10 G for recovery will then provide us with a braking acceleration due to drag of 1 G.
Assuming a 300 m radius circular pull-out, this would result in a specific energy loss equivalent to 471 m. The specific energy gain at the same time due to gravity is 300 m, so the aircraft is hardly slower at the end than it was in the beginning.
(For comparison: The specific kinetic energy of the Fw 190 at 800 km/h is equivalent to 2520 m.)
Regards,
Henning (HoHun)
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Kweassa,
In the report of the FW190A5 vrs the F4U/F6F there was including in the report and breakdown of the aircraft restrictions place card found in the cockpit when the airplane was captured. These cards were common in WW2 fighters of all types.
It says (IAS of course)
466MPH from 10,000ft down
426 from 10,000ft to 16,400ft
360MPH from 16,400 to 25,000FT
It is not like the wings would fall off or anything but virtical diving at 490MPH at less than 10,000ft IRL would be suicide in any WW2 aircraft. At that speed straight down pull out at low alt would be almost impossible without blacking out, augering or exceeding the G limits of the A/C and really pulling your wings off. Hence the limitation.
For comparison
at 10,000FT
P-47 500MPH the Brits had it at 520MPH
P-51 505MPH
F4U-1 472MPH at 3.5 g's
P-38L 420MPH with dive flaps add 10-15MPH
(http://www.geocities.com/slakergmb/1fa60b00.jpg)
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I've read in several book (can't remember which exactly - was a long time ago, but I think Closterman mentions it also) about spitfires breaking a wing in a very high G turn.
Bozon
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Here is a chart I produced for an article some time ago, it relates the dive angle and dive speed to the pull out altitude needed for recovery.
As an example, the red dotted lines show that for a 45 degree dive at 500mph with a 6g pull out, the recovery would need to be initiated at 1300ft.
Hope that helps.
(http://www.badz.pwp.blueyonder.co.uk/Files/Images/pullout.jpg)
Badboy
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F4U,
Just a note. It's "placard", not "place card".
I was wondering what you were talking about there for a minute.
It comes from having little embossed plates in the cockpit where the pilot can read them and be reminded of limitations.
A common one is "Intentional spins Prohibited", for example, stuck on the instrument panel.
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Wow badboy! That's exactly the sort of data I was looking for!
Thank you! :)
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Originally posted by Kweassa
Wow badboy! That's exactly the sort of data I was looking for!
Thank you! :)
No problem.
I also have sets of similar curves for other load factors, and 6g is admittedly a tad conservative for an emergency dive recovery. The aircraft mentioned in the original thread would almost certainly have suffered some form of structural damage, but you can see from the diagram below that even a relatively safe 8g recovery brings the minimum altitude down to around 900ft.
(http://www.badz.pwp.blueyonder.co.uk/Files/Images/pullout2.jpg)
Hope that helps...
Badboy
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Hi Badboy,
>Hope that helps.
Hm, actually it confuses me. What's the physical background of your curves? Using the calculation I outlined initially, I get a pull-out radius of 2088 ft for a 500 mph/8 G recovery, which should be equal to the pull-out altitude in a 90° dive. Your chart indicates 1000 ft more, though.
I'm not surprised you don't agree with my simplistic calculation, but I'd have expected the accurate value to be lower than my estimate, not higher :-)
(The reason would be the energy loss due to pulling Gs, as pointed out by Bozon.)
Regards,
Henning (HoHun)
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Originally posted by HoHun
Hi Badboy,
I'm not surprised you don't agree with my simplistic calculation, but I'd have expected the accurate value to be lower than my estimate, not higher :-)
Yep, that's because I posted the chart without checking it. The calculations were actually in knots and because the chart was originally produced for the recovery from a dive bomb delivery it also included a frag pattern allowance.
In this chart I have removed the safe frag clearance altitude and corrected for mph. For a 90° dive at 500mph, your figure is correct. Well spotted.
(http://www.badz.pwp.blueyonder.co.uk/Files/Images/pullout3.jpg)
Sorry for any confusion!
Badboy
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Terms and terms! :)
What's the differece between knots and mph? How do they convert?
Also, what does "frag clearance" mean?? :o
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Originally posted by Kweassa
Terms and terms! :)
What's the differece between knots and mph? How do they convert?
Also, what does "frag clearance" mean?? :o
The Knot in correct usage denotes a speed of one nautical mile per hour. Nautical miles are defined as the average length of one minute of latitude on the earths surface. There are two values for the knot, one is exactly 6080ft and the other is 1852m or about 6076ft. This is the international value, and the one I was using. The statute mile is of course 5280ft so the conversion factor is just 6076/5280, or approximately 1.151 statute miles to the nautical mile.
The frag distance is illustrated in this diagram, along with the equation for the radius. If you have any questions about how that equation was derived, just ask. However, the curves in my diagram, take into account factors that aren't included in that equation, for example, the reduction in g when the aircraft becomes g limited.
(http://www.badz.pwp.blueyonder.co.uk/Files/Images/pullout4.jpg)
Hope that helps.
Leon "Badboy" Smith