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I have trouble understanding why the ball (on the turn & slip indicator) falls into the turn e.g. drop to the left when aircraft is rolled left, when in an uncoordinated turn (slipping turn). What are the forces that goes into this?

From reading up online, many sources quote that in a slipping turn, the aircraft is banked too much for the rate of turn. So, the horizontal lift component is greater than centrifugal force. As a result, the ball drops into the turn since there is less g force to "centralise" the ball.

Isn't centrifugal force suppose to be equal and opposite of the horizontal component of lift. How can being in a slipping turn make the centrifugal force lesser? Essentially, what is so special about slipping turn can allow the centrifugal force to change?

Normal, Slipping, and Skidding Turns

Image from: http://avstop.com/ac/flighttrainghandbook/forcesinturns.html

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    $\begingroup$ Thank you for your replies. As much as I agree that the links you shared have their merits, I am still unable to pinpoint how exactly a slipping turn can have an effect on the centrifugal force. $\endgroup$ – Flightsimrightnow May 4 at 7:07
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    $\begingroup$ I fail to understand how this answer does not answer your question. You may elaborate that. $\endgroup$ – Manu H May 4 at 7:11
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    $\begingroup$ Please pardon me but I have trouble relating that particular answer to my question. I am under the impression that the centrifugal force generated is EQUAL and opposite to the horizontal component of lift. As such, I cannot relate to why the centrifugal force can be changed. $\endgroup$ – Flightsimrightnow May 4 at 7:38
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    $\begingroup$ @Manu H many of these "answers" only serve to confuse a very simple concept. $\endgroup$ – Robert DiGiovanni May 4 at 11:53
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    $\begingroup$ "many of these "answers" only serve to confuse a very simple concept." YES!!! I only wish I would have been logged on in time to vote to close this earlier. Good grief already... $\endgroup$ – Michael Hall May 4 at 15:11
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I have trouble understanding why the ball (on the turn & slip indicator) falls into the turn ... in an uncoordinated turn (slipping turn). What are the forces that goes into this?

The ball is displaced to one side because the aircraft is being allowed to fly sideways through the air. The nose is not aligned with the instantaneous direction of the flight path, but rather is yawed to point to the left or right of the instantaneous direction of the flight path.

As a result, the airflow is striking the side of the aircraft and generating an aerodynamic sideforce, which is what causes the ball to displace to the side.

There are many reasons why an aircraft might tend to fly yawed slightly toward the left or the right in a turn. Most of these reasons tend to promote a slip, rather than a skid. Other than the most obvious reason of a significant, intentional deflection of the rudder by the pilot, these reasons include the curving nature of the relative wind during turning flight and its effects on tail surfaces (also described as "aerodynamic damping"), and the fact that the outboard wingtip must move faster, and thus tends to generate more drag, than the inboard wingtip. In an aircraft with a propeller, P-factor can also play a role, tending to promote either a slip or a skid depending on the direction of turn. The setting of the rudder trim may also be factor. When the bank angle is changing, adverse yaw from the deflected ailerons and from the rolling motion itself plays a major role, strongly tending to promote a slip when the bank angle is increasing and a skid when the bank angle is decreasing. In all cases, the cure is for the pilot to use the rudder as needed to compensate for these effects and align the nose of the aircraft with the instantanous direction of the flight path, so the nose is pointing straight into the "relative wind" and the airflow is not hitting the side of the fuselage.

Note that "having too little or too much lift in relation to the bank angle" or "having too little or too much centrifugal force in relation to the bank angle" are not causes of slips or skids. Rather, they are results of slips or skids. The causes are the aerodynamic effects listed above, creating a yaw torque that makes the aircraft fly slightly sideways through the air, despite the generally stabilizing effect of the vertical fin. More on this later.

So why does the slip-skid ball get displaced off center when the aircraft flies sideways through the air? When the airflow hits the side of the fuselage, this generates an aerodynamic sideforce, acting orthogonal (perpendicular) to the flight path in the direction that is roughly parallel to the wingspan, This sideforce points toward the high wingtip in a slip, and toward the low wingtip in a skid. This real aerodynamic force has been completely omitted from the diagram of the slipping turn attached to your question. This is a major flaw in the diagram.

This aerodynamic sideforce contributes to the net aerodynamic force generated by the aircraft, so that the net aerodynamic force vector is no longer pointing straight up" in the aircraft's reference frame, i.e. parallel to the vertical fin.

The slip-skid ball always tends to position itself in the curved glass tube at the point in the tube that is exactly perpendicular to the net aerodynamic force vector. If the ball is positioned at any other point in the tube, it will "feel" an apparent force to the left or right, and will shift position. If the net aerodynamic force is not pointing "straight up" in the aircraft's reference frame, then the ball will not position itself at the "bottom", i.e. at the center, of the curved glass tube.

Another way to look at the situation is to consider the apparent "load" acting on the ball, due to the combined effect of weight and "centrifugal force". The ball always tends to position itself in the curved glass tube at the point in the tube that is exactly perpendicular to the "load" vector. If the "load" vector is not pointing "straight down" in the aircraft's reference frame, then the ball will not position itself at the "bottom", i.e. at the center, of the curved glass tube.

The key thing to understand is that the apparent "load" felt by the aircraft structure and contents, including the pilot and the slip-skid ball, is nothing more than the mirror image of the real aerodynamic force vector generated by the aircraft. This is explored in more detail in a several answers to the related ASE question "Forces "felt" by pilot, G-meter, inclinometer--are they the aerodynamic forces generated by the aircraft, or the sum of weight+centrifugal force?".

The diagram is misleading because it suggests that some mysterious effect, presumably somehow related to turn rate, yet seemingly unrelated to any real aerodynamic force, is affecting the amount of "centrifugal force" generated by the aircraft or acting on the aircraft. This is false. The difference in the magnitude of the "centrifugal force" vector between a "coordinated" turn and a "slipping" turn is due to the way that a sideslip generates a real aerodynamic sideforce vector, as described above.

While it is in fact true that the "load" acting on the ball can be said to be equal to the vector sum of weight and "centrifugal force", this is not a concept that has a lot of explanatory power, unless we understand that the "centrifugal force" is intimately related to the real aerodynamic forces generated by the aircraft. The "load" vector, which can be said to be the vector sum of weight and "centrifugal force", is nothing more than the mirror image of the real aerodynamic force generated by the aircraft. The "load" vector is always equal and opposite to the vector representing the real aerodynamic force generated by the aircraft.1

An extreme case of a slipping turn is a forward slip or sideslip (these are actually exactly the same thing, just aimed differently in relation to the target), where the aerodynamic sideforce from the slip is sufficient to bring the turn rate all the way to zero, creating a linear flight path. Whenever the flight path is linear, the slip-skid ball doubles as bank angle gauge. The most extreme case of a non-turning (linear) sideslip is vertically-banked sustained knife-edge flight, like we often see at an airshow. In this case the aerodynamic sideforce vector from the airflow striking the side of the fuselage (plus an additional component due to the engine thrust line being aimed above the horizon) is supporting the entire weight of the aircraft, and the wing's lift vector is zero. In this case, the "load" vector is simply equal and opposite the aerodynamic sideforce vector from the slip (including the component due the engine thrust), because aerodynamic sideforce vector and the net aerodynamic force vector are one and the same. The net aerodynamic force vector is equal to the weight of the aircraft but acting in the upward direction, while the "load" vector is equal to weight of the aircraft, acting in the downward direction.

While established in a steady-state turn at a constant turn rate and airspeed, trimmed for hands-off flight, if we then pull aft on the stick to "load up" the wing with "extra" lift to boost the G-load to a higher-than-normal value for the bank angle, or if we push the stick forward to "unload" the wing to decrease the lift vector and drop the G-load to a lower-than-normal value for the bank angle, the turn rate will immediately change, and so will the horizontal component of the "centrifugal force" vector. The diagram above might tend to lead us to think that the ball will immediately shift off-center. This is not the case.

In a situation like this, the flight path will curve up or down, introducing a vertical component of "centrifugal force". The vector sum of weight and centrifugal force will still be "straight down" in the aircraft's reference frame, and the ball will stay centered.

Or to look at it from a much simpler perspective-- when we pull the stick aft or push the stick forward to change the wing's angle-of-attack, we are increasing or decreasing the magnitude of the lift vector, but we aren't introducing any aerodynamic sideforces-- we aren't yawing the fuselage to fly sideways through the air. Because the net aerodynamic force vector still acts "straight up" in the aircraft's own reference frame-- parallel to the direction that the vertical fin is pointing-- the ball still stays centered.

(For simplicity, we're ignoring the forward or aft components in the net aerodynamic force vector-- the components that wouldn't show up in a head-on view of the aircraft. The basic explanation here remains the same regardless of whether thrust is exactly in balance with drag, or not.)

Of course, "loading" or "unloading" the wing in this manner during a turn will also drive a change in airspeed, as the flight path curves up or down. For an aircraft of a given weight at a given bank angle, there is only one value of the lift vector (G-loading) that will yield a steady-state turn at constant airspeed.2

From reading up online, many sources quote that in a slipping turn, the aircraft is banked too much for the rate of turn.

From the content above, you'll see that this isn't always true. It can be true given certain constraints, but it is not a concept with a lot of explanatory power.

As noted above, if we are established in a stabilized turn, and then we pull the stick back, or push the stick forward, to boost or decrease the lift vector and G-loading, the turn rate immediately changes, but the ball does not shift off-center.3

The idea that (for a turn at a given airspeed) "in a slipping turn, the aircraft is banked too much for the rate of turn" is only true given the constraint that there is zero upward or downward (skyward or earthward) curvature in the flight path. And the only way4 that (for a given airspeed) we can have a mismatch between the bank angle and turn rate, while not allowing the flight path to curve up and down, is that we are applying the rudder to expose the side of the fuselage to the airflow and generate an aerodynamic sideforce, or we are failing to apply the rudder as needed to cancel out other aerodynamic torques that are tending to expose the side of the fuselage to the airflow. If we do this, while adding or subtracting power as needed, we can indeed create the following situations. Starting from a stabilized coordinated turn at a constant airspeed and altitude, we can--

a) vary the turn rate and turn radius while holding altitude, airspeed, and bank angle constant

b) vary the bank angle and while holding altitude, airspeed, and turn rate and radius constant

The ball will deflect off center in both of these sitations. It is equally true to say that the ball goes off-center because the centrifugal force vector is no longer correctly matched to the bank angle, or because the load vector (the vector sum of centrifugal force and weight) no longer points straight "down" in the aircraft's reference frame (i.e. parallel to the vertical fin), or because the net aerodynamic force vector no longer points straight "up" in the aircraft's reference frame (parallel to the vertical fin), or because the pilot is using the rudder in a way that exposes the side of the fuselage to the airflow, which generates an aerodynamic sideforce.

The pilot's rudder usage is the key to centering the slip-skid ball, or to intentionally driving it off center. The sideways force we "feel" in a slip is the real aerodynamic force generated by the air striking the side of the fuselage, which changes the direction of the net aerodynamic force vector, and its mirror image, the "load" vector. Explanations centered around some sort of "balance" or "imbalance" between bank angle and turn rate, or bank angle and lift force, or bank angle and G-loading, are misleading, and have little explanatory power, and certainly do not apply to aerobatic flight. These explanations tend to obscure, rather than illuminate, what is really going on.

Some of these sorts of explanations are well suited to a car driving on banked track, or a bobsled going down a banked track, but not to flight, where the trajectory is not constrained to have a specific turn radius, and also is not constrained to have no upward or downward (skyward or earthward) curvature.5

Note that in flight, generally speaking, changing airspeed while holding the bank angle constant creates no tendency toward a slip or skid, though the turn rate and radius will both vary. As long as the net aerodynamic force vector continues to point "straight up" in the aircraft's reference frame, there will be no slip or skid. Though again, we can come up with specific contrived situations where varying the airspeed while holding the bank angle constant does cause a slip or skid-- for example if the turn rate and radius are constrained to remain constant as the airspeed is changed. Again, this can only happen if the pilot applies the rudder to expose the side of the fuselage to the airflow, generating an aerodynamic sideforce.

I have trouble understanding why

This is not surprising, because you have been provided with faulty explanations, as well as faulty diagrams. The diagrams attached to your question have omitted the aerodynamic sideforce vector caused by the airflow striking the side of the fuselage. The diagrams attached to your question give the impression that the net aerodynamic force vector generated by the aircraft is the same in all 3 cases (coordinated flight, slipping flight, and skidding flight), when it actually is not. The diagrams attached to your question give the impression that the "load" vector can somehow be something other than the mirror image of the net aerodynamic force vector, when it actually cannot be anything other than the mirror image of the net aerodynamic force vector.

To read more about the specific errors in the vector diagrams attached to your question, see the related question What is missing from these diagrams of the forces in slips and skids? and answer What is missing from these diagrams of the forces in slips and skids? .

Other related ASE questions or answers:

(Q) Forces "felt" by pilot, G-meter, inclinometer--are they the aerodynamic forces generated by the aircraft, or the sum of weight+centrifugal force?

(A) Forces "felt" by pilot, G-meter, inclinometer--are they the aerodynamic forces generated by the aircraft, or the sum of weight+centrifugal force?

(Q) What does the balance ball actually indicate?

(A) What does the balance ball actually indicate?

The topic of accuracy of the diagrams illustrated in your question has also been discussed on Physics Stack Exchange. See--

(Q) Is this vector diagram of the forces at play in turning flight correct?

(A) Is this vector diagram of the forces at play in turning flight correct?

Footnotes

1 -- For simplicity, we're assuming that the slip-skid ball is located near the aircraft's CG, or aircraft's yaw rotational rate is constant, or both. A change in the yaw rotational rate will create an apparent centrifugal or inertial force that displaces the slip-skid ball to one side if it is located far forward or aft of the CG, and this displacement does not reflect any real aerodynamic force component. For example, if the yaw rotational rate is increasing toward the left, a slip-skid ball located far in front of the CG would tend to be displaced toward the right, and a slip-skid ball located far aft of the CG would tend to be displaced toward the left. These second-order effects are beyond the intended scope of this answer, and are also beyond the scope of what pilots are generally able to notice in actual flight.

2 -- For simplicity, we're overlooking the fact that magnitude of the lift vector associated with a steady-state coordinated turn at some particular bank angle and some particular airspeed and angle-of-attack is very slightly different when we are descending or climbing, than when we are maintaining altitude (with respect to the airmass in all cases). For modest dives and climbs, this is a very minor effect and definitely not something that someone still learning about the basics of slips and skids should concern themselves with, but in the interest of accuracy, this point should not be completely omitted. To learn more about the "unloading" of the lift vector in a descent or climb, see What produces thrust along the line of flight in a glider? and Does lift equal weight in a climb?

3 -- Surprisingly, Wolfgang Langewiesche's classic flight-mechanics-for-pilots book "Stick and Rudder" (1944), which contains a lot of good material on the physics of turning flight, does make the claim (see pages 205, 219-220, and 223-226) that while excess back pressure on the stick does not cause a slip or skid, too little back pressure on the stick can cause a slip, with the ball deflected to the low side of the turn, which should be corrected by increasing back pressure rather than by applying inside rudder. This is probably the only faulty idea in the entire book. The 3rd edition of the book "Modern Airmanship" (1966, edited by Neil Van Sickle, Major General USAF) has similar content on pages 308-309. By the 8th (1999) edition of "Van Sickle’s Modern Airmanship", edited by John F. Welch, Lewis Bjork, and Linda Bjork, this content had been deleted, while the remaining content on the physics of turning flight had been retained (page 441). For more, including specific quotes from these books, see https://web.archive.org/web/20180905112047/http://aeroexperiments.org/critiques.shtml .

4 -- Strictly speaking, it's not exactly true that the only way that (for a given airspeed) we can have a mismatch between the bank angle and turn rate, while not allowing the flight path to curve up and down, is to deflect the rudder to the side in a way that exposes the side of the fuselage to the airflow. But it's a good first approximation. For a more nuanced view, we have to consider the effect of the sideforce generated by the rudder itself. For example, we could shut down one engine on a conventional piston twin and apply the rudder just enough to keep the fuselage exactly aligned with the airflow. Since the sideforce from the rudder itself is not zero, the ball will be slightly deflected (toward the working engine) when the nose of the aircraft is pointing straight into the relative wind. If we bank into the good engine as needed to cancel all turning tendency, we'll end up flying along in a straight line (zero turn rate) while banked slightly toward the good engine, as discussed in detail elsewhere on ASE, such as here, here, and here (link to be added).

5 -- Speaking of bobslebs on curved tracks, "Hang Gliding Training Manual" and "Performance Flying" by Dennis Pagen both include extensive content-- including an illustration of a bobsled on a track-- purporting to show how, as a pilot enters a turn while flying a hang glider, if he or she doesn't immediately "load up" the wing with an adequate pitch input, the glider will slip sideways through the air, toward the low wingtip. Also the idea that an intentional reduction of angle-of-attack and lift and G-loading while turning will make the glider slip sideways through the air toward the low wingtip. See for example pp. 128-129 in "Hang Gliding Training Manual", and page 45 in "Performance Flying". A similar idea appears on page 70 of Peter Cheney's "Hang Gliding for Beginner Pilots" (3rd edition, 1997). I've checked these ideas in flight with a yaw string, in hang gliders as well as conventional aircraft, and found no basis for them. For more on the content in these books, see https://web.archive.org/web/20180905112047/http://aeroexperiments.org/critiques.shtml .

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – Federico May 10 at 7:10
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Because of the insufficient speed and pull for the bank angle of the aircraft .For level turn you must pull up not bank ,but first you need to bank some degrees and have enough speed to maintain the G required for the bank angle .If you pull lower than the needed G's for bank angle your aircraft nose will dive ,if you pull more G's the nose rise above horizontal line.Now there are two types of skidding turns (inside the turn or drifting outside of turn)

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