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I've reviewed quite a few questions on this forum including

Forces in a slipping turn

is-this-vector-diagram-of-the-forces-at-play-in-turning-flight-correct

what-does-the-balance-ball-actually-indicate

but I don't believe they have answered this specific question.

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    $\begingroup$ Btw, if this pertains to helicopters, the sideforce from the tail rotor is significant and can cause the slip-skid ball to be off center even when the yaw string is perfectly centered (fuselage streamlined to airflow). Pertains to the paragraph in my answer starting with "All the above is really just a first approximation, but it is a very good approximation for most situations in actual flight." So it appears that in a helicopter in cruising flight, the most efficient situation is to have the yaw string centered but slip-skid ball slightly off center, meaning that the a/c is slightly banked. $\endgroup$ – quiet flyer Apr 4 at 13:17
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    $\begingroup$ Of course it goes without saying that any time the flight path is linear, the slip-skid ball doubles as a bank angle gauge. $\endgroup$ – quiet flyer Apr 4 at 13:21
  • $\begingroup$ PS I'm not a helicopter pilot, just thinking through the physics-- $\endgroup$ – quiet flyer Apr 4 at 13:30
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It is the same thing that makes you slide sideways in your seat going around a corner in your car.

Or the same thing that would cause you to slide sideways the other way if you were driving straight ahead on a side hill.

Don't overthink it... the ball simply responds to the same forces you feel.

ADDENDUM:

“I agree the comparisons are spot on if you are only considering a coordinated turn, but I'm asking about a steady level turn in which the aircraft is slipping/skidding; which I don't think the comparison can easily be extended to?”

I actually think it is the other way around. Consider the following:

CAR - Unless it is travelling on a banked racetrack, at precisely the speed the banked turn was designed for, a car will never be able to make a “coordinated turn”. There will generally always be a centrifugal sideforce.

PLANE - Once adverse yaw at the turn initiation has been countered, and when established in a constant angle of bank turn, most (positively stable) aircraft will default to a coordinated turn.

In case it helps further, let’s take the aircraft “uncoordinated turn” to a couple logical extremes:

  1. Wings Level, unbanked turn, using rudder only. Think about this as it compares to turning a car on level ground. This is an exaggerated skidding turn, with all the centrifugal force pulling you (and the ball!) horizontally, or sideways.

  2. Straight ahead, level flight in a side slip. The aircraft nose is tracking straight ahead and not turning, while the wings are banked. This is an exaggerated slipping turn, (minus actually turning...) and is just like the example of driving your car on a side hill. In this case, gravity is pulling you (and the ball!) vertically straight down, which from the aircraft's frame of reference, is a bit sideways.

The ball is a "dumb" instrument. Nothing more than ball in a tube reacting to centrifugal force and gravity. You could accomplish the same thing by holding a full glass of water between your knees: If the inside knee gets wet it is a slipping turn, if the outside knee gets wet it is a skidding turn.

If these explanations are inadequate the only thing left to suggest is that you find a local flight school that offers orientation flights and ask the pilot to demonstrate for you.

If that suggestion isn't practical or affordable, try this simple experiment: Get a round mixing bowl and a marble. The round bowl simulates the curve of the glass tube. Put the marble in the bowl and hold it at arms length and begin turning in place. (if you think you might get dizzy and fall against the furniture, wear a helmet!) Your goal is to adjust the angle of the bowl to keep the marble in the center. Note that if you tip the bowl too much either way the marble will start to roll up the sides.

It isn't a perfect comparison because it doesn't simulate yaw, but can you appreciate the correlation? Please let me know if any further explanation is needed...

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  • $\begingroup$ Thanks for the reply @Michael Hall! But I don't see how this answers my question as the dynamic of a car (ground friction) are different than the dynamics of an airplane (aerodynamics)? $\endgroup$ – eball Apr 2 at 13:11
  • $\begingroup$ True that some "dynamics" are different, but other fundamental forces are identical. This is one of them, and the comparison is entirely valid. Centriifugal force and gravity, that's it. As I cautioned, don't overthink this just because it is an airplane, and because other explanations include graphs and diagrams... $\endgroup$ – Michael Hall Apr 2 at 15:01
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    $\begingroup$ @eball Replace ground friction with the side force from a tilted wing and the comparisons are spot on. +1 $\endgroup$ – Peter Kämpf Apr 2 at 17:00
  • $\begingroup$ @MichaelHall, I agree the comparisons are spot on if you are only considering a coordinated turn, but I'm asking about a steady level turn in which the aircraft is slipping/skidding; which I don't think the comparison can easily be extended to? $\endgroup$ – eball Apr 2 at 17:07
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    $\begingroup$ @eball, please see my addendum. $\endgroup$ – Michael Hall Apr 2 at 18:10
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A simple answer: the ball is forced to move along an arc in a plane. Let's imagine the center of that arc as the point where the resultant of the sum of the gravitational and inertial forces is applied. Now, you project that resultant on the plane of that arc. The intersection of the arc and the projected resultant is exactly where the ball lies, at any time or circumstance...

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  • $\begingroup$ See my second-to-last paragraph for similar ideas (but not explained so concisely), which itself was inspired by something in eball's answer, namely the concept that the ball always sits at the point in tube that is "normal" to the lift vector that would be required in a coordinated turn at that same velocity and turn rate or turn radius-- $\endgroup$ – quiet flyer Apr 3 at 11:58
  • $\begingroup$ (now third-to-last paragraph) $\endgroup$ – quiet flyer Apr 3 at 12:07
  • $\begingroup$ I hope this eventually becomes the "accepted" answer, very nicely put. $\endgroup$ – quiet flyer Apr 3 at 12:10
  • $\begingroup$ Wow, @xxavier, that is a really concise/correct way to explain it! You could also say "Let's imagine the center of that arc as the point where the resultant of the sum of the external forces ($\bf L$ and ${\bf L}_{slip}$) is applied." $\endgroup$ – eball Apr 3 at 14:10
  • $\begingroup$ @quietflyer, I would make this the accepted answer but it doesn't address where ${\bf L}_{slip}$ comes from...tho I guess that's irrelevant if you think of it from the point of view xxavier did above (i.e. the gravitational/inertial forces instead of the external forces) $\endgroup$ – eball Apr 3 at 14:13
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Assume the aircraft is trying to perform a steady, level, coordinated turn. If the airplane is slipping, excess outward lateral force (${\bf L}_{slip}$) will be generated. To be coordinated, the bank angle must be decreased or the rudder into the turn increased.

Because the ball will position itself such that it is normal to the sum of the external forces in a coordinated turn (here simply the lift vector $\bf L$, it should sit in the center, in a coordinated turn (similar to a banked turn in a car where the normal force applied by the road replaces the lift force).

In a slip, the ball will be to the left of center because it rests at the point where the surface is normal to the Lift vector required for a coordinated turn at the same turn radius and airspeed (equivalent to the sum of the lift vector $\bf L$ and ${\bf L}_{slip}$) as indicated in the diagram.

The old adage "step on the ball" applies here.

The below diagrams should help clarify:

enter image description here

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  • $\begingroup$ Thanks for the edits @Robert DiGiovanni!! I saw you were an active contributor in the other threads I linked so I value your input. But why would the airplane generate insufficient lateral outward force? Looking at the diagram, wouldn't it generate excess outward lateral force because the fuselage acts like a wing at an angle of attack of $\beta$ (which would need to be balanced by a larger inward force in a turn, provided by a larger bank angle $\phi$)? -Thanks! $\endgroup$ – eball Apr 2 at 13:15
  • $\begingroup$ the "ball" inclinometer is in the aircraft reference. Wing does pull the plane sideways, but needs help to center the ball. Thats what the rudder does. Remember they are teaching shallow turns, and adverse yaw of the ailerons also needs to be counteracted. "Step on the Ball". Adding more bank just makes it harder to center, as the center is now more "uphill". $\endgroup$ – Robert DiGiovanni Apr 2 at 14:04
  • $\begingroup$ Ooo, I think we may be talking about this from the opposite point of view. I'm trying to figure out how to fly with sideslip (and it's effect on the ball), not how to recover. So If you are in a steady slip, you need to decrease the angle of bank to return to a coordinated turn. But if you are in a coordinated turn and want to slip, you need to increase the angle of bank. Right? (I believe they way my answer above is currently written disagrees with the diagrams below it haha.) Thanks! $\endgroup$ – eball Apr 2 at 14:33
  • $\begingroup$ yes, increase the bank and/or push the rudder the other way. That's how you do a side slip (against a cross wind), but please make sure you have adequate airspeed and expert instruction as flying uncoordinated can have a bad ending. What you do with the controls depends on what you want the plane to do. $\endgroup$ – Robert DiGiovanni Apr 2 at 14:38
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    $\begingroup$ A suggestion-- in the diagrams, the apparent inertial force "centrifugal force" (which is a pseudoforce) is equal and opposite the net centripetal force (turning force), which is equal to the vector sum of the (leftwards) horizontal component of the lift vector plus the (rightwards) horizontal component of Lslip. $\endgroup$ – quiet flyer Apr 3 at 0:06
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The ball is off-center in a slip or skid because the bank angle is "wrong" for the turn rate and airspeed, and thus the resultant of "centrifugal force" plus weight does not point "straight down" in the aircraft's reference frame.

While true, this statement isn't terribly useful from a pilot's point of view, since we normally don't vary the bank angle without also changing the turn rate, so we normally wouldn't try to center the ball by adjusting the bank angle. (An exception would be if we're performing a standard-rate turn by reference to a turn rate indicator or turn coordinator, especially if we're using the old "needle-ball-airspeed" technique, where pilots were advised to keep the turn rate constant with the rudder, while using the ailerons to center the slip-skid ball.)1

An exactly equivalent-- but more useful-- statement is that the ball is off-center whenever the net aerodynamic force generated by the aircraft has a sideways component in the aircraft's reference frame, rather than acting "straight up" in the aircraft's reference frame.

Note that the "centrifugal" force we're speaking of here is actually a pseudoforce. It is the apparent inertial force generated by the turn. It is exactly equal in magnitude and opposite in direction to the actual net centripetal force that is causing the turn in the first place. In a steady-state coordinated turn, this centripetal force is simply the horizontal component of the lift from the banked wing, while in an uncoordinated turn, we have to add or subtract an additional sideways force that is caused by the fuselage moving sideways through the air.

The analogy of a car going around a banked track can help us understand all this. In this case, substitute the force exerted by the track on the tires for the net aerodynamic force generated by the aircraft. At some combination of turn rate and speed, the banked ground will be pushing "straight up" on the tires in the car's reference frame, and a slip-skid ball would be centered. In such a case, the tires are "feeling" no sideways loading. This is like an airplane in a coordinated turn, where the net aerodynamic force acting on the aircraft is simply the lift force from the wing, acting "straight up" in the aircraft's own reference frame.

However, it's important to recognize that the car analogy only goes so far. After all, the track is constrained to have a certain turn radius regardless of the car's speed, and that's not true in actual flight.

It's easy to see why at all speeds except for one, the bank angle will be "wrong" for the turn rate as a car goes around a banked track. But why would the bank angle ever be "wrong" for the turn rate and airspeed in actual flight? Or to put it another way, why would the net aerodynamic force acting on the aircraft ever be anything other than the lift force from the wing, acting "straight up" in the aircraft's own reference frame?

Beware of explanations that invoke "centrifugal force" without giving sufficient attention to the real, tangible, aerodynamic forces generated by the aircraft.

At least one aviation author (writing for pilots of ultralight aircraft) has actually suggested that flying at the "wrong airspeed for the bank angle" will make the aircraft slip or skid. This is completely wrong, unless we artificially constrain the turn radius to be fixed, which is a very odd constraint in the context of flight.

So in actual flight, how can it happen that the bank angle can be "wrong" for the turn rate and airspeed?

Because the aircraft is being allowed to fly sideways through the air-- i.e. the nose is not being kept pointed in the same direction as the aircraft is actually travelling through the air at any given instant.

As a result, the airflow is striking the side of the fuselage and making a real, tangible, aerodynamic sideforce, acting to the "left" or the "right" in the reference frame of the aircraft and pilot. The slip-skid ball "feels" the combined effect of this sideforce, and the wing's lift force.

Or to put it another way, this real, tangible aerodynamic sideforce adds or subtracts to the turn rate, so that the turn rate is no longer "right" for the bank angle and airspeed, so the slip-skid ball goes off-center.

The ball will always displace in the "upwind" direction relative to the airflow that the fuselage is feeling. It will move toward the side of the fuselage that the airflow is striking.

(An extreme case is sustained, vertically-banked, linear "knife edge" flight, where the wing is unloaded to the zero-lift condition, and the airflow striking the side of the fuselage must support the entire weight of the aircraft, apart from whatever component is supported by the thrust vector. The ball will be fully deflected, to the "earthward" corner of the curved glass tube.)

So it's equally valid to say that the slip-skid ball will be displaced off-center whenever the bank angle is "wrong" for the turn rate and airspeed, or to say that the slip-skid ball will be displaced off-center whenever the airflow is striking the side of the fuselage because the nose of the aircraft is not aligned with the actual direction of travel and the airflow is hitting the side of the fuselage. The two statements are equivalent--they are two different "sides of the same coin". Pilots who understand that, understand what the ball is really telling them.

Once we understand this, it becomes clear why in aircraft with no propwash over the nose-- such as sailplanes-- a yaw string (a tuft of yarn attached to the outside of the canopy, on the aircraft centerline, in the pilot's field of forward vision) is a perfectly acceptable substitute for a slip-skid ball. Any discrepancy between the direction the aircraft is pointing and the direction the aircraft is moving will make the "yaw string" blow sideways.

Whenever the pilot uses the rudder as needed to keep the nose of the aircraft pointing directly into the airflow, so that there is no airflow striking the side of the aircraft, the ball (and the pilot) only "feel" the lift generated by the wing. Since the wing's lift force always acts "straight up" in the aircraft's reference frame, the ball will always be centered in such a case, regardless of the bank angle angle or airspeed. In such a case, the turn rate will automatically be "correct" for the bank angle and airspeed.

Note that if we change the shape of the fuselage, we also change the amount of deflection of the slip-skid ball that we'll see for some given sideways angle of airflow. A slender streamlined fuselage will create less sideways aerodynamic force, and less displacement of the slip-skid ball, than a big boxy one.

The biggest deficiency in diagrams like the ones accompanying this related ASE question is that they don't specifically show the aerodynamic force created by the airflow striking the side of the fuselage in a slip or skid, and thus leave the reader with no understanding of what is really going on in a slip or a skid.

All the above is really just a first approximation, but it is a very good approximation for most situations in actual flight. To take the conversation to a deeper level, we can recognize that it's actually possible for the rudder itself to create a significant sideforce--which will affect the turn rate, and will also affect the slip-skid ball--even if the fuselage is completely streamlined to the airflow. That's why the most efficient flight is not always obtained with the slip-skid ball completely centered. This comes into prominence when we are dealing with a failed engine on a twin-engine aircraft, because the rudder will be strongly deflected. But for all practical purposes, as long as we are not dealing with a huge asymmetry in thrust, it's a very good approximation to assume that whenever the ball is off-center, it's because the pilot is not keeping the nose pointed in the same direction as the aircraft is actually travelling, and thus is allowing the airflow to strike the side of the aircraft.

Please keep in mind that the "airflow" we're speaking of here has nothing to do with the external, meteorological wind. Rather, it is the "relative wind" created by the aircraft's own motion through the airmass. When the aircraft is not pointing in exactly the same direction as it is actually travelling through the airmass, this "relative wind" has a sideways component in relation to the fuselage. The pilot fixes this by applying the rudder as needed to keep the aircraft pointing in the same direction as it is actually travelling. The rule is to "step on the ball" to bring it back to center.

We could go on to discuss the reasons why many aircraft-- especially slow-flying, long-spanned aircraft like sailplanes-- tend to require the pilot to hold a touch of inside rudder during a turn to keep the slip-skid ball centered. But that is better saved for another question. For now, we'll simply note that while the vertical fin does tend to keep the nose of the aircraft pointing directly into the airflow, like a weathervane, sometimes it needs a little extra help from the rudder.

What if we are maneuvering in three dimensions rather than just two? The concept of the slip-skid ball going off-center whenever the bank angle is "wrong" for the turn rate and airspeed is really only valid when there is no vertical acceleration. For example, near the top of a wingover, the bank angle may briefly approach vertical and the airspeed may drop quite low, yet given appropriate rudder usage, the ball can remain centered. A more complete statement would be that the slip-skid ball will go off-center when there is a mismatch between the bank angle, turn rate, airspeed, and vertical acceleration component. The slip-skid ball will position itself in the tube to be aligned with the vector sum of weight, the horizontal "centrifugal force" pseudoforce generated by the turn, and a vertical (upward) "centrifugal force" pseudoforce generated by the earthward curvature in the flight path. But even in this more complex case, it's still true to say that if the aircraft is kept aligned with the instantaneous direction of travel so the airflow is not allowed to strike the side of the fuselage, the slip-skid ball will stay centered.

Here's an interesting point originally brought out in another answer (and now also in yet another)-- we can imagine that the ball represents an imaginary pendulum, suspended from a point located at the center of the radius of curvature of the glass tube. For turns where we are only maneuvering in two dimensions (as opposed to the wingover example above), if we varied the bank angle while holding the airspeed and turn rate constant, we'd notice that the position of the pendulum (or the position of the ball) relative to the horizon would stay the same. The ball always moves to the part of the tube that is "normal" to the vector sum of the weight vector and the apparent "centrifugal force" (pseudoforce) vector. This also means that the ball always moves to the part of the tube that is "normal" to the vector sum of upward lift and net centripetal force. And this also means that the ball always moves to the part of the tube that is "normal" to the direction of the lift vector that would be required for a perfectly coordinated turn at the current airspeed and turn rate. So we can think of the ball as telling us what bank angle would be required for a perfectly coordinated turn at the current airspeed and turn rate. This brings us back around to the opening line of this answer-- the concept that in a slip or skid, the bank angle is "wrong" for the turn rate and airspeed. In theory, in a slip, we can center the ball by reducing the bank angle while holding the turn rate and airspeed constant.

But unless we're performing a standard-rate turn by reference to a turn rate indicator or turn coordinator, this isn't a terribly useful concept in actual flight, since any change in bank angle will normally also drive a change in turn rate. Rather than thinking of adjusting the bank angle to match the airspeed and turn rate, thus centering the ball, it's much more useful to recognize that the bank angle will always be "correct" for the airspeed and turn rate as long as the airplane is not being flown sideways through the air. That's why, practically speaking, the rudder is almost always the best tool for centering the slip-skid ball.

In fact, in the few instances where we choose to use the ailerons rather than the rudder to center the ball, the main reason that works is not because we are changing the bank angle, but rather because the ailerons are generating an "adverse yaw" torque that has the effect of a rudder input in the opposite direction. Example: we're holding a thermal turn in a sailplane and we're suspecting that a steeper bank angle might help us to stay in the thermal better-- the ball accidentally moves to outside (high side) showing a skid-- we know that applying outside (high side) rudder would center the ball, but also will reduce the bank angle due to coupling between yaw, slip, and roll-- but we don't want to reduce the bank angle as we might lose the thermal-- so we apply inside aileron instead. The main reason that this centers the ball is actually not that the bank angle is increasing, but rather that the ailerons are creating an "adverse yaw" torque and we aren't doing anything with the rudder to counteract it. So the nose swings away from the direction that we've moved the aileron control stick, bringing the nose into better alignment with our actual direction of travel, and increasing the bank angle as an extra bonus. In actual practice, in an aircraft with a high wing or dihedral or sweep, if we want to make a yaw adjustment (i.e. change the sideslip angle) without introducing a roll torque, we actually need to move the rudder one direction while also moving the ailerons slightly in the opposite direction. There are situations where it will best suit our purposes to use only the rudder ("step on the ball") or only the ailerons ("move the stick against the ball") for a correction, but the former is much more common than the latter, and the latter works mainly due to "adverse yaw" from the ailerons, not as a direct result of the actual change in bank angle.

Footnotes--

  1. As Wolfgange Langewiesche noted back in 1947 in his classic book "Stick and Rudder", this very antiquated "needle-ball-airspeed" technique is arguably rather flawed. It would fail completely on any aircraft that didn't either a) exhibit substantial coupling between yaw, sideslip, and roll, or b) generate substantial aerodynamic sideforce during a sideslip or skid. It's better to control to the turn rate with the ailerons, while using the rudder as needed to center the slip-skid ball-- this will work in any aircraft. Nonetheless, this old technique is likely what led to a turn rate indicator (with included slip-skid ball) sometimes being a called a "turn and bank indicator", as if the indication of the ball somehow had something to do with the bank angle.
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  • $\begingroup$ thanks for the detailed reply! I'm currently working on helicopter simulations and have never actually flown a real aircraft so the controls are not intuitive to me. But your explanation seems correct to me. And I was actually looking for those diagrams you said you'd get to eventually! haha. Thanks again for your input. $\endgroup$ – eball Apr 3 at 14:04
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    $\begingroup$ Sometimes it’s hard to know whether you are answering for a teenager who’s never taken a hard corner, unbelted, on a bench seat, or a computer programmer looking for equations... (doubly confusing if you don’t know what “unbelted” or “bench seat” means!) $\endgroup$ – Michael Hall Apr 3 at 16:03
  • $\begingroup$ Haha totally agree!! Probably would have been helpful for me to make that clear in my question. $\endgroup$ – eball Apr 3 at 17:20
  • $\begingroup$ @eball as you are working on simulations of a special breed of aircraft, the helicopters, you shoud know that a helicopter pilot that has flown an entire flight (with a helicopter) so that the ball remained fully centered, is superhuman 😃 $\endgroup$ – Jpe61 Apr 4 at 6:26
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    $\begingroup$ There's a duplicated word in this sentence, but I'll wait till I have other changes to make before editing-- " Since the wing's lift force always acts "straight up" in the aircraft's reference frame, the ball will always be centered in such a case, regardless of the bank angle angle or airspeed. " $\endgroup$ – quiet flyer Apr 4 at 16:34
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As the OP is not a pilot (but asking a very good question), it remains to be said that flight is 3 dimensional and adequate vertical lift is always required to do a ... "steady level turn".

So now we may discuss the virtues of flying the plane, not the instrument.

It so happens that increasing bank, or slipping, is an undesirable turning technique because it requires more pitch (or more speed) to maintain adequate vertical lift. You now have a double whammy of airspeed reducing drag factors. What happens to a pilot at night or in the clouds? They start pulling back on the stick more, losing airspeed and rapidly entering a spiral.

Skidding flight also creates a higher drag condition, and also carries the risk of tip stalling the slower wing.

This is why "low and slow" planes benefit by staying coordinated, but exactly what is this?

For cars, trucks, and planes alike it is the direction of inertial resistance to the vector sum of gravitational and aerodynamic acceleration directly towards the bottom of the vehicle.

For a turn of a given radius and bank, the controlling factor is speed. This works for cars and certainly for trucks.

With aircraft, bank is rarely given, but radius and speed can be (when flying a landing pattern curcuit). When speed is given, we match bank (ailerons), and "steering" (rudder), to hold course in a coordinated (lowest drag) manner while also providing adequate lift for level flight.

So, pilots do not choose (even though it is theoretically possible) to control coordinated flight by changing bank. Far better to reduce drag with rudder by stepping on the ball. As drag will always be higher in turns than straight flight, increasing power to maintain airspeed is also not a bad idea, especially if your plane is very draggy.

So it follows that adequate airspeed is crucial for safe flight and lower drag helps maintain airspeed, so slipping (or skidding) in level turns is out (not safe).

These techniques do have value in adjusting glide angles for landing but should not be attempted without understanding the risks.

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  • $\begingroup$ You're a little late to the party... Good answer, but now that we know he's writing software for a helicopter simulation I'm not sure this is helpful. $\endgroup$ – Michael Hall Apr 3 at 22:10
  • $\begingroup$ @Michael Hall well, I felt a little responsible for all this, but hey, would this not apply to overbanking a helicopter too? $\endgroup$ – Robert DiGiovanni Apr 3 at 23:10
  • $\begingroup$ Helicopters are a whole different animal. I only flew one once, but tail rotors and power changes magnify what us fixed wing pilots experience WRT the ball by 10x... $\endgroup$ – Michael Hall Apr 3 at 23:40
  • $\begingroup$ @RobertDiGiovanni, why is it more efficient to step on the ball than to change bank? Is this because to change bank at a given speed/turn radius, you have to increase AOA which is more draggy than using rudder to gain the inward force required for the turn? $\endgroup$ – eball Apr 5 at 10:47
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    $\begingroup$ @eball the actual difference will vary from aircraft to aircraft, but the answer is yes. With increase bank one must increase pitch to maintain vertical lift...more drag. Ball is not centered. Also more drag. The gist of your idea makes sense, as the horizontal lift vector of a banked wing will produce a lot of lateral force. But here, it is acceleration that holds the ball "up", and the drag of a wing moving sideways will not allow you to accelerate for long in one direction. The rudder keeps changing that direction. Acceleration from 0 (constantly) is easiest. $\endgroup$ – Robert DiGiovanni Apr 5 at 19:48

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