More Camel torque please

[SCTusk]

Thanks HiTech, I'll try to restrict myself to referencing the points in the previous post, that way if I go off the rails you should be able to point out the errors in a way I can follow.

At this point I would like to address the following question: Could rotary powered WW1 aircraft turn better (either at a higher rate or shorter radius) to the right than to the left as a result of gyroscopic effect?

In reference to points 5 and 6 in my previous post, it can be seen that a different condition occurs when the aircraft turns to the left (as opposed to the right) :
 
In a left hand 45 degree banked turn the gyroscopic effect will cause a pitch up and a yaw to the right. As the aircraft is banked at 45 degrees some amount of pitch up is required anyway, but it should be noted that if the effect were large enough some pitch down input may be required. At this time I will assume that all that is required in this instance is a reduced pitch up input to maintain the turn. The yaw to the right must be countered with some amount of left yaw input in addition to the left yaw input normally required in the absence of gyroscopic effect.

So to outline the differences between similar co-ordinated left and right turns under gyroscopic effect :

The left turn requires more left rudder than usual and less pitch up (or possibly even pitch down under extreme effect).
The right turn requires left (opposite) rudder and pitch up.

Consider the conditions for a tighter turn, which would require a higher angle of bank and more input to counter the increased gyroscopic effect. Assuming that sufficient airspeed is maintained (especially on the wings on the inside of the turn) the limiting factor to increasing the left turn is the limit of rudder travel. Once this limit is reached the turn cannot be made tighter as the gyroscopic effect will cause the aircraft to yaw to the right, which goes against the turn. Any further pitch up input increases the yaw to the right (due to gyroscopic effect) negating the effect of increasing pitch in the turn.

The limiting factor to increasing the right turn however is not the limit of rudder travel. The rudder is being held to the left to counter the yaw to the right, so if the rudder reaches the limit of travel in this case the aircraft will yaw right, which goes with the turn. Put another way, we actually want more yaw to the right as we tighten the turn. Additional pitch up input will increase the yaw to the right, so that the turn can be tightened provided the limit of elevator travel is not reached (in reality the aircraft is more likely to stall a wing before this occurs). At this stage the right turn will no longer be co-ordinated due to the left yaw input being at maximum yet insufficient, but with the error in co-ordination favouring the turn rather than going against it.

Since the rudders on these aircraft are known to have been insufficient (I believe there was some misunderstanding of the basic design principles at the time?) I suspect that the limit of rudder travel might have been reached quite quickly in the presence of a substantial gyroscopic effect. My contention therefore is that the rotary engined aircraft were capable of tighter turns to the right than to the left.

If I have made this argument correctly there is still the matter of degree, not only in general but in relation to the Camel versus the Dr1. However there seems little point in pursuing those questions without your acceptance of this initial premise.

[hitech]

First you are asking a slightly different question then originally spoke of, you are asking can a gyro slow a left turn more then slowing the right turn, VS can a gyro speed up a turn.

Ill give you a more detail analysis later but your first major error is this.

Consider the conditions for a tighter turn, which would require a higher angle of bank and more input to counter the increased gyroscopic effect. Assuming that sufficient airspeed is maintained (especially on the wings on the inside of the turn) the limiting factor to increasing the left turn is the limit of rudder travel. Once this limit is reached the turn cannot be made tighter as the gyroscopic effect will cause the aircraft to yaw to the right, which goes against the turn. Any further pitch up input increases the yaw to the right (due to gyroscopic effect) negating the effect of increasing pitch in the turn.

And specially

effect will cause the aircraft to yaw to the right, which goes against the turn.

Yes on yaw right and up but the torque is not against the turn, it is 90 degrees to the turn or up, it is in no way against the turn. As you run out of rudder (assuming this is the case, it would be very plane specific and that is a different topic) you would simply be slipping the plane in the turn. And this slip would happen if turning either direction creating drag and hence slowing the turn both left and right.

2nd the limiting factor in the turn is MAX AOA (I.E. stall) not the rudder.

HiTech

[hitech]

Ok lets first describe some terms and how an aircraft turns.

Turn/revolution (to move about a circle) think of this as an airplane moving about a circle always pointing north.

Rotate (to pivot about an axis) picture this as an airplane not moving but spinning like a top.

How an airplane turns ,as the plane banks to the left, the lift vector is pointed side ways, this force to the left accelerates the airplane to the left and the airplane is now moving in a slightly different direction and hence the wind is traveling at a different angle across the plane.

Now since the wind is from a different direction it pushes the the tail of the air plane creating a torque and rotating the plane.

This process continues and the plane turns and pivots as it travels about the circle.

So please note very carefully what really TURNED the airplane was the horizontal component of lift.

Now since we banked the lift vector the plane started descending causing the pilot to pull the stick back, causing a torque on the plane , rotating the plane creating a larger AOA, and creating more total lift, enough until the up component of the lift vector = the weight of the plane.

As the plane banks more, more total lift is required  and generated for level flight by pulling on the elevator also creating more horizontal force and hence accelerating the plane to the side more. Now for any bank angle the total up force will be 0 (gravity - lift). The total torque on the plane will also be 0 because if we have  an net torque or net force on any object by definition it will accelerate its rotation or its translation.

The horizontal component of the lift is not 0 and accelerating the plane in the horizontal, 90 degrees to its direction of travel.

Now onto the gyro problem.

The plane is Turing a circle parallel to the ground.

The gyro is is spinning around the tangent of the circle, As the gyro moves/ translates around the circle it is also  rotated as described above.

This rotation creates a torque either up or down. (basic gyro principle) So far I am fairly sure you agree.


Now if we were doing a flat 0 deg bank turn (not reality) all this up or down torque would have to be offset by use of the elevator.

Now if we were flying a 90 bank angle (again not reality) all the up or down torque would be controlled with the rudder.

Now at any bank angle in between 0 & 90 the up or down torque is controlled with a combination of rudder and elevator but to maintain the same rotation rates (note not turn rate, the rotation rate simply must match the turn rate) and hence must create  a net torque of 0 on the plane. Now if the gyro was not there again the net torques would have to be zero.   Since the gyros torque is up, the only thing that can be added is a net torque in the opposite direction. If we add any other net torque we are accelerating the rotation and this would either spin the plane on its tail or cause it to stall, because in a steady state turn all torques must sum to 0.

To view this view the plane in a 45 deg bank,you would have a 45deg force from the rudder and a 45 deg force from the elevator like legs on a saw horse. The horizontal forces must = each other and the vertical force * the arm must = the torque created by the gyro.

Hence we have not created any net extra horizontal force and thus the plane can not turn faster.

Now the other piece in involved. As I stated very early on the only thing that could possibly change the turn rates is the net drag on the plane. If for some reason (I have not been able to think of one, but I am not positive) more drag is applied turning left then turning right it do to the gyro it would effect the turn rate, simply because increasing speed increases turn rate.

HiTech

[Ghosth]

Bravo HT  <S>

And thank you sir.

[SCTusk]

Thanks HiTech. I have two problems with your previous posts. I know you must be very busy and this is just one of many items on your to-do list, so I will presume a 'slip of the tongue' but I must still point to the errors so my apologies in advance. Firstly you neglected to mention thrust. Lift and drag are effects brought about by thrust, therefore are subordinate to it. Second you refer to my mention of adverse yaw in a turn as not being against the turn, but rather 90 degrees to the turn. But you are referencing a point along the line of lift directly above the centre of the turn. This point is not the centre of the turn. The centre of the turn is in the same plane as the turn, i.e. directly below the point you are referencing and not somewhere along the line of lift. As I will demonstrate this is vital in understanding the significance of adverse yaw. 

In a turn the line of thrust would usually be approximately tangential to the arc of the turn and acting in the direction of the turn. If an effect were to cause the thrust to be directed as in say, a right yaw in a left turn, then that is an adverse yaw, i.e. away from the centre of the turn due to the thrust now working against the turn, or if you prefer moving you away from the centre of the turn. If however the thrust was directed towards the centre of the turn instead of away from it, as in say, a right yaw in a right turn, that is the opposite of an adverse yaw, or you might say the yaw is assisting the turn due to the thrust moving you closer to the centre of the turn.

If I could demonstrate with a hand experiment; left arm straight out to the side then swing forward 45 degrees, left hand represents centre of turn circle. Right arm straight out to the front, palm flat down, hand represents aircraft in level flight viewed from the rear. Add 45 degrees of left bank. So far so good. Now add 45 degrees of right yaw. You can clearly see that there is no way the thrust line is going to help you move around the centre of the turn circle (left hand). Now yaw 90 degrees back to the left and observe that although the 'aircraft' is in a nose down attitude, the thrust line is now positioned so as to assist the turn.

But here's the really fascinating part, which I initially missed due to over simplification in my previous post. I will now consider secondary effects in the aforementioned left and right gyroscopic turns:

1. Left turn induces gyroscopic effects of right yaw (adverse yaw) and pitch up. Secondary effect of right yaw is bank right (adverse bank). Pilot induces left bank to counter, inducing secondary effect of ailerons i.e. right yaw (adverse yaw). A total of three adverse effects in what I think you will agree is a closed loop of cause and effect. There seems to be nothing the pilot can do to prevent the aircraft skidding out of the intended turn.

2. Right turn induces gyroscopic effects of right yaw and pitch down. Secondary effect of right yaw is bank right. If pilot corrects with left bank secondary effect of ailerons induces right yaw. No adverse effects.

So it seems clear that if the gyroscopic effect is significant, the aircraft will be unwieldy in a left hand turn, but will turn right quite readily.   

[hitech]

SCTucks, 1st adverse yaw is typically refereed to as the yawing effect from ailerons do to differential drag but that is just a definition.

But what you are missing one big thing, the plane is in the exact same position as if it would not have gyro torques because the torque was corrected with a combination of rudder and elevator and hence the prop is not outside the turn and would be in the same position relative to the turn.

2nd even if the torque is uncorrected with controls and the plane is left to slip you are including 1 force and ignoring the force apposing it creating the 0 torque sum. In essence you seem to be hung up on rudder torques/forces working for you with out looking at elevator forces/torques working against you.

There seems to be nothing the pilot can do to prevent the aircraft skidding out of the intended turn.

This has me completely lost because it is obvious in almost any plane, (all prop planes have gyro forces) that the pilot can fly a non skidding turn. So why is a rotary any different except in the size of forces & torques? So are you claiming a plane can never fly a coordinated turn?

To summarize, the plane is in the same position relative to the tangent of the circle with or with out gyro torques. The pilot simply moved the controls to hold it in the same position.

So it seems clear that if the gyroscopic effect is significant, the aircraft will be unwieldy in a left hand turn, but will turn right quite readily.

Again read previously I have stated it is much easier to turn to the right then the left, this is obvious to anyone flying the rotary engined planes, but that is a different question then the turn rates. And is primary do to the planes tendency of slowing vs speeding up when flying uncoordinated.

HiTech

[Baumer]

One area I am beginning to investigate is how the WW1 aircraft were rigged at the time. Given that neither the F.1 or the Dr.1 had any trim controls the initial rigging would effect how the plane responds in flight.

My original assumption was that the planes would be rigged for cruising speed. However, given that takeoff and landing were very tricky, perhaps most planes were rigged to make low speed handling better. Or perhaps they chose a compromise between the two.

Now I know you liked my trim control setup HiTech at the last CON, so it's pretty easy for me to test.
 

[Pongo]

What is clear is why few developers engage in discussions with the public.

[Charge]

"What is clear is why few developers engage in discussions with the public."

It is also clear that if the game has survived more than ten years, with players active all those years, it is also recommended.

-C+

[FLS]

One area I am beginning to investigate is how the WW1 aircraft were rigged at the time. Given that neither the F.1 or the Dr.1 had any trim controls the initial rigging would effect how the plane responds in flight.

My original assumption was that the planes would be rigged for cruising speed. However, given that takeoff and landing were very tricky, perhaps most planes were rigged to make low speed handling better. Or perhaps they chose a compromise between the two.

Now I know you liked my trim control setup HiTech at the last CON, so it's pretty easy for me to test.
 

Baumer as you know the point of rigging is to lesson the pilot work load. Since you spend most of your time at cruise speed and altitude it makes sense that you rig for that condition. You're going to be active on the controls when taking off and landing regardless of the speed you rigged for.

[SCTusk]

Thanks again HiTech, you state your case clearly (whereas perhaps I struggle a little) so that I can attempt to address your objections specifically.

the plane is in the exact same position as if it would not have gyro torques because the torque was corrected with a combination of rudder and elevator and hence the prop is not outside the turn and would be in the same position relative to the turn.

You would know the figures whereas I can only suggest that the maximum torque applicable with the elevators would significantly exceed the maximum torque applicable with the rudder. As the gyroscopic effect produces a  torque equal to the inducing torque but at 90 degrees to it, this suggests that in hard turns the rudder would in fact not be capable of correcting the induced yaw from the initiatiating action of the elevators. 

even if the torque is uncorrected with controls and the plane is left to slip you are including 1 force and ignoring the force apposing it creating the 0 torque sum. In essence you seem to be hung up on rudder torques/forces working for you with out looking at elevator forces/torques working against you.

I am not ignoring the opposing force creating your 0 torque sum, I am ignoring your torque sum. The whole premise of your equations is based on a point on the line of lift above the centre of the turn circle, not on the centre itself (i.e. on the same plane as the circle). I presume you employ this method in the simulation. In reality I'm sure you are aware that an aircraft in a left turn with right yaw must move away from the actual centre of the turn (unless the movement along the line of lift exceeds the movement along the line of thrust?) thus widening the turn as indicated in the hand experiment of my previous post. If that lacked clarity then imagine the following situation:

If you are turning to the left around a tree at tree height so that the tree is at the centre of the turning circle and allow the aircraft to go nose high and yaw right it becomes immediately obvious that the aircraft is pointing more than 90 degrees away from the tree, therefore must move further away from it.

I would think we could just allow that a skidding turn can not be a tight and efficient turn. Surely nobody would seriously argue this point. As you say, there would be increased drag for one thing. Also the nose will be high (both from the pitch up and the right yaw) slowing the horizontal movement of the aircraft around the turn as the aircraft attempts to climb.

Obviously all this depends on the degree of gyroscopic effect. I have assumed a significant degree in the discussion so far, so that the effect and its' result can be visualised more easily. If the degree of the effect in a hard turn proves small enough to be easily corrected in the turn, then my argument is invalid.

[hitech]

SCTusk you are starting to change the discussion.

From the out set we stated we are speaking a general case not a specific case. You are now making assumptions that the plane can not fly coordinated and then really are changing the question to 3 questions.

1. Does non coordinated  flight slow a turn.
We both agree yes.

2. Can a gyro produce enough force to be able to fly none coordinated.
A very plane specific question.
We would both agree that we can create a plane and condition that this is the case.

But these 2 questions are very very different then the statement a gyro makes your plane turn faster to the right. Which is the claim you made at the outset.

3. Exactly how and in what directions would this non coordination be in both left and right turns.

If the above is your premis, I have no desire to discuses the very plane specific items required for this.
If you believe a plane flying coordinated in a turn will turn more quickly right do to gyro torques , I will be happy to continue the discussion.

Also in this discussion if you arrive at only very minor force differences such as the different moment arms of the elevator and the rudder then again I have no desire because at this level of force many many other things must be considered and the precision we are starting to speak of no longer can be spoken of in a general since and would again be very plane specific.

Any way regardless of the above here is an analysis I have done of your post.

You would know the figures whereas I can only suggest that the maximum torque applicable with the elevators would significantly exceed the maximum torque applicable with the rudder. As the gyroscopic effect produces a  torque equal to the inducing torque but at 90 degrees to it, this suggests that in hard turns the rudder would in fact not be capable of correcting the induced yaw from the initiating action of the elevators. 

You have multiple premise problems here.

1. A gyro does not produce a torque = to the torque applied to it in a 90 deg direction. It only applies a % of torque that is calculated by rpm * moment and mass of the entire object. I assume one you think about it, this should be obvious, because other wish even a mass spinning 1 1 rpm would produce a torque 90 degrees to the applied torque.

2. The 2nd piece you are missing is that how much the rudder and elevator must applied are only = to each other at a 45deg bank. In all other case the amount required are related by a sin cos functions. As I stated earlier in a flat turn all elevator is needed, in a 90 deg bank all rudder is needed.

3. Using your argument you are stating you can not fly a max g loop with out skidding.

4. The forces need to generate the torques are very speed specific. So you could fly coordinated at one speed but slower you could not.

If you are turning to the left around a tree at tree height so that the tree is at the center of the turning circle and allow the aircraft to go nose high and yaw right it becomes immediately obvious that the aircraft is pointing more than 90 degrees away from the tree, therefore must move further away from it.

Think about what you just said, if the plane is continuing to move away from the center of a circle it can never return to the same point hence it is not flying circle.

Also note I am not saying that a thrust vector can not point out side the circle,and produce a force apposing the turn.Obviously if you allow the plane to slip it will turn slower, for many reasons not just the thrust vector.  I am saying that if you are in a coordinated turn it will be in the same position  regardless of gyro torques.

I am ignoring your torque sum.

If you do not believe that the sum of torques have to be 0, then we must back up and discuss this, because with out that premises we can not really begin. It is a basic technique used in problem solving and stems for the simple formula F = M * A, so if you are left with a torque or force after summing all torques and forces, the object MUST be accelerating in some direction or rotation.

The whole premise of your equations is based on a point on the line of lift above the centre of the turn circle, not on the centre itself

Your assumption would be completely false assume.

The core idea of simulation is very simple and is physics 101. You can take all torques and forces on any object and reduce them to 1 force vector and 1 torque attitude.

then F = m * a for both translation and rotation. Rinse repeat.

So  thrust direction is very simple, what direction is the plane pointing, add thrust at the prop point in the direction the plane is pointing + any engine angle offsets.

And hence any portion of the thrust that is pointing out side the turn will oppose the horizontal component of lift. Note this same effect in a normally coordinated turn or slow flight adds to the lift vector.

[SCTusk]

Salute again HiTech, I agree that the discussion is drifting off the issue, and will gladly 'kick in some corrective rudder'

If you believe a plane flying coordinated in a turn will turn more quickly right do to gyro torques , I will be happy to continue the discussion.

I can conditionally commit to that, as I'll explain later.

But these 2 questions are very very different then the statement a gyro makes your plane turn faster to the right. Which is the claim you made at the outset.

Actually in my OP I included a quote from Wikipedia or similar which made that claim. I was careful not to make that claim myself, although I still think it a possibility, provided we are careful to define the conditions. If a may refer back to this point later, as there is a more pressing matter:

1. A gyro does not produce a torque = to the torque applied to it in a 90 deg direction. It only applies a % of torque that is calculated by rpm * moment and mass of the entire object. I assume one you think about it, this should be obvious, because other wish even a mass spinning 1 1 rpm would produce a torque 90 degrees to the applied torque.

I think this may be the focal point of our problem. The resultant torque does indeed occur at 90 degrees to the applied torque at any rpm, that is a fundamental characteristic of the phenomena. If I can refer you to an extract from a magazine article on Seafires (you can find similar statements from other sources) :

"Any applied force which changes the axis position of a gyroscope causes the axis to move 90° to the applied force and in the direction of rotation"

ref http://www.auf.asn.au/magazine/seafires1.html#takeoff_swing

and from Wikipedia:
"If the speed  of the rotation and the magnitude of the torque are constant the axis will describe a cone, its movement at any instant being at right angles to the direction  of the torque."

and again from Wiki:

"If the rotating body is symmetrical  and its motion unconstrained, and if the torque on the spin axis is at right angles to that axis, the axis of precession will be perpendicular to both the spin axis and torque axis."

ref http://en.wikipedia.org/wiki/Gyroscopic_precession#Torque-induced

In addition, and although I cannot find confirmation of this, the formulae seem to indicate that the output torque always = the input torque. I suspect that can only be true under 'normal' circumstances, as it doesn't seem right for instance, in the case of a small mass and a large input torque. It appears to be one of those Newtonian action=reaction things. But I'm definitely in over my head with the math at this level.

I've had my son check this (BEng & BSc Major in Pure Math) and he agrees, I say this only to make it clear that I've done about all I can to verify, I could still be in error. But the terms we have been using here (input torque and output torque) seem not to be in common use, rather just the term 'gyroscopic torque', which also suggests they are considered to be the same thing. In any event, the formulae should provide the correct output torque, but I would be interested to know if you are getting less output than input. Of course you could be referring to the gyroscopic effects as calculated on the engine in-situ, and so slightly forward of the centre of mass of the aircraft and therefore subject to torques at angles other than simple 90 degrees off the spin axis?  

I hesitate to add more at this point due to the significance of this misunderstanding by one or the other of us (genuinely perplexed)

[hitech]

I think this may be the focal point of our problem. The resultant torque does indeed occur at 90 degrees to the applied torque at any rpm, that is a fundamental characteristic of the phenomena. If I can refer you to an extract from a magazine article on Seafires (you can find similar statements from other sources) :

I agree the torque is always 90 degrees.

In addition, and although I cannot find confirmation of this, the formulae seem to indicate that the output torque always = the input torque. I suspect that can only be true under 'normal' circumstances, as it doesn't seem right for instance, in the case of a small mass and a large input torque. It appears to be one of those Newtonian action=reaction things. But I'm definitely in over my head with the math at this level.

As is obvious to you something spinning at a slow rate produces almost zero gyro effect I.E. torque 90.

I suspect that can only be true under 'normal' circumstances.

There is not any "NORMAL" circumstance with physics the equations are the same for any rate.

I believe you have answered you own question.



The problem lies in that we are not really speaking in and out torques but deal with a torque generates a 90 rotation rate. Or conversely in the case of a plane rotation rate generates a torque.

The key here is this line and equation in wiki .

http://en.wikipedia.org/wiki/Gyroscope

Under a constant torque of magnitude τ, the gyroscope's speed of precession ΩP is inversely proportional to L, the magnitude of its angular momentum:

This line simply is saying with more moment you get less 90 deg rotation rate for the same input torque.

Or conversely with more momentum you get more torque 90 deg up or down for the same turn rate.

But the problem we are speaking of stems from this statement of yours which I completely missed the real problem in 2 Places.

The first was in your original definitions.

4. The reactive torque is instantaneous to the initiating torque.

While this is a little correct it is very not exactly correct. Yes there is an instant reactive torque but that torque would be growing with time.

and this creates the error here.

You would know the figures whereas I can only suggest that the maximum torque applicable with the elevators would significantly exceed the maximum torque applicable with the rudder. As the gyroscopic effect produces a  torque equal to the inducing torque but at 90 degrees to it, this suggests that in hard turns the rudder would in fact not be capable of correcting the induced yaw from the initiating action of the elevators. 

The toque creates a 90 deg rate or a rate creates a 90 deg torque.  But if my memory serves me, you will not find equations directly of torque to torque because  the directions of the torque would have to be continually changing .

Think of it this way as you start the plane rotation you input a torque and the plane stares accelerating its rotation. This rotation is generating a torque 90 degrees. As the rotation speed increases the torque increases, but the assumption we are making is that the torque is corrected with controls and hence we are accelerating rotation in the same plane/axis.

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

[Angus]

Camel:
Imagine welding/fixing a shaft to the inside of the drum of a 200 hp washing machine. The weight would be the same as of the camel's engine. The shaft is fixed to the aircraft on one side, and the back of the (rotating) washing machine is fixed to the airscrew. This is basically how it works.
Inline engines etc, - the whole mass of the engine is fixed to the aircraft, the shaft is pushed around by the pistons, and turning just the airscrew.
Is there no difference in the force that wants to throw your craft to the side? If it is and has been posted, I apologize, but would appreciate the point of correction.