Load Factor and Forces in a Turn

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chapfo
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Load Factor and Forces in a Turn

Post by chapfo »

I'm referring to this picture: http://avstop.com/ac/flighttrainghandbook/image54p.jpg
(Or this if the other one is down: http://www.free-online-private-pilot-gr ... g_turn.gif)

I'm reviewing my PGI notes for a Class 4 renewal and this confusion came up.
So based on my understanding, there seems to be two methods of reasoning/understanding of load factors and forces in a turn. (I capitalized FUGAL and PETAL to avoid miscommunication.)

Initially during my PPL, I've understood it like this:
  • *In Straight and Level, LIft=Weight.
    *In a banked attitude, TotalLift is inclined so that VerticalLift<Weight.
    *Aircraft starts to lose altitude.
    *HorizontalLift (CentriPETAL) pulls the aircraft to the inside of the turn.
    *CentriFUGAL force (Inertia) adds to Weight and gives a minor Resultant Load.
    *But since we want to maintain altitude, we need to pitch up to increase TotalLift so that VerticalLift=Weight.
    *TotalLift has increased and so has HorizontalLift (CentriPETAL), which also increases CentriFUGAL force.
    *This greater CentriFUGAL force plus Weight gives a moderately larger Resultant Load.
    *That's why load factor increases exponential from 0 to 90 degrees bank
Also based on that reasoning, if I'm not maintaining altitude, a 60 degree bank wont give me a load factor of 2.

During my Instructor Rating, my Class 1 explained it to me like this:
  • *Lift is inclined to create CentriPETAL.
    *CentriFUGAL gets created due to CentriPETAL.
    *Resultant Load increases due to combination of Weight and new CentriFUGAL.
    *Load factor increases and tends to push nose down.
    *Back pressure is required to increase TotalLift and to maintain altitude.
I thought the decrease in VerticalLift 'pushes' the nose down and not because of Resultant Load.

Are they both right?
Which method should I explain it as a Class 4 instructor?
It made sense at the time (18 months ago) when the Class 1 explained, but now the order of the forces occuring doesnt seem to make sense to me.

Thanks!
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photofly
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Re: Load Factor and Forces in a Turn

Post by photofly »

I like your original explanation better.

But to thoroughly unconfuse yourself, you need to brush up your basic physics.

VERTICAL FORCES

Unbalanced forces cause accelerations, not fixed velocities. So when you roll into the bank and tilt your lift force away from the vertical, the vertical forces become unbalanced, and you start to accelerate downwards. If you keep the same pitch attitude - the downward motion all by itself increases your angle of attack, increasing the lift, bringing the vertical forces back into balance and stabilizing your descent at a constant rate. The forces were only unbalanced for the brief fraction of a second it took you to pick up a steady rate of descent.

If the vertical forces stayed unbalanced for very long you would within a few seconds pick up a vertical speed of tens or hundreds of knots downwards - a rate of descent of thousands and thousands of feet per minute, increasing. Clearly this doesn't happen, even if you don't pull back on the yoke, demonstrating that the vertical forces don't stay unbalanced for very long. So your initial analysis of "unbalanced" forces is correct, but only for a fraction of a second.

That doesn't make it wrong, or useless as an explanation, but you have to know in what context it applies - and the answer is, only briefly.


HORIZONTAL FORCES : CENTRIPETAL VS. CENTRIFUGAL

Secondly - the debate about centripetal vs. centrifugal forces has been going on for centuries. The real answer about which force is correct depends entirely and only on your frame of reference.

In the (arguably*) inertial frame in which the air is stationary, which is a natural frame of reference to look at the behaviour of an airplane, an airplane moving in a curved path - like one that's turning - needs a sideways force to create the sideways acceleration that results in that curved path. This force acts on the airplane in the direction of the centre of the turn, and is a very real force, generated by the sideways-acting part of the lift. It's called the centripetal force. In this frame of reference there is *no* centrifugal force.

In the frame of reference in which the aircraft cabin is stationary (which is a natural frame of reference to consider the behaviour of the occupants of the airplane - like the passengers, and the coffee in their coffee cups) whenever the aircraft flies in a curved path the passengers and their coffees "feel" a force pushing them to the outside of the turn. There are two ways to explain this phenomenon. One is to acknowledge the aircraft-centric frame of reference isn't inertial - that is, Newton's second law (F=ma) isn't obeyed in a rotating frame of reference. The second way is to *invent* a new pseudo-force, so that Newton's law appears to be obeyed, by "creating" an extra "F". The extra force that you have to create to make Newton's law apply in a rotating frame of reference acts *away* from the centre of the turn, and is called a centrifugal force. In a rotating frame there is no centripetal acceleration, and no centripetal force. The centrifugal force is entirely imaginary. It was invented because it's what your body appears to experience, and it has hung around because it makes the equations simpler.


From a flight training point of view, it's extremely unfortunate (actually, it's just plain ignorant and wrong) that the Flight Training Guide shows one diagram with both a centripetal and a centrifugal force acting at the same time on the airplane. It's trivial to see this is incorrect physics: if both forces applied to the same object, they would cancel each other out, and the airplane would continue in a straight line. Which force is "correct"? Either, depending on your frame of reference. But you can't have both at the same time.




Finally - and I'm sorry it's such a long answer - don't try to mix into this reasons why the nose "drops". Explanations of forces in turns work just as well for stones being whirled around on the ends of strings as they do for airplanes, and stones don't have wings, noses or pilots. The reasons for the airplane attempting to change pitch when you bank into a turn are best looked at from a point of view of pitch stability in general, which is a different topic.


*note that this frame of reference isn't really inertial at all. It's actually everywhere accelerating "downwards" at close to 9.81ms^-2. This is a very inconvenient fact for analyzing the motion of small nearby rigid bodies so we brush it under the carpet by inventing a pseudo-force we call "the force of gravity" so we can pretend to ourselves that Newton's laws are obeyed in our own stationary frame.
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Last edited by photofly on Tue Jan 30, 2018 4:30 am, edited 1 time in total.
DId you hear the one about the jurisprudence fetishist? He got off on a technicality.
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Re: Load Factor and Forces in a Turn

Post by photofly »

OK I lied - one more point. If both your Class 1 instructor and the TC official who wrote the Flight Training Manual can both misunderstand and mis-explain the forces in a turn quite so badly, it's obvious that not knowing this stuff clearly doesn't stop anyone advancing in their instructing career or working for government.
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Re: Load Factor and Forces in a Turn

Post by AuxBatOn »

The only reason lhe load factor increases in a non-linear fashion (but not really exponential) has to do with angles. In order to maintain a constant altitude, the vertical lift needs to remain the same as the aircraft weight. In a banked turn, vertical lift equals total lift times cosine of the bank angle. Knowing that load factor is defined as total lift divided by aircraft weight, we can easily demonstrate that load factor=1/cos(bank angle).
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photofly
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Re: Load Factor and Forces in a Turn

Post by photofly »

1/cos rises faster than any exponential function, as the angle approaches 90° from below - it's unbounded for a finite value of its argument, which no exponential function is.
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Re: Load Factor and Forces in a Turn

Post by AuxBatOn »

Sure, but it is not an exponential function.
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Re: Load Factor and Forces in a Turn

Post by chapfo »

Thanks photofly for your detailed explanation!

So a better picture to use would be the same picture, but remove centriFUGAL force?
But then how would I explain the increase in load factor without centriFUGAL force? Load factor is pretty much Weight+CentriFUGAL (with the appropriate sin and cos) isnt it?
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Re: Load Factor and Forces in a Turn

Post by photofly »

First pick your frame of reference, and then stick to it. The two frames of reference that make sense in this scenario are the “air-stationary“ frame where the aircraft flies a curved path; and the aircraft-stationary frame, where the aircraft is fixed in the frame and the frame itself is turning. Either will do to explain the increase in load (which is real).

In the air-stationary frame, the airplane (in a level turn) and all the parts of it and everything in it all move in a curved path, and also all are supported against gravity. The vector sum of the force required to resist gravity and the force supplied to make it move in a circle (the centripetal force) means each object pushes against its support by more than its own weight. And hence “feels” heavier.

In the aircraft-stationary frame (which is rotating, non-inertial, and has a centrifugal pseudo-force) nothing is accelerating, so all forces must balance and cancel out. The force exerted by an object on its support is the vector sum of its gravity weight, plus the force required to oppose the centrifugal pseudo-force.
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