FlyByWire22
Well-Known Member
Another noob question, but do big jets or regionals for that matter still require use of rudder when making turns? Just curious due to the previous feet on the floor comments...
Turning tendencies on jet engines?
- Thread starter beasly
- Start date
dasleben
That's just, like, your opinion, man
Yes, they do. However, in normal ops the yaw dampers coordinate the jet to the point that manual input of the rudder is not normally required.
SpiraMirabilis
Possible Subversive
Even on turboprops I prefer flying with the rudder trim and keep my feet just sort of laying on the rudders, except for landing.
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Cessnaflyer
Wooooooooooooooooooooooooooooooo
The ATR goes bananas when you try flying with rudder.
Autothrust Blue
Welcome aboard the Washington State Ferries
Aileron plus coordinated rudder, just like every other airplane you've ever flown...
Cessnaflyer
Wooooooooooooooooooooooooooooooo
No the yaw damper system is not the best. We were told it was made for smaller aircraft and ones that drive the rudder directly.
The Dash on the other hand is so amazing to hand fly. The perfect balanced controls with just the right amount of feedback.
Autothrust Blue
Welcome aboard the Washington State Ferries
Oh yeah ours is pretty crappy too.
I can't speak much to jet operations, but maybe I can give you some insight into a single engine prop aircraft on initial roll out. The predominant effect is typically that of the slipstream, as it's effects increase as forward velocity decreases. The faster you go the more the spiral is stretched, like stretching a slinky, and the less chance it has to impact the rudder. In other words, slipstream will be a dominating effect at low air speeds.
Torque is a roll and plays little part, gyroscopic effects exists only when AOA is changing, and p-factor can be ignored at low AOA. Since on roll out AOA is not changing and is very low, both p-factor and gyroscopic effects can be ignored. Further, pitching up causes a gyroscopic right yawing tendency.
So what does all this mean as it applies to a jet? I'd suspect a jet has very little, if any issues with turning tendencies for three reasons: Jet's accelerate the slipstream as accelerating air is how they produce thrust, minimizing the most dominant effect on roll out. Many jets have a slipstream that does not impact the vertical tail at all, completely neutralizing any possibility of slipstream effect. Finally, as the previous paragraph points out, the other effects in a propeller aircraft can be ignored for roll out considerations; I suspect that holds true for a jet as well.
I leave you with a quote from AFNA on the bottom of page 294, speaking to control requirements for the rudder:
AFNA said:
fish314
Well-Known Member
Great points shdw. I hadn't even thought about slipstream effect, but your absolutely right. Since the air leaving the engines on a jet doesn't hit the rest of the airplane (on most jets), no slipstream.
And as for p-factor, even if the jet is at an angle of attack, the air goes through an inlet nozzle before it hits the fan blades, which takes some of that angle of attack out before the air hits the fan blades. So no (or negligible) p-factor.
Regarding the other question about requiring the rudder in a turn, another point to consider is that a lot of jets use spoilers rather than ailerons to turn (or they use spoilers AND ailerons). With spoilers, the down-moving wing has increased drag from the spoiler-- which negates adverse yaw. So that's another reason why one might need little to no rudder to turn the aircraft.
TFaudree_ERAU
Mashin' dem buttons
I've been to type school on three different jets. Not once have I heard the words "you really gotta watch the (left/right) turning tendencies on this one...they're worse than anything else you've ever flown."
In other words, if they're happening, I'm certainly not feeling it, and even more certainly couldn't care less if there's some minute amount occurring.
In other words, if they're happening, I'm certainly not feeling it, and even more certainly couldn't care less if there's some minute amount occurring.
FM had turning tendencies but those were from crappy rigging.
TFaudree_ERAU
Mashin' dem buttons
Nonsense. That was the most straight flying, excellent performing airplane ever manufactured.
/sarcasm
Yes, in props anyway, it is the dominant effect during the roll out. Arguably, I've heard torque causing higher friction on the left main could be dominating. I don't posses the mathematical prowess to run the numbers to verify this, but I believe the idea to be a silly one. Especially since I've numerous text stating slip stream or spin recovery as the critical design condition for rudder size.
Do you happen to have anything that says otherwise with regard to torque effect causing more friction on the left main tire? Does this even seem plausible?
Thank you in advance and apologies beasly for hijacking your thread for a moment here.
Not in a single.
Apologies for not clarifying that in my latest post. Vmc is typically the design requirement in a twin, yes. In a single it's usually slipstream or spin recovery.I was assuming multi's with the thread title.Not in a single.
fish314
Well-Known Member
No, not off the top of my head and it doesn't make any sense to me that it would either.
TUCKnTRUCK
That guy
Most turbine engines use counter rotating components to minimize torque furring acceleration.
Airflow through a turbine engine passes through stators, which minimize if not stop all rotation of flow, and then immediately prior to the combustion chamber it passes through a diffuser ring which turns the airflow 90 deg or so in a short space to decrease speed, and increase pressure.
The net result of the whole deal means that the thrust on the top/bottom are equal.
On the bypass air section, the same stators reside between the fan sections, so airflow in generally the same
Outside of that, the nacelle and tail cone are part of the airframe design, and should in theory slow the airflow, and present a uniform pressure surface just in front of the compressor.
In short, a modern turbine should not have any left or right turning tendency inherently. Installation angle, or airflow issues outside of the engine may affect this.
Airflow through a turbine engine passes through stators, which minimize if not stop all rotation of flow, and then immediately prior to the combustion chamber it passes through a diffuser ring which turns the airflow 90 deg or so in a short space to decrease speed, and increase pressure.
The net result of the whole deal means that the thrust on the top/bottom are equal.
On the bypass air section, the same stators reside between the fan sections, so airflow in generally the same
Outside of that, the nacelle and tail cone are part of the airframe design, and should in theory slow the airflow, and present a uniform pressure surface just in front of the compressor.
In short, a modern turbine should not have any left or right turning tendency inherently. Installation angle, or airflow issues outside of the engine may affect this.
inigo88
Composite-lover
shdw and TUCKnTRUCK have mostly answered this, but I wanted to clear up something about torque.
You got the first part right, but I think you're combining torque and "spiraling slipstream" together in the second part. Like you say, torque is literally just Newton's 3rd law: "Every applied force experiences an equal and opposite force," or "Every action has an equal and opposite reaction." The moving propeller blade surface hits a bunch of air molecules (fluid particles), which collectively push back with the same amount of force. You see this most noticeably in a helicopter, where the main rotor spins one way and the helicopter is being twisted the opposite way due to the "reaction forces" acting on the blades. Hence, the pedals on the floor which control the AOA of the tail-rotor are called "anti-torque pedals," because the tail rotor is creating a sufficient amount of thrust in the direction the tail boom would spin to keep it from spinning. Now I'm not a helicopter pilot, but I believe this is also why you slam the collective all the way down during a tail-rotor failure - because the lower the AOA of the rotor blades is, the less surface area there is exerting a force against the air molecules, and thus the less they push back with reaction forces and try to spin the helicopter.
In an airplane, torque is just a twisting moment strictly limited to the longitudinal (roll) axis, so it'll wear your left tire and wheel bearing out faster due to friction against the ground, but it's not going to try and weathervane you to the left like slipstream does (as shdw pointed out).
This doesn't make much sense to me, because I think you're implying that the air has some sort of memory (if you stop the spin of the air later then it could cancel out the torque from before - which wouldn't be true). The fluid particles of air are simply moving through the engine so fast, that by the time a blade hits a particle and "feels" a reaction force, that original particle is already long gone out the back. Stator blades are always positioned after each compressor and turbine blade, and I've always heard that is done not only to build up the air pressure but also mainly to line the air up so its free stream velocity vector meets the compressor/turbine blade at the most efficient AOA possible. In other words, the angle of the stators relative to the air departing the blades of the previous stage maximizes the efficiency of the next stage. I'm sure you've seen ceramic/titanium turbine blades on newer engines covered in all the little cooling holes, so that kind of efficiency is obviously paramount in modern jet engine design.
I think you are correct about the net torque effect on the engine being zero, but it's not because the air comes out straight. Here's what I think happens:
1.) Air initially enters the 1st compressor stage of the engine (assuming turbojet, not turbofan, the stages are all more or less the same size). The 1st stage compressor is spinning clockwise, so the equal and opposite reaction force tries to spin the engine nacelle counter-clockwise.
2.) The resulting air is spinning in a clockwise slipstream as it exits the 1st stage compressor blades, and hits the 1st stage stator blades. The stators stop the motion of the air in that direction (deflecting it 90 degrees or so). Air pushes against the stators (connected to the nacelle) in the clockwise direction, so the stators push back with an equal and opposite force in the counterclockwise direction.
The compressor blades and stators end up looking like this from a side view: / \ / \ / \ / \ with blue being a compressor blade and red being a stator blade (hence the ~90 degree change in direction each time).
From the example above, if we assume counter-clockwise = negative clockwise (CCW = -CW), and sum up all the axial longitudinal moments happening in that first stage in #1 & #2, we should get.
Total moment = CW + (-CW) + CW + (-CW) = 0
So I think we're onto something here, and my best guess - and I should stress that I don't know for sure - is that each time a compressor or turbine blade tries to twist the engine nacelle one direction through torque, a stator blade twists the engine nacelle back in the opposite direction, so that the net result is nearly zero.
I think whatever torque imbalance is left over isn't felt by the airplane very much, mainly due to the larger mass moment of inertia a big jet aircraft has over a small single. This is the same reason a Cessna gets tossed around more than say an A380 in an equivalent amount of crosswind.
If anyone bothers to read this far I commend you. Let me know if this sounds reasonable, because I'm confident about the definition of torque but not 100% on the turbine explanation. And apologies in advance if this is what you meant all along, and we're simply arguing semantics.
fish314 said:
You got the first part right, but I think you're combining torque and "spiraling slipstream" together in the second part. Like you say, torque is literally just Newton's 3rd law: "Every applied force experiences an equal and opposite force," or "Every action has an equal and opposite reaction." The moving propeller blade surface hits a bunch of air molecules (fluid particles), which collectively push back with the same amount of force. You see this most noticeably in a helicopter, where the main rotor spins one way and the helicopter is being twisted the opposite way due to the "reaction forces" acting on the blades. Hence, the pedals on the floor which control the AOA of the tail-rotor are called "anti-torque pedals," because the tail rotor is creating a sufficient amount of thrust in the direction the tail boom would spin to keep it from spinning. Now I'm not a helicopter pilot, but I believe this is also why you slam the collective all the way down during a tail-rotor failure - because the lower the AOA of the rotor blades is, the less surface area there is exerting a force against the air molecules, and thus the less they push back with reaction forces and try to spin the helicopter.
In an airplane, torque is just a twisting moment strictly limited to the longitudinal (roll) axis, so it'll wear your left tire and wheel bearing out faster due to friction against the ground, but it's not going to try and weathervane you to the left like slipstream does (as shdw pointed out).
This doesn't make much sense to me, because I think you're implying that the air has some sort of memory (if you stop the spin of the air later then it could cancel out the torque from before - which wouldn't be true). The fluid particles of air are simply moving through the engine so fast, that by the time a blade hits a particle and "feels" a reaction force, that original particle is already long gone out the back. Stator blades are always positioned after each compressor and turbine blade, and I've always heard that is done not only to build up the air pressure but also mainly to line the air up so its free stream velocity vector meets the compressor/turbine blade at the most efficient AOA possible. In other words, the angle of the stators relative to the air departing the blades of the previous stage maximizes the efficiency of the next stage. I'm sure you've seen ceramic/titanium turbine blades on newer engines covered in all the little cooling holes, so that kind of efficiency is obviously paramount in modern jet engine design.
I think you are correct about the net torque effect on the engine being zero, but it's not because the air comes out straight. Here's what I think happens:
1.) Air initially enters the 1st compressor stage of the engine (assuming turbojet, not turbofan, the stages are all more or less the same size). The 1st stage compressor is spinning clockwise, so the equal and opposite reaction force tries to spin the engine nacelle counter-clockwise.
2.) The resulting air is spinning in a clockwise slipstream as it exits the 1st stage compressor blades, and hits the 1st stage stator blades. The stators stop the motion of the air in that direction (deflecting it 90 degrees or so). Air pushes against the stators (connected to the nacelle) in the clockwise direction, so the stators push back with an equal and opposite force in the counterclockwise direction.
The compressor blades and stators end up looking like this from a side view: / \ / \ / \ / \ with blue being a compressor blade and red being a stator blade (hence the ~90 degree change in direction each time).
From the example above, if we assume counter-clockwise = negative clockwise (CCW = -CW), and sum up all the axial longitudinal moments happening in that first stage in #1 & #2, we should get.
Total moment = CW + (-CW) + CW + (-CW) = 0
So I think we're onto something here, and my best guess - and I should stress that I don't know for sure - is that each time a compressor or turbine blade tries to twist the engine nacelle one direction through torque, a stator blade twists the engine nacelle back in the opposite direction, so that the net result is nearly zero.
I think whatever torque imbalance is left over isn't felt by the airplane very much, mainly due to the larger mass moment of inertia a big jet aircraft has over a small single. This is the same reason a Cessna gets tossed around more than say an A380 in an equivalent amount of crosswind.
If anyone bothers to read this far I commend you. Let me know if this sounds reasonable, because I'm confident about the definition of torque but not 100% on the turbine explanation. And apologies in advance if this is what you meant all along, and we're simply arguing semantics.