Aces High Bulletin Board
General Forums => Aircraft and Vehicles => Topic started by: Zacherof on April 12, 2013, 05:18:58 PM
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Does anyone have a
compilation of comparisons regarding the amount of stress different aircraft wings can take? Also what speeds, degrees, and gforces have as an affect to limiting manuevers?
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Simple answer - No. How to get there? 1.) get the design Gross weight for which stress calculations were performed, 2.) nail down the Structural design load factors - in general all US applied Stress = Yield for Design max and 1.5 Yield for Ultimate/Failure... for US the numbers were generally 8g Limit/12g Ultimate. P-51H was an exception as it was 7.5 for 8000 pounds, 11.25 for Ultimate.
I think that USN operated to same Spec. The key is that you can assume 8/12 G as the boundary but the harder assumption is "at which GW".
For example the XP-51 was stressed at 8/12 for 8,000 GW - which was full fuel and guns, but the same GW for the P-51B was closer to 9100 - so it WAS NOT stressed at 8g/12 as it was already 1100 pounds greater GW. It was a Limit Load of 8000/9100 X 8,000 pounds --- 7.03G limit load.
The conditions that the design engineer looked at was high G/ZHigh AoA flight in a turn, the Pull out for dives - and next (if the guys were really good and had good aero load guesstimates - was the asymetrical load manuever like slow or snap rolls or a rolling dive manuever to evade.
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I have the data for the Mossie somewhere. I'll see if I can dig it out when I have time.
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If you can find it that wod be really awesome. And thanks for pointing me in the right direction :salute
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The amazing thing is, they appear to never have tested how much those wings actually bent in an actual horizontal turn with a low-wing nose-puller... To this day even...
I've been looking for those wing in-flight turn stress tests for years now... And no, dive-pull-outs don't count, and that is what they do if they do it: I checked... This is because in a dive, the propeller is, in large part, unloaded by the extra dive speed...
For those who know my thinking about this, I think 3 G of turning at full power could equal 6 G of wing stress on these WWII fighter types, and the pilot limit of 6-7 Gs could mean 9-10 Gs of actual wing stress: Well within the ultimate failure limit...
Those who claim this is a violation of physical laws seem to confuse force and energy... I remember reading the claim (elsewhere) that this is is like claiming a perpetual motion machine... This shows you how clueless are those who claim this violates physical laws...
Unlike energy, yes you can get more force out of something than you put in... If you don't think this is true, you don't understand the difference between force and energy...
Gaston
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That is rather interesting. When inposted this I thought for sure someone might have this info.
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Zacherof Sir,
There are lots of folks that likely know the answer to your question.
Some of them reside 50-60 miles north of your reported location.
If they won't let you onto the base, look for a cantina in the vicinity of Rosamond that has many old pictures of early test pilots and test vehicles on the walls.
Pose your question there (when the flight-suit clad folks show up). . . if they don't know, they will likely take your name and number and figure it out...a bunch of inquisitive squirrels that lot!
Good Hunting!
- Rodent57
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That is perhaps the best advice I've been given on BBS :salute
Theres actually quite a few bases. Edwards(might be able to go on), march AFB,
and miramar. Now if I can find a way to get there... :bhead
Wonder if those chumps at china lake would let me on... :noid
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Spitfire was designed to specification no. F 7/30:
Structural Strength
(a) The strength of the main structure when carrying the load specified in paragraph 3, plus 100 lb shall not be less than as specified hereunder
Load factor throughout the structure with the centre of pressure in its most forward position 9.0.
Load factor for wing structure with the centre of pressure in its most backward position in horizontal flight 6.0
Load factor in terminal nose dive 1.75
Inverted Flight
(1) Load factor at incidence corresponding to the inverted stall and with C.P. at 1/3 of chord 4.5
(2) Load factor at incidence appropriate to steady horizontal inverted flight and at the maximum speed of horizontal normal flight 4.5
(b) The alighting gear must be able to withstand an impact at a vertical velocity of 10 feet per second and at this velocity the load on the alighting gear must not exceed three times the fully loaded weight of the aircraft.
(c) When subject to the impact forces on alighting, as specified above, the load factor for the alighting gear must not be less than M/3, and for the remainder of the structure not less than 1- 1 /2. The load factor for the structure and the attachment fittings of the alighting gear must always be greater than that for the alighting gear itself by the margin indicated above.
(d) The maximum weight per wheel of the aircraft in pounds must not exceed 12 times the product of the wheel and tyre diameters in inches with the aircraft carrying the full load specified above.
(c) The above factors are to be determined by the approved official methods as published by the DTD and the detail requirements given in A.P.970 are also to be satisfied. With a view to minimising the risk of flutter, attention should be given to the recommendations of A.P. 1177, particularly as regards the static balance of ailerons.
(0 The wing is to be sufficiently rigid to withstand satisfactorily any torsional or other loads which may be encountered during service operations.
(g) Ribs (both mainplane and tail unit) are required to develop, on test, factors 20% greater than those specified for the aircraft as a whole.
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Of course for all fighters Rule 27 of the fighter pilots creed apply: “The aircraft G-limits are only there in case there is another flight planned for that particular airplane. If subsequent flights do not appear likely, there are no G-limits.”
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Of course for all fighters Rule 27 of the fighter pilots creed apply: “The aircraft G-limits are only there in case there is another flight planned for that particular airplane. If subsequent flights do not appear likely, there are no G-limits.”
:rofl
:aok
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Spitfire was designed to specification no. F 7/30:
Interesting in that there is no delineation between Design Limit (usually within elastic/Yield stress) and Ultimate (Failure and in US 1.5xLimit)
The Load factor in Terminal dive should be the Q load Does 1.75 seems to reflect that? But, if so what is the Design Q? Is it then a reflection of max positive G load to be undertaken in a terminal dive?
Next question - was F 7/30 a spec for the MkI at Mk I design GW or updated to reflect the increase in GW for say Mk IX or XIV?
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Spitfire was designed to specification no. F 7/30:
It was F.10/35.
F.7/30 was for Blackburn F.3, Bristol Type 123, Bristol Type 133, Gloster Gladiator, Gloster SS.19, Hawker P.V.3, Supermarine Type 224, Westland F.7/30
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Supermarine Type 224 was also to be called Spitfire, hence there sometimes being confusion.
drgondog,
Yes, there were increases in strength for the Spitfire's wing, though I don't recall when they showed up. Mk V or VIII perhaps.
Gaston,
It would do you a wonder of good to invest the time and money in some physics and aerospace engineering classes.
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My mistake.
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Interesting in that there is no delineation between Design Limit (usually within elastic/Yield stress) and Ultimate (Failure and in US 1.5xLimit)
The failure load of steel and aluminum is typically 1.5 times the yield load.
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The failure load of steel and aluminum is typically 1.5 times the yield load.
:aok Correct,
Curiously since the airframe business I was in when I was a structures guy dealt with such materials ( and early composites and titanium forgings) that might be the reason that was our standard in US.. for normal flight loads and cycles.. greatly reduced for critical components in high frequency reversible loads such as helicopter engine mounts, etc,
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Supermarine Type 224 was also to be called Spitfire, hence there sometimes being confusion.
drgondog,
Yes, there were increases in strength for the Spitfire's wing, though I don't recall when they showed up. Mk V or VIII perhaps.
Gaston,
It would do you a wonder of good to invest the time and money in some physics and aerospace engineering classes.
Karnak - does that mean the load profiles in the above spec were at Ultimate?.. wow, if so - unless that was Late in the development cycle when GWs had pretty well max'ed out...
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I'm fairly certain they are limit loads. While British specifications typically were a bit lighter on the limit load (btw. P-51H was designed after the British specs, thus the slightly lower limit load), they weren't THAT different. The Jerries also typically designed their fighters to an 8G limit load.
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I seem to remember an old discussion here years ago about the P-38's limit load. Having been designed as an high alt interceptor it only had a limit load of 6G, if my memory serves me. Also someone here thought it was the cause of the death of an American ace in the Pacific.
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I seem to remember an old discussion here years ago about the P-38's limit load. Having been designed as an high alt interceptor it only had a limit load of 6G, if my memory serves me. Also someone here thought it was the cause of the death of an American ace in the Pacific.
I would tend to doubt that Lockheed set a limit/ultimate load spec - but as in the case of the XP-51 designed to 8/12G Limit and ultimate load factor for 8000 pounds GW, that limit load reduced to 6.4/9.6 at 10,000 pounds for the P-51D. The P-51H was 11 Ultimate and 7.4 Limit at 8000.
I suspect, because I can't prove otherwise, that the Load Factors were based on YP-38 basic weight which was 1600 pounds less than the basic weight of the P38J... by the time the full internal combat load was attained it grew another 1500 pounds.. and the limit load dropped dramatically to the "6G" range.
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Speed also plays a factor.
You can't ignore the force that is pulling the wings backwards because of your speed.
That has to be factored in and it complicates things greatly concerning at what G loading you will experience wing failure.
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There are other factors.. Planes such as the P-38 & Typhoon were among the 1st to really
encounter compressibility effects in high speed dives.. with tails falling off & uncontrollable
terminal dives.. both an aero & fatigue stress dynamic..
The P-38 needed & got 'dive flap' spoilers that enabled control to be regained..
but remained fundamentally low Mach/Vne limited.
The Typhoon needed the Tempest wing to really solve its issues..
[ but received the beneficial thinner Tempest tail-plane in the late-production versions]
Typhoon Vne/dive speed limit of 520mph IAS @ 10,000ft ..
..was still ~100mph higher than the P-38 though.
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The amazing thing is, they appear to never have tested how much those wings actually bent in an actual horizontal turn with a low-wing nose-puller... To this day even...
I've been looking for those wing in-flight turn stress tests for years now... And no, dive-pull-outs don't count, and that is what they do if they do it: I checked... This is because in a dive, the propeller is, in large part, unloaded by the extra dive speed...
For those who know my thinking about this, I think 3 G of turning at full power could equal 6 G of wing stress on these WWII fighter types, and the pilot limit of 6-7 Gs could mean 9-10 Gs of actual wing stress: Well within the ultimate failure limit...
Those who claim this is a violation of physical laws seem to confuse force and energy... I remember reading the claim (elsewhere) that this is is like claiming a perpetual motion machine... This shows you how clueless are those who claim this violates physical laws...
Unlike energy, yes you can get more force out of something than you put in... If you don't think this is true, you don't understand the difference between force and energy...
Gaston
Wait...what???
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Unlike energy, yes you can get more force out of something than you put in... If you don't think this is true, you don't understand the difference between force and energy...
Don't you hold with those fangled laws of thermodynamics then Gaston?
Quite right, how is the perpetual motion machine coming along? :banana:
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TEST:
Go to dueling arena, or training arena, I forget which... take off from the '30k base'... it will drop you in flight, then dive the plane: get to high speed, pull back!!!
I did this long ago, most planes break their wings, but some don't! Long lost list, need to make new!
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For your amusement http://youtu.be/OhwLojNerMU (http://youtu.be/OhwLojNerMU)
From my reading you don't often get much warning from flutter induced failure. The film clips are often super slow mo and an excited control surface can in reality just fly off in a half second.
Here is a flutter analysis of a famous amateur build airplane. What is interesting is that in investigating the flutter characteristics of adding more fuel capacity they found a configuration that lowered flutter speed to 160 knots, (this is a 300 mph airplane.)
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Thanks for the responses guys. I hink I have what I want now :salute
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wing stress is over rated! ;)
http://www.youtube.com/watch?v=SXQ5uoMBnuE
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Don't you hold with those fangled laws of thermodynamics then Gaston?
Quite right, how is the perpetual motion machine coming along? :banana:
What? Do you mean to say you can't get more force out of something than you put in? Oh ok, that was not what you meant. Thank God....
Gaston
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What? Do you mean to say you can't get more force out of something than you put in? Oh ok, that was not what you meant. Thank God....
Gaston
The first law of thermodynamics says it. I merely agree with it.
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The first law of thermodynamics says it. I merely agree with it.
Shida, I always thought if you used a tool like a pulley or lever and fulcrum you could get more effort put out than was required to put in! Heavens knows aircraft are full of pulleys and levers...... :D
:rofl :rofl :rofl
You engineer types,always looking at the fine print and never seeing the big picture!
:salute
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What? Do you mean to say you can't get more force out of something than you put in? Oh ok, that was not what you meant. Thank God....
Gaston
Name something that yields more energy than is put into it.
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I seem to remember an old discussion here years ago about the P-38's limit load. Having been designed as an high alt interceptor it only had a limit load of 6G, if my memory serves me. Also someone here thought it was the cause of the death of an American ace in the Pacific.
It wasn't what caused McGuire's crash.
ack-ack
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Name something that yields more energy than is put into it.
I didn't say more energy, I said more FORCE... What is bending the wings on a constant speed turning aircraft (especially at a constant speed, so at the lower speed end) is purely a FORCE, and that has nothing to do with energy...
That you can get more FORCE out of something than you put in has been established since Archimedes.
All those who claimed that my idea, which was that a light aircraft with bigger wings might possibly bend its wings with more force, at the same G load, than a heavier aircraft with smaller wings at the same G load (not as an established fact, but just as something theoretically possible and allowed by physical laws), and then went on to claim that this violated physical laws, then these people simply showed they did not understand the difference between force and energy...
Although this apparently has never been measured in horizontal turns for low-wing nose-driven aircrafts (horizontal turns which I think are fundamentally different in dynamics than dive pull-outs), my basic idea that a Spitfire may, under horizontal turn circumstances, have a heavier wingloading than a FW-190A, is allowed by basic physical laws...
As far as I know, all the wing bending that was definitely tested on low-wing nose-driven types is static ground wing bending tests...
To simplify, the nose position of the thrust, and the assymetrical incoming air of a turn, both open the door (in my opinion) to in-flight leverages in a horizontal turns that would not really show up in a dive pull-out (which dive pull-outs I am told are the only way wing-bending measurements in flight are actually taken: Don't ask me why).
It is not clear to me that even dive pull-outs wing-bending measurements in flight were ever done with nose-pulled low-wing monoplanes, since those tend to be old or low-cost aircrafts.
In-flight wing-bending measurements are a very expensive and very uncommon thing for small or old aircrafts, apparently... Ask any Warbird operator (as I have): It is not even on their radar screen...
Once again, it is part of the most basic laws of physics that you can get more force out of something than you put in... If the speed in the turn is not decaying, as with sustained speed turns of around 3 Gs, then you are dealing only with pure force bending the wings.
Zero energy at play, except for what is burning away in the fuel tank...
And with these basic physical facts established, some first hand observations finally start to make sense:
RCAF John Weir: "A Hurricane was built like a truck, it took a hell of a lot to knock it down. It was very manoeuvrable, much more manoeuvrable than a Spit, so you could, we could usually outturn a Messerschmitt. They'd, if they tried to turn with us they'd usually flip, go in, at least dive and they couldn't. A Spit was a higher wing loading..."
"The Hurricane was more manoeuvrable than the Spit and, and the Spit was probably, we (Hurricane pilots) could turn one way tighter than the Germans could on a, on a, on a Messerschmitt, but the Focke Wulf could turn the same as we could and, they kept on catching up, you know."
Gaston
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Force is energy.
As to your laughable "theory", as has been explained, you'd need something like 30 degrees of flex to accomplish what you claim happens. You don't need instruments to see that, it can be observed from the ground.
You claim that Gs can be different for different aircraft doing the same circle at the same speed and physics simply says that isn't possible. To pick your two pet bogeymen, a Spitfire and an Fw190 doing 250mph in a 750ft radius circle will both be under exactly the same G forces.
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That you can get more FORCE out of something than you put in has been established since Archimedes.
No, it hasn't. You don't even understand simple machines do you?
Shida, I always thought if you used a tool like a pulley or lever and fulcrum you could get more effort put out than was required to put in! Heavens knows aircraft are full of pulleys and levers...... :D
This is a disaster Morfiend! Why did you have to ruin our conspiracy against Gaston. Someone call HTC immediately and fess up. Tell them to remodel the 190, it should out turn a Spitfire after all since the 190 has more levers in it :old:
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No, it hasn't. You don't even understand simple machines do you?
This is a disaster Morfiend! Why did you have to ruin our conspiracy against Gaston. Someone call HTC immediately and fess up. Tell them to remodel the 190, it should out turn a Spitfire after all since the 190 has more levers in it :old:
Are you sure about that,the FW's use electically controlled flaps and IIRC also for tail trim so it might actually have fewer levers and pulleys than a spit,TBH I'd have to count them all and compare but it has to be the reason the spit can out turn the FW's.... :devil
:salute
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You do, you gave me that doc back in 2005. I'll try to find it.
I have the data for the Mossie somewhere. I'll see if I can dig it out when I have time.
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I seem to remember an old discussion here years ago about the P-38's limit load. Having been designed as an high alt interceptor it only had a limit load of 6G, if my memory serves me. Also someone here thought it was the cause of the death of an American ace in the Pacific.
Tommy McGuire stalled while pulling a high G turn with his externals still attached.
So his GW was very high and 6G in a turn at near corner speed put him into an accelerated stall (probably) and at max GW it was much higher weight than the original design Limit Load...
The standard for US structural engineers was 8G Limit and 12G Ultimate for the original design gross weight for symmetrical load conditions. The P-38 (like the P-51) grew significantly and as a result the Limit and Ultimate G loading decreased inversely proportional to the increase in GW over the original. When a Mustang grew from 8000 pound design GW the Limit was 8G (~stress close to elastic peak). when it grew to say 10,000 pounds it would have reduced to 6.4G.
If you can find a P-38 pilot handbook the V-N diagram will show you the max speed to G plot - all based on design GW and max CL without flaps.
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Are you sure about that,the FW's use electically controlled flaps and IIRC also for tail trim so it might actually have fewer levers and pulleys than a spit,TBH I'd have to count them all and compare but it has to be the reason the spit can out turn the FW's.... :devil
:salute
:cry :aok
I once tried to think like Gaston, just as an experiment. Look what I came up with to explain how a Spitfire could out-turn a 190:-
http://bbs.hitechcreations.com/smf/index.php/topic,330942.msg4451230.html#msg4451230
:banana: :banana:
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Force is energy.
As to your laughable "theory", as has been explained, you'd need something like 30 degrees of flex to accomplish what you claim happens. You don't need instruments to see that, it can be observed from the ground.
You claim that Gs can be different for different aircraft doing the same circle at the same speed and physics simply says that isn't possible. To pick your two pet bogeymen, a Spitfire and an Fw190 doing 250mph in a 750ft radius circle will both be under exactly the same G forces.
I never claimed the Gs were different, I claimed the forces applied to the wings were different, and not proportional to wingloading, between different aircraft types for the same G.
Are you saying 3 Gs produces 15° of flex?
As to force is energy...: Did you and nrshida ever finish high school?
You guys pretend to be knowledgeable in flight physics, and you don't even know the difference between FORCE and ENERGY????!!!!
Jeez, some high schools are really no longer doing their work...
Ask any engineer what the difference between force and energy is... Grown-ups will do as well...
Gaston
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I never claimed the Gs were different, I claimed the forces applied to the wings were different, and not proportional to wingloading, between different aircraft types for the same G.
Are you saying 3 Gs produces 15° of flex?
As to force is energy...: Did you and nrshida ever finish high school?
You guys pretend to be knowledgeable in flight physics, and you don't even know the difference between FORCE and ENERGY????!!!!
Jeez, some high schools are really no longer doing their work...
Ask any engineer what the difference between force and energy is... Grown-ups will do as well...
Gaston
Your are right Gaston. It's just not good enough! :mad:
We should be reported to The Scientific Inquisition for heresy, for believing in those satanic works the laws of thermodynamics and other dastardly writings. It's Karnak's fault, he lead me astray. :eek:
Never mind high school, I shall be sending back both my Bachelor of Science degree and my Master's degree certificates at once and asking for a refund :old:
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Uhh, I'll try...
Force = mass x acceleration ?
force x distance = work ?
Work per time = energy ?
E=mgh ? (Mass x gravity x height)
E=.5 mv2 (1/2 mass x velocity squared)
Not sure which type of force or energy...
Uhh, so did our space shuttle crash or blow up?
To answer the question:
Energy = work/time = (force x distance)/time ?
Do I get a star? :D
Extra credit!
Force = Energy x time / distance ?
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We can't create or destroy energy.
We can only manipulate it, or change its form.
So... we never get complete energy transfer because each time we change it, we lose some.
Efficiency = (energy we put in - energy we got out) / energy we put in.
For kinetic energy ( 1/2 mass x velocity squared)... our velocity is losing some E to air resistance.
Air resistance has to do with the frontal area and shape of the plane.
Air resistance also has to do with friction created from the 'surface tension' or 'boundary layer' of air going past the plane. Wax your plane! This is why large wing area usually means slower, but turnier. And small wings are usually faster and less turny.
Also air resistance has to do with lift, the more lift created, the more forward energy is transfered away. Don't turn so much!
For potential energy (mgh) or (mass x gravity x height) we are fighting against gravity!
So diving down and climbing up, we constantly lose energy to gravity. Our Work = force x distance too, distance is much futher when bnz'ing to just flying straight. Energy = work/time.
Wing loading is plane weight / wing area.
Power loading is engine power x wing loading.
Why an f14 will out turn a zeke? F14 is heavier, but much more wing area and more powerful engines.
Hence the more powerful 109s should be faster than spitys! And still able to turn with them in a full power turn, but not in a power off turn, hmm, depends on the ratios.
For wing stress, forces / energy can get in sink (be in phase with) with other forces.
Example: planes engine, frame, wings, and control surfaces - like elevators, each as sparate components could all reach a state of 'harmony' so flabbing in the same dirrection at the same time, cooperatingly overstressing, and... something breaks, like the elevators. With speed, flabbing gets more energy from air passing.
And I've thoroughly confused me self now. Someone correct all me mistakes pls!
Some stuff I don't understand, why didn't th Ki team just make stronger elevators?
Same with the yak3 team. Could have to do with the mach cone squishing stuff just at the right spot to break it?
Same question about 109 (and all other planes) compression. Change strength and shape till fixed?
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Other thing you guys are talking about is 'statics' of trusses (frames)
Look up 'statics'... NOT statistics btw.
Its about sheer and bending moment diagrams.
Put a 'load' on a system/framing... you get sheer and bending.
Sheer moment diagrams are for when frames have opposing forces and want to 'snap'...
Bending is when the frame does just that, bends.
Also torq = perpendicual force x distance. <-- why spity's got their wing tips cut to roll better, that little bit of area was FAR from the frame, and apparently it was significant enough to chop some off.
Yes different areas, shapes, and weights get different results from same evenly distributed loads - air, when pulling Gs.
'Mechanics of materials', different types of materials break at different times, some are brittle, some are ductile (bendable).
Need this stuff just right, thought I read yaks broke wings on early ones because the rivits, put in hot, and the holes something so too tight, brittle also, sheering happened under G's. But this is some of what happens when trying to copy stuff and not knowing how it was assembled.
Boss coming, hasta la visatas, from me phone!
Editting, yeah 109s had all the aces! 190 pilots 2nd place! 262, Komets, V1 and V2's, salamander too! luft tech was uber!
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I am so lost, and it my thread :cry
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I am so lost, and it my thread :cry
don't read anything Franz posts and you will be fine...
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Ok, someone said about 'levers' and "pullys.'
THIS IS TO GET MORE FORCE OUT THAN PUT IN.
The topic is 'Mechanical Advantage.'
Back to this:
Work = force x distance
:banana: vs :noid
Suppose, Mr Banana, wants to use a lever to lift a Rock .
He puts a 'fulcrum' pivot point between, like a teeder-todder
We have work done by Mr Banana, and the work done to the rock.
The Distances, D, is each distance to the fulcrum pivot point.
Work for banana will be W(b) = F(b) x D(b)
Work on the rock will be W(r) = F(r) x D(r)
Ideally (not realistically) conditions would be no work lost... sooo
Work done by Mr Banana would equal work done on the Rock.
W(b) = W(r)
or
F(b) x D(b) = F(r) x D(r)
If f(b) is small, but the D(b) is far, and d(r) is short then the F(r) is large.
Hmm
f(b) x D(b) = F(r) x d(r).
Put the fulcrum nearer to the rock.
Same as a hydraulic lift.
Small pressure on large area = big pressure on small area.
THIS IS TO GET MORE ENERGY THAN PUT IN.
Ok, only way to do this is to find energy that was already stored.
Energy = mass x gravity x height.
Work = force x distance
Energy = work / time = force x distance / time.
Coyote vs Rock vs Roadrunner
Coyote puts a little force to push a round rock up a tall mountain, by a not-steep shallow road, but far distance over a long time, and now has a lot of E=mgh, big H, over a long time.
AT THIS POINT, he can now put only a small energy to send the rock falling down fast and smash stuff with a lot of E.
Stored Energy.
If Coyote FOUND a rock at high alt, then he could technically get more out than HE put in.
The E was already there.
Extreme example:
Kinda like a nuclear bomb, a little bit of energy to release what was already there a LAGE ammount of atomic energy..
Not so extreme example:
A 30mm tater hitting something, igniting by impact heat(?) and then a large release of chemical energy xfered into pressure and heat - an explosion.
Yes you can get more Energy out than you put in, if it was already there or pre stored.
So that proves it, p38 Js should sux! haha just kidding. :banana:
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From a "get more energy out than you put in" standpoint we have to include the energy from plate tectonics that raise the rock up there or the energy of the explosives that was created during the manufacturing process of that shell to be energy that "you" put in.
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Energy = work / time = force x distance / time.
Work per time = energy ?
Once you introduce time, you are talking about power, so to be correct you need to replace the word energy in the quotes above with the word power.
Hope that helps.
Badboy
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Eeeek, you might be right, I need to put actual units in, instead of just titles, to see how it works out. :cry
To be continued! :(
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Uhh, I'll try...
Force = mass x acceleration ?
force x distance = work ?
Work per time = energy ?
E=mgh ? (Mass x gravity x height)
E=.5 mv2 (1/2 mass x velocity squared)
Not sure which type of force or energy...
Uhh, so did our space shuttle crash or blow up?
To answer the question:
Energy = work/time = (force x distance)/time ?
Do I get a star? :D
Extra credit!
Force = Energy x time / distance ?
No STAR for you.
Power = D * F / T (like 1 HP = 550 ft lb per second)
And Since D * F = Work
Work per time I.E. Work/time = power.
From above P * T = D * F
Since Energy E = P * T (like WATT HOURS or KWH)
E = D * F
hence Energy = Work
While Gaston is correct in that Force and distance can be converted to and from each other via levers or pulleys while still maintaining the Same Energy
Since W = E and W = F * D take any distance you want and solve for the force.
His conclusion from the above fail to take into account things like the direction of a force. And many people seem to think of G's as a force, it is not, it is a unit of acceleration.
HiTech
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Will the real caoder please stand up!
Ok, yes difficult to type science on computers, let alone a phone.
Can't realy type equations let alone graphs with vectors, or free-body diagrams, let alone ridgid-body diagrams.
But the real question is... drum roll...
WHY DO 190s ALWAYS FLY PULLING UP??? :huh
WHY DO 410s ALWAYS FLY NOSING DOWN??? :headscratch:
How to aim the Great German Luftwaffe fighter or bomber-intercepters' weapons with such issues?!?!
Yes STAR for me! and no STAR for you! :ahand
Fix combat trim please!
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All this palaver and magical thinking reminded me of something I always wondered about with wings. It is normal to have a wing designed to stall inboard first so that ailerons remain effective as long as possible. So, if you are pulling 6g in your 2000lb P-Whatzit, just nibbling at a stall, does that mean that the 12000lbs of force holding you up is now further out along your wing thereby raising the bending moment on the spar compared to a uniformly distributed beam model?
As you pass through the stall is there a spike in bending moment that is sharply different from loading your wing with sandbags?
Anybody?
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Out of my league here but I'll try.
Single engine prop, the prop would still be putting some airflow for lift even at zero speed.
And the flaps are always 'inboard' so seems the ailerons would be left out as 2nd place anyways.
Twin engine planes, same thing - flaps on both sides of engines for most planes?
Ailerons are outside towards ends of wings so their area gets better leverage.
Torq = force x perpendicular distance. Imagine using a wrench to take off a bolt, push from outer side of the lever.
For the spike... hmm, not sure the topic...
change in position over time = speed
change in speed over time = acceleration
change in acceleration over time = "the jerk" ~ Mr Bell lol
so the 'spike' might be the same as 'jerk'... a change in acceleration (over a very short time.)
Could be that the metal of the wings has some elasticity and gets in phase with the g's on the plane?
Like just as the g's come off the wings, the metal unloads and wobbles/slingshots back?
Or the disruption of airflow from the g's which was smooth thru corner, then flutters as the angle of attack changes and gets smooth again?
Ok I'm lost!
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As you pass through the stall is there a spike in bending moment that is sharply different from loading your wing with sandbags?
Yes, as the pressure distribution on the wing changes the bending moments and thus the stress values change. Spikes in the wing stress are also due to the rate at which those changes take place. Sandbags are a static load, wing loading is dynamic. For example, a suddenly applied load will cause stress values double that of static loads.
You can test this yourself by carrying out a simple experiment. If you take a set of kitchen scales and place an object on the scales very slowly and very gently. You should see the needle on the scales rise slowly until it reaches the weight of the object. Now start again, but this time place the object in contact with the pan on the scales so gently that the needle hardly deflects, then just let it go. That's a suddenly applied load and the needle will momentarily spike at a value double the weight of the object. The needle will normally oscillate for a short time (depending on how much friction is causing damping in the scales) before settling back at the weight of the object.
Aeroelastic effects can damp the stress response, and under certain conditions an oscillatory response can lead to flutter and rapid load onset and thus large magnification of the stress.
Regards
Badboy
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I have sometimes wondered if the floppy wings of early Spitfires actually helped in delaying stall by slightly bending under heavy load to add to wash-out that was already built in the wing.
-C+
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No STAR for you.
Power = D * F / T (like 1 HP = 550 ft lb per second)
And Since D * F = Work
Work per time I.E. Work/time = power.
From above P * T = D * F
Since Energy E = P * T (like WATT HOURS or KWH)
E = D * F
hence Energy = Work
While Gaston is correct in that Force and distance can be converted to and from each other via levers or pulleys while still maintaining the Same Energy
Since W = E and W = F * D take any distance you want and solve for the force.
His conclusion from the above fail to take into account things like the direction of a force. And many people seem to think of G's as a force, it is not, it is a unit of acceleration.
HiTech
Well indeed it looks like power is a more correct term than my use of the term energy: Energy was explained to me by an engineer as force over distance over time, as opposed to force over distance being work, which I misremembered as just "force"... He did not use the term power in the conversation we had, for some reason, but it was twenty years ago... At least now I know the exact terms...
G is a unit of acceleration, but the momentum of the fuselage pressing down on the wings, in a sustained turn, is work: The turn being sustained speed means time disappears from the equation beyond the gas tank: It is just force over distance.
In any case, the accepted wingloading theory, which in my opinion is unverified for WWII-type low wing monoplane nose-driven fighters (I have yet to find any evidence of wing bending measurement ever being taken during horizontal turns, as opposed to dive pull-outs, but would love to be shown some...), this currently accepted theory is different from my theory -that the FW-190A exerts less work on its wings than the Spitfire- because it assumes the distances between the various forces are that same between types.
First of all, I find it suspicious that the assymetrical inflow of air into the prop disc is not taken into account: From what I understand, this is assumed to be negligible... I don't see what is implausible about a hundred pounds of force being required for each degree of angle of attack increase, when forcing a flat disc to follow a curvature... Over the whole thrust of the disc, a hundred pounds per degree is very little...
The only argument against this (made to me by an aeronautic engineer) is the following: At a modest sustained turn 7° AoA, it would imply 700 lbs of force at the nose, which would require a similar force on the elevators to maintain just 7° (maybe just 500 lbs given the tail being longer than the nose, a ratio of about 1.5 to one on a P-51).
Obviously the pilot is not figting such huge forces in his elevators... The error everyone is making is to assume that the movement of the whole aircraft makes, while starting to pivot nose up, is initially just a pivoting action around its CG. I think the pivot point is very briefly closer to the prop, and this means the whole aircraft is, for a micro-second, more sweeping tail-down into the angle rather than pivoting, which action could shift the CL in front of the CG.
The CL shifting in front of the CL means that, as the turn tightens, the further pivoting action is actually helped by the scissor action of the CL being in front of the CG: That is why the pilot cannot feel the resistance of the prop in his elevator...
You can actually see the effect in the very claims Hitech has made of his own flying experience: He has confirmed 6G dive pull-outs in a P-51D at speeds well below 300 mph.
Yet, when the Society of Experimental Test Pilots tested in 1989 the minimum speed to reach 6 Gs in actual horizontal turns, the minimum speed the P-51D needed to reach 6 G horizontally without stalling was 320 mph: Why is 6 G so easy to reach below 300 mph in dive pull-outs compared to genuine horizontal turns?
Well at least the question must be asked before it can even be answered... The answer could be that, while diving, the prop disc is unloaded by air hitting the front of the blade at a greater rate than propulsion: This could tip the balance to nullify the effect of the prop getting assymetrical incoming air...
Suggestions that the SETP were somehow distorting the tests, which they bothered to make at great expense, doesn't seem too convincing to me... Sorry, but I'll take their word over yours...
So if, for the sake of argument, we accept that through incoming air assymetry the prop is putting, through nose length leverage, a greater downward force on the wings, for this to actually work at bending the wings more, then the wings themselves also have to come up with more force coming up, for the whole thing to stay up...
I think there is a way for the wing to generate correspondingly greater lift: If the CL does shift forward, there is something going on with the airflow around the wing... The fact that no wing bending tests in actual horizontal turns has surfaced is pretty telling: I would have expected those to be easily found...
The point I wanted to make about the Force/Energy issue is that there is no physical laws against getting more "work" out of something than you put in: A simple lever will do it... The assumption here is that, in the case of these aircrafts, the direction of the "work" creates no leverages that could play around with the outcomes: Until that assumption is backed by actual wing bending measurements in actual flight while actually turning horizontally, this is an assumption with no science behind it...
And the fact that a supposed 50% advantage in wingloading remains completely invisible in actual live combat is particularly ludicrous: I have five separate examples of FW-190As out-turning Spitfires in slow speed sustained turns (right where my theory predicts the greatest advantage), two of them general pilot statements as to an overall FW-190A advantage in that area, and so far I have never found slow-speed sustained turn examples of the contrary in actual combat. That's five to zero. A mighty discrete 50% wingloading advantage...
Yes test and general pilot lore tells the opposite, but somehow when an actual specific combat happens, it is always the opposite that comes up... But in these things remember this: Any out-turning event for the FW-190A counts, while for the Spitfire it has to be demonstrably low-speed and sustained turns, with no diving or high speed just before: By this accounting it is five to zero for the Fw-190A so far...
Gaston
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And the fact that a supposed 50% advantage in wingloading remains completely invisible in actual live combat is particularly ludicrous: I have five separate examples of FW-190As out-turning Spitfires in slow speed sustained turns (right where my theory predicts the greatest advantage), two of them general pilot statements as to an overall FW-190A advantage in that area, and so far I have never found slow-speed sustained turn examples of the contrary in actual combat. That's five to zero. A mighty discrete 50% wingloading advantage..
How many examples are there of Spitfires out turning Fw190s in low speed sustained turns? Have you looked?
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Well indeed it looks like power is a more correct term than my use of the term energy: Energy was explained to me by an engineer as force over distance over time, as opposed to force over distance being work, which I misremembered as just "force"... He did not use the term power in the conversation we had, for some reason, but it was twenty years ago... At least now I know the exact terms...
G is a unit of acceleration, but the momentum of the fuselage pressing down on the wings, in a sustained turn, is work: The turn being sustained speed means time disappears from the equation beyond the gas tank: It is just force over distance.
In any case, the accepted wingloading theory, which in my opinion is unverified for WWII-type low wing monoplane nose-driven fighters (I have yet to find any evidence of wing bending measurement ever being taken during horizontal turns, as opposed to dive pull-outs, but would love to be shown some...), this currently accepted theory is different from my theory -that the FW-190A exerts less work on its wings than the Spitfire- because it assumes the distances between the various forces are that same between types.
First of all, I find it suspicious that the assymetrical inflow of air into the prop disc is not taken into account: From what I understand, this is assumed to be negligible... I don't see what is implausible about a hundred pounds of force being required for each degree of angle of attack increase, when forcing a flat disc to follow a curvature... Over the whole thrust of the disc, a hundred pounds per degree is very little...
The only argument against this (made to me by an aeronautic engineer) is the following: At a modest sustained turn 7° AoA, it would imply 700 lbs of force at the nose, which would require a similar force on the elevators to maintain just 7° (maybe just 500 lbs given the tail being longer than the nose, a ratio of about 1.5 to one on a P-51).
Obviously the pilot is not figting such huge forces in his elevators... The error everyone is making is to assume that the movement of the whole aircraft makes, while starting to pivot nose up, is initially just a pivoting action around its CG. I think the pivot point is very briefly closer to the prop, and this means the whole aircraft is, for a micro-second, more sweeping tail-down into the angle rather than pivoting, which action could shift the CL in front of the CG.
The CL shifting in front of the CL means that, as the turn tightens, the further pivoting action is actually helped by the scissor action of the CL being in front of the CG: That is why the pilot cannot feel the resistance of the prop in his elevator...
You can actually see the effect in the very claims Hitech has made of his own flying experience: He has confirmed 6G dive pull-outs in a P-51D at speeds well below 300 mph.
Yet, when the Society of Experimental Test Pilots tested in 1989 the minimum speed to reach 6 Gs in actual horizontal turns, the minimum speed the P-51D needed to reach 6 G horizontally without stalling was 320 mph: Why is 6 G so easy to reach below 300 mph in dive pull-outs compared to genuine horizontal turns?
Well at least the question must be asked before it can even be answered... The answer could be that, while diving, the prop disc is unloaded by air hitting the front of the blade at a greater rate than propulsion: This could tip the balance to nullify the effect of the prop getting assymetrical incoming air...
Suggestions that the SETP were somehow distorting the tests, which they bothered to make at great expense, doesn't seem too convincing to me... Sorry, but I'll take their word over yours...
So if, for the sake of argument, we accept that through incoming air assymetry the prop is putting, through nose length leverage, a greater downward force on the wings, for this to actually work at bending the wings more, then the wings themselves also have to come up with more force coming up, for the whole thing to stay up...
I think there is a way for the wing to generate correspondingly greater lift: If the CL does shift forward, there is something going on with the airflow around the wing... The fact that no wing bending tests in actual horizontal turns has surfaced is pretty telling: I would have expected those to be easily found...
The point I wanted to make about the Force/Energy issue is that there is no physical laws against getting more "work" out of something than you put in: A simple lever will do it... The assumption here is that, in the case of these aircrafts, the direction of the "work" creates no leverages that could play around with the outcomes: Until that assumption is backed by actual wing bending measurements in actual flight while actually turning horizontally, this is an assumption with no science behind it...
And the fact that a supposed 50% advantage in wingloading remains completely invisible in actual live combat is particularly ludicrous: I have five separate examples of FW-190As out-turning Spitfires in slow speed sustained turns (right where my theory predicts the greatest advantage), two of them general pilot statements as to an overall FW-190A advantage in that area, and so far I have never found slow-speed sustained turn examples of the contrary in actual combat. That's five to zero. A mighty discrete 50% wingloading advantage...
Yes test and general pilot lore tells the opposite, but somehow when an actual specific combat happens, it is always the opposite that comes up... But in these things remember this: Any out-turning event for the FW-190A counts, while for the Spitfire it has to be demonstrably low-speed and sustained turns, with no diving or high speed just before: By this accounting it is five to zero for the Fw-190A so far...
Gaston
This is your finest piece so far. Outstanding :banana:
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there is no physical laws against getting more "work" out of something than you put in
Gaston
No STAR for you.
Power = D * F / T (like 1 HP = 550 ft lb per second)
And Since D * F = Work
Work per time I.E. Work/time = power.
From above P * T = D * F
Since Energy E = P * T (like WATT HOURS or KWH)
E = D * F
hence Energy = Work
HiTech
NOTE ENERGY = WORK
So gaston again goes against rule #1.
by stating
there is no physical laws against getting more "work"
per above proof you could substitute "Energy" for "Work" out of something than you put in[/i]
Good luck with that perpetual motion machine you keep trying to say is possible.
HiTech
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This all seems waaaay over complicated. All your gears and levers don't make it take less energy to accomplish a given unit of work. If the opposite were true than ultimately you could get an infinite amount of work out of any amount of energy.
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NOTE ENERGY = WORK
So gaston again goes against rule #1.
by stating
per above proof you could substitute "Energy" for "Work"
Good luck with that perpetual motion machine you keep trying to say is possible.
HiTech
We need a Gaston forum,just to see how much work you can get out of a little energy!
:salute
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Good luck with that perpetual motion machine you keep trying to say is possible.
He's not into perpetual motion, he just wants the 190 to out turn the Spitfire.
Have you ever considered making a "Dastardly and Muttley in Their Flying Machines" arena? Then you could model a Fw190Gs - the Gaston special - a stripped out A-5 with a helicopter rotor for a prop hub, then he'd teach us all a lesson with prop disk theory :old:
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Hi Gaston
It is good to see you posting again, I always enjoy reading your comments and I see you are trying to come to terms with some of the scientific concepts involved. However, I notice you have not yet fully succeeded in that, and there are some minor issues that I would like to help clarify for you.
The point I wanted to make about the Force/Energy issue is that there is no physical laws against getting more "work" out of something than you put in: A simple lever will do it... The assumption here is that, in the case of these aircrafts, the direction of the "work" creates no leverages that could play around with the outcomes:
In that passage of text you are making a very small mistake, you just have one single word wrong, but unfortunately it changes the meaning of what you are saying entirely. Change one word and your statement is correct.
In your text quoted above, if you swap the word "work" for "force" and read it again it will be correct. I assume that is what you actually meant to say.
It is an easy mistake to make, for example work is calculated as force times distance, and has similar units to leverage which is also force times distance, but one of them is a measure of energy and the other is just a force acting at a distance causing leverage, and is sometimes called a bending moment, or just a moment. Using leverage to get out more force than you put in is perfectly correct and is known as a mechanical advantage. However, if you change the word force and replace it with work, as you have done, your statement is not just wrong, it is profoundly wrong and breaks the laws of physics and thermodynamics.
I very much enjoy reading your theories. So forgive me for being blunt, but in that one passage of text there was just a single wrong word, but it changed what you were saying from being potentially correct, to being completely wrong. The problem is that every other passage of text contains similar and more numerous errors. Each mistake small by itself, but having a similarly negative impact on what you are saying to the point it is almost all complete and utter nonsense.
I hope that does not sound too harsh, because personally, I am always delighted to see you post on these boards, and I look forward to reading your analysis on this subject. However, if you are still a little fuzzy regarding the distinction between the terms, please accept the following explanation.
Energy comes in many forms, and "work" is simply a term used to describe a form of mechanical energy. When a force moves a distance, work is said to be done and it is a form of energy. In terms of energy, the time doesn't matter, you could move a force over some distance very slowly or very rapidly, the energy, or work, will be the same. When you introduce time, you are then talking about power. If you want to move a force some distance very slowly you don't need much power, if you want to move the same force the same distance very quickly you need more power.
It is worth pointing out that although we call mechanical energy work, energy is only involved when the force actually moves some distance. This should not be confused with bending moments, which has the same units of force times distance, but if a force is exerted at some distance from a point it is a description of the bending effect, and despite having the same units as work it is very different to work or energy. Energy can be involved, but only when the force or bending effect causes some form of movement, linear or rotation. The confusing thing is that most forces do cause movement, even if it does not move the object it acts on as a whole, it may just cause internal movement within the object, in the form of strain or deflection and thus energy is almost always involved when forces are applied.
I know that appears to complicate matters, but despite that I hope I have at least clarified the terms as they are normally used. I appreciate how frustrating it must be trying to communicate your theories when the rest of us are using technical language correctly. I feel sure that if you are able to make some progress in using those terms in the same way, we will all be able to enjoy your fascinating theories more fully.
At least, I hope something here has been helpful.
Regards
Badboy
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Bad boy you are a prince, it is always a pleasure to read your postings.
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How many examples are there of Spitfires out turning Fw190s in low speed sustained turns? Have you looked?
What do you think? Did you ever see one? I really would like to see one, as in 17 years of looking I found only the opposite...
Which kind of explains why John Weir says it has a heavier wing loading...
Gaston
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In your text quoted above, if you swap the word "work" for "force" and read it again it will be correct. I assume that is what you actually meant to say.
It is an easy mistake to make, for example work is calculated as force times distance, and has similar units to leverage which is also force times distance, but one of them is a measure of energy and the other is just a force acting at a distance causing leverage, and is sometimes called a bending moment, or just a moment. Using leverage to get out more force than you put in is perfectly correct and is known as a mechanical advantage. However, if you change the word force and replace it with work, as you have done, your statement is not just wrong, it is profoundly wrong and breaks the laws of physics and thermodynamics.
I very much enjoy reading your theories. So forgive me for being blunt, but in that one passage of text there was just a single wrong word, but it changed what you were saying from being potentially correct, to being completely wrong. The problem is that every other passage of text contains similar and more numerous errors. Each mistake small by itself, but having a similarly negative impact on what you are saying to the point it is almost all complete and utter nonsense.
I hope that does not sound too harsh, because personally, I am always delighted to see you post on these boards, and I look forward to reading your analysis on this subject. However, if you are still a little fuzzy regarding the distinction between the terms, please accept the following explanation.
Energy comes in many forms, and "work" is simply a term used to describe a form of mechanical energy. When a force moves a distance, work is said to be done and it is a form of energy. In terms of energy, the time doesn't matter, you could move a force over some distance very slowly or very rapidly, the energy, or work, will be the same. When you introduce time, you are then talking about power. If you want to move a force some distance very slowly you don't need much power, if you want to move the same force the same distance very quickly you need more power.
It is worth pointing out that although we call mechanical energy work, energy is only involved when the force actually moves some distance. This should not be confused with bending moments, which has the same units of force times distance, but if a force is exerted at some distance from a point it is a description of the bending effect, and despite having the same units as work it is very different to work or energy. Energy can be involved, but only when the force or bending effect causes some form of movement, linear or rotation. The confusing thing is that most forces do cause movement, even if it does not move the object it acts on as a whole, it may just cause internal movement within the object, in the form of strain or deflection and thus energy is almost always involved when forces are applied.
I know that appears to complicate matters, but despite that I hope I have at least clarified the terms as they are normally used. I appreciate how frustrating it must be trying to communicate your theories when the rest of us are using technical language correctly. I feel sure that if you are able to make some progress in using those terms in the same way, we will all be able to enjoy your fascinating theories more fully.
At least, I hope something here has been helpful.
Regards
Badboy
Quote: "it may just cause internal movement within the object, in the form of strain or deflection and thus energy is almost always involved when forces are applied."
Does this inclusion of energy means that my notion of some aircraft types bending their wings differently through different-than-expected leverages actually break physical laws?
I think if you think my theory is entirely an issue of leverages, you are misunderstanding it:
There is two parts to my theory: Only one part involves leverage: That is the part where the prop assymetry and the nose length comes in, tilting the thrust axis down.
The other part involves a change in the airflow around the wing: That has nothing to do with leverages, except for the issue of moving the Center of Lift in front of the CG.
If you understand anything about my theory, you will be able to say why that move of the CL in front of the CG is essential...
From what Hitech said, you can get more energy out than you put in...: This is his quote:
"While Gaston is correct in that Force and distance can be converted to and from each other via levers or pulleys while still maintaining the Same Energy
Since W = E and W = F * D take any distance you want and solve for the force.
His conclusion from the above fail to take into account things like the direction of a force. And many people seem to think of G's as a force, it is not, it is a unit of acceleration."
I don't understand his objection that I don't take into account the direction of the force.
In any case, there are several ways to easily disprove my theory:
1-My theory depends on the frontal engine power level affecting the wing's bending moment on powerful low-wing types during a horizontal turn: Show any data that shows no such correlation between engine power, on these types, and the the wing bending exists...
2-My theory depends on the minimum speed to reach 6 Gs on WWII fighters to be different in a dive pull-out than on a horizontal turn: If they are the same, then the theory doesn't work. So far the SETP test definitely leans my way, with 320 mph on a P-51 at METO, and close to their respective maximum level speed on all the other types: Same thing accross the board in other words...
3-Find in-flight wing bending data, in horizontal turns, on these WWII types, that matches theoretical calculations...
(I've been told unfortunately that such data is usually gathered form dive pull-outs, if at all...)
4- Not a final proof, but if some Spitfires in combat could ever be found to gain in low-speed multiple level turns on a FW-190A, that could be an indication as to what kind of hole that 50% wingloading advantage is hiding... It is not absolute, because a lot of these FW-190As were armored-up to fight bombers from Spring 1944, but it would still be interesting...
Gaston
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From what Hitech said, you can get more energy out than you put in...: This is his quote:
"While Gaston is correct in that Force and distance can be converted to and from each other via levers or pulleys while still maintaining the Same Energy
Since W = E and W = F * D take any distance you want and solve for the force.
Sigh,
a complete lack of understanding that I NEVER said you can get more energy out then you put in. In the quote you have, I said Energy or work remains constant and do not change.
But for any amount of work, you can covert distance to force or oppositely force and distance. It is like saying 4 times 4 is the same as 16 times 1 or 2 times 8 or 8 times 2. They all = 16 units of energy.
HiTech
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Hey, y'all never know. This could be the science standard of the future generations that none of us have yet to understand.
Ya never know! :old:
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The confusing thing is that most forces do cause movement, even if it does not move the object it acts on as a whole, it may just cause internal movement within the object, in the form of strain or deflection and thus energy is almost always involved when forces are applied.
Badboy
Yes but there is something inherently not relevant, it seems to me, to wing bending in sustained turns in what you say above:
While the wing is in the process of bending there is movement until the effort has reached its peak: Let's say 3 G. For that peak-preceding period, there is indeed wing movement, there is in fact a part of energy, as you say, because there is movement from a point A to a point B, most visibly for the wingtips at least...
If the wingtips are raised fast, or slow, this will affect the amount of energy present: This is irrelevant to the discussion of sustained turns...
To maintain the wing bent at the same value, there is no longer any wing movement, therefore energy, or "work", is no longer present, except in a "potential" form, stored in the elasticity of the wings.
You are pointing out something that is not relevant throughout the turn, for the sustained part of the turn: Once the peak has been reached there is no longer any wing movement, and therefore no longer any "work".
And, once again, depending on leverages, you can get more force out than you put in, which is what I have been saying all along... Your insistence there is energy involved in maintaining the wings bent seems to me to muddy the waters: The only reason there is "potential energy" stored here is that the wing are not infinitely rigid... We are discussing physics, not materials...
The biggest problem of my theory, which you consistently fail to point out, is that the described leverages forces are all downward.
The real problem with my theory is that it implies far greater wing lift forces than anything previously assumed to counter-act this, with minimal loss of speed (I think in part because the prop assymetry is actually an increase of thrust on the inside of the turn: Prop blade efficiency becomes greater inside the turn: Again, not an increase of energy, but a simply a reduction of inefficiency).
Again, the greater than assumed lift forces would be an increase in lift efficiency. No energy or "work" involved here, if the sum of all the added downward forces reduce movement to zero, which of course they do...
That this increase of wing lift efficiency is unknown from wind-tunnel tests is very easy to understand: No powered flight, no control surface-induced change in attitude, and, I'll bet, no real replication of curved flight airflow...
The only way to really know if these extra forces are present is to measure the wing bending in level turns.
Guess what? No such test has surfaced for WWII types, except that the procedure is seen as a modern high-tech luxury, exclusively done by pulling out of dives...
Another problem with the theory is that it implies that the CL moves forward of the CG.
This forward CL movement would, in my opinion, not affect directional stability for an obvious reason: Given what I described by now, you should easily understand why...
Gaston
P.S. Hitech did not address what he meant about me not taking into account the direction of the forces.
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Gaston I am trying to understand what you are saying. In simplified language are you saying that in a 3g horizontal turn some of the thrust from the engine/prop acts against the direction of lift such that the wings must support a greater load than 3 times the weight of the aircraft?
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Gaston I am trying to understand what you are saying. In simplified language are you saying that in a 3g horizontal turn some of the thrust from the engine/prop acts against the direction of lift such that the wings must support a greater load than 3 times the weight of the aircraft?
No he's saying a Fw190 should out-turn a Spitfire. :lol
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No he's saying a Fw190 should out-turn a Spitfire. :lol
Because the Spitfire is taking 6 Gs of acceleration in a 3 G turn due to its wings flexing and the Fw190A is taking 3 Gs of acceleration in a 3 G turn due to its wings not flexing.
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Because the Spitfire is taking 6 Gs of force in a 3 G turn due to its wings flexing and the Fw190A is taking 3 Gs of force in a 3 G turn due to its wings not flexing.
Ah. Pure genuis. I'm totally convinced.
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If the wingtips are raised fast, or slow, this will affect the amount of energy present: This is irrelevant to the discussion of sustained turns...
Gaston
P.S. Hitech did not address what he meant about me not taking into account the direction of the forces.
I should have said "Power" instead of "Energy" in the above sentence, since time is involved...
Also in a turn, I suggested all the forces are equal to zero, which is true, but I forgot to add that if there is no Power "into" the turn, then there is no turn... I was wrong to say it this way, and there is of course Energy and Time involved in a sustained turn, not just Force: Force is holding the wings bent, but Power moves the aircraft up into the turn continuously while turning...
I think I have figured out what Badboy and all the others fail to understand: You guys assume that for a heavy small wing aircraft to have more "Power" available to turn, vs a light big wing aircraft, it requires "added Power"...
There is no "added Power" in my theory: To think that is to show you haven't read much of what I said...
In my theory, all the nose-driven prop types lose turning Power compared to what is theoretically available. They are all losers, so there is no excess turning Power...
This is why, despite a lesser climb rate and acceleration, and a minuscule 5% wingloading advantage (at 33.1lbs per sq. ft., compared with the Spitfire XIV's 35.1 lbs per sq. ft), the Vampire Mk I gains 25% in turn time per turn vs the Spitfire Mk XIV... (It reverses a tailing Spitfire XIV in four 360° turns)
Let me re-iterate this: An aircraft with less Power-to weight (thus with less acceleration and climb rate) and a 5% wingloading advantage, will display a 25% advantage in turn time in low-speed sustained turns over a Spitfire XIV... How does that square with your calculations?
The reason for the above is that all the low-wing nose-driven prop types (of some weight and power at least) lose turning Power compared to what is available to similar straight-wing jets. My contention is that some of the prop types lose less, some of the prop types lose more. There is nothing that involves "extra" turning Power here...
The real problem, with my idea, is that there is a requirement for an extra lift Force to compensate for the huge lift Force loss needed to stabilize the nose thrust location higher vs the trajectory. Once the nose is lifted to the new AoA, there is no movement to keep it there: Just Forces.
The problem of this extra lift Force is that it seems to mean there is more lift Force available in turning flight than what is seen in wind tunnels: This extra Force can't come from nowhere: For a wing to maintain easily a powered nose at a more upward angle, there is apparently extra Force to be found in curved airflow, just like your arm will find extra Force pulling on a pulley... How it does that is the real problem...
If the object is too heavy and the pulley doesn't lift it, your effort remains just Force, but that doesn't mean you don't lose your available physical Power pulling on it... Or that you would not lose more available "Power" for applying the same Force with a smaller pulley... Well the Spitfire has a smaller pulley, so it loses more lifting Power than the FW-190A, which doesn't mean the FW-190A ever "gained" anything...
The "curved airflow pulley" is here to help hold the nose up into the turn, which contributes no Power to pulling the aircraft up into the turn: On the contrary it subtracts from it... I don't think there is any physical law that says that a Force cannot reduce the effective motion of a given Power...
The Fw-190A certainly doesn't gain 25% per turn on a Spitfire Mk XIV, so it would still get creamed by the Vampire... Its turn Power potential is way below the Spitfire Mk XIV, but it just doesn't lose turn Power potential, to waste it on Force, as much as the Spitfire does.
Gaston
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I think the answer to how the theoretical structure known as a skyhook has finally been discovered. It is now obvious to me it must work on the "curved airflow pulley" principal.
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DH Vampire is a jet, Spit XIV is a piston engined fighter. How are you calculating power to weight ratio for each and at what speed are we talking about?
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Also in a turn, I suggested all the forces are equal to zero, which is true, but I forgot to add that if there is no Power "into" the turn, then there is no turn... I was wrong to say it this way, and there is of course Energy and Time involved in a sustained turn, not just Force: Force is holding the wings bent, but Power moves the aircraft up into the turn continuously while turning...
Gaston
You may wish to tell glider pilots about this theory so they make sure to never try and turn.
HiTech
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You man wish to tell glider pilots about this theory so they make sure to never try and turn.
HiTech
:rofl :rofl :rofl
But what about the tow rope? It has to input some power to the glider.....
When I was a youngster my Dad would say "you cant shoot pool with a rope" I think I finally understand what he meant.
:salute
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Hey Gaston, I see that your grasp of the mechanics of flight are as solid as they've ever been.
For those interested, here are a few extracts from the tests, via a post here a couple of years back
Turning Circles: The Vampire I is superior to the Spitfire XIV at all heights. The two aircraft were flown in line astern formation. The Spitfire was positioned on the Vampire's tail. Both aircraft tightened up to the minimum turning circle with maximum power. It became apparent that the Vampire was able to keep inside the Spitfire's turning circles. After four or five turns the Vampire was able to position itself on the Spitfire's tail so that the deflection shot was possible. The wing loading of the Vampire is 33.1lbs per sq. ft. compared with Spitfire XIV's 35.1 lbs per sq. ft.
...The Vampire will outmanoeuvre the Spitfire type of aircraft at all heights, except for initial acceleration at low speeds and in rolling.
... The Spifire XIV used in the comparison trial was a fully operational aircraft fitted with a Griffon 65, giving 2,015 h.p. at 7,500 ft. Vampire I had an operational take-off weight of 8,800 pounds, powered by a de Havilland Goblin 2 turbojet, developing 3,000 pounds static thrust."