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flyjet787
2nd Feb 2010, 17:51
Hello people,

How do you explain which engine is Critical for a Twin Engine Propeller aircraft during a T/o run on the ground.

I am aware of the factors (assymentric etc.) that determine a critical engine when the aircraft is in air. But since assymetric factor does not occur during a T/o run I am really confused as to what factor determines a critical engine during T/o run.

Thnak you,
fly

Kirks gusset
2nd Feb 2010, 18:19
Flyjet 787, May I suggest you look at the previous posts on VMCG and you may find a wealth of explanations and theory on this subject.
I am curious as to why you might consider there is no assymetric factors on the take off run? If we fail an engine you will still get yaw.
KG

bobrun
2nd Feb 2010, 18:23
But since assymetric factor does not occur during a T/o run

Are you sure of that? I always thought there was no difference from when an aircraft is flying.

FE Hoppy
2nd Feb 2010, 18:24
Or to save time,

Which ever engine has the down going prop blades closer to the fuselage.

vikrant soni
2nd Feb 2010, 19:10
Flyjet 787

sir,
Thrust asymmetry WILL OCCUR if an engine fails on a twin, whether in flight or on ground.
The effects of asymmetry are opposed by the forces of friction between wheels and ground, but is WAY TOO SMALL to consider.

As far as the critical engine is concerned, it is always the engine, the loss of which will most adversely affect the directional stability of the aircraft. In conventional props it is the left engine.

Vicky :O

737ngpilot
2nd Feb 2010, 21:16
Asymmetrical yaw



http://upload.wikimedia.org/wikipedia/commons/7/76/Criticalengine1.jpg (http://en.wikipedia.org/wiki/File:Criticalengine1.jpg) The operating right-hand engine will produce a more severe yaw towards the dead engine, thus making the failure of the left-hand engine critical

bfisk
2nd Feb 2010, 23:02
I think the answer the thread starter may be looking for is the upwind engine.

During the takeoff roll, before the aircraft is flying, there is really no P-factor to talk of (assuming the propeller disc plane of rotation is perpendicular to the runway), thus the thrust line will be more or less through the centre of the propeller disc. Factors such as slipstream may still be a factor, though.

However, since "critical engine" is define as the engine whose failure would result in the most adverse effects on the aircraft's handling and performance, one could argue that the upwind engine is critical. Obviously the vertical fin would, in a crosswind, add to the yawing effect of the downwind engine, while it would counteract the yawing from the upwind engine, if the downwind engine was to fail.

Vmcg, as certified and described, does not allow for a crosswind. However, the speed at which the aircraft is directionally controllable on the ground in case of an engine failure, will in fact vary with crosswind component.

flyjet787
3rd Feb 2010, 08:00
Ok. Firstly could you please explain why D2 lies farther away from the fuselage as compared to D1 during a take off roll on the ground (not when in the air).

As far as i know the assymetric blade effect or P-factor is when they down going peopeller travels through a greater distance than the up going propeller i.e the down going prop produces more thrust than the up going one hence you have different thrust lines. But this normally occurs only after a take off or for tail draggers during a take off run. I think the assymetric factos that I am refering to does not happen for a normal multi engine prop aircraft during a take off roll.

(please correct me if I am wrong)

Thanks.

flyjet787
3rd Feb 2010, 08:03
I am not saying that there would'nt be a yaw. There will be a yaw towards the failed engine. But how do you determine which engine is critical when the aircraft is on the ground.

flyjet787
3rd Feb 2010, 08:13
P-factor, also known as asymmetric blade effect and asymmetric disc effect, is an aerodynamic (http://en.wikipedia.org/wiki/Aerodynamics) phenomenon experienced by a moving propeller (http://en.wikipedia.org/wiki/Propeller) with a high angle of attack (http://en.wikipedia.org/wiki/Angle_of_attack) that produces an asymmetrical ceter of thrust

Causes:
When an aircraft is in straight and level flight at cruise speed, the propeller disc will be normal (i. e. perpendicular) to the airflow vector. As airspeed decreases and wing angle of attack increases, the engines will begin to point up and airflow will meet the propeller disc at an increasing angle, such that horizontal propeller blades moving down will have a greater angle of attack and relative wind velocity and therefore increased thrust, while horizontal blades moving up will have a reduced angle of attack and relative wind velocity and therefore decreased thrust. (Vertical blades are not affected). This asymmetry in thrust displaces the center of thrust of the propeller disc towards the blade with increased thrust, as if the engine had moved in or out along the wing. The engine with the down-moving blades towards the wingtip produces more yaw and roll than the other engine, because the moment (arm) of that engine's thrust about the aircraft center of gravity is greater. Thus, the engine with down-moving blades towards the fuselage will be critical, because its failure will require a larger rudder deflection by the pilot to maintain straight flight.


So this is the factor that normaly determines which engine is critical when the aircraft is in the air. So what determines a critical engine on the ground.??

737ngpilot
3rd Feb 2010, 16:40
When one engine becomes inoperative, a torque will be developed which depends on the lateral distance from the center of gravity (C.G.) to the thrust vector of the operating engine multiplied by the thrust of the operating engine. The torque effect attempts to yaw the aircraft's nose towards the inoperative engine, a yaw tendency which must be counteracted by the pilot's use of the flight controlls. Due to P-factor, the right-hand engine typically develops its resultant thrust vector at a greater lateral distance from the aircraft's C.G. than the left-hand engine. The failure of the left-hand engine will result in a larger yaw effect via the operating right-hand engine, rather than vice-versa. Since the operating right-hand engine produces a stronger yaw moment, the pilot will need to use larger control deflections in order to maintain aircraft control. Thus, the failure of the left-hand engine is less desirable than failure of the right-hand engine, and the left-hand engine is critical.

flyjet787
3rd Feb 2010, 17:51
"Due to P-factor, the right-hand engine typically develops its resultant thrust vector at a greater lateral distance from the aircraft's C.G. than the left-hand engine."

P-factor: also known as asymmetric blade effect and asymmetric disc effect, is an aerodynamic (http://en.wikipedia.org/wiki/Aerodynamics) phenomenon experienced by a moving propeller (http://en.wikipedia.org/wiki/Propeller) with a high angle of attack (http://en.wikipedia.org/wiki/Angle_of_attack) that produces an asymmetrical ceter of thrust

According to the above statement P-factor is experienced only by a moving prop with HIGH ANGLE OF ATTACK but the propeller does not have a high angle of attack during a take off run.

I understand what you have explained in your post. But what you are saying hold good when the aircraft/propeller are at high angles of attack which then cuases the down going prop to produce more thrust than the up going one. But all this doesn't occur on the ground.

thepotato232
3rd Feb 2010, 18:24
Flyjet:

You're right. P-Factor does not exist when the aircraft is at a precise zero degree AoA. It receives far too much blame for yaw tendencies during the groundroll portion of the takeoff phase. The real culprits behind yaw in this stage for a nosewheel-equipped airplane are the corkscrew-shaped propwash striking the empennage, torque produced by the engine itself (Newton's Third Law), and, as has been mentioned, the crosswind direction. P-Factor and gyroscopic precession are of primary concern for taildraggers in this stage.

The short answer to your question is that - on the ground - there is no "critical engine" as it relates to P-Factor. If the engine fails in a twin and you're on the ground, you have two options: Continue the takeoff roll until airborne (at which point P-Factor and the critical engine become major issues), or chop the power and stay on the ground. The wisdom of each course of action is beyond the scope of this discussion.

This website did a great deal to broaden my understanding of yaw during my CFI days: Yaw-Wise Torque Budget [Ch. 8 of See How It Flies] (http://www.av8n.com/how/htm/yaw.html)

EDIT: It's also worth noting that most nosewheel airplanes are designed such that the propellers are not at a precise zero degree AoA while the airplane is on the ground, so P-Factor still exists. It is negligible in this stage, however, compared to the other factors listed.

I hope that, between my disjointed explanation and the nigh-impenetrable wall of text on the other side of the link, you've got a satisfactory answer.

737ngpilot
4th Feb 2010, 00:30
according the to Airplane Flying Handbook issued by the FAA, P- factor is an issue during take off roll....

Airplane Flying Handbook - Google Books (http://books.google.com/books?id=V3SZXFWuCIgC&pg=PT59&lpg=PT59&dq=P+factor+on+take+off+roll&source=bl&ots=HkPJGihDru&sig=IdYHtBVZihUuY8iRds4bd19QkFA&hl=en&ei=aPhpS9LzFY7gNdGhpIIG&sa=X&oi=book_result&ct=result&resnum=7&ved=0CBgQ6AEwBg#v=onepage&q=P%20factor%20on%20take%20off%20roll&f=false)

Pugilistic Animus
4th Feb 2010, 00:39
....and dealing with assymetry:)

thepotato232
4th Feb 2010, 01:48
The FAA puts forward a lot of aerodynamic literature that is either oversimplified or flat-out incorrect. Just reference the ongoing battle to teach Newtonian principles in initial aerodynamic theory courses in addition to the FAA-sanctioned over-application of Bernoulli's Principle. Long story short, starting in the 1950's, a lot of these complicated principles were dumbed down to "our level" on the basis that we're pilots, not mathematicians, and we don't need to (or simply can't) understand the full story. When you go full Math Nerd and really get into the physics in play, you find yourself needing to unlearn a lot of The Sacred Truths as taught in ground school.

I'm in serious danger of throwing this discussion way off topic, so to get back to what the OP was asking: The existence of P-Factor is predicated on the propeller operating in an AoA other than zero. A propeller producing thrust at a zero degree AoA will not create P-Factor. A propeller producing thrust at very-near-to-zero degree AoA will produce P-Factor, but the asymmetry is negligible compared to the other factors cited. The term "Critical Engine" doesn't really apply on the ground the same way it does in the climbout.

fly_antonov
4th Feb 2010, 03:08
Apparently, the question isn' t as boring as it looks.



"Due to P-factor, the right-hand engine typically develops its resultant thrust vector at a greater lateral distance from the aircraft's C.G. than the left-hand engine."

P-factor: also known as asymmetric blade effect and asymmetric disc effect, is an aerodynamic (http://en.wikipedia.org/wiki/Aerodynamics) phenomenon experienced by a moving propeller (http://en.wikipedia.org/wiki/Propeller) with a high angle of attack (http://en.wikipedia.org/wiki/Angle_of_attack) that produces an asymmetrical ceter of thrust

According to the above statement P-factor is experienced only by a moving prop with HIGH ANGLE OF ATTACK but the propeller does not have a high angle of attack during a take off run.



During the take-off run, the fixed-pitch propeller has the lowest TAS, thus also the highest angle of attack.

Propeller angle of attack is the angle between the propeller' s chord and the relative airflow, the latter being a combination of the propeller' s movement in the air and the relative air from the movement of the airplane.

Next, the P-factor is caused by a positive angle between the propeller disc' s axis and the relative airflow of the airplane.
The down-going blade will have a higher angle of attack than the up-going blade because of how the airplane' s relative airflow bounces against them.
Also, the relative airflow vector will be larger, adding to the equation.


http://www.boundvortex.com/images/pfactordownup.png


The real culprits behind yaw in this stage for a nosewheel-equipped airplane are the corkscrew-shaped propwash striking the empennage, torque produced by the engine itself (Newton's Third Law), and, as has been mentioned, the crosswind direction. P-Factor and gyroscopic precession are of primary concern for taildraggers in this stage.

I don' t agree. The corkscrew effect is irrelevant on twin engined airplanes.

Also, the torque produced by the Newton' s 3rd law itself only affects single-engine airplanes. On a twin engine, they more or less compensate themselves because the left engine will try to twist the left wing counter-clockwise from the propeller/engine' s axis, thus try to push the fuselage up, the right engine will also try to twist the wing counterclockwise but this time the fuselage being on the other side, it will try to pull the fuselage down.
The resulting roll torque from this is insignificant to create any additional weight on any of the main wheels, meaning that the wheel drag will be approximately equal and that no yaw will result.

The problem of the critical engine on the ground for twins is solely a P-factor problem, itself caused by the postive tilt of the propeller and/or the angle of the propeller disk during rotation.
It is not that significant (or non-existing if there is no tilt by design) when the nose gear is still on the ground but it is significant during rotation and until the airplane sticks off.

When does it make a big difference?

100kts.....V1.. BANG BANG BONG (left engine torquemeter drops)..CONTINUE (apply right rudder to make-up for the thrust assymetry)... reading XXX kts nearing V2 of XXX kts..rotate... (additional correction for P-factor with rudder = more drag!!)....(due to impaired performance and additional drag, more time between rotation and unsticking)...minimum V2 is XXX kts.

After the rotation, the airplane with only one engine, and all the added drag from the rudder and the windmilling engine could take several seconds to unstick and this is where the P-factor is going to hurt your performance most.

This portion of the take-off run is why the V1 must be higher than Vmcg (minimum control speed ground using aerodynamic controls only, mainly the rudder). The critical engine scenario affects Vmcg calculation and as a result also the V1 calculation.

The critical engine awareness is an important factor for a safe post-V1 continue call. It defines your error margin and the level of precision you need to operate at. You should also know that you can expect to need a little more rudder input during rotation.

In any case, this is basic theory and you should always operate your aircraft in accordance with the procedures as described in your airplane manual.

flyjet787
4th Feb 2010, 06:31
@232 and antonov. Thanks for your help guys.

Pugilistic Animus
4th Feb 2010, 15:25
The FAA puts forward a lot of aerodynamic literature that is either oversimplified or flat-out incorrect. Just reference the ongoing battle to teach Newtonian principles in initial aerodynamic theory courses in addition to the FAA-sanctioned over-application of Bernoulli's Principle. Long story short, starting in the 1950's, a lot of these complicated principles were dumbed down to "our level" on the basis that we're pilots, not mathematicians, and we don't need to (or simply can't) understand the full story. When you go full Math Nerd and really get into the physics in play, you find yourself needing to unlearn a lot of The Sacred Truths as taught in ground school.



pilots ain't that smart;..... the USA invented it all and the whole world follows the Avigation Act of 1926 wheter or not they think so:ooh:

nobody knows more about this stuff than the US government

a.....Newton's law ---wing pushes air down; air pushes wing up

b. the aerodynamic forces and moments can be computed from Bernoulii's relation F = MA ,...no more :zzz:

PA

thepotato232
4th Feb 2010, 17:56
The corkscrew effect is irrelevant on twin engined airplanes.

Also, the torque produced by the Newton' s 3rd law itself only affects single-engine airplanesTrue when both screws are turning, but the original question was regarding the causes of yaw when one engine fails (besides the obvious asymmetrical thrust).
That's one of the best explanations of P-Factor I've ever seen, but in this instance, we are discussing specifically the stage of the takeoff when the angle of the propeller disc axis and the relative airflow is quite near to perpendicular. As you mentioned, P-Factor becomes a primary controllability issue once AoA is increased and liftoff is achieved. While still on the ground roll, however, the factors I listed (torque, propwash and x-wind) will have a larger effect than P-Factor on a twin engine aircraft that has suddenly and unexpectedly become a single.

fly_antonov
6th Feb 2010, 20:59
True when both screws are turning, but the original question was regarding the causes of yaw when one engine fails (besides the obvious asymmetrical thrust).



You have a point there.

Basically the effects of torque act in opposite directions on the fuselage, one acts upwards, one downwards. They compensate eachother during normal operation on the ground, but indeed the engine failure case is different as the engines become the center of a left roll axis.
I got messed up about that.

So let' s discuss the engine/prop' s torque' s effects in determining a critical engine on a twin, on the ground.

In a few words: different explanation, same conclusion being little to no factor for critical engine.

If you' re not interested in the analysis, jump to the last paragraph.

Take an imaginary heavy twin engined turboprop with a very long wingspan and engines very close to the root. Due to the central positions of the engine, you can assume that it will behave like a single-engine airplane if an engine fails, including the torque creating a left roll moment, adding weight on the left side of the airplane and to the main gears on that side, increasing drag on that gear and yawing the airplane to the left.
Now take the same turboprop and place the engines at the tips of the wings. The engine/propeller combination produces the same torque for a given configuration and specific air data.
Torque is T = r x F, r being your arm length, F being the force.

(note: supposing that the wing-fuselage angle is 90, sin teta = sin 90 = 1, thus sin teta factored out)
Now the arm length between the engine and the fuselage is significantly longer than the engines at the root configuration. So for the same torque, the force acting on the fuselage will be alot smaller to compensate for the longer arm. As a result, the left roll moment will be reduced significantly.
Also the torque' s effects will be relative to the size/weight of the airplane.

If you put 2 x 11000hp TP400 turboprop engines on a piper seminole, you're going to have a take-off run like a rocket but if an engine fails, your entire airplane will start rotating around the engine, even on the ground.
If you want to imagine the effects, just compare it to losing a wing during flight, but the same effects happening to you on the ground.

If you put 2 x 180hp O-360 engines on an ATR-72, you could get to move the ATR-72, and if you have an endless runway, you could make it take-up a certain speed. If an engine fails at any point then, it wouldn' t matter because it would have no roll nor yaw effects to the airplane.

The effects can be discussed for hours but in the end they would differ so much depending on the airplanes considered due to several factors such as the actual torque, the distance of the engine on the wing from the CG and other factors like air density.


In general, for twin aircraft of average thrust to weight ratios, due to the lateral moment arm length to the center of gravity on the Y/pitch axis, the effects of such engine/prop torque that ultimately cause a higher drag on the main gears is barely perceivable, due to the arm length reducing the perceivable force on fuselage and landing gear and also because the force would act in a more balanced way on both sides of the main gear compared to a single engine aircraft with an engine's torque axis running between the main gears, segregating the force more unequally onto the left main gear.

Pugilistic Animus
7th Feb 2010, 18:36
I had never known flight to be so complicated,...new ideas everyday:zzz:

Lester:E

FullWings
7th Feb 2010, 19:17
There's a definition of a "critical engine" if you take off in a multi-engined aircraft: it's the one that causes you the most problems if it stops for whatever reason.

There's another definition, which I always thought depended on whether the props/fans/whatever rotated in the same or contrary directions. If they are going round the same way, then you'll hit the limits of yaw control sooner with a failure on one side than the other (the side depending on whether you're clockwise/anticlockwise) due to torque reaction.

In a twin, you've effectively become a single with an offset propulsion unit - depending on which way it's going round, the secondary effects will add to or oppose the yaw generated by the other power plant failing, thus leading to one engine being more "critical" than the other in terms of yaw containment.

That's how I always understood it but am ready to be corrected...