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Consider a pilot sitting at the CG of an aircraft, with instruments also located at the CG of the aircraft. (Feel free to consider other cases as a bonus, but the core of the question is aimed at this simple case.)

Consider the apparent force "felt" by the pilot, and measured by a traditional-style panel-mounted G-meter, when the aircraft is upside-down at the top of a loop. Should this force considered to simply be an expression of the wing's lift vector, plus any other aerodynamic force components that the aircraft is generating? Does gravity make a contribution to this force? Does "centrifugal force" make a contribution to this force? Is the force "felt" by the pilot, and the G-meter, due to the combined effects of gravity and "centrifugal force"? Can more than one of these things be true at the same time?

Consider the apparent force "felt" by the pilot, and determining the deflection of the inclinometer (slip-skid ball), during a coordinated turn, or during an uncoordinated turn. Is this force simply an expression of the net aerodynamic force generated by the aircraft? Does gravity make a contribution to this force? Does "centrifugal force" make a contribution to this force? Is the force "felt" by the pilot, and the inclinometer, due to the combined effects of gravity and "centrifugal force"? Can more than one of these things be true at the same time?

Feel free to substitute "acceleration" for "force" in your reading of the question, if you feel it makes it a better question.

Basically this question is seeking to reconcile the view that a pilot only "feels" the aerodynamic force generated by an aircraft, and the view that gravity and / or centrifugal force also play a role in the forces "felt" in flight. Can these two views be reconciled? If so, how? Or is only one of these two views correct?

A thought problem might that better inform the question -- consider an astronaut in orbit. Is he or she feeling both centrifugal force and gravity? Or neither? Or is this purely a manner of convention, i.e. either answer is accurate?

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Can more than one of these things be true at the same time?

Yes. It is only a matter of the point of view, or, more precisely, the reference system.

Consider a trivial case first: The aircraft sits on the ground, engine off. The pilot will feel his weight pressing into the seat even in the absence of aerodynamic forces. What he feels is gravity, and the force is the product of his mass and gravitational acceleration.

Now do the same with the aircraft flying in level flight. The forces acting on the pilot are still the same, so it is again gravity what he feels. Only that now the aircraft is not supported by its landing gear and the ground, but by wing lift (and the ground again).

Next step: The pilot flies a dive, followed by a 2g pull-up, meaning that the g-meter reads 2 gs. Again we look at the moment when the aircraft is level, but now it flies on a curved flight path. This adds a centrifugal load of 1g to the 1g of gravity and the pilot feels twice the weight force that he felt before. The nitpickers will now say this is not weight, because weight is the product of mass and gravitational acceleration, but our pilot couldn't care less at the moment. It hurts in his buttocks, that's what counts.

And neither the g-meter nor the pilot could tell gravity apart from centrifugal force – they all feel the same and only the sum of all is felt as a single sensation.

Now switch reference systems and look at the flow around the wing. Speed and angle of attack combine to produce a lift force of twice the weight the aircraft had in level flight. This is pure aerodynamics, no gravity or centrifugal load involved. Strictly speaking, the aerodynamic force on the wing is even a bit higher in order to compensate for the higher tail downforce resulting from trim to overcome pitch damping. Looking at the full aircraft as one object, its total lift force is now twice as high as in level flight.

This force gets transmitted to the pilot seat and the instrument panel and is felt there as a resistance against the acceleration. Depending on your point of view, either the pilot is accelerated into his seat or the seat pushes against the pilot. Actio equals reactio.

Next switch: The same curved flight path, but now at the moment when the aircraft is inverted. Now centrifugal and gravitational acceleration cancel each other, the g-meter reads zero and the pilot feels weightless. In order to do so, the pilots needs to push the stick forward and reduce wing lift to about zero, or wing lift would pull the aircraft down. At this moment we have no aerodynamic contribution to the vertical forces felt by both the pilot and the g-meter.

Now the pilot pushes the stick even more forward so his flight path becomes straight and level again, only with the aircraft in an inverted position. The g-meter now reads -1 g, the force of the harness pressing on the pilot's shoulders is again his weight times gravitational acceleration, but what allows the harness to support the pilot is the lift of the wing the harness is attached to. Change your reference system, and it is again aerodynamics that the pilot feels. Let him wiggle the stick and the change in aerodynamics is immediately felt on his shoulders.

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  • $\begingroup$ @quietflyer: No problem, I'm open for any suggestions. My only concern is the ever growing length of my answers … $\endgroup$ – Peter Kämpf Apr 19 at 4:59
  • $\begingroup$ All above is based on Newton.. .But Newton was wrong. Using Newton to explain aircraft motion only makes the issue unnecessarily complicated. The only forces we "feel" in an aircraft are the forces generated by the engines, and by the aerodynamic pressure of the air on the surface of the airframe. Everything else is engineering self-abuse done to explain away the math because we are trying to measure everything in a non-inertial (accelerated) frame of reference. $\endgroup$ – Charles Bretana Apr 25 at 2:43
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    $\begingroup$ If you believe you can actually feel the force of gravity, then please explain why the feeling goes away when you push forward on the stick and unload? Do you believe that gravity actually has been turned off by the forward motion of the stick? Is it not so much simpler that what you "felt" was simply the lift on the wings, (not Gravity). And when it goes away, you no loner feel it. $\endgroup$ – Charles Bretana Apr 25 at 2:46
  • $\begingroup$ @Charles Bretana, what we feel as gravity is actually resistance to gravity. I.e., when something is stopping us. A body in free fall is certainly experiencing gravity, but it feels like weightlessness. $\endgroup$ – Michael Hall Apr 25 at 5:40
  • $\begingroup$ @Charles Bretana yes, the earth and the moon rotate around the sun, which rotates around our galaxy, which is moving away from other galaxies.... But everything on earth acts in the earth frame of reference. The effect of gravity is like acceleration, but, in the earth frame, at 1G, we are not accelerating relative to earth. A full explanation of "gravity" awaits us all, but in an airplane it is simply balanced with vectors of acceleration. $\endgroup$ – Robert DiGiovanni Apr 25 at 6:21
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It's the acceleration of the object laterally due to a change from a straight path to a curved one. You're just along for the ride. If you are spinning a ball around on a string on a vertical plane, I think you could look at it as the lift vector being the string. As the ball passes the bottom of the arc, you have centrifugal force (the ball being forced to move in an arc) plus gravity applying tension on the string; the string, or the lifting force, is subjected to the weight of the ball plus centrifugal force imposed by moving in an arcing path. Both the plane and you as pilot feel your weight plus the magnification of your weight due to the acceleration created my moving in an arc, when at the bottom of the loop, and feel (and the plane's structure subjected to), typically 3+ Gs down (Gravity + acceleration off a straight path adding to it).

And at the top, it's centrifugal force MINUS gravity. In the spinning string/ball, if the rotation is too slow, centrifugal force becomes less than gravity, the string goes slack and the ball falls. In the airplane, if the rotation is too slow, or you don't hold the correct arc with the elevator, you go zero G, the wings aren't making any lift at all (slack string) and may fall out of the top of the loop if you try to make the wing fly inverted to hold the loop's arc (this is how pilots get into inverted spins), or at minimum your loop becomes D shaped and the judges mark you down or the airshow crowd boos.

So what the pilot feels is gravity, plus or minus the acceleration created by moving in a non-linear path. At the bottom of a loop, apparent gravity is below, but magnified. At the top of the loop, apparent gravity is above, because lateral acceleration imposed by the arcing movement is more than gravity (at the top of a loop you normally feel maybe one quarter to half a G because the centrifugal acceleration is about 1.2-1.5G total; if you barely make it over the top and end up at only 1G of centrigual acceleration, it's fully cancelled out by gravity and you will feel weightless and your're right on the edge of falling out - best not push especially if you don't have an inverted fuel and oil system).

When skidding, it's just the same acceleration applied laterally, which sums with gravity to provide an new "apparent" gravity. Spin the ball on the string horizontally, but slowly so that the string's movement describes a cone. The angle of the string is the "apparent" gravity felt by the ball, and is also what your perceive as "down" when skidding in an airplane. Bank the plane to an angle appropriate to the rate of change in direction, and now the apparent gravity is aligned with the airplane's vertical axis and the glass of water on the instrument panel sits level to the panel as if you were stationary and everything is right with the world.

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    $\begingroup$ "at the top of a loop you normally feel maybe one quarter to half a G" – well, I always try to round the loop at the top instead of flying a flight path looking like a standing egg. When I did this in an open cockpit Stearman once, the flight instructor in the back seat desperately cried "pull, pull!" because he forgot to close his pockets. I don't mind the negative gs at the top of the loop. $\endgroup$ – Peter Kämpf Apr 18 at 14:37
  • $\begingroup$ If you're doing loops in a clip wing Cub with the fuel tank and wire gauge in the nose, and you push at the top to keep it round, you get a gasoline windshield wash to boot. $\endgroup$ – John K Apr 18 at 16:35
  • $\begingroup$ Not every airplane is made for this – I love to do parabolas and let a coin hover over the palm of my hand. In some cases, the engine sputters and stops when doing this. $\endgroup$ – Peter Kämpf Apr 18 at 19:28
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The answer lies in a breakdown of the term aerodynamic. It may be obvious, but I will say it anyway: "Aero" is the Latin word for air.

So is it possible for the human body to perceive an aerodynamic force? Absolutely. Leaning into a gale force wind, sticking your hand out the car window, skydiving, even the sensation of a light breeze across your face are all examples of feeling an aerodynamic force.

Absent direct contact with outside air though, other perceived forces in a moving vehicle are not, (by definition) aerodynamic. When you are in an airplane, (or car, or train...) you are not directly feeling aerodynamic force, other than from the vents. You are effectively sealed off and shielded from the outside forces acting on the airframe.

That's why you can walk down the aisle of an airliner to use the lavatory without fighting against 500 knots of drag.

Gravity is the primary force perceived by the human body that we are most familiar with. In a moving object, an additional force may be perceived when the direction of travel is changed. What you are feeling is your momentum being restrained. This is called centrifugal force, and anybody who has ridden in a car has felt it going around a corner.

Many would argue that centrifugal force isn't a "real" force. From an engineering standpoint I accept this, and am not going to argue that point here. But it is easily perceived, demonstratable, and offers the best explanation of G force to the lay person.

And it is real enough: It is real enough to keep that water inside the bucket you swing over your head. It is real enough to keep your body pinned to the wall when the floor drops on the spinning Gravitron ride at the county fair. It is real enough to be misperceived as (vertical) gravity in a shallow turn when visual cues are absent. And it is real enough to cause the fighter or acrobatic pilot to black out at 7 Gs if countermeasures are not used.

If it enhances your understanding of "real" forces to negate centrifugal force for computation purposes then please disregard my explanation. But if you are trying to relate what pilots feel in the aircraft to more common everyday experiences, then maybe this answer will offer some simple clarity.

In summary, the aerodynamic forces the aircraft generates produce the centrifugal force that, (added to earth's gravity) the body experiences.

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  • $\begingroup$ The specific problem with this answer is the last sentence. The net centrifugal force is not the result of only the aerodynamic forces. For why this is relevant see chat comments chat.stackexchange.com/transcript/message/54138356#54138356 and chat.stackexchange.com/transcript/message/54138386#54138386 . $\endgroup$ – quiet flyer Apr 20 at 13:22
  • $\begingroup$ The body feels earth's gravity, AND any lateral Gs produced by the aircraft turning. You do not subtract out gravity. If there is something factually incorrect, or if you have a better way of phrasing that last sentence then please offer a concise suggestion. But I am not going to let this drag into another protracted circular discussion with you. My whole intent with this answer was to steer away from overthinking something that is fundamentally simple and relatable to other things non-pilots are able to observe. In other words, no math. $\endgroup$ – Michael Hall Apr 20 at 15:01
  • $\begingroup$ If the last sentence of the answer were corrected to be true (i.e. centrifugal force is produced by aerodynamic effects PLUS GRAVITY) , it would show that saying the body feels centrifugal force plus gravity is inherently circular and while not false, is really just equivalent to saying the body feels only the aerodynamic effects. Explained in more detail in chat permalinks above. $\endgroup$ – quiet flyer Apr 20 at 15:15
  • $\begingroup$ From a practical standpoint the following are equivalent: "the aerodynamic forces the aircraft generates produce the centrifugal force that, (added to earth's gravity) the body experiences" = "centrifugal force is produced by aerodynamic effects PLUS GRAVITY". When the floor drops on the Gravitron ride the horizontal centrifugal force pins you to the wall, but you still feel vertical gravity from the earth. $\endgroup$ – Michael Hall Apr 20 at 15:49
  • $\begingroup$ ... and is was just the kind of nit-noid circular discussion I was hoping to avoid. Please, let's stop saying essentially the same thing in subtly different ways. $\endgroup$ – Michael Hall Apr 20 at 15:50
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The sum of inertial forces in the body frame of the aircraft is equal magnitude and opposite direction of the sum of all external (which for aircraft means aerodynamic forces and thrust) forces acting on the aircraft.

  • Per second law of motion, $a = \frac{F}{m}$. Since the aircraft is not accelerating in a reference frame attached to itself, the external and inertial forces must be balanced in it. They act as action and reaction pair regarding objects inside the aircraft like the slip-skid ball or the pilot.
  • Per general relativity, gravity is locally indistinguishable from acceleration of the reference frame and behaves as an inertial force for all purposes. Therefore external forces means just the aerodynamic forces and thrust (propeller thrust can be included in aerodynamic forces, but rocket thrust shouldn't be and for jet thrust it's somewhat unclear).

So saying you feel the aerodynamic forces or you feel gravity and centrifugal forces is equally valid and only a matter of point of view.

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  • $\begingroup$ "Gravity is indistinguishble", it is certainly included in the "4 forces" to yield 0 acceleration steady state. "Thrust", no need to discriminate, it is a force acting on the aircraft for force vector analysis, particularly, as in many aircraft, it is angled down and to the right. Let's not forget p forces for props, and, yes, gyroscopic forces (especially for single engine jets). $\endgroup$ – Robert DiGiovanni Apr 18 at 22:46
  • $\begingroup$ @RobertDiGiovanni, the principle of general relativity states: “effects of uniform gravity are indistinguishable from effects of acceleration of the frame of references”. That is the defining postulate of General Relativity, in its entirety. It means that the role of inertial reference frames from Newtonian mechanics goes to reference frames in free fall, and gravity should be treated as an inertial force (which are no longer called fictitious, because general relativity talks about all reference frames, and in those where they exist the inertial forces are real). $\endgroup$ – Jan Hudec Apr 18 at 23:27
  • $\begingroup$ @RobertDiGiovanni, in the reference frame of Earth, the gravity is obviously included in the “4 forces” that yield 0 acceleration in steady state. But the reference frame of Earth is in constant acceleration upward at $1g$ (it's actually gravity, but it is equivalent), and an accelerometer measures that $1g$ in it. $\endgroup$ – Jan Hudec Apr 18 at 23:31
  • $\begingroup$ @RobertDiGiovanni,… That reference frame is also rather uninteresting in this discussion, because it has some, but not all of the inertial forces. Either we use the body reference frame, where inertial forces—including gravity, because gravity is an inertial force—balance external forces, or we use the free-falling reference frame, where there are no inertial forces, but the aircraft is accelerating upward at $1g$ in it most of the time. Many things are simpler with the right choice of reference frame. $\endgroup$ – Jan Hudec Apr 18 at 23:35
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    $\begingroup$ @RobertDiGiovanni, yes, IN THE EARTH FOR (Frame of Reference), "what you are left with is gravity". But you do NOT feel this. What you feel is the force of the lift on the wings. "Gravity" is the result of the frame translation tensor you would use to Translate answers derived on one FOR to another FOR when the two FORs in question are accelerating with respect to one another, To make things balance you need to apply an artificial "force" (which is unfelt) to make things balance. Take a look at en.wikipedia.org/wiki/Inertial_frame_of_reference $\endgroup$ – Charles Bretana Apr 26 at 15:02
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The only forces felt by a pilot are the aerodynamic forces from the pressure of the atmosphere on the surface of the aircraft, and the thrust of the engines. Contrary to generally accepted concepts, Gravity is not a “force”. It is just an artificial construct we use to make the Math work out because we are generally measuring the motion of things (like an aircraft), in an accelerated frame of reference (FOR). In fact, if the aircraft was in free fall (Zero AOA and zero Lift) then the pilot would feel only the force of the engines and the force of aerodynamic form drag. When in “Level” flight, the pilot feels an additional “G” of force pushing on his butt because the airframe is generating additional Lift (an aerodynamic force perpendicular to the flight path vector and the earth) sufficient to maintain a constant altitude. In a turn, the pilot does not feel “Centrifugal” force. “Centrifugal” force is another artificial construct or "fictitious force" we use because of measuring things in non-Inertial (accelerated) FORs. A rotating FOR is an accelerated FOR. What the pilot feels is the additional aerodynamic force on the airframe necessary to turn the aircraft (change it’s direction of motion) If you roll into a bank and do not pull harder on the stick/yoke, you will not generate any additional lift and you will not feel any additional force. Explaining this away by resorting to “Centrifugal” force, and then thinking about it as balancing some other fictitious “Centripetal” force just adds unnecessary complexity. All forces you feel are due to the aerodynamic forces pushing the surface of the airframe (and the engine).

In fact, if you read a bit about Gravity, you will learn that Newton's theory of Gravity, where it is represented as a "force", although it generates almost correct answers in 99.999% of the cases we are familiar with, is in fact absolutely wrong. Einstein showed, and proved, that gravity is the effect of mass distorting the curvature of space-time, in such a way as to cause everything traveling relative to an inertial (non-accelerated) frame of reference to "appear" to travel in a curved path.

In the FOR of the earth, "what you are left with is gravity". But you do NOT feel this. What you feel is the force of the lift on the wings. "Gravity" is the result of the frame translation tensor you would use to Translate answers derived on one FOR to another FOR when the two FORs in question are accelerating with respect to one another, To make things balance you need to apply an artificial or fictitious "force", (which is unfelt), to make things balance. Take a look at Inertial_frame_of_reference. In particular, look at the third paragraph, to wit:

In a non-inertial reference frame in classical physics and special relativity, the physics of a system vary depending on the acceleration of that frame with respect to an inertial frame, and the usual physical forces must be supplemented by fictitious forces.
* My Italics *

and the last sentence:

Another example of such a fictitious force associated with rotating reference frames is the centrifugal effect, or centrifugal force.

One of the coolest representations of this concept is in a book about Gravitation and shows how the paths of a thrown baseball, a bullet, and a beam of light, traveling from one point to another, will have exactly the same curvature when graphed in space-time, (in a one-G accelerated frame of reference),as opposed to their path in ordinary 3-D space. If these were all graphed in a zero-G (inertial) frame of reference (Free-Fall) they would all appear to be traveling in an absolutely straight line.

enter image description here

Bottom line. All of the complexities and fictitious artificial "forces" we use to explain the forces acting on objects in motion are only necessary because we are measuring motion in some linear or rotationally accelerated frame of reference. We need to come up with additional fake "forces" like gravity, and centrifugal force, to explain things because we are attempting to measure acceleration in a frame of reference (the earth) that is ALREADY ACCELERATING! The only real forces acting on the aircraft are the force of the air pushing on the aircraft (aerodynamic forces) and the thrust of the engine[s].

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – Federico Apr 28 at 10:31
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Throughout this answer we'll use "feel" or "felt" to mean the force or acceleration that is perceived by the pilot and transmitted through the aircraft structure as stresses or strains. In a "weightless" situation, the net force or acceleration is not zero, but the "felt" force or acceleration is zero.

It is valid to argue that the force or acceleration "felt" by the pilot, and registered by the instruments, in flight is simply the net aerodynamic force or acceleration generated by the aircraft. To get the actual net force acting on the aircraft, we have to add gravity to the aerodynamic forces. In other words, the "felt" force or acceleration is the net force or acceleration acting on the aircraft, minus the force or acceleration due to gravity.

Examples--

1) Straight and level upright flight-- net acceleration zero, net force zero, "felt" acceleration 1-G in the upward direction, force "felt" by the pilot is simply equal to his weight, acting in the upward direction. This force is the aerodynamic lift force generated by the wings, transferred through the aircraft structure to the bottom of the seat and then to the pilot's body.

2) Aircraft inverted at the top of a loop-- again the acceleration and force "felt" by the pilot are simply the result of the net aerodynamic force generated by the aircraft. If the wings are generating positive lift, the pilot and G-meter will feel positive G's, and if the wings are generating negative lift, the pilot and G-meter will feel negative G's. If the wings are generating zero lift, the pilot and G-meter will feel zero G's (weightlessness).

3) Steady-state coordinated turn-- the net aerodynamic force generated by the aircraft is simply the wing's lift vector, which contains no lateral (sideways) component in the aircraft's reference frame. The slip-skid ball is centered. Meanwhile the net force including gravity is the purely horizontal force vector-- the centripetal force vector-- that drives the turn. This force vector does have a lateral (sideways) component in the aircraft's reference frame, yet the slip-skid ball is centered.

4) Turn with some sideslip-- the nose of the aircraft is allowed to point toward the outside of the turn. The airflow impacts the side of the fuselage and generates an aerodynamic sideforce. Now the net aerodynamic force generated by the aircraft does include a lateral (sideways) component, so the slip-skid ball rides toward the inside of the turn, and the pilot's body tends to lean toward the wall of the cockpit that is on the inside of the turn.

Yet if we take our airplane and bolt it to a roller coaster and run it through loops, flat turns, etc, the pilot still "feels" apparent forces, and the G-meter and the inclinometer still give indications. What are causing these apparent forces? (Assume for simplicity that the roller coaster is in a vacuum.)

We can say that these apparent forces are entirely due to the forces exerted by the rails of the roller coaster upon the wheels of the roller coaster, with gravity playing no role. This is analogous to saying that the forces "felt" in flight are entirely due to the aerodynamic forces created by the aircraft, with gravity playing no direct role.

Yet we can also say that these apparent forces are entirely due to the combined effects of gravity, and of "centrifugal force". Obviously, if we don't start a loop in the roller-coaster with enough speed to maintain positive G's all the way around, we'll hang upside-down in the seatbelts as we go over the top.

The key to reconciling these viewpoints is to understand exactly what "centrifugal force" is. It is the apparent force created by a curvature in a vehicle's trajectory. It is not a real force. It is basically just the mirror image of the real forces that are acting to curve a vehicle's trajectory. For (almost) every centripetal force acting to drive a curvature in a vehicle's trajectory, we can say the occupants "feel" an apparent "centrifugal force" acting in the other direction. When we do a high-speed turn to the left in a car on flat ground, the tires generate a real force toward the left, and we say we perceive a centrifugal force to the right, and are thrown against the right wall of the inside of the car. If we want to understand which direction a slip-skid ball would deflect, we could consider the real centripetal force from the tires, or the apparent "centrifugal force" generated by the resulting turn rate, but not both. If we consider both forces, we end up with zero net force-- they would cancel each other out-- predicting that the slip-skid ball would not deflect at all. This would be wrong. We can't mix and match apparent "centrifugal force" with real centripetal force. We have to use one or the other.

In this context, we're not using "apparent" to mean "the felt component of". We're just using "apparent" to remind ourselves that "centrifugal force" doesn't actually exist.

We stated that "For (almost) every centripetal force acting to drive a curvature, we can say the occupants 'feel' a "centrifugal force" acting in the other direction". Why "almost"? Because we can't "feel" any centrifugal force that is caused by a curvature in a vehicle's trajectory that is due to gravity. (Consider the forces felt by an astronaut in orbit.)

Why is gravity such a special force? Forces from wings and forces from wheels must be transmitted through the aircraft structure and to the exterior of the pilot's body. As a result, stresses and strains are created in the aircraft structure and in the pilot's body. In contrast, gravity exerts an equal acceleration per unit mass on every molecule of the aircraft structure, and on every molecule of the pilot's body, creating no stresses or strains, i.e. no displacement of one part of the structure or body relative to another part. Likewise, unless we consider the outer housing of the slip-skid ball or the G-meter to be rigidly fixed in space, gravity has no tendency to displace any element of either of those instruments in relation to any other element. For example, gravity has no tendency to pull the inclinometer ball to one side within the glass tube, even if the tube is tilted relative to the earth. (We're assuming that our instrument is much smaller than planetary in scale, so that tidal effects may safely be ignored!)

We've defined "centrifugal force" as an apparent force that is the mirror image of the net centripetal force acting to drive a curve in a vehicle's trajectory. If the force of gravity is not "felt", then what is the "felt" component of that apparent "centrifugal" force?

Since gravity is already "built in" to the net centripetal force determining a vehicle's trajectory, it follows that an apparent upward force opposite to gravity is already "built in" to the apparent "centrifugal" force that is paired with that centripetal force. This means that the "felt" centrifugal force is equal to the total centrifugal force plus gravity. Just as the "felt" centripetal force is equal to the total centripetal force minus gravity.

(Note-- it's really only the component of gravity that acts perpendicular to the flight path that is "built in" to the centripetal force vector. For simplicity, in the discussion of "centripetal force" in this answer, we'll assume that we're looking at the vehicle's trajectory at a point where the trajectory is exactly horizontal rather than rising or falling, and thus the full weight vector acts perpendicular to the flight path. Alternatively, we could think in terms of a three-dimensional vector sum rather than a two-dimensional one.)

The key point here is not to argue about whether "centrifugal force" actually exists or not. Rather the key here is to recognize that we're starting from a situation where the exact trajectory is known, and then we're calculating the centripetal (or centrifugal) force that is involved in creating this trajectory, and then we're subtracting away (or adding) gravity to get the "felt" component of that centripetal (or centrifugal) force.

In essence, "centripetal force" is just another way to say "net force", except that we're discarding components that act tangential rather than perpendicular to the flight path. In the context of flight, the aerodynamic forces and the force of gravity are both built in to the concept of "centripetal force". "Centrifugal force" is an apparent force or pseudoforce that is equal and opposite to the real "centripetal force".

In the case of a roller coaster, the "felt" component of the centripetal force would be the force exerted by the tracks on the wheels, and the "felt" component of the centrifugal force would be equal and opposite. In the case of an aircraft, the "felt" component of the centripetal force would be the aerodynamic force created by the aircraft, and the "felt" component of the centrifugal force would be equal and opposite. Whether we want to say that the pilot, G-meter, and inclinometer are responding to the "felt" component of the centripetal force or the "felt" component of the centrifugal force is really just a matter of convention, but only the former actually exists.

So all these three things are true:

1) "Felt" force = aerodynamic force = net force minus gravity

And

2) "Felt" force = net centripetal force minus gravity

And

3) "Felt" loading (the sensation of apparent weight due to being accelerated by a force) = ( net "centrifugal" force plus gravity ) / mass

We're taking the convention here that the felt "loading" acts in the opposite direction to the felt "force" or felt "acceleration", just as the apparent "centrifugal" force acts in the opposite direction to the actual centripetal force.

Whether we want to work in terms of "centripetal" or "centrifugal" forces is really just a matter of convention. Only the "centripetal" forces are actually real. To get the "felt" component, we subtract gravity in the former case, and we add gravity in the latter case. Because of the way that gravity can't actually be "felt", yet it is already "built in" to the centripetal and "centrifugal" forces.

Note also that by specifying "centripetal" (or "centrifugal"), we're saying that we're limiting the force and acceleration components of interest to those acting perpendicular (orthogonal) to the flight path, and disregarding those acting parallel (tangent) to the flight path. The first equation in the list of three above does not have that constraint.

Often one sees diagrams in flight training manuals that try treat "centrifugal force" as just one component out of several forces acting to curve the flight path. Many diagrams have been published, including by the FAA, that illustrate the deflection of the slip-skid ball in turning flight as being due to a force balance between "centrifugal force", and "gravity", and lift, or between "centrifugal force" and lift. (Oddly, the real aerodynamic sideforce caused by the fuselage flying sideways through the air is invariably omitted.) At first this seems intuitively right-- if you are bringing gravity into the picture rather than just purely considering centrifugal force alone, then why not bring lift into the picture too? But this approach just doesn't work. The effect of the lift vector is already included in the centrifugal force vector. To explain the forces "felt" in a coordinated or uncoordinated turn, we need to work in terms of the real aerodynamic forces, or we need to work in terms of the balance between "centrifugal force" and gravity, which is essentially the same as working in terms of the balance between centripetal force and gravity. A diagram combining gravity, centrifugal force, and lift just can't hold water.

If the flight path is fully constrained, then the "centrifugal force plus gravity" (or "centripetal force minus gravity") approach works well. For example, if the flight path is known to be linear, then centrifugal force is known to be zero, and the slip-skid ball works as a bubble level, telling us which way is down.

But in many situations, the flight path is not constrained to be linear. Also, unlike the case of the roller coaster, the turn radius is free to vary as the lift force is varied, and the flight path is free to curve up and down in the vertical plane as well. That is why it is so useful to take the viewpoint the forces "felt" in flight are simply the real aerodynamic forces generated by the aircraft, with gravity and "centrifugal force" playing no direct role.

Consider for example this related ASE answer about why a glider experienced a negative G-load at the top of the loop. If we imagine the loop to be constrained to follow a pre-defined track with as set radius like a roller coaster, we could imagine that the explanation is simply that we didn't start with enough speed to generate enough "centrifugal force" to counteract gravity and "make it all the way around" without hanging loose in the belts. But recognizing that the flight path is not constrained to follow a pre-defined track, it is more useful to recognize that when we "felt" a negative G-loading, the wing must have been flying at a negative angle-of-attack and generating negative lift. Which raises the question, "Why was the wing generating negative lift-- with the stick full aft? Why didn't the angle-of-attack stay positive, as the pilot was intending to command, even if the resulting trajectory did not result in a perfectly round loop?" The negative-lift situation is in fact related to the low airspeed at the top of the loop, but for reasons that are much more subtle than the fixed-track roller-coaster analogy would suggest.

The idea that a coordinated turn features a "balance" between centrifugal force and gravity may be technically true, but it has no explanatory power, especially in the context of a flight path that is free to vary in three dimensions as lift and other variables are varied, rather than being constrained to run along a pre-existing track as in the case of a roller coaster. It doesn't reveal that the key factor in a "coordinated" turn is simply that the fuselage is streamlined to the airflow, so that the aircraft is generating no aerodynamic sideforce and thus has no aerodynamic force component in the lateral (sideways) direction in the aircraft's reference frame. It can lead one to believe that in a sideslip, an abnormally low turn rate is somehow causing the aircraft to slide sideways through the air, rather than understanding that the aircraft is being forced or allowed to fly sideways through the air which is creating aerodynamic sideforce (as well as drag), which is in causing a reduction in the turn rate. It can lead one to believe that a pilot should somehow be making adjustments in angle-of-attack to get the "right" lift force for the airspeed and bank angle, or else the aircraft will go sliding off toward the inside of the turn. (Footnote to be added.)

In short, the "centrifugal force plus gravity" approach is often not the most useful way to look at the problem when the flight path is free to vary rather than constrained to be some particular shape like a straight line, or a perfectly circular loop of pre-defined radius, or a round horizontal circle with a pre-defined turn radius and bank angle or airspeed. Unless we are back-engineering the "felt" forces (i.e. the aerodynamic forces) from a saved track log that was recorded with tremendous precision, the "centrifugal force plus gravity" approach doesn't have much application. To know what the "centrifugal" force is, we have to know what the trajectory is, and to know what the trajectory is, we usually have to know the forces acting on the aircraft, including the aerodynamic forces, in which case we already know the answer to question of what will be the result of "centrifugal force plus gravity".

The "centrifugal force plus gravity" approach is generally better suited to understanding the forces felt in race cars travelling along tracks of fixed bank angles and fixed turn radii, or the angle taken by a rope of fixed length tied to bucket full of water swung round in a circle, than to the forces "felt" in an airplane whose path is free to vary in three dimensions. But when applied properly, it is not an invalid way to look at the mechanics of flight.

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  • $\begingroup$ Sorry, tried to answer in chat, blanked out again. Easy to fly a constant G circle in the horizontal. In the vertical loop, you would have to vary speed and in a perfect circle, which would involve variable thrust. Notice the difference, in the horizontal circle, the gravity vector does not change, in the vertical, one hangs from the harness at top without pulling. $\endgroup$ – Robert DiGiovanni Apr 18 at 18:51
  • $\begingroup$ See also chat room for this topic -- chat.stackexchange.com/rooms/107177/… $\endgroup$ – quiet flyer Apr 25 at 12:00
  • $\begingroup$ No such thing as "actual" force!!!!- all forces are the result of a calculation done in some specific frame of reference. And the answer you will get from the calculation depends completely on which frame if reference you do the calculation in. For measurements of forces, the accelleration that the frame of reference is experiencing will change the answer. Its exactly analogous to calculating your position.... Until you specify the frame of reference, no answer is meaningful. Asking where are we? And getting the answer 90 miles, tells you nothing. You have to know where position 0 is. $\endgroup$ – Charles Bretana Apr 25 at 23:21
  • $\begingroup$ For measurements of forces, the acceleration that the frame is experiencing will shift the answer you get from any force calculation. And because the frame we.commonly use (fixed to the earth), is not a zero-G inertial frame, it is accelerating upward at 32ft/sec2, any measurements are shifted by this amount. It's exactly the same as if you measured it in a rocket in outer space accelerating at one G. $\endgroup$ – Charles Bretana Apr 25 at 23:26
  • $\begingroup$ @CharlesBretana , you have a lot of good insights and I don't necessarily disagree with anything you say. I might be just tackling the problem from a little lower level, more aligned to "common sense" or the forces that we seem to perceive in everyday life, than you are. $\endgroup$ – quiet flyer Apr 25 at 23:28

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