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The Cetn

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How does the Aircraft continue to turn when the both the Horizontal component of lift and the centrifugal force are equal?

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  • $\begingroup$ thebackseatpilot.com/pages/…. $\endgroup$
    – Nish
    Oct 21, 2023 at 19:23
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    $\begingroup$ Re the link above-- wow, the diagram labelled "Left bank, no back pressure" makes even worse errors. Implies that the amount of back pressure you apply affects the vertical component of lift but not the horizontal component. Fails to recognize that if true, this would create a skid (net aerodynamic force pointed to inside of turn, not "straight up", in a/c ref frame.) Completely bogus. In the hang gliding world an argument is sometimes made that inadequately pitching the nose up while turning creates a slip rather than a skid. That's bogus too. Slip/skid is about yaw not pitch. $\endgroup$ Oct 23, 2023 at 14:59
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    $\begingroup$ Try it-- enter a turn and change the back pressure. You'll see little effect on the ball. $\endgroup$ Oct 23, 2023 at 14:59
  • $\begingroup$ Just noticed that this question is a duplicate of aviation.stackexchange.com/questions/38040/… . Some well-thought-out answers were posted there. $\endgroup$ Oct 23, 2023 at 15:16
  • $\begingroup$ Related -- aviation.stackexchange.com/q/79338/34686 $\endgroup$ Oct 23, 2023 at 15:19

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How does the Aircraft continue to turn when both the Horizontal component of lift and the centrifugal force are equal?

Those pictures are not 100% correct.

To enter and keep a turn, a force toward the centre of the curve has to be created. In an airplane this is achieved by tilting the lift laterally, like in the following picture (source) where the airplane is turning left (as seen from the front):

 Airplane in a turn

The vertical component of the lift balances the weight out while the horizontal component keeps the airplane in a turn. This horizontal component is called centripetal force. The higher the centripetal force is, the steeper the turn is.

End of the story.

So what about the centrifugal force? Let's make an everyday comparison with what happen in car that accelerates. Due to the traction force the car gets accelerated forward. But what you experience as a driver/passenger is actually a backward force (aka inertia) pushing you against the seat. This is exactly the same as for our airplane: the centripetal force makes it bend leftward but what you experience is actually an opposite, rightward, centrifugal force. Those two forces have the same magnitude but opposite direction. Anyway is only the centripetal force that keeps the airplane in a turn. The centrifugal force is the inertia "felt" by the pilot/passenger and acting on all the non-lifting surfaces of the airplane.

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The forces in the first pictures would be as shown if you are observing situation from reference frame connected to the turning airplane. Then, in this reference frame, the airplane is not moving and therefore neither turning. Other forces, gravity and interaction with air molecules remains in place, so in order to make the description consistent, you need to add centrifugal force.

If you observe the situation from ground, there is no centrifugal force (because this fictitious/inertial force appears as a consequence of applying laws of motion in rotating reference system) and horizontal component of lift makes observed trajectory circular.

Edit2: The later two images are not complete. Forces has to be always in balance if viewed from airplane's reference frame. They are missing part of aerodynamic force generated by side-wise motion through air, which is definition of slip and skid (you can call it drag or airframe-lift). See @quietflyer's answer for more details on this.

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  • $\begingroup$ Re "(if it should be a view from airplane's reference frame, CF and horizontal lift has to match always)" -- well, not if "horizontal lift" is only meant to include the horizontal component of the wing's lift and not the horizontal component of the aerodynamic sideforce from the airflow striking the side of the fuselage. See my answer for more-- $\endgroup$ Oct 23, 2023 at 14:18
  • $\begingroup$ @quietflyer: True, I have realized this later, but had not found time to add it to my answer. $\endgroup$
    – Martin
    Oct 28, 2023 at 13:48
  • $\begingroup$ All fictitious, (non-inertial, psuedo forces), simply represent the acceleration of the frame of reference you are doing your analysis in. Doing your analyis in any accelerating, non-inertial frame invalidates Newton's kaws, UNLESS you add in an extra, fictitious force to represent the acceleration of the frame itself. $\endgroup$ Oct 28, 2023 at 14:17
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The centrifugal force is produced by the act of turning. If you weren't turning, it would be zero. If it were zero, you would have an imbalance of forces which would cause an acceleration that would cause you to turn.

Properly, the centrifugal force is not really a force, it is an acceleration.

$a=V^2/r$

Recall that we can write $F=m\,a$ where forces balance out accelerations. So, the horizontal component of lift needs to balance out (or it causes) the acceleration in the turn.

As @Martin said, the second two images are bogus and the source that generated them doesn't understand what is going on.

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    $\begingroup$ I do think "force is balancing out acceleration" formulation is misleading. Force is causing acceleration. In any reference frame. The problem with centrifugal force is that it is result of choosing accelerating frame of reference. So you could say, in some sense (this particular) force is "balancing out" acceleration of frame of reference (even if it is bit unusual wording), but frame of reference is no real object. For any physical object, force is causing acceleration, not "balancing" it. $\endgroup$
    – Martin
    Oct 22, 2023 at 11:14
  • $\begingroup$ @Martin Fair point -- I thought about wording things differently, but then I started to think about the acceleration due to gravity -- which we have in (what we consider) an inertial frame -- but that we often freely replace with a force (weight). I meant balance out to really imply algebraic equivalence. And although the force undoubtedly causes the turn, when analyzing the problem, you may well observe the turn in order to then infer the force. $\endgroup$ Oct 22, 2023 at 15:44
  • $\begingroup$ Acceleration causes Gravity, not the other way round, or perhaps a better way of stating it is that they are the same thing. Think about it, how do we measure, or detect acceleration? With a G-meter. When we let Gravity accelerate us, (in free-fall), a g-meter will indicate zero. $\endgroup$ Oct 28, 2023 at 11:50
  • $\begingroup$ I shouldn't have brought gravity into it. It is a fundamental force of nature that is a body force (acts on all the mass simultaneously) rather than a point, pressure, or shear force that we're more used to thinking about. Gravity is a force field. If we are free to fall, then we accelerate at a fixed rate (due to gravity). However, the acceleration does not cause it (the way an accelerating frame causes a force). For if we can't fall, we still feel the force of gravity. So a mystical force field causes the body force of gravity F=mg. $\endgroup$ Oct 28, 2023 at 17:11
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Centrifugal force is a fictitious force, i.e., it is not real, it is added by engineers only to make the F=ma equation work in what is called a non-inertial or accelerated frame of reference. They can also be referred to as Inertial or Psuedo Forces. It is the same concept as trying to do F=ma for motion in a room on a rotating space station that has artificial gravity due to the rotation. In that frame of reference, the cup of coffee on the table is not accelerating, it is motionless. To make the F-ma equation work, (to make the force F and the acceleration be zero we need to include a fictitious Centrifugal force to oppose and balance the centripetal force. Other forces which fall into this category include Coriolis force and Gravity.

That is exactly what is going on for the diagrams you are looking at for an aircraft in a steady state turn.

What is not addressed, (and rarely explained in basic flight training), is that Newton's rules of motion (F=ma for example), are only valid in an inertial or non-accelerated frame of reference. Unless you are in free-fall (experiencing zero acceleration), you are in a non-inertial or accelerated frame of reference, and some fictitious force must be added to your analysis to compensate for the acceleration of the frame of reference in which you are doing your analysis.

In addition, it must be remembered that everything is relative. Not just velocity. Most of us will agree that all velocities must be measured relative to some reference velocity. How fast is the earth moving? well, there is no correct answer unless you also specify "with respect to what"? What is not generally understood is that the same concept also applies to acceleration. Sitting in your chair in your living room, we are in fact in an accelerating frame of refence (as we feel acceleration), upwards. That is why Einstein's famous "thought experiment" is so illuminating. In it he explains that there is no way that someone can tell the difference between that situation, and being in a room in a spaceship in deep space that is accelerating with a rocket engine thrust at 32 feet per second squared. These two situations are not just similar, they are equivalent, (identical). In both cases you are in a non-inertial, accelerating frame of reference, and a fictitious force must be added to any analysis to compensate for the intrinsic acceleration of the frame of reference to make any F=ma analysis equation balance properly.

Heree is a graphic from Misner, Wheeler & Thorne's Gravitation that illustrates this: enter image description here

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    $\begingroup$ Note: centrifugal force is for most purposes as real as for example gravity. The fact that it appears because of choice of frame of reference does not make its effects less real. (You are actually telling just this in the last paragraph mentioning indistinguishably of gravity force and accelerating-frame forces.) $\endgroup$
    – Martin
    Oct 22, 2023 at 11:10
  • $\begingroup$ Actually, Gravity, as understood in the physics world, is also a "fictitious" force, as it is, according to General Relativity, (now pretty much accepted, verified and considered to be the current "correct" view of what gravity is), is not a force at all, but just our perception of the straight-line path in space-time taken by objects that are not accelerated at all, because we are looking at them in an accelerated, non-inertial frame of reference, exactly as when we might look at the path of a dropped object in a rocket ship accelerating from thrust. $\endgroup$ Oct 23, 2023 at 14:38
  • $\begingroup$ And saying that the effects are more or less real or not real, when the choice of which frame of reference you choose to look at them in totally changes how you see them is a bit foolish. It betrays a lack of understanding of what the issue is. Bottom line, If you can feel the force, it's real, if you can't then it is fictitious. Consider this, the only time we say we can "Feel" Gravity is when we are not accelerating. When we are accelerating due to gravity, (free fall), then we don't feel it. If this doesn't bother you a bit, then you don't understand this. $\endgroup$ Oct 23, 2023 at 14:44
  • $\begingroup$ There are countless videos explaining this on the internet. Here's a few... youtube.com/… youtube.com/… youtube.com/… $\endgroup$ Oct 23, 2023 at 14:47
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    $\begingroup$ And another quick point here. Your intuitive comment about "anything we can directly experience" also betrays the essence of this misconception. You do NOT experience the force of gravity. Even before physicists understood that gravity is not a force, but just the curvature of space-time, when they thought that it was a force, they understood that this gravitational force was not, and could not, be felt by anyone, as it affected every atom in your body equally. (except for the edge case represented by extremely strong tidal forces - also fictitious). $\endgroup$ Oct 27, 2023 at 3:27
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Those diagrams, versions of which have been reproduced in many different flight training "ground school" materials, including some published by the FAA, are extremely misleading, and contain errors and omissions.

Slips and skids are not characterized by an "imbalance" between the pseudoforce called "centrifugal force", and the horizontal1 component of the net aerodynamic force generated by the aircraft.

Rather, slips and skids can be said to be characterized by an imbalance between the pseudoforce called "centrifugal force", and the horizontal component of the wing's lift force. (This isn't the simplest or most intuitive definition of a slip or skid, but it is a valid one.)

Since the pseudoforce called "centrifugal force" is actually just a mirror image of the horizontal component of the net aerodynamic force generated by the aircraft, the above statement is really just an overly complicated way of saying that in a slip or skid, something other than the wing is generating an aerodynamic force which has a horizontal component.

That "something" is the fuselage. In a slip or skid, the airflow is striking the side of the fuselage, which generates a real aerodynamic sideforce, oriented perpendicular to the wing's lift vector. Because the fundamental defining characteristic of a slip or skid is that the nose of the aircraft is not aligned with the actual direction of travel through the airmass, in the yaw axis.

If the aerodynamic sideforce generated by the airflow striking the side of the fuselage were included in the diagrams attached to the question, the horizontal component of the net aerodynamic force generated by the aircraft would be exactly equal in magnitude (and opposite in direction) to the "centrifugal force" vector in every case.

If the vectors labelled "HCL" are supposed to represent the horizontal component of the net aerodynamic force generated by the aircraft, including the horizontal component of the aerodynamic sideforce generated by the airflow striking the side of the fuselage, then they are drawn incorrectly. For a given bank angle, the horizontal component of the net aerodynamic force generated by the aircraft is larger in a skidding turn than in a coordinated turn, and is smaller in a slipping turn than in a coordinated turn. If the vector labelled "total lift" is supposed to represent the total aerodynamic force generated by the aircraft, including the sideforce contribution from the airflow striking the side of the fuselage, then the coordinated turn is the only case where the "total lift" vector should be exactly "square" to the wingspan. In all cases the "centrifugal force" vector should be a mirror image of the vector representing the horizontal component of the net aerodynamic force generated by the aircraft.2

On the other hand, if the vectors labelled "HCL", "VCL", and "Total Lift" are only supposed to represent the horizontal, vertical, and total components of the wing's lift vector, then the diagrams become very confusing because the vector representing the aerodynamic sideforce generated by the airflow striking the side of the fuselage has been entirely omitted. In this case the vectors labelled "centrifugal force" should still appear as described in the above paragraph, but in the slipping and skidding cases, the vector labelled "HCL" will no longer be a mirror image of the vector labelled "centrifugal force". For a given bank angle, the wing's total lift vector, and therefore the horizontal and vertical components of the wing's lift vector, are all slightly smaller in a slipping turn than in a coordinated turn, because the aerodynamic sideforce generated by the airflow striking the side of the fuselage supports a small portion of the aircraft's weight.3 Similarly, in a skidding turn, the wing's total lift vector, and therefore the horizontal and vertical components of the wing's lift vector, are all slightly larger than in a coordinated turn, because the aerodynamic sideforce generated by the airflow striking the side of the fuselage contains an earthward component that must be opposed by the wing's lift vector. These differences should be apparent in the vectors labelled "VCL" and "Total Lift" as well as the vectors labelled "HCL", and the vector labelled "Total Lift" should be "square" to the wingspan in all three cases.

Obviously the diagrams would be greatly improved by changing to an airmass-based reference frame rather than an aircraft-based reference frame, so that the "centrifugal force" vector could be entirely discarded, and by also including by the aerodynamic sideforce vector generated by the air striking the side of the fuselage. There's no need to break things into horizontal and vertical components-- just show the wing's lift vector and the aerodynamic sideforce vector from the airflow striking the side of the fuselage. These two vectors are oriented perpendicular to each other. In a coordinated turn, there is no airflow striking the side of the fuselage, so the aerodynamic sideforce vector is zero, so the net aerodynamic force acts "straight up" in the aircraft's reference frame. In a slip or a skid, the aerodynamic sideforce vector is not zero, and so the net aerodynamic force does not act "straight up" in the aircraft's reference frame. End of story. (If desired, an "apparent load" vector could be added, which would always be the mirror image of the vector sum of the wing's lift vector and the aerodynamic sideforce vector from the airflow striking the side of the fuselage. Only in the case of the coordinated turn, where aerodynamic sideforce is zero, would the "apparent load" vector be exactly "square" to the wingspan. And now-- keeping in mind that the weight vector makes no contribution to the "apparent load" vector-- we understand why the inclinometer ball behaves as it does in slipping, skidding, and coordinated turns. And we've come to this understanding without ever invoking some sort of hypothetical "imbalance" between "centrifugal force" and some other force. It all boils down to the question of whether the airflow is, or is not, striking the side of the fuselage and generating an aerodynamic sideforce vector.)

This concept was also explored in this related, highly upvoted question-- What is missing from these diagrams of the forces in slips and skids?

But as to your specific question:

If in a coordinated turn, the horizontal lift vector is equal to the Centrifugal force. Then how is the aircraft still turning?

Because when we include the pseudoforce called "centrifugal force" in our vector diagrams, we are using an aircraft-based reference frame rather than an airmass-based reference frame or ground-based reference frame. Since the aircraft can't accelerate relative to itself, the net force in the aircraft-based reference frame will always be zero, whether the aircraft is turning or not. In the aircraft-based reference frame, the fact that the centrifugal force vector exists at all is actually evidence that the aircraft is turning. But for teaching purposes, using the aircraft-based reference frame (and therefore including the "centrifugal force" vector whenever the flight path is not linear) is arguably an inferior approach to using the airmass-based reference frame (and therefore not including any "centrifugal force" vector.)

Footnotes:

  1. In this answer, when we talk about "horizontal", we mean as seen looking at the aircraft in a head-on view. We're not referring to fore-and-aft forces such as thrust and drag.

  2. Note that in the diagrams attached to the question, the illustrator chose to make the "total load" vectors in the "skidding turn" and "coordinated turn" identical, and to make the "HCL" (horizontal component of lift) vectors in the "slipping turn" and "coordinated turn" identical. That's all kind of random and makes no sense.

  3. For an extreme case, consider sustained linear "knife-edge" flight at a 90-degree bank angle-- here the aerodynamic sideforce from the airflow striking the side of the fuselage is doing all the work of supporting the aircraft's weight, and the wing's lift vector is zero.

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  • $\begingroup$ Re "In a coordinated turn, there is no airflow striking the side of the fuselage, so the aerodynamic sideforce vector is zero, so the net aerodynamic force acts "straight up" in the aircraft's reference frame." -- maybe "in the aircraft's coordinate system." would be a better way to end this sentence. The reference frame we are using here is that of the airmass, not the aircraft. (Hence no need to include "centrifugal force".) $\endgroup$ Oct 26, 2023 at 8:46
  • $\begingroup$ Regardless of whether we are taking the approach outlined in the 7th paragraph ("If the vectors labelled "HCL" are supposed to represent the horizontal component of the net aerodynamic force generated by the aircraft..."), or the 8th paragraph ("On the other hand, if the vectors labelled "HCL", "VCL", and "Total Lift" are only supposed to represent the horizontal, vertical, and total components of the wing's lift vector..."), it makes no sense whatsoever for the HCL vector in the "skidding" case to be smaller than the HCL vector in the "coordinated" case. It should be larger. $\endgroup$ Oct 26, 2023 at 13:52
  • $\begingroup$ Likewise, at least if following approach outlined in 7th para, in the skidding case the Total Lift vector should be tilted further to the inside of the turn than in the coordinated case, not closer to vertical. And the Total Load vector should reflect this, being a mirror-image of the Total Lift vector. This set of diagrams really is completely flawed. $\endgroup$ Oct 26, 2023 at 13:55
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I'm gonna go out on a limb and suggest that all of these answers seem to be answering a different question than "why does the airplane turn" when the bank is such that the acceleration in the airplane's rest frame points straight down (a coordinated turn). A ball on a string swung over your head moves in a circle without any aerodynamic forces. So there's something else going on.

Part of the problem is conflating turn = "move in a circle" with turn = yaw.

You can have a banked airplane where the wing lift vector points off to one side but the airplane doesn't turn. Every PPL candidate has to demonstrate a slip to landing, which is exactly that - the rudder input exactly opposes the yaw tendency created by the horizontal relative wind component.

What that highlights is that the yaw is caused by the side-force of the relative wind on the tail. The horizontal relative wind at the tail is caused by the horizontal lift component created by the bank. In a constant rate turn the rate of yaw must match the rate of turn, but that doesn't mean the airplane's nose has to be pointed along the tangent of the turning circle.

If the apparent "g" in the airplane frame is not straight down, that will cause a horizontal wind component on the tail which will yaw the plane. If the horizontal "g" points to the inside of the turn, the effective wind will yaw the nose to the inside of the turn & vice versa. That's why you can make a turn without using any rudder - the "g" to the inside will create enough yaw to keep the plane turning. But it's less aerodynamic than if you get all the yaw from having inside rudder input instead of depending on inside "g". So "coordinated turn" = "apparent g straight down".

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  • $\begingroup$ Comments have been moved to chat; please do not continue the discussion here. Before posting a comment below this one, please review the purposes of comments. Comments that do not request clarification or suggest improvements usually belong as an answer, on Aviation Meta, or in Aviation Chat. Comments continuing discussion may be removed. $\endgroup$
    – Ralph J
    Oct 26, 2023 at 21:18
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Newton’s first law. If the forces are perfectly in balance then the aircraft will continue to turn at a constant rate.

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    $\begingroup$ No, that's not true. With forces in balance (as seen from the given reference frame), the movement will continue along straight line with constant speed. $\endgroup$
    – Martin
    Oct 21, 2023 at 20:36
  • $\begingroup$ An object moving at a constant velocity is not accelerating, all forces considered. Notice, in a turn, fuel input (power) must be increased, due to increased drag. But once in the turn, fuel burn is constant. $\endgroup$ Oct 22, 2023 at 0:24
  • $\begingroup$ @Martin ok the principle is more succinctly described by conservation of angular momentum but if you consider the aircraft as a collection of individual components then 3rd law applies; the frame of reference is the aircraft as a whole, it ‘sees’ the universe rotating around it and there’s no resultant force that would change it. $\endgroup$
    – Frog
    Oct 22, 2023 at 20:00
  • $\begingroup$ Anytime you attempt to analyze a scenario within, or with reference to, a non- inertial or accelerated Frame of reference, (one in which a g-meter will show any non-zero value), you must include some non-inertial or fictitious forces that represent the acceleration of the frame of reference itself or Newton's laws will not work - they will no longer be valid. $\endgroup$ Oct 28, 2023 at 12:07

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