parsifal
Colonel
I agree with NJ....the viewpoints of the protewgonists are well known....done to death really. i say we move along and discuss something new....
Spitfire IX v. FW 190A
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-Well... I have corresponded with several aeronautic engineers so far for over two years, and to date none of them has explained to me how:
1- You can curve the trajectory of an aircraft without compressing its thrust on one side...
2- Then, none of them has explained to me how compressing the thrust on one side will not induce a "slant" to the thrust, no matter how small... One aeronautic engineer has agreed, on the "Aces High" forum, that the thrust "slant" probably does occur, but is very small in amplitude... (Halleluia!)
3- If the thrust slanting does occurs on the curving of the trajectory (slanting the forward direction of the thrust towards the outside of the turn since the extra pressure that is applied by the wing angle of attack rotation is to pull the the top of the prop disc back, not the bottom), then there is also no way that the wing's increased angle of attack does not shorten the length of a line paralell to the trajectory between the wing's pivot point level and the top end of the prop disc... In effect, the nose gets shorter when viewed from the center of the turn...
4- If 3 is true, then there is no way that an increasing wing angle of attack does not require "pulling back" on the thrust if the thrust "traction" is from the nose... (The effect is much less on rear propulsion because the thrust source is going down, not up, and is also going forward as the aircraft rotates to the higher angle of attack: No real fighting of the thrust, or very little)
5- If 4 is true, then there is no way that pulling back on about half or more of the prop disc does not require defeating the entire amount of thrust existing there...
6- 5 remains true even if the amount of top prop disc half pull-back is so small a microscope cannot measure it... ANY, I repeat, ANY amount of pull-back requires defeating ALL the thrust there to move the top of the prop back even a micrometer compared to the trajectory...
If it is 2000 lbs pulling in the upper prop disc half, then it is 2000 lbs you pull before you can even start thinking about that first millionth of a degree of angle of attack increase...
The elevator action does not bear all that load to beat the prop: It may provide 100 lbs of extra force for it: The tilting of the wings creates the upward wing leading edge drag that multiplies the elevator effort to finally beat the prop with little actual pilot-elevator effort. Say 100 extra pounds of effort on the elevators leads to 2100 pounds of extra effort on the wings, 2000 lbs of it being by the wings: total extra wing load due to prop disc: 2100 pounds. (Not a big deal out of the 50 000+ pounds total at very high Gs, but at a sustained 3 Gs a P-51D will be down to say 27 000-30 000 lbs of wing load, at wich point the 2100 lbs will start to be a significant advantage if they could be reduced to, say, 500 lbs by downthrottling...
Hence downthrottling from WEP to well below METO could lead to the prop load being much less on the wingload...
Think about it: All these effects are clearly irrelevant to tail propulsion, since obviously no "pulling back" on the thrust occurs when you increase the wing angle of attack: Do you really think basic common sense allows for there to be NO difference in that regard for nose traction?
Do you really think there can be no fundamental handling differences between being pulled from the nose and pushed from the tail?
I have been discussing this for over two years now, on various boards with several aeronautic engineers, and so far not one has advanced the slightest rebuttal to the points above, except to deny there is any slanting of the thrust, as if there was no compression of the thrust in a curve...
In other words, to them, the thrust is exactly the same forced in a curve or going straight... It is not altered in any way by the curvature...
Without the thrust slanting compared to the aircraft's attitude, there is, of course, no "shortening" of the nose thrust location-to-wing distance. Basically their argument is that the aircraft's nose stays the same length in a turn!
Not compared to the trajectory it doesn't...
Just because that is an integral part of their schooling, does that mean you have to buy it?
If they cannot calculate the 6G "Corner Speed" within 50-80 MPH of actual data at similar weight, just how much do you think they know about the esoteric issue of fighting at the limit on long outdated fighters?
Gaston
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Oooops... Note I said downthrottling could reduce the prop disc load on the wings down from 2100 lbs to 500 lbs: The reason I made what looks like an exaggeration in reduction (1600 lbs: But the real exaggeration is in the three-quater reduction in thrust...
) is that I usually assume for my example that half the prop disc load (the pulled-back upper disc half) is at 1000 lbs of thrust, so that is why I halved it to 500 lbs downthrottled...
I think that, more realistically on a WWII single engine fighter, the engine thrust on the upper disc half could be cut, by downthrottling, from 2000 lbs at the prop disc half at WEP, this giving say 6000 lbs of force at the wings (this much larger number coming from the nose lenght and the thrust center height leverage, giving a right-angle compound leverage above and ahead of the wing's center of lift), which then allows a more realistic one-third power reduction from WEP: A one-third reduction from 2000 lbs to 1400 lbs of thrust at the prop disc half, and thus of one-third from 6000 lbs to 4000 lbs at the wings, which is still about 2000 lbs less wingloading (this being achieved through downthrottling alone) out of a rough 30 000 pounds of total force in a 3G sustained turn... (About a 7% reduction in total wing load: Possibly a significant edge in sustained turns.)
Since the elevators do originate all the 60 000 lbs of total force existing in a P-51D doing a 6G turn, and doing this with only a few hundred pounds of force at the starting point (the tail), such an ability to apply massive forces through leverage and the wing leading edge's "raised drag" is in fact not as wild as it seems...
I hope it is clear I don't intend these numbers as actual precise values; this is just to give a better sense of "scale" to what I mean...
I have heard very disparate numbers as to what the total prop disc thrust actually is...
The nose length, and the thrust center's height above the wing's center of lift, both do give a very long compound right angle leverage to the nose thrust, which is why I think the nose length matters a lot...
Gaston
) is that I usually assume for my example that half the prop disc load (the pulled-back upper disc half) is at 1000 lbs of thrust, so that is why I halved it to 500 lbs downthrottled...I think that, more realistically on a WWII single engine fighter, the engine thrust on the upper disc half could be cut, by downthrottling, from 2000 lbs at the prop disc half at WEP, this giving say 6000 lbs of force at the wings (this much larger number coming from the nose lenght and the thrust center height leverage, giving a right-angle compound leverage above and ahead of the wing's center of lift), which then allows a more realistic one-third power reduction from WEP: A one-third reduction from 2000 lbs to 1400 lbs of thrust at the prop disc half, and thus of one-third from 6000 lbs to 4000 lbs at the wings, which is still about 2000 lbs less wingloading (this being achieved through downthrottling alone) out of a rough 30 000 pounds of total force in a 3G sustained turn... (About a 7% reduction in total wing load: Possibly a significant edge in sustained turns.)
Since the elevators do originate all the 60 000 lbs of total force existing in a P-51D doing a 6G turn, and doing this with only a few hundred pounds of force at the starting point (the tail), such an ability to apply massive forces through leverage and the wing leading edge's "raised drag" is in fact not as wild as it seems...
I hope it is clear I don't intend these numbers as actual precise values; this is just to give a better sense of "scale" to what I mean...
I have heard very disparate numbers as to what the total prop disc thrust actually is...
The nose length, and the thrust center's height above the wing's center of lift, both do give a very long compound right angle leverage to the nose thrust, which is why I think the nose length matters a lot...
Gaston
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Glider
Captain
So you say that you have discussed this with a number of engineers and found one, on a web site (with no evidence that they are an engineer) who says it probably does occur but the difference in very small and as a result, this proves all your theories.
It's a bit like your theory that the Fw190 can turn as well as if not better than a Spitfire (which this thread is all about) based on one combat report. The fact that all the test flights between the Fw190 and the Spit, and the vast majority of the combat reports that exist say something else are therefore null and void.
You may want to get support i.e. qualifications and experience of the engineer before you nail your flag to the mast. I was given some training in aerodynamics as a five year apprentice with the FAA, but do not claim to be an aeronautical aerodynamic engineer. But if I had to choose between your totally unsubstantiated theories and that is all they are, theories, and the experience of all the engineers of all nations then I know which I will pick.
It's a bit like your theory that the Fw190 can turn as well as if not better than a Spitfire (which this thread is all about) based on one combat report. The fact that all the test flights between the Fw190 and the Spit, and the vast majority of the combat reports that exist say something else are therefore null and void.
You may want to get support i.e. qualifications and experience of the engineer before you nail your flag to the mast. I was given some training in aerodynamics as a five year apprentice with the FAA, but do not claim to be an aeronautical aerodynamic engineer. But if I had to choose between your totally unsubstantiated theories and that is all they are, theories, and the experience of all the engineers of all nations then I know which I will pick.
claidemore
Senior Airman
Which (Glider) brings us back to the original premise of the discussion.
All discussions and conversations tend to deviate (evolve?) from the original topic, whether online or face to face, and I personally do not see a problem with using the P51 and data and tests involving it as evidence to support or disprove a theory about turn ability of the FW190. Both planes are after all subject to the same laws of physics and there is limited data available on any one type.
I have found the discussion thus far entertaining and enlightening, in fact it is one of only two threads I have been following for the past few weeks.
Whether or not either protaganist ever 'sees the light' is irrelevant for me, as I will make my own decisions based on the evidence and arguments presented.
All discussions and conversations tend to deviate (evolve?) from the original topic, whether online or face to face, and I personally do not see a problem with using the P51 and data and tests involving it as evidence to support or disprove a theory about turn ability of the FW190. Both planes are after all subject to the same laws of physics and there is limited data available on any one type.
I have found the discussion thus far entertaining and enlightening, in fact it is one of only two threads I have been following for the past few weeks.
Whether or not either protaganist ever 'sees the light' is irrelevant for me, as I will make my own decisions based on the evidence and arguments presented.
From the "Report of Comparative Combat Evaluation of Fw-190 A/4 (should be A-5/U4) Airplane":
In general it [Fw 190] is considered to be an excellent interceptor-type airplane which is at a disadvantage against airplanes designed for the purpose of "in-fighting".
...
In view of the fact that the Fw-190 can outrun the F4U-1 and the F6F-3 in a 160 knot or faster climb, the best solution in offense is for the F4U-1 and F6F-3 to get the Fw-190 to close with them so that advantage can be taken of the superior maneuverability, provided, of course, that any initial advantage in altitude is not sacrificed merely for the sake of closing.
In general it [Fw 190] is considered to be an excellent interceptor-type airplane which is at a disadvantage against airplanes designed for the purpose of "in-fighting".
...
In view of the fact that the Fw-190 can outrun the F4U-1 and the F6F-3 in a 160 knot or faster climb, the best solution in offense is for the F4U-1 and F6F-3 to get the Fw-190 to close with them so that advantage can be taken of the superior maneuverability, provided, of course, that any initial advantage in altitude is not sacrificed merely for the sake of closing.
drgondog
Major
Oooops... Note I said downthrottling could reduce the prop disc load on the wings down from 2100 lbs to 500 lbs: The reason I made what looks like an exaggeration in reduction (1600 lbs: But the real exaggeration is in the three-quater reduction in thrust...
T=ma. In equilibrium T=D: Present your model and explain your assumptions.
I think that, more realistically on a WWII single engine fighter, the engine thrust on the upper disc half could be cut, by downthrottling, from 2000 lbs at the prop disc half at WEP, this giving say 6000 lbs of force at the wings (this much larger number coming from the nose lenght and the thrust center height leverage, giving a right-angle compound leverage above and ahead of the wing's center of lift), which then allows a more realistic one-third power reduction from WEP: A one-third reduction from 2000 lbs to 1400 lbs of thrust at the prop disc half, and thus of one-third from 6000 lbs to 4000 lbs at the wings, which is still about 2000 lbs less wingloading (this being achieved through downthrottling alone) out of a rough 30 000 pounds of total force in a 3G sustained turn... (About a 7% reduction in total wing load: Possibly a significant edge in sustained turns.)
In sustained Turn, T=D, L=W (in vector format); present your model and explain assumptions
Since the elevators do originate all the 60 000 lbs of total force existing in a P-51D doing a 6G turn, and doing this with only a few hundred pounds of force at the starting point (the tail), such an ability to apply massive forces through leverage and the wing leading edge's "raised drag" is in fact not as wild as it seems...
In a banked turn the forces generated on the aircraft are applied by a.) spanwise pressure distribution along the wing, as well as similar loads applied to both Rudder and Elevator. The pressure distribution of the 'up' aileron wing is slightly different from the 'down' aileron wing. In the banked state the aircraft is experiencing asymmetric loading to maintain flight equilibrium in a steady turn.
I hope it is clear I don't intend these numbers as actual precise values; this is just to give a better sense of "scale" to what I mean...
'scale' absent metrics or values derived mathmatically is meaningless.
I have heard very disparate numbers as to what the total prop disc thrust actually is...
T=K*eta*Bhp/Vel + Exhaust Thrust (and in case of Mustang + 'Meridith Effect thrust).. in pounds of thrust
K= conversion factor based on Velocity units ---> 550 if Velocity in Mph
eta=propeller efficiency - generally between .8 and .86 depending on altitude, Bhp, prop diameter, ratio of engine rpm to prop rpm, etc, etc
Bhp= brake Horsepower at the throttle setting and altitude density of operation
The nose length, and the thrust center's height above the wing's center of lift, both do give a very long compound right angle leverage to the nose thrust, which is why I think the nose length matters a lot...
It matters more relative to center of mass.
Gaston
Think, then Do the math, then pause and re-think
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parsifal
Colonel
drgondog
Major
Claidmore - I think you can very safely assme that I will not 'see the light' and embrace the 'new religion' of 6G sustained turns at 'corner velocities of 320mph' while embracing the mysteries of 'prop disk and long nose leverage while downthrottling at stall speed' forces of the universe.
On the other hand I did resort to putting a Mickey Mouse hat, with taped antenna and tinfoil sail, on my head to improve my receptivity to the ether and such input that may improve my comprehension..
Stay tuned..
Glider
Captain
claidemore
Senior Airman
Question about curved trajectory: Is a curved trajectory in flight actually a series of straight trajectories connected by imperceptable small turns? A ball on a string makes a smooth curve (I think), but an object in free flight tends to want to go straight?
This post (#339) in another thread got me thinking.
http://www.ww2aircraft.net/forum/aviation/bf-109-vs-p-40-a-16213-23.html#post748543
Or is the curve simply not smooth and constant?
This post (#339) in another thread got me thinking.
http://www.ww2aircraft.net/forum/aviation/bf-109-vs-p-40-a-16213-23.html#post748543
Or is the curve simply not smooth and constant?
drgondog
Major
Here is a jewel - a true beauty that approaches a Casey Stengal rupturing of both the English language and priciples of Aerodynamics...
From Gaston on ubi.com What is the P-51D's Corner Speed in Il-2? - Topic Powered by Social Strata
No. What you are saying is that the stall hierarchy relationship between aircraft types at "medium" speeds stays the same going down to stall speed (I don't really contest that "stable" relationship for speeds much higher than "medium"; say 350 MPH+: It is the higher prop load relative to wing lift in "medium" speeds that mixes things up).
Here is why I think the stall in sustained turns from low speed to "medium" speeds is altered differently, from prop aircraft type to prop aircraft type: Higher speeds mean higher power, and propeller traction gains a NEW leverage when the nose is raised, either through a raising of the center of thrust, or a lowering of the airframe"s center of drag vs the prop center of thrust, and this new leverage taxes the WINGLIFT.
The center of thrust is inevitably offset upward compared to the airframe center of drag by the act of raising the nose. That offset can only create a vertical lever, and that vertical lever will inevitably push downward on the center of the wing lift: In effect increasing the WINGLOADING in ways peculiar to a specific aircraft type.
The only way that this extra wingloading could be avoided (besides downthrottling) would be to keep the drag center and the prop disc"s center of thrust PERFECTLY aligned when the elevator is deflected: Impossible on a prop-traction fighter. This is irrelevant on jets because there is nothing ahead of the wing that needs to be kept in perfect alignment.
Also the lack of a jet thrust "disc" means the "thrust center" cannot move upward and "ouside" the disc center under the pressure of a nose-up angle change, creating leverage; there is no lateral extension (the prop blade) for the thrust center to move on to...
Either through fuselage/wing shape or the shortness of the nose, WWII fighters do not tax their wing lift in the same way in "medium-speed" turns, where the prop disc load relative to the wing lift is the greatest.
I think the nose lenght probably plays a big role in mixing things up at those "middle" speeds."
From Gaston on ubi.com What is the P-51D's Corner Speed in Il-2? - Topic Powered by Social Strata
No. What you are saying is that the stall hierarchy relationship between aircraft types at "medium" speeds stays the same going down to stall speed (I don't really contest that "stable" relationship for speeds much higher than "medium"; say 350 MPH+: It is the higher prop load relative to wing lift in "medium" speeds that mixes things up).
Here is why I think the stall in sustained turns from low speed to "medium" speeds is altered differently, from prop aircraft type to prop aircraft type: Higher speeds mean higher power, and propeller traction gains a NEW leverage when the nose is raised, either through a raising of the center of thrust, or a lowering of the airframe"s center of drag vs the prop center of thrust, and this new leverage taxes the WINGLIFT.
The center of thrust is inevitably offset upward compared to the airframe center of drag by the act of raising the nose. That offset can only create a vertical lever, and that vertical lever will inevitably push downward on the center of the wing lift: In effect increasing the WINGLOADING in ways peculiar to a specific aircraft type.
The only way that this extra wingloading could be avoided (besides downthrottling) would be to keep the drag center and the prop disc"s center of thrust PERFECTLY aligned when the elevator is deflected: Impossible on a prop-traction fighter. This is irrelevant on jets because there is nothing ahead of the wing that needs to be kept in perfect alignment.
Also the lack of a jet thrust "disc" means the "thrust center" cannot move upward and "ouside" the disc center under the pressure of a nose-up angle change, creating leverage; there is no lateral extension (the prop blade) for the thrust center to move on to...
Either through fuselage/wing shape or the shortness of the nose, WWII fighters do not tax their wing lift in the same way in "medium-speed" turns, where the prop disc load relative to the wing lift is the greatest.
I think the nose lenght probably plays a big role in mixing things up at those "middle" speeds."
drgondog
Major
Claidmore - a ball on a string is a classic example of the 'centrifugal force' construct whereas the curvilinear flight about a center is an example of Centripetal force.
String/Ball wishes to go 'straight' but the tether exerts a constant force making the velocity vector change in direction but not magnitude.
Aircraft in a perfect circular path, banked turn, constant speed also wishes to follow an 'independent' path but it also is thwarted by inward direction force making the velocity vector change direction but not magnitude
Both are described and modelled mathmatically as F= d/dt (MV)
You probaly have done the Calculus thing so I won't be pedantic about this differential expression with respect to time.
Short answer - in 'real life' and in the calculus, each increment of time will cause the velocity vector to change in direction but the physical world sees only a continuous circle about a fixed point..Differential calculus however is a minute change in direction for a minute change in time, with both change intervals approaching zero - meaning the math says that you have an infinite series of straight line segments, attached, one at a time, each with a very slight change in direction but with equal magnitude.
Please don't get offended if you are conversant with Calculus. If not, visualize the first 'approximation' if we integrated over six intervals as a hexagon, then 8 intervals an octagon, and so on until you have a polygon with so many sides that it is indistinguishable from a circle.
In a circular path about a fixed point and constrained by a tether, the force magnitude is constant, the radius of the circle is constant. If the 'tether' is gravitational force, the radius is also constant and the calculation for 'r' and "V' and 'omega=rate of turn' is straightforward.
There is no difference between an infinite polygon to express the path and a circle for an aircraft under perfect thrust, perfect control, perfect altitude maintenance, perfect aerodynamic behavior of the free body aircraft in a perfect atmosphere, no outside interference from turbulence/gusts/20mm impact, etc.
An aircraft in a banked turn without Yaw, experiences a Lift vector perpendicular (normal) to the plane of the wing, and would have a component equal and opposite of the Weight vector pointing to the center of the earth, and a 'lateral force' pointing inward to the center of rotation.
Strictly speaking the downward component of negative lift from the Elevator and positive lift component from the Rudder must be added to Weight to determine the required vertical component of the total Lift Vector.
BTW - NONE of the discussions we have all enjoyed or cursed for turning manuever calculations have ever introduced the contribution of the tail forces into the model. They would not be ignored in a sophisticated computer model
Summary - virtually all the models of interest in Performance predictions resuly from the above equation which is simply expressing the forces as change in momentum with respect to time.
This thread is about the Spitfire IX and the Fw 190A. Lets try and keep it that way.
I am having some trouble with this thrust slanting and prop disc.
With reference to the fuselage datum line, the Fw190A has a slightly negative angle for the engine and ~+6 degree for the wing and a nominal +2 degree for the stab.
In normal level flight I would think this gives the Fw190A a slight nose down attitude.
With reference to the fuselage datum line, the Fw190A has a slightly negative angle for the engine and ~+6 degree for the wing and a nominal +2 degree for the stab.
In normal level flight I would think this gives the Fw190A a slight nose down attitude.
drgondog
Major
Milosh - I further encourage you to have a 'lot of trouble' with 'thrust slantin and prop disk' as presented in the above narrations.
I have done a bit of thinking about the forces involved in the nose lenght leverage, and I came up with this, as it seems to me a drop of 7% in total wingload would not cover a drop of power of 30%...
We can take as a basis what Karhila has said about the Me-109G-6 (this is all very relevant to the FW-190A vs Spitfire as you'll see...):
virtualpilots.fi: 109myths
Full quote:"I learned to fly with the "Cannon-Mersu" (MT-461). I found that when fighter pilots got in a battle, they usually applied full power and then began to turn. In the same situation I used to decrease power, and with lower speed was able to turn equally well."
" When the enemy decreased power, I used to throttle back even more. In a high speed the turning radius is wider, using less speed I was able to out-turn him having a shorter turning radius. 250kmh seemed to be the optimal speed."
- Kyösti Karhila, Finnish fighter ace. 32 victories.
Ok. So he quotes an "optimal" downthrottled sustained turn speed of 160 MPH for the Me-109G-6, which allowed him to "Turn equally well" as one who throttled up to WEP: Optimal sustained turn speed at full power for the Me-109G-6 is usually quoted as around 220 MPH in Fin tests: So "equally well" means, say, the same 3 G (no decimal for the sake of simplicity) sustained at both speeds: 160 MPH and 220 MPH: All of it 3 Gs...
The radius of 160 MPH, to be equal in Gs to 220 MPH, must be a whole lot smaller, which makes up some if not most of the speed lost by downthrottling.
Let's take a lower figure of 2000 lbs prop disc total, and split it in half for the upper disc load: 1000 lbs, just for the sake of simplicity.
Let's assume the height of the wing's center of lift is one foot below the center of thrust when the angle of attack is raised to turn: Nose up, and slightly slanted thrust all in one...
Now we agree I hope that the elevator's effort is multiplied by the wing's leading edge drag as it is raised, so that the 3 G total of 21 000 lbs total wing lift is not borne by the elevator directly (I have it from an aero engineer that the load on the tail is only a few hundred pounds at most to generate those 21 000 lbs at 3Gs in the end).
So the upper prop disc load (1000 lbs) is on a vertical lever that is one foot high pulling forward at a right angle to another lever that is say, 12 foot long, on the Me-109G, to the point of rotation as the angle of attack is increased. (I think there is some thrust center migration into the upper disc half on nose-up too, but let's keep the total to one foot above the wing's center of lift.)
I will assume, for the sake of argument, that the point of rotation to change the angle of attack is also one foot behind the wing's center of lift, to give the wing's lift-drag center some leverage to raise the nose, and thus tilt the prop back compared to the trajectory...
With the wing's center of lift one foot lever vs the angle of attack change's point of rotation, what is the leverage ratio to the nose?: I don't know what the vertical one foot right angle at the nose precisely does, except that I am sure it gives a downward pull to the thrust, this at the end of a 12 foot lever...
The critical point is, what is the lever available to the wing's center of lift compared to the (unclear location to me) actual point of rotation as the angle of attack is raised?: If it is one foot, I would say it is 12 to 1 in favor of the nose thrust upper disc half load: So it takes, right from the start, 12 000 lbs of extra lift on the wing's center of lift to even begin tilting back the prop compared to the trajectory, as changing the wing's angle of attack requires...
Mind you, the prop is not pushing straight down with the 1000 lbs top disc half, put to pull the top disc half backward vs the trajectory against its thrust, you do have to beat 1000 lbs from a twelve foot lever with a one foot lever...
If the point of rotation is .5 foot behind the wing's center of lift, then 24 000 lbs of force would be required of the wing's center of lift... Let's again keep it at one foot for simplicity...
If you reduce the throttle by one third to 700 lbs, that 12 000 lbs becomes 8400 lbs... A drop of 3600 lbs... 3600 lbs on a total Me-109G wing load at 3 G of 21 000 lbs: That's a 17.1% drop...
But if you add the prop disc and wing load totals it becomes 33 000 lbs of total effort, of which 3600 lbs is down to a still reasonable 10.9% of drop in total force when a 30% reduction in prop thrust is done.
I am not sure for now if the prop disc load is beaten by moving the center of lift forward for greater than one foot leverage at a greater cost in drag, or if simply the force is added to the total wingload, or a mixture of both... I'll just add totals for now...
A 17.1 % or 10.9% drop in the total wingload would certainly better allow sustaining the same 3Gs at 160 MPH just like full power would allow it at 220 MPH... Especially when you consider the much smaller 160 MPH turn radius which makes up a lot of the lost speed by being smaller in circumference...
So you can see here how what Karhila says makes perfect sense...
Same thing with the Spitfire Mk IX being defeated at 7450 lbs by a FW-190A-8 at 9460 lbs...
Here's the FW-190A8's 4300 kg weight in a document:
http://www.wwiiaircraftperformance.org/fw190/fw190-a8-climb-13nov43.jpg
I picked the lower weight, since all of the numerical values I present here are simply a very rough demonstration of the potential scale of the issue. Spitfire IX weight is the heaviest because that is all I could see when I saw the summary test outline text originally, and I don't want to do over the whole calculations again...
http://www.spitfireperformance.com/spitfire-lfix-ads.jpg
Let's assume a nose lenght of 12 feet on the Spitfire (to center of lift), meaning 12 000 lbs to lift the nose by the wing's center of lift theoretical one foot lever...
At 3G we get 22 350 lbs total wingload for the Spitfire Mk IX.
For the FW-190A-8 we get 28 380 lbs. But the nose is only, say conservatively (I would think 7 feet is possible even for the later ones, but again, only a rough demonstration here...), 8 feet: "Only" 8000 lbs extra to lift the nose...
So the actual total wingload at "1000 lbs upper half disc" at full WEP power for both (assuming same power for simplicity again) is:
FW-190A-8: 28 380 lbs + 8000 lbs: 36 380 lbs
Spitfire Mk IX: 22 350 lbs + 12 000 lbs: 34 350 lbs.
Look at the margin: 2030 lbs: 5.5 % of margin...
Now what happens if the FW-190A-8 downthrottles in turns, Which I have heard at least one Western FW-190A-8 ace say is the only way to fight with it, and that he feared NO other fighter doing this...
700 lbs at the disc (30% downthrottled) with 8 foot of lever: 5600 lbs: 28 380 lbs + 5600 lbs:
Result: 33 980 lbs for the FW-190A-8... Vs 34 350 lbs for the MK IX...
Now you might say the Spitfire's wing was larger and thus had more lift: I heard quite the opposite: That it was a very sleek wing with a high Mach numbers, but with poor lift capabilities relative to its size, so that at high speed (300 MPH +) you could only move the stick top back 3 quaters of an inch before the wings warned of stall quite severely... What the Spitfire did have was an uncanny ability to maintain 3 axis control even while the wings were stalling, allowing "accross the circle" high angle of attack shots while stalling; great offensively, but a terrible feature defensively because of the lost speed...
The only real question is: If lifting the nose to tilt the prop back compared to the trajectory requires up to 12 000 lbs of force from the wings, is this achieved by shifting the center of lift more forward at the cost of more drag, or is this actually simply more load on the wing the way I calculated it? Likely a mixture of both...
For the record, I think no Spitfire at full power could sustain turns with a downthrottled FW-190A-8, but the Mk IX was the closest, perhaps very close. It is clear from combat accounts, and the above calculations, that the Spitfire Mk XIV was particularly hopeless against a FW-190A-8 in sustained turns, and this is correlated by at least one Mk XII and one Mk XIV account I have read... (Not even remotely such a thing for the Me-109G of course...)
Wartime Mk XIV pilots complained that the Mk XIV was handling more like a P-51D Mustang, to which the joking reply, long post-war, was "And that's a bad thing?" Well, not a very good thing, since the Merlin P-51D is about even, or barely better, than a Me-109G... And needs downthrottling in spades to really beat it, as the Hanseman account clearly demonstrates...:
http://www.spitfireperformance.com/mustang/combat-reports/339-hanseman-24may44.jpg
Again, early Razorback P-47Ds displayed some parity with earlier FW-190A-6s to the left, which leads me to think the Razorback P-47D could slightly out-turn some, maybe all, Spitfire marks in sustained left turns, but not of course in unsustained radius...
The P-47D of course crushed the Me-109G in sustained left turns like it wasn't even funny, and, again, the Luftwaffe captured testing by KG 200 fully correlates this: "The P-47D out-turns our Bf-109G" (Source: On Special Missions, KG 200)
Overall the picture seems pretty coherent to me...
Gaston
We can take as a basis what Karhila has said about the Me-109G-6 (this is all very relevant to the FW-190A vs Spitfire as you'll see...):
virtualpilots.fi: 109myths
Full quote:"I learned to fly with the "Cannon-Mersu" (MT-461). I found that when fighter pilots got in a battle, they usually applied full power and then began to turn. In the same situation I used to decrease power, and with lower speed was able to turn equally well."
" When the enemy decreased power, I used to throttle back even more. In a high speed the turning radius is wider, using less speed I was able to out-turn him having a shorter turning radius. 250kmh seemed to be the optimal speed."
- Kyösti Karhila, Finnish fighter ace. 32 victories.
Ok. So he quotes an "optimal" downthrottled sustained turn speed of 160 MPH for the Me-109G-6, which allowed him to "Turn equally well" as one who throttled up to WEP: Optimal sustained turn speed at full power for the Me-109G-6 is usually quoted as around 220 MPH in Fin tests: So "equally well" means, say, the same 3 G (no decimal for the sake of simplicity) sustained at both speeds: 160 MPH and 220 MPH: All of it 3 Gs...
The radius of 160 MPH, to be equal in Gs to 220 MPH, must be a whole lot smaller, which makes up some if not most of the speed lost by downthrottling.
Let's take a lower figure of 2000 lbs prop disc total, and split it in half for the upper disc load: 1000 lbs, just for the sake of simplicity.
Let's assume the height of the wing's center of lift is one foot below the center of thrust when the angle of attack is raised to turn: Nose up, and slightly slanted thrust all in one...
Now we agree I hope that the elevator's effort is multiplied by the wing's leading edge drag as it is raised, so that the 3 G total of 21 000 lbs total wing lift is not borne by the elevator directly (I have it from an aero engineer that the load on the tail is only a few hundred pounds at most to generate those 21 000 lbs at 3Gs in the end).
So the upper prop disc load (1000 lbs) is on a vertical lever that is one foot high pulling forward at a right angle to another lever that is say, 12 foot long, on the Me-109G, to the point of rotation as the angle of attack is increased. (I think there is some thrust center migration into the upper disc half on nose-up too, but let's keep the total to one foot above the wing's center of lift.)
I will assume, for the sake of argument, that the point of rotation to change the angle of attack is also one foot behind the wing's center of lift, to give the wing's lift-drag center some leverage to raise the nose, and thus tilt the prop back compared to the trajectory...
With the wing's center of lift one foot lever vs the angle of attack change's point of rotation, what is the leverage ratio to the nose?: I don't know what the vertical one foot right angle at the nose precisely does, except that I am sure it gives a downward pull to the thrust, this at the end of a 12 foot lever...
The critical point is, what is the lever available to the wing's center of lift compared to the (unclear location to me) actual point of rotation as the angle of attack is raised?: If it is one foot, I would say it is 12 to 1 in favor of the nose thrust upper disc half load: So it takes, right from the start, 12 000 lbs of extra lift on the wing's center of lift to even begin tilting back the prop compared to the trajectory, as changing the wing's angle of attack requires...
Mind you, the prop is not pushing straight down with the 1000 lbs top disc half, put to pull the top disc half backward vs the trajectory against its thrust, you do have to beat 1000 lbs from a twelve foot lever with a one foot lever...
If the point of rotation is .5 foot behind the wing's center of lift, then 24 000 lbs of force would be required of the wing's center of lift... Let's again keep it at one foot for simplicity...
If you reduce the throttle by one third to 700 lbs, that 12 000 lbs becomes 8400 lbs... A drop of 3600 lbs... 3600 lbs on a total Me-109G wing load at 3 G of 21 000 lbs: That's a 17.1% drop...
But if you add the prop disc and wing load totals it becomes 33 000 lbs of total effort, of which 3600 lbs is down to a still reasonable 10.9% of drop in total force when a 30% reduction in prop thrust is done.
I am not sure for now if the prop disc load is beaten by moving the center of lift forward for greater than one foot leverage at a greater cost in drag, or if simply the force is added to the total wingload, or a mixture of both... I'll just add totals for now...
A 17.1 % or 10.9% drop in the total wingload would certainly better allow sustaining the same 3Gs at 160 MPH just like full power would allow it at 220 MPH... Especially when you consider the much smaller 160 MPH turn radius which makes up a lot of the lost speed by being smaller in circumference...
So you can see here how what Karhila says makes perfect sense...
Same thing with the Spitfire Mk IX being defeated at 7450 lbs by a FW-190A-8 at 9460 lbs...
Here's the FW-190A8's 4300 kg weight in a document:
http://www.wwiiaircraftperformance.org/fw190/fw190-a8-climb-13nov43.jpg
I picked the lower weight, since all of the numerical values I present here are simply a very rough demonstration of the potential scale of the issue. Spitfire IX weight is the heaviest because that is all I could see when I saw the summary test outline text originally, and I don't want to do over the whole calculations again...
http://www.spitfireperformance.com/spitfire-lfix-ads.jpg
Let's assume a nose lenght of 12 feet on the Spitfire (to center of lift), meaning 12 000 lbs to lift the nose by the wing's center of lift theoretical one foot lever...
At 3G we get 22 350 lbs total wingload for the Spitfire Mk IX.
For the FW-190A-8 we get 28 380 lbs. But the nose is only, say conservatively (I would think 7 feet is possible even for the later ones, but again, only a rough demonstration here...), 8 feet: "Only" 8000 lbs extra to lift the nose...
So the actual total wingload at "1000 lbs upper half disc" at full WEP power for both (assuming same power for simplicity again) is:
FW-190A-8: 28 380 lbs + 8000 lbs: 36 380 lbs
Spitfire Mk IX: 22 350 lbs + 12 000 lbs: 34 350 lbs.
Look at the margin: 2030 lbs: 5.5 % of margin...
Now what happens if the FW-190A-8 downthrottles in turns, Which I have heard at least one Western FW-190A-8 ace say is the only way to fight with it, and that he feared NO other fighter doing this...
700 lbs at the disc (30% downthrottled) with 8 foot of lever: 5600 lbs: 28 380 lbs + 5600 lbs:
Result: 33 980 lbs for the FW-190A-8... Vs 34 350 lbs for the MK IX...
Now you might say the Spitfire's wing was larger and thus had more lift: I heard quite the opposite: That it was a very sleek wing with a high Mach numbers, but with poor lift capabilities relative to its size, so that at high speed (300 MPH +) you could only move the stick top back 3 quaters of an inch before the wings warned of stall quite severely... What the Spitfire did have was an uncanny ability to maintain 3 axis control even while the wings were stalling, allowing "accross the circle" high angle of attack shots while stalling; great offensively, but a terrible feature defensively because of the lost speed...
The only real question is: If lifting the nose to tilt the prop back compared to the trajectory requires up to 12 000 lbs of force from the wings, is this achieved by shifting the center of lift more forward at the cost of more drag, or is this actually simply more load on the wing the way I calculated it? Likely a mixture of both...
For the record, I think no Spitfire at full power could sustain turns with a downthrottled FW-190A-8, but the Mk IX was the closest, perhaps very close. It is clear from combat accounts, and the above calculations, that the Spitfire Mk XIV was particularly hopeless against a FW-190A-8 in sustained turns, and this is correlated by at least one Mk XII and one Mk XIV account I have read... (Not even remotely such a thing for the Me-109G of course...)
Wartime Mk XIV pilots complained that the Mk XIV was handling more like a P-51D Mustang, to which the joking reply, long post-war, was "And that's a bad thing?" Well, not a very good thing, since the Merlin P-51D is about even, or barely better, than a Me-109G... And needs downthrottling in spades to really beat it, as the Hanseman account clearly demonstrates...:
http://www.spitfireperformance.com/mustang/combat-reports/339-hanseman-24may44.jpg
Again, early Razorback P-47Ds displayed some parity with earlier FW-190A-6s to the left, which leads me to think the Razorback P-47D could slightly out-turn some, maybe all, Spitfire marks in sustained left turns, but not of course in unsustained radius...
The P-47D of course crushed the Me-109G in sustained left turns like it wasn't even funny, and, again, the Luftwaffe captured testing by KG 200 fully correlates this: "The P-47D out-turns our Bf-109G" (Source: On Special Missions, KG 200)
Overall the picture seems pretty coherent to me...
Gaston
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Jabberwocky
Staff Sergeant
Walter Wolfrum (137 victories, all in 109sn for JG52)
"The P-47 wasn't so bad because we could out turn and outclimb it, initially. [...] The P-51 was something else. It could do everything we could do and do it much better. First off, it was hard to recognize. Unless you saw it from the side, it looked like a Me 109. This caused us trouble from the outset. We would see them, think they were ours and then the damned things would shoot us full of holes. We didn't like them at all! " .
Hansotto Nehls (8. 11./JG 300)
".. At low to medium altitudes the P-51s were agile and could pull tight turns but the AS-engined Messerschmitt was generally superior over 6,500 metres due to the lower wing loading. Our Messerschmitts largely out-performed the P-47. I would often re-trim during a dogfight but had to be careful not to let too much speed bleed off in the turn. Combat was usually very brief.."
"The P-47 wasn't so bad because we could out turn and outclimb it, initially. [...] The P-51 was something else. It could do everything we could do and do it much better. First off, it was hard to recognize. Unless you saw it from the side, it looked like a Me 109. This caused us trouble from the outset. We would see them, think they were ours and then the damned things would shoot us full of holes. We didn't like them at all! " .
Hansotto Nehls (8. 11./JG 300)
".. At low to medium altitudes the P-51s were agile and could pull tight turns but the AS-engined Messerschmitt was generally superior over 6,500 metres due to the lower wing loading. Our Messerschmitts largely out-performed the P-47. I would often re-trim during a dogfight but had to be careful not to let too much speed bleed off in the turn. Combat was usually very brief.."
Glider
Captain
Shame that the test flights and combat reports say something else, isn't it.
drgondog
Major
Gaston - IF an attacking fighter like a 109 or 51 or 47 entered combat with a slower opponent, or one that initiated a turn to evade - then if the attacking fighter wished to pull deflection, it will initiate a pursuing curve.
Different fighter pilots like Duane Beeson would never downthrottle..If they could get a deflection shot, they simply used their superior energy to zoom past and above - then assess whether to return for another pass.
Some Would downthrottle to reduce speed, decrease their turn radius to seek deflection, and make a decision whether to try to out turn their opponent or break off the fight and use the remaining superior energy to gain altitude and position.
The last batch would reduce speed in the turn and try for the deflection shot all the way down to stall speed - and live or die based on a.) ability to do so, or b.) having a wingman take it from another position.
Giving up all available superior energy to reduce turn radius was risky if the fight was between an Fw 190 or an Me 109 and a Spit IX and risky, but less risky against a P-51B/D
You have constantly referenced Encounter reports written by 'winners'.
Different fighter pilots like Duane Beeson would never downthrottle..If they could get a deflection shot, they simply used their superior energy to zoom past and above - then assess whether to return for another pass.
Some Would downthrottle to reduce speed, decrease their turn radius to seek deflection, and make a decision whether to try to out turn their opponent or break off the fight and use the remaining superior energy to gain altitude and position.
The last batch would reduce speed in the turn and try for the deflection shot all the way down to stall speed - and live or die based on a.) ability to do so, or b.) having a wingman take it from another position.
Giving up all available superior energy to reduce turn radius was risky if the fight was between an Fw 190 or an Me 109 and a Spit IX and risky, but less risky against a P-51B/D
You have constantly referenced Encounter reports written by 'winners'.
- Status
