More Camel torque please

[SCTusk]

For a moment there I thought we might have found a solution But it seems we can agree on the physics both in terms of the aerodynamics and the gyroscopic effect. I can see that your knowledge of both undoubtably exceeds my own, so if there is any substance to my claims we must be close to resolving the issue. I hope so because I'm sure your time is valuable, and I think perhaps I have taken too much of it already. I do find it encouraging that we can engage in a discussion like this in good humour, and not take the outcome too seriously <Salute>

And here is where I believe you are correct about how toque in = torque out. The rate of turn generates a 90 deg torque. When we add torque apposing this torque we are also adding a  rotation rate via gyro  opposite  our current rotation rate. This then require a torque = to the 90 torque maintain the rotation rate of the turn.

Now I have not done the math on this piece so it could be incorrect but simply knowing how all net toques must be 0 (this is not debatable) in a stable turn I would think it is correct.

HiTech

The presumption that we would correct the gyroscopic torque is only justified in a left turn (which is the reason we agreed that the left turn would incur an energy penalty). In the right hand turn following the right bank, we need only pitch up to induce sufficient right yaw for a co-ordinated turn (but here I presume enough gyroscopic effect to negate the usual requirement for right rudder).

Note that in this situation there is no pitch down, as the gyroscopic effect does not 'build' on itself. If it did, the result of the right yaw would be pitch down, the result of the pitch down would be left yaw, and we would be back where we started. This clearly does not occur so it seems we can avoid the pitch down, at least in theory.

So I think you will agree that given enough gyroscopic effect, to turn right we need only bank right and pitch up. And this is in fact the method outlined in the anecdotal evidence.

Several questions remain unanswered; and you have set me on a difficult path if I am to defend a premise which originated elsewhere. By this I refer to your challenge to prove that gyroscopic effect makes it possible to turn faster to the right in a co-ordinated turn. I have two issues with that, the first being that at no point did I make this claim (I did however provide several conditional quotes which indicated an unusually fast turn to the right, and I did say that the right turn would be significantly faster than the left).

The second  issue is the constraint of co-ordination, as my understanding from the anecdotal evidence is that these turns were deliberately unco-ordinated, so that the effect on right yaw was allowed to go largely uncorrected (much like an aerial powerslide) or possibly that the rudder authority was not sufficient to fully correct it.

That said, I would certainly enjoy attempting to defend it, given the understanding that I am not certain at this time what the outcome will be. The conditions I mentioned in my previous post would be encapsulated in the following question:

(see next post, exceeded character limit lol)

[SCTusk]

Could the Camel turn faster to the right due to gyroscopic effect than could be expected if it was fitted with a conventional engine of equivalent power and weight?

This is something that you could certainly test in simulation, provided the gyroscopic effect is modelled accurately (by simply switching it on and off while testing). So I will put down some ideas as a general outline of this new question:

As we all know in physics (Newtonian physics, Einstein opened a real can of worms beyond the scope of this debate) you can't get something for nothing. So any idea that the gyroscopic effect somehow 'powers' the turn seems invalid. For one thing, there is inertia in the spinning mass which must be overcome. I did some quick experiments with a gyroscope and obtained the following (admittedly rough) data:

Time to spin to a stop from an approximately equal rpm:

No applied torque, gyro spinning freely undisturbed -                                 3 mins 55 secs
Continual applied torque, gyro free to rotate under precession -                  4 mins 05 secs
Continual applied torque, gyro prevented from rotating under precession -    1 min  45 secs

Given some margin for error we can see that by allowing the precession to occur unchecked the energy in the spinning mass dissipates normally, i.e. as when the gyro is undisturbed. But when the precession torque is corrected (the spinning mass prevented from precessing) the energy in the spinning mass dissipates much more rapidly.

This certainly has implications for our infamous left turn with all its' corrective measures (energy drain on engine) but more importantly it demonstrates that the energy to overcome any resistance to the resultant torque originates in the spinning mass, which clearly loses energy when the resultant torque meets resistance. So this is not 'something for nothing', but a valid 1:1 transfer of energy from one axis to another through the spinning mass.

If we now imagine an aircraft with poor rudder authority but more than sufficient elevator authority, equipped with a rotary engine such that a right hand turn might be attempted by simply banking right and pitching up, it seems possible that the resultant gyroscopic torque (right yaw) could indeed occur with more energy than by mere application of rudder, due to the transfer of energy from the 'strong' axis to the 'weak' axis via the gyroscopic effect. If we imagine that same aircraft to have most of its' mass located in very close proximity to the C of G, and consider the effects on the turn of that design in addition to the gyroscopic effect, we can begin to imagine how rapidly this aircraft might turn to the right. No doubt you possess the data on maximum pitch and yaw torques for the Camel, so you can probably evaluate this theory easily. 

If that same aircraft was fitted instead with a conventional engine, the entire mass concentration around the C of G would need to be changed due to the unique compact design of the rotary engine. This makes any comparison of turning with and without gyroscopic effect difficult, but still possible by flight testing in simulation.

So in answer to this new question, I would say 'yes' for the following reasons:

1. If the Camel had been fitted with a conventional engine, the entire design would have been different and would be unlikely to turn as rapidly or tightly.
2. If a conventional engine of equal size and power had been available, poor rudder authority would have resulted in a less rapid right turn (and yet a more rapid left turn).

Therefore a conditional 'yes', because at the time there was no other way to achieve this turn rate other than with gyroscopic effect from a small engine situated close to the C of G. The designers at the Sopwith plant seem to have designed the aircraft specifically for this ability following on from their experience with the Pup. The Fokker Dr1 may very well have been an attempt to produce an aircraft of similar performance, but I believe the attempt failed for a number of reasons, not least of which that the Dr1 failed to make as effective use of the gyroscopic effect due to less spinning mass, mass distribution and less power.

HiTech, you clearly want to model flight accurately and just as clearly have the necessary knowledge and skills to achieve the best possible outcome. But models are models and there are bound to be limitations. For instance, I have an old gliding habit of coming in too high on finals (better high than low deadstick lol) and sideslipping to wash off the extra alt, but I have not been able to find a simulation which allows me to do this (other than RoF, which has other issues I can not live with). Anyone following me in must think I am drunk (often true lol) or a terrible pilot (hopefully not entirely true) as I stubbornly swing from one side to the other while the FM modifies the slip and forces me off track. I heave on the stick and kick hard on the rudder in the forlorn hope that your zero torque sum will come to my aid but alas, I can only enjoy the sensation briefly before the rudder authority fails and I end up being forced to swing it across the other way.

So I think you would agree with me that while the highest level of authenticity is sought, there must always be shortcomings. The goal in the end is surely entertainment, and I doubt you would recommend that someone with high hours on an AH Mustang for instance should climb into the cockpit of a P51 at the next airshow and strut their stuff. So if entertainment is a key factor, perhaps I could suggest that issues such as this might be dealt with in less clinical fashion. What has developed in this instance into a rather cold scientific debate might instead be approached from the point of view where historical evidence carries as much weight as raw data. Just a few thoughts, and I've probably drifted off topic (apologies).

Certainly neither of us has flown either a Camel or a Dr1 (and I for one hope never to do so lol). Therefore I think we must look closely at the science, but also make use of whatever evidence we can obtain from historical notes, flying models and the very few replica aircraft still flying. I've contacted the warden of one such aircraft but as yet had no reply.

I do know my campfire etiquette, and bringing a tin of beans to your fire does not give me the right to mess with it. I can enjoy its' warmth and if invited maybe poke it with a stick, but I must always fart downwind and never pee on it. So thanks for being both patient and helpful, I trust my enquiries were not too much of a nuisance. I merely hoped to make a suggestion in the first instance, and perhaps a small positive contribution in the second.

Again conforming to good etiquette, I think you should have the last word (answers to any questions notwithstanding).

Salute

[hitech]

1st I am still not sure this statement of mine is correct.

nd here is where I believe you are correct about how toque in = torque out. The rate of turn generates a 90 deg torque. When we add torque apposing this torque we are also adding a  rotation rate via gyro  opposite  our current rotation rate. This then require a torque = to the 90 torque maintain the rotation rate of the turn.

Anyone other physics guys out there know if this is correct of not?

_______

I can not see why leaving a plane in uncoordinated flight would be of any use. Either way the torque has to be stopped, only you are doing it more with fuselage then  with rudder, which would create more drag then using rudder and hence slow the turn.

n very close proximity to the C of G, and consider the effects on the turn of that design in addition to the gyroscopic effect, we can begin to imagine how rapidly this aircraft might turn to the right.

Why would the distance an engine is away from the CG have any effect on turn rates? Remember a turn is a translational  movement not rotational , hence it is not effected by moments. And once the plane is rotating at the turn rate it requires no additional torque to maintain the turn. A shorter plane I.E. shorter tail moments would then require more force to offset the gyro and hence more drag.

1. If the Camel had been fitted with a conventional engine, the entire design would have been different and would be unlikely to turn as rapidly or tightly.
2. If a conventional engine of equal size and power had been available, poor rudder authority would have resulted in a less rapid right turn (and yet a more rapid left turn).

These are 2 strange case, how about we fit the camel with the same HP and weight engine, but a conventional turning prop, I.E. less moment. Would it turn better or worse in a right turn. I would think less.

No applied torque, gyro spinning freely undisturbed -                                 3 mins 55 secs
Continual applied torque, gyro free to rotate under precession -                  4 mins 05 secs
Continual applied torque, gyro prevented from rotating under precession -    1 min  45 secs

I would think this is almost all do to friction, but really this is a very minor force . It appears a large difference but off the top of my head you are looking at less then 1% of the power produced by the engine.And then you are taking the difference of 1% or 0.5% of the power. An easy way to think of it is how much energy did you put into the system to make the gyro spin. It took maby 1.5 secs to pull the string.
So now what percentage of that pull would it take to maintain the RPM of the gyro given your above numbers?

But models are models and there are bound to be limitations. For instance, I have an old gliding habit of coming in too high on finals (better high than low deadstick lol) and sideslipping to wash off the extra alt, but I have not been able to find a simulation which allows me to do this (other than RoF, which has other issues I can not live with). Anyone following me in must think I am drunk (often true lol) or a terrible pilot (hopefully not entirely true) as I stubbornly swing from one side to the other while the FM modifies the slip and forces me off track. I heave on the stick and kick hard on the rudder in the forlorn hope that your zero torque sum will come to my aid but alas, I can only enjoy the sensation briefly before the rudder authority fails and I end up being forced to swing it across the other way.

You have me completely confused a slip would be left rudder forward stick and right aileron and the plane continues in a straight ahead path, just with more drag.Your description seems to be back stick.Slips are one of the most common hot landing maneuvers used in AH. And why you would run out of rudder authority In a slip I have no idea.


And finally if you are going to analyze which way a turn  would be faster, you can not just analyze the gryo forces. You must also look at other components that can create an asymmetrical turn such as slip stream ,Pfactor, engine angle offsets. And this is why I said I have no desire to have that discussion because the level of precision needed for that discussion is far out side the realm of the bbs.

HiTech

[Baumer]

I am still reviewing my understanding of the math behind various gyro torques, but I did find a very helpful website to do the math.

http://www.gyroscopes.org/math.asp

So far the calculations at this website correlate to the examples in my Physics text book. So I am reasonably comfortable that the website is mathematically accurate. For me the calculation I am most interested in is at the bottom and refereed to as "reaction couple".

<S> Baumer

[Yeager]

Its all very interesting but lets talk about the REAL freak of the bunch.  That darned Dr1.  What the heck is that manoever where it stalls sideways and then snap recovers in the reverse direction.  It's as if there is no fuselage to provide drag....it reminds me of a box kite falling sideways that suddenly reverses direction.  Very odd to be fighting one of these Dr1 when it pulls that move....some folks have gotten scarey good using that move. 

[Baumer]

To me this issue applies to both aircraft, not just the Camel. Now I can understand it having different implications based on other aerodynamic factors between the two (F.1 and Dr.1) but I think the gyroscopic forces (torques) need to be understood better by everyone. Or in my case, I would like to look at them and understand how it applies mathematically to an aircraft no matter what model it is.

[hitech]

Howdy Baumer.

I am still reviewing my understanding of the math behind various gyro torques, but I did find a very helpful website to do the math.

http://www.gyroscopes.org/math.asp

So far the calculations at this website correlate to the examples in my Physics text book. So I am reasonably comfortable that the website is mathematically accurate. For me the calculation I am most interested in is at the bottom and refereed to as "reaction couple".

<S> Baumer

I assume you realize the reaction couple is just stating a torque is create 90 deg to a rotation velocity? I.E. the basic thing that makes a level circle move you nose up or down?

The computation of that force is really quite simple and I assume you already understand the equations posted on that page, amazingly the force couple equation exactly match AH'.

So what are you interested in?

HiTech

[Yeager]

is it safe to say that a spinning engine from ww1 is in fact a force field generator similar to what was used on Tie Fighters?

[Baumer]

The part I still don't understand properly is how long that torque is present. I think that reaction couple torque is present as long as the nose of the plane is turning along the horizon (assuming a coordinated turn), is that correct?

Also since the roll axis is aligned with the engines axis (gyro axis) rolling in to the turn, dose not change the vector of the reaction couple torque (again assuming a flat coordinated turn). Is this last statement correct?

And one last statement just to make sure I understand this properly.

In the last data fields the "distance between bearings" that would be the distance from the center of mass of the engine, to the center of gravity, correct?

Thanks for taking the time to help me understand the physics better HiTech.

[hitech]

The part I still don't understand properly is how long that torque is present. I think that reaction couple torque is present as long as the nose of the plane is turning along the horizon (assuming a coordinated turn), is that correct?

As long as the plane is turning (technically the rotation part of the turn , not the translation piece, I.E. the plane could be standing still) it will be generating the torque.

2nd. Assuming you are turning a flat circle. I.E. parallel with the ground. The torque will always be up or down relative to the GROUND depending on turn direction.

[/quote]
Also since the roll axis is aligned with the engines axis (gyro axis) rolling in to the turn, dose not change the vector of the reaction couple torque (again assuming a flat coordinated turn). Is this last statement correct?

It does not change the direction relative to the world / circle, but it does change the direction relative to the plane as it rolls. I.E in a 45deg left bank turn the torque will be on a 45deg line between your right wing and rudder. I.E. some pitch, some yaw.

And one last statement just to make sure I understand this properly.

In the last data fields the "distance between bearings" that would be the distance from the center of mass of the engine, to the center of gravity, correct?
[/quote]

Yes, that part is simply translating a torque which is defined as 2 opposing forces each 1 ft apart (when using ft/lb as torque unit) to a force with a different moment arm as you know Torque = force * distance.

Thanks for taking the time to help me understand the physics better.

The pleasure is mine, I love thinking about this stuff from different perspectives.

HiTech.


[/quote]

[bustr]

HiTech,

I've been reading as much as I can find on the Internet about the F1. No one gives any specifics about the effect of the rotary engines torqe other than anecdotally and the statement of legend, "gyroscopic effect". No physics explanations as far as I can find. But, a common theme is the specific results the designer wanted to achieve. The engine, pilot, guns and fuel tank all cramed into the first 7ft of the airframe. The tail heavyness. The propensity to suddenly stall or spin if forward speed was not maintained. Specificly a strong nose high climb to the left or a nose down turn to the right.

This almost makes me wonder if the 270dgr left hand turn by turning right was really taking advantage of the right hand nose down tendancy to snap the plane down right and recover around to the left. After all one of the things that killed many pilots was its rapid stall and spin.

After all of your research to create the F1 in AH, what is your personal take on the planes design and how the design would allow skilled pilots to trade nose for tail to the right or seem to turn faster to the left by performing a right hand manuver?

[hitech]

bustr: I have no idea of the story you are speaking of. I'm the physics guy, pyro is the history guy.

HiTech

[bustr]

Derned if I could find any physics discussions to explain the rotary engines 270dgr right turn in the F1 other than the aircraft was designed to rapidly stall and spin on purpose to the right. All references point out the extreme departure from the previous rotary engined Triplane and Pup's solid stability to a purpose driven design that was unstable. Why didn't the Pup, Triplane, or even Dr1 have the magical gyroscopic effect in some degree? If you look at the wing area for the Triplane and F1 they are equal with similar engine horse power.

F1 - 231 ft\sq....130hp rotary
Tri - 231 ft\sq....130hp rotary
Pup - 254 ft\sq....80hp rotary
Dr1 - 201 ft\sq...100hp rotary

Makes me consider that skilled pilots learned to take advantage of making it spin out of control to the right in a nose down attitude. You have modeled how quickly the F1 stalls based on the reading I've done around the internet. I have yet to find any pilot comments from the era specific to "x,y,z here is how you use the F1's purpose designed instabilities" to make impossible looking turns to the right so as to turn inside of the Hun. Is it possible the genius of the Camel is not the gyroscopic effect from the engine but the specific design instabilities to make use of what was known about torqe at the time?

[SCTusk]

As indicated I'm happy to answer specific questions but feel I've had my time on the 'soapbox', so I'll attempt to reply without adding any further argument if possible. I found only two direct questions (the others were I think rhetorical) so here goes, if I need to elaborate it is only in the interests of clarity:    

Why would the distance an engine is away from the CG have any effect on turn rates?

In a balanced turn, none. But the whole premise of the Camels' rapid right turn is based on the idea of a skidding gyroscopic turn. e.g. If you imagine an aircraft suddenly rotated 90 degrees to the right yet which somehow maintains its' attitude and structural integrity (there are all kinds of issues with this, but bear with me) then at some point (after some altitude loss) thrust will regain authority, the aircraft will regain flying speed and continue at 90 degrees to the original direction. Clearly this is an extreme case with any number of failure issues but somewhere between this example and a co-ordinated turn there exists a workable skidding turn, with effectiveness and extent determined by aircraft design and pilot skill. No doubt the entry speed would be high so that loss of airspeed over the inside wing would not cause a stalled condition, and the angle of bank probably about 30 degrees. In the Camel (and to a lesser extent in other rotary engine aircraft) the torque required to rotate the aircraft rapidly was provided by the gyroscopic effect induced by a powerful and rapid pitch torque working on a spinning mass which was close to the C of G, which as you know would allow the aircraft to rotate (yaw) more rapidly than if the engine had been further from the C of G.

The amount of skid which could take place while maintaining control is dependent on more factors than I would care to resolve, but providing the airframe maintained integrity and the pilot avoided the spin, the Camel should have been capable of changing direction more rapidly than any other aircraft of the time. I have seen a Pitts Special performing a similar manoeuvre, but that was likely done mainly with rudder, whereas the Camel had relatively poor rudder authority and was reliant on gyroscopic effect transferred from the pitch axis for the rapid rotation.

So now what percentage of that pull would it take to maintain the RPM of the gyro given your above numbers?

I included the test results mainly to demonstrate that the energy for any work done came from a valid source. Like yourself I doubt that the drain on the engine would be particularly significant to the debate. The data was obtained by rapidly and repeatedly applying torque of unknown value so as a means of determining specific energy loss data it was less than useful.

[SCTusk]

I am an idiot  . Sorry to break my own rules regarding only answering questions but I thought you wouldn't mind reading that.

I've allowed my loathing of the Dr1 to obscure something important; just tried the manoeuvre in the Tripe and achieved a full 360 degree turn in about 6 seconds, abit messy but the Dr1 isn't my usual mount. I'm sure someone with appropriate experience can do better. So the effect is there, and it works, but appears not to be achievable with the Camel.

Its all very interesting but lets talk about the REAL freak of the bunch.  That darned Dr1.  What the heck is that manoever where it stalls sideways and then snap recovers in the reverse direction.  It's as if there is no fuselage to provide drag....it reminds me of a box kite falling sideways that suddenly reverses direction.  Very odd to be fighting one of these Dr1 when it pulls that move....some folks have gotten scarey good using that move. 

Thanks Yeager, you put your finger right on the problem <Salute>

This requires a change of perspective.

First it seems that HiTechs' gyro model is alive and well (which explains your astonishment HiTech <Salute> hehe).
Second and unless I'm particularly clumsy (or have very poor controls/control setup) the Camel isn't quite set up right for the effect. Probably something to do with lift coefficient, not my forte, a minor tweak perhaps? Anyway I can't master it, but the Dr1 seems fairly well behaved through the turn with a gentle touch.

To fly the manoeuvre try 110mph, right bank approx 30 degrees with initial right rudder to commence the turn and gentle pitch up. Move quickly to left rudder (lots to max) and continue pitch up but not so much as to overly rattle the stall horn. Try to maintain 30 degree bank throughout, if it gets away from you it accelerates rapidly to inverted.

I have to ask, with so many Dr1 pilots about, why didn't someone speak up about this? Sneaky little monkeys