Draining E in turns

[Crumpp]

Well, I don't put much weight to that site; the author uses same Cd0 for all Spitfire models despite radiator changes and Cd0 seems to be value which actually is total Cd in the Lednicer's paper.

Yes he mentions the radiator changes.  It is not just the radiator.  It's the windshield is too far sloped on the Spitfire and the boundry layer seperation for the cooling system occurs just inside the ducting from what I read, not in the entrance.

You can see a late Mk Spitfires radiators here:
http://www.flightjournal.com/fj/store/viewissue.asp?issueid=BSFG
Down at the bottom.

And your Cl values are not true for 1 g and your drag values are not true for Lednicer's flat plate areas at 300 mph near sea level (real values are somewhere around 5000 N).

For the Cl calcs, the only thing that changes is "r" and the rest is easy.  Make up a velocity and plug it inot the formula.  I cross-referenced my "r value with the constant you use and it behaved exactly as it should.  Close but a fraction less dense because the constant is figured at a lower tempature.   I did not use the flat plate areas.  I used the wet area.

The math is the math Gripen.  You run it through the formula and it spits out the numbers.  Again we are missing some of the Details.

You are assuming the A/C is at full throttle (horsepower and exhaust thrust).  In order to fly at that Airspeed all of the planes must throttle back.

 

Well, I don't put much weight to that site; the author uses same Cd0 for all Spitfire models despite radiator changes and Cd0 seems to be value which actually is total Cd in the Lednicer's paper.

He does the same for the FW-190.  His calculations apply to an FW-190A8 and a Spitfire IX LF from what I read.  You keep thinking the radiator problem was fixed.  That is Davids point.  It was not.

How come you keep trying to discredit this guy?  Do you think he is on an agenda to make his favourite fighter better?  All human beings make mistake, but I would have to say this guy is much more qualified than anyone here.  Is anyone here published in one of the leading homebuilt aircraft organizations in the world periodicals?  Does anyone here work for a engineering firm specializing in fluid mechanics and flow problems?  Is anyone here an aeronautical engineer?

Calculations at our level are fine for very simple, simple problems.  Reading to much into them however is dangerous and can lead to wrong conclusions.  It's the same thinking as:

If you live in New Jersey you are an American.  Therefore all Americans are from New Jersey.

It still does NOT explain why the fundatmental forces in high speed turn do not act, at least according to the Flight instructor and PhD., like it claimed in this forum.  The CL does not change from level flight much.  Yet the claims here is that it does.  I can see where it would be needed to change the vector.  Once the vector is changed however the plane will follow the new vector until acted upon by other forces.

Nor does it explain why our planes self recover from a steep bank.

Well, in last thread someone called me butthead, lawyer etc. And somehow you concluded that my statement that Clo of the wing profile must be tested in the wind tunnel means that 3D to 2D conversion is impossible according to me. And you also came up with numbers which you could not explain despite requested.

I am not going to play the he said she said game.  Anybody that is interested can simply read the thread.  In my book, you have blown your credibility on several points.  If you want to rediscuss that the feel free to start a new thread.  

http://www.hitechcreations.com/forums/showthread.php?s=&threadid=125882

Getting back to the subject.


If the spitfire is so efficient in the High Speed Turn then why are they not being used in the Unlimited class racers?  The later Mk Spitfire certainly should be able to compete.  Yet we see the heavier wingloaded and less draggy aircraft like the P51 being used.

Crumpp

[gripen]

Originally posted by Crumpp
Yes he mentions the radiator changes.  It is not just the radiator.  It's the windshield is too far sloped on the Spitfire and the boundry layer seperation for the cooling system occurs just inside the ducting from what I read, not in the entrance.

This has nothing to with credibility of the J22 site. The author there does simple and clear errors in his analyses as noted above.

Originally posted by Crumpp
For the Cl calcs, the only thing that changes is "r" and the rest is easy.

If it is so easy why you can't do it?

Originally posted by Crumpp
I did not use the flat plate areas.  I used the wet area.

Do a little math: Multiply the wet area with the Cdwet, just like you do in the drag formula. What is the number you get?

Originally posted by Crumpp
I am not going to play the he said she said game.  Anybody that is interested can simply read the thread.  In my book, you have blown your credibility on several points.  If you want to rediscuss that the feel free to start a new thread.  

Actually it's you who started to talk about that thread and now you are blaming me for that.

gripen

[Crumpp]

I'm not sure where you thought I was saying Cl increases in a turn. It would only increase if the pilot increased the AOA to pull a tighter turn.

That is where their rub lays.  He does not constantly change angle of attack.  Aircraft move on using the laws of vector motion.  

First Newtons Laws of Motion:

http://www.grc.nasa.gov/WWW/K-12/airplane/newton1.html

http://www.grc.nasa.gov/WWW/K-12/airplane/newton1a.html

http://www.av8n.com/how/htm/motion.html#sec-straight-line-motion

Aircraft manuver using the laws of vector motion.  Each vector has a manitude and direction.



http://www.grc.nasa.gov/WWW/K-12/airplane/newton2.html

Each vector has a quality:

http://www.grc.nasa.gov/WWW/K-12/airplane/vectors.html

Once you input the controls, the A/C changes to a new vector and flies along it.  If you max out the control inputs at the begining of the turn then you have no more input into the vector.  In fact in a maximum turn it is not uncommon for the pilot to end up using maximum rudder and opposite alerion to stop the aircrafts natural tendancy to tighten the turn (overbanking tendancy) and dive.

It is interesting to combine this with what we learned about long-tail slip effect (section 8.9) --- in the slow, steeply banked turn in the glider, you would be holding nearly full inside rudder (to prevent the long-tail slip) and nearly full outside aileron (to counteract the overbanking tendency). If you are not expecting this, it will appear very strange. You are holding completely crossed controls, yet the turn is perfectly coordinated. You can confirm this by observing that the slip string is perfectly centered.

Once you change vector with maximum A/C control input the Aircraft will stay on that vector until control input is changed.  When you enter the turn the relative wind changes and while you are on that vector the wing act pretty much like it is in level flight except for assymetrical lift.  You will suffer a Cdi hit only when the lift goes up to input the vector change.  The instant the forces balance your back in level flight where parasitic drag and speed take over traveling on a new vector.  You are assuming the Cl goes up throughout the turn.

This is assuming of course full control inputs and full power which is how 99 percent of the break turns we see in AH are conducted.

BTW.  I emailed several aeronautical engineers to find out exactly why the Cl does not go up much in the turn just to confirm.

Remember we covered this before and you said you got it? Cl does not vary with speed. It varies with AOA.

Huh?  All I did was post this link I believe.  You can see that it does go down by calculating it.

http://www.av8n.com/how/htm/aoa.html#fig-ias-cl

If the airspeed goes down, the coefficient of lift must go up. This relationship is illustrated in figure 2.13.

Cl had a direct relationship with AoA but an inverse relationship to velocity.

Crumpp

[dtango]

I give up.  I've spent far too much energy on the topic.  

You are throwing information out all over the place trying to act like you know what you're talking about when infact you don't.

Now you're saying that I said things that I didn't even say - e.g. I never said Cl increases in a turn by definition.

Also you're inferring that you have better access to knowledge than any of us here.  Have you considered that I speak with aero engineers or that some of the folks posting here have aero backgrounds?

And lastly your memory is slipping.  Look back over the thread regarding the Cl discussion and see what we said about it.

I'm done.

Tango, XO
412th FS Braunco Mustangs

[Crumpp]

If it is so easy why you can't do it?

I did it and posted the results.  

You are throwing information out all over the place trying to act like you know what you're talking about when infact you don't.

No, actually I don't claim to know all that much about it except for the fact I am getting my pilots license.  

I am just trying to reconcile what folks are saying on this BBS with what aerodynamic experts say.

Also you're inferring that you have better access to knowledge than any of us here. Have you considered that I speak with aero engineers or that some of the folks posting here have aero backgrounds?

Sure and I have enjoyed our discussions.  I am not inferring anything. Just seeking clarification.  Sorry that offends you.

 

Well, from the Fw 190A-1 onwards all the A-series Fw 190s had same wing area and span so these all had same aspect ratio 6,02 (18,3 m2 and 10,5 m). Some early A-0s had smaller wing which had aspect ratio 6,07 (14,9 m2 and 9,51 m).

I would have to Email you a copy of the original documentation.

This whole thing boils down to:

1. It either like is being claimed here and CL goes up dramatically throughout the turn hence Cdi has a large effect on turn forces

or

2.  It is like the PhD. claims - The Cl does not go up in a turn and therefore Cdi doesn't factor in as much as the parasitic drag does in our turn forces.

3.  There is an unknown element which is missing from the puzzle.


Crumpp

[Crumpp]

Well, from the Fw 190A-1 onwards all the A-series Fw 190s had same wing area and span so these all had same aspect ratio 6,02 (18,3 m2 and 10,5 m). Some early A-0s had smaller wing which had aspect ratio 6,07 (14,9 m2 and 9,51 m).

BTW your correct.   Area and Span did not change  The cord did though.

 

The maximum distance between the upper and lower surfaces is called the thickness. Often you will see these values divided by the chord length to produce a non-dimensional or "percent" type of number.

The wing was made thicker to provide more room for the outboard MG 151's.  

Crumpp

[dtango]

My absolute last reply for clarity -

To turn your CL has to increase ABOVE what is necessary for level flight.

To maintain a turn your CL has to remain above what is necessary for level flight.  This does not mean CL is increasing in a turn.  This does mean that CL has to remain above what is needed for level flight.

To determine contribution of CDi to drag in a turn even with nose low just simply evaluate the CD = CD0 + CDi relationship.  You said you've done the math which showed you the relative difference between planes in a level turn.  The relationship doesn't change much in a nose low situation.

Best of luck understanding the concepts.

Tango, XO
412th FS Braunco Mustangs

[Crumpp]

To turn your CL has to increase ABOVE what is necessary for level flight.

Right you have to change the vector.  The Cl jumps up after you bank and pull back on the yoke.


The PhD sent me a reply.

Basically he says it depends on the control inputs.  If you max out or hold the same control input then the Cl is the same as level flight.  If you have more yoke and you pull it then the CL will go up again.

From what I understand then:

Your stick inputs in the turn will control the amount of Cdi in the turn as long as you don't let your speed drop to much.  Just like the Cl the Cdi changes based on the angle of attack to the relative wind.

The relationship doesn't change much in a nose low situation.

Except that the heavier plane will begin to add it's mass to thrust, right?

Crumpp

[gripen]

Originally posted by Crumpp
I did it and posted the results.  

Why your results are not right then? Why don't you post your step by step calculation so we can see what goes wrong?

Originally posted by Crumpp
BTW your correct. Area and Span did not change The cord did though.

The aspect ratio is a simple relation between the wing span and the wing area.

Regarding drag in the turn here is a little quick and dirty comparison at different g loads calculated with Lednicer's flat plate values. Speed is 483km/h ie  300mph and e factor is same for all planes (0,8).

Spitfire IX 3400 kg
1g 5524 N
2g 6481 N
3g 8074 N
4g 10305 N
5g 13174 N
6g 16680 N

P-51B 4200 kg
1g 4716 N
2g 6162 N
3g 8573 N
4g 11947 N
5g 16286 N
6g 21589 N

Fw 190A 4000 kg
1g 5340 N
2g 6854 N
3g 9377 N
4g 12910 N
5g 17452 N
6g 23003 N

The values show how drag of the Spitfire increases at much lower rate than in the other two planes.

gripen

[Crumpp]

Why your results are not right then? Why don't you post your step by step calculation so we can see what goes wrong?

There is nothing wrong with my calculations.

Gripen first you calculate the density.  

Use the equation of state with the measured values since we are sea level and don't have to account for tempature and pressure changes at altitude.

Tempature - 65*F

Pressure - 14.696psi

http://www.grc.nasa.gov/WWW/K-12/airplane/atmos.html

r= .00163029

Then work the Cl formula in English units.

The aspect ratio is a simple relation between the wing span and the wing area.

Yes and I have said of have original documentation that shows an aspect ratio change.  And it is not in the V5.  It is in the FW-190A6.

Regarding drag in the turn here is a little quick and dirty comparison at different g loads calculated with Lednicer's flat plate values.

Please show me HOW you are applying the force of G's to the Airframe.  The G forces the plane experiences are not the same as what the pilot experiences in a turn.  The plane is in its flight envelope and is simply flying on a new vector.  It only experiences G's for a few seconds at the begining of the vector change.  The pilot, who has no thrust or lift, is trying to move on a different vector and experiences different forces.

If that was the case then calculate the Cl and compare it to the max Cl for the Spitfire.  If Clmax is 1.1 like Izzy posted then the Spitfire is incapable of flying a vector that loads the Cl with more than a couple of G's.

Crumpp

[hitech]

Please show me HOW you are applying the force of G's to the Airframe.  The G forces the plane experiences are not the same as what the pilot experiences in a turn.  The plane is in its flight envelope and is simply flying on a new vector.  It only experiences G's for a few seconds at the begining of the vector change.  The pilot, who has no thrust or lift, is trying to move on a different vector and experiences different forces.

If that was the case then calculate the Cl and compare it to the max Cl for the Spitfire. If Clmax is 1.1 like Izzy posted then the Spitfire is incapable of flying a vector that loads the Cl with more than a couple of G's.

Crumpp, email this to your Phd friend, let him try explain it to you.


HiTech

[Crumpp]

Actually Hitech,

I already went over to some buddies of mine that are fighter pilots.  I work practically on top of an Air Force Base.

Talk about a bunch of guys who love to talk theory about airplanes.  Well it started a "heated discussion" among too guys with a Master's in Aeronautical engineering...

Not attempting to educate anyone here just spelling it out how they explained it.

I misunderstood the "G" application and the plane does experience G's in a Break turn.  It's not flying a vector and is constantly changing its vector due to the nature of the turn.

The forces are totally different from a coordinated turn.

On the rest of the turn:

The short answer they ended up giving me was...It Depends.

All on the stick inputs.  

There is no doubt that when the Cl goes up the Cdi goes up much more so than parasitic drag.

If the pilot makes a coordinated turn then the Cl does not change much at all.  He makes a break turn and keeps cranking back then yes:

1.  The plane loses lots of altitude at once.  In fact they usually make a quick climb then do the break turn.

2.  And break turns put the brakes on fast and will kill your speed no matter what the wing loading of the plane.  Because of your rapid speed loss the angle of attack does constantly change and the plane does experience constant G's.  

Break turns are purely defensive moves according to them.  Any pilot that started a fight off with one was putting himself in hurt locker energy wise.  It was definitely NOT the preferred method to use at the merge!!

So between High induced drag and low parasitic drag, in the break turn the induced drag does become the dominant force.  It is better to have higher parasitic drag and lower induced drag for break turns.  In level speed and zoom climb it is better to have the lower parasitic drag.

However NOBODY should be want to make a break turn except for immediate survival.

BTW I just noticed:

According to Gripens Calculations the Spitfire and FW-190 keep about the same ratio of mass to force throughout with the FW-190 pulling out ahead the higher the G's.  The clear winner in that line up is the P51!  It has less force acting on it and more mass.  The Spitfires appears much lower but since it is a much lighter plane it does not need as much force to put the brakes on.

Crumpp

[hitech]

2. And break turns put the brakes on fast and will kill your speed no matter what the wing loading of the plane. Because of your rapid speed loss the angle of attack does constantly change and the plane does experience constant G's.

That is it depends.

AOA Might be constantant and G's decresing with a loss in speed,
AOA And G Might be constantant and altitude is lost.
G Might be constant and AOA Increases as speed is lost

[Crumpp]

That is it depends.
AOA Might be constantant and G's decresing with a loss in speed,
AOA And G Might be constantant and altitude is lost.
G Might be constant and AOA Increases as speed is lost

Yep,

Any of those scenarios though the fighter conducting the break turn ends up with a lot less energy than he started the turn with.

It Depends....Engineers Number 1 answer to everything!!


Crumpp

[gripen]

Originally posted by Crumpp
There is nothing wrong with my calculations.

Actually there is quite a lot of wrong in your calculations.

Originally posted by Crumpp
Gripen first you calculate the density.  

Use the equation of state with the measured values since we are sea level and don't have to account for tempature and pressure changes at altitude.

Tempature - 65*F

Pressure - 14.696psi

http://www.grc.nasa.gov/WWW/K-12/airplane/atmos.html

r= .00163029

Let's calculate it first right.

Density formula from the NASA site, note the units:

r = p / [1718 * (T + 459,7)]

where:

r = density (slugs/cu ft)
p = pressure (lbs/sq ft)
T = temperature (*F)

So we have:

T = 65*F
p = 14,696 psi = 2116,224 lbs/sq ft

And the density calculation:

2116,224 / [1718 * (65 +459,7)] = 0,002347618 slugs/cu ft

Then Cl calculation for the Fw 190, again note the units:

Cl = L / (A * .5 * r * V^2)

where:

L = Lift (lbs)
A = Wing Area (sq ft)
r = Density (slugs/cu ft)
V = Speed (ft/s)

So we have:

L = 4272 kg = 9418 lbs
A = 18,3 m2 = 197 sq ft
r = 0,002347618 slugs/cu ft
V = 300 mph = 440 ft/s

And the Cl calculation:

9418 / (197 * 0,5 * 0,002347618 * 440^2) = 0,210394

So what went wrong in you your calculation? Obivioysly you have used wrong pressure unit for density calculation because using the psi value results:

0,0000163029

Funny thing is that you have accidentally or purposedly multiplied the result with 100 to get sensible result.

And similar error can be found from the Cl calculation; because your result was:

FW-190 Cl = .651727567

we can easily see that you have used mph value instead ft/s. It's very probable that you have done similar errors also in the drag calculation too.

Originally posted by Crumpp
Yes and I have said of have original documentation that shows an aspect ratio change.  And it is not in the V5.  It is in the FW-190A6.

Aspect ratio is simply:

AR = S^2 / A

where:

AR = aspect ratio
S = wing span
A = wing area

So if the wing span and the area were unchanged (as you claimed above), there should be no difference in the AR.

Originally posted by Crumpp
Please show me HOW you are applying the force of G's to the Airframe.

Hm... Actually we are just interested about needed lift and that is simply:

L = g * w * 9,81

where:

L = lift (N)
g = g load
w = weight of the plane (kg)
9,81 = normal acceleration (m/s2)

Originally posted by Crumpp
If that was the case then calculate the Cl and compare it to the max Cl for the Spitfire.  If Clmax is 1.1 like Izzy posted then the Spitfire is incapable of flying a vector that loads the Cl with more than a couple of G's.

Well, the Cl values are naturally part of the calculation and therefore allready calculated:

Spitfire IX 3400 kg (483 km/h):
1g Cl=0,135
2g Cl=0,269
3g Cl=0,404
4g Cl=0,539
5g Cl=0,674
6g Cl=0,808
7g Cl=0,943
8g Cl=1,078

Regarding the Clmax value of the Spitfire, this has been discused earlier and RAE data gives Clmax 1,36 in glide and 1,89 at full power. The NACA values are quite questionable due to the condition of the tested plane and the testing methods.

gripen