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The power of the engines will apply the exact same torsional force to the plane independent of the weight. However, the acceleration of the object (plane) itself is inversely correlated to the mass of the plane multiplied by that force. Therefore, the heavier the plane, the less rotational acceleration will occur.
That is correct but:
- The issue here is not the linear acceleration (and in particular the lateral acceleration which is who will move the plane laterally, away from the RWY center line and towards the grass. The lateral acceleration would also diminish with the increase in weight, except that the lateral force that provides that acceleration will increase with the weight (normal force, weight-on-wheels) compensating the previous effect.
- Since the wheels will basically provide all the necessary lateral acceleration to keep the wheels from skidding sideways as the plane yaws, the question is how quickly the plane will yaw, that is the angular acceleration. You already said that the torque due to the asymmetric thrust is independent of the weight. The other factor is the moment of inertia. I've already mentioned that, while the moment of inertia tends to increase with the weight, that is not necessarily the case: Take 5000 lb of fuel from the right wing tank, 5000 from the left wing, and 5000 from the tail. Now put 30000 lb in the center wing tank. More mass, less moment of inertia.
--- Judge what is said by the merits of what is said, not by the credentials of who said it. ---
--- Defend what you say with arguments, not by imposing your credentials ---
The way you word this seems to assume that the plane remains pointing parallel to the runway and slides sideways on all the wheels. I am actually assuming that the rotational force only slides the nose wheels resulting in a change of angle of the plane and the force pushing the plane into the grass is the forward force and subsequent acceleration (no longer parallel to the runway), not a sideways one.
If it is indeed the alignment of the plane with the runway that determines where it ends up (vs a lateral slide) the only thing that matters is the rotational force. The net rotational force (I will call it force rather than torque so we keep the terms clear) will be the torsional force generated by the asymetric thrust MINUS the lateral friction force provided by the nose wheel. The counteracting force of wheel friction increases approximately linearly with weight. Therefore the front wheels will provide more counteracting force to the asymetric engines. Even if it can't counteract the full force, the amount of movement over time (acceleration) will also be a lot slower because the mass is greater so if a slide occurs it will be much slower (rotating around the centre of mass) and depending on the length of the run, may still keep you off the grass. You are right that moving the center of mass will affect the calculation, but it is more complicated than you imply. There are two torsional forces acting on the plane around the centre of mass: The asymetric thrust and the nose wheel friction so you have two moments of inertia coming into play.
That is exactly the same that I am assuming with an added condition: that the main wheels don't skid sideways, so the plane turns like a car. So the yaw acceleration together with the forward speed is what will determine how soon the plane will end on the grass. I am assuming that, during this turn, the main wheels will provide enough lateral traction (and hence lateral acceleration) to make the plane turn without skidding. What may be confusing is: What lateral acceleration if the plane never moves laterally, always forward, just while turning. Well, it is the same than when you make a rock spin over/around your head holding it with a cord (a conical pendulum). Acceleration is any change in the speed VECTOR, so even if the rock is traveling at constant "speed", the speed vector is changing because the direction of the speed vector is changing. And the acceleration that makes the speed change in direction (not in magnitude) is the lateral acceleration (also called radial, normal or centripetal acceleration).
You are messing with Physics with me, you are looking for troubleIf it is indeed the alignment of the plane with the runway that determines where it ends up (vs a lateral slide) the only thing that matters is the rotational force. The net rotational force (I will call it force rather than torque so we keep the terms clear) will be the torsional force generated by the asymetric thrust MINUS the lateral friction force provided by the nose wheel. The counteracting force of wheel friction increases approximately linearly with weight. Therefore the front wheels will provide more counteracting force to the asymetric engines. Even if it can't counteract the full force, the amount of movement over time (acceleration) will also be a lot slower because the mass is greater so if a slide occurs it will be much slower (rotating around the centre of mass) and depending on the length of the run, may still keep you off the grass. You are right that moving the center of mass will affect the calculation, but it is more complicated than you imply. There are two torsional forces acting on the plane around the centre of mass: The asymetric thrust and the nose wheel friction so you have two moments of inertia coming into play.
So a few points here. The first one is that we agree in the big picture.
I don't know why you want to call a torque force instead of torque to keep it clear, but ok, I follow you (in fact the proper name would be moment, the moment of the thrust, the moment of the friction force of the nose wheel).
The third thing is that I neglected the nose-wheel friction, like if the nose-wheel would caster freely (which is not the case)
Why I did this simplification? Well, to begin with Vmca is defined taking into account aerodynamic means only, so it looked fair to keep that assumption.
Then, a really good model would be extremely complicated. The weight on the nose wheel depends not only on the weight but also on the position of the CG and the lift of the wing and horizontal tail and the pitching moment of the wing. The same would be true for the main wheels so, at a somehow high speed, the hypothesis that both you and I did that the plane will not skid can't hold true because the weight on the main wheels can be reduced to a level where a lateral skid WILL happen. Then, there are other means to control the plane. Asymmetric braking, idling the engines that did not fail (if you want to maximize the correction power, in the 474 you would reduce thrust only on the engines on the "good" side and will leave the "good" engine next to the failed engine to keep producing thrust. Neither you or me took any of this into account, which is going toward the worst case scenario. So I just removed the friction on the nose wheel.
Finally, if you still want to keep the friction on the nose wheel, it is not fair to say ONLY that with a higher weight you have more friction. If the wight is added with the same distribution that it already was, then the moment of inertia will also increase so you will need more lateral force (more friction) to produce the same yaw angular acceleration. And not even mention what would happen if the added weight is because you put fuel in the fin tank. The CG moves back and the weight on the nose wheel will actually diminish, maybe by more than the wight you added. So, depending on the distribution of the added weight, the power of the nose wheel to make an angular acceleration may increase, remain the same, or diminish with added weight.
Also, the addition of wight, if it changes the moment of inertia (which it typically will), it will also change the the yaw angular acceleration (speed of the change in direction) produced by the same "rotational force". So what you said "If it is indeed the alignment of the plane with the runway that determines where it ends up (vs a lateral slide) the only thing that matters is the rotational force" is just not correct. It is the "rotational force AND the moment of inertia. In the same way you have F=m*a, you also have M=I*α, where M is the net moment of the forces (or the rotational force as you like to call it, equivalent to the force F in F=m*a), I is the moment of inertia (the resistance of an object to change its rotational speed when a moment is applied, equivalent to m the mass in F=m*a), and α is the angular acceleration (equivalent to a in F=m*a). Form F=m*a you can solve for a and get a=F/m, and in M=I*α you can solve for α and get α=M/I.
The best argument to say, if the engine is going to fail at one given speed when all engines were set at full thrust, a higher weight will help you prevent end up in the grass, is that, in general (but not necessarily), a higher weight also means a higher moment of inertia, and since the moment of the asymmetric thrust will be the same in the high and low weight scenarios, that asymmetric thrust will cause a slower angular acceleration in the high-weight case, giving a bit more time to the pilot to react (reduce thrust in the other engines, compensate with differential brakes and nose wheel, etc...) before a departure from the paved surface becomes unavoidable.
--- Judge what is said by the merits of what is said, not by the credentials of who said it. ---
--- Defend what you say with arguments, not by imposing your credentials ---
Les règles de l'aviation de base découragent de longues périodes de dur tirer vers le haut.
We're definitely on the same page. The reason I preferred to stick with force vs torque is that if I visualize the formula it is easier to be able to add/subtract two things that are of the same unit.
The one thing I learned in engineering, is that when a factor was negligible or close to zero, it is good to eliminate it to simplify things. We are far less exact than the mathematicians, physicists or economists.
My statement was incorrect as you pointed out. I should have stated: "... the only thing that will result in movement off the pavement will be the net rotational force". You are correct in that the net rotational force will result in movement (or acceleration) based on the moment of inertia which is directly related to the mass. I also missed the rudder impact on the net torsional force in this simplified calculation, a torsional force directly related to airspeed.
I did make the assumption that the main wheels would not slide laterally. I based that entirely on the assumption that in order to introduce a lateral slide, you would need to have a significant perpendicular force on the plane that was directly aligned with the CG. I couldn't see one occurring unless there was a massive cross wind. Even immediately prior to full gear liftoff -- and under asymetric thrust I suspect a quick transition of force would be better -- I don't think the rear wheels are providing much lateral force against anything other than wind just like in a normal takeoff.
The question of the amount of weight on the nose gear did cross my mind but I guessed it would be substantial, until the time where it was lifted by the pilot beginning the rotation. If a plane is resting on 3 points, I imagine they are going to make sure the CG is somewhere in the middle of the triangle because otherwise they risk an unstable resting position if the CG was too close to the line between the main gear. I assumed it would be too easy to topple in the case of a sudden weight shift etc. I don't really know where the balance is though, those were my guesses. Based on that guess though, I think the nose wheel friction plays a significant role in the result, though you are right, the amount would be highly determined by the location of the CG, an outcome determined by the weight distribution.
You were right to point out that as the plane rotates and the nose lifts that counteracting force will disappear and the only thing counteracting the torsional force from the asymetric thrust would be the rudder and I think in the case of asymetric thrust, the takeoff would look a little like a plane taking off in a cross wind the difference being the plane would not float straight down the runway (facing slightly sideways) but would move directly forward in the direction of the nose.
So yes, I agree with you, even the simplified problem is still a set of variables which would act on the three main torsional forces acting in the scenario. To summarize:
-- Nose wheel friction will generally be higher with a heavier plane, but might not be depending on the loading of the aircraft and the resulting center of mass in relation to the 3 wheels and will change as the pilot introduces rotation and applies lift to the wing surfaces for takeoff
-- Rudder force will be zero at first and will increase as the plane accelerates
-- Asymetric force from the engines will be constant with both airspeed and unaffected by weight
The amount of rotational movement -- angular acceleration as you stated -- (like the turning like a car) will be determined by when in the run there is a net non-zero torsional force governed by the moment of inertia as you stated. The net movement being inversely proportional to the weight of the plane.
I suspect with those variables (some of which change as the plane accelerates down the runway) you could model the outcomes for a variety of weights, weight distributions, and pilot inputs.
Fun exercise.
And you know who you're sharin your opinion with? If he weren't that far away from here, he'd be my choice for the flight instructor that you'll trust. I wouldn't have accepted a jetphotos online friendship with him if that weren't the case.
But after all, I am not a friend of all to much brain or calculation in the air. Decisions have to be made FAST after you've left the ground.
Nice to read you again, Schwartz.
A new year, for all of us. But (not only) for me, it'll be a special year. Four decades in life, still this winter. And almost ten years here on this brilliant platform.
Aviation enthusiast since more than 30 years. Almost a decade here on this platform.
Here we are. I think that I'll never regret the choice of my jetphotos prototypes. MD-80s are and have been so very uncommon in Germany, at least since I am an aviation enthusiast, I really regret that.
If we talk about my home airport, there only 1 airline comes to my mind which in the past here operated MD-80s. And who else if not the jetphotos platform will tell me in a second if I am right...
Man, I should be right if we talk about my home airport. Scandinavian airlines. Here is the evidence:
SK-MD82 at Dusseldorf Rhein Ruhr International airport
SK stopped to fly that beautiful bird in 2013. But I don't think that the reason was bad t/o computation. I respect the pilots from Copenhagen everytime I see them, nowadays without MD-80.
Btw, Bobby tried to do what I told him in late August? Back then I thought that he and me overfly the 1,000 post hurdle at the same time. But probably that'll not happen.. But that's ok.
PS: Iberia also once owned MD-80s before they've become an "Airbus only airline" (!). But I just can't remember if they also flew their MD-80s to DUS. That's a question which you can ask the jetphotos database.
Last edited by LH-B744; 09-04-2017 at 10:33 PM. Reason: Iberia
A new year, for all of us. But (not only) for me, it'll be a special year. Four decades in life, still this winter. And almost ten years here on this brilliant platform.
Aviation enthusiast since more than 30 years. Almost a decade here on this platform.