# What is missing from these diagrams of the forces in slips and skids?  See these images that are widely reproduced in many different on-line ground school materials.

A) Is the magnitude of the wing's lift vector illustrated correctly in each of the three cases? Should it really be the same size in each case?

B) What real aerodynamic force generated by the aircraft has been completely omitted from the second two cases, but should be included to make the diagrams more comprehensible? Specifically, to explain why the vector labelled "load" is not the same in all the figures?

Assume that the aircraft is maintaining a constant altitude and airspeed regardless of whether the turn is coordinated, slipping, or skidding.

(The same diagrams could apply to gliding flight as well, in which case we would assume that the aircraft was maintaining a constant airspeed and a constant descent rate in relation to the surrounding airmass.)

The intent of the question is to address a major flaw in the diagrams, not to nitpick small errors on the part of the artists. The second diagram is the best one to focus our attention on, because the bank angle is clearly drawn to be identical in every case, and the horizontal and vertical components of the lift vector are clearly drawn to be identical in every case.

A word about the images included here-- both images are widely reproduced in many different on-line ground school materials. For example, the first image appears as Figure 3-21 from this "Aerodynamics in Flight" section from an on-line ground school. For another example, see page 12 of this document. It may have originally been published in the FAA's "Pilot's Handbook of Aeronautical Knowledge". The second image appears as figure 5-35 on page 5-24 of the FAA's "Pilot's Handbook of Aeronautical Knowledge" (2016 edition). It also may be found in various on-line ground school materials-- see for example this one.

• I think it is very incomplete because you have to consider force vectors introduced by the sideways fuselage orientation and the offset thrust line. This means in the slipping turn you have lateral forces acting on the fuselage and the thrust line acting outward on the lateral axis, and vice versa for skidding. Now to quantify that, not sure. – John K Apr 25 '20 at 17:28
• – Michael Hall Apr 26 '20 at 6:12
• Please don’t cross-post almost exact questions across the network. Either this question or the one on Physics should be deleted. – dalearn May 12 '20 at 16:39
• @dalearn -- I am considering what to do next -- I do not intend to delete the one from ASE but maybe will delete the other one -- physics.meta.stackexchange.com/questions/12902/… – quiet flyer May 15 '20 at 16:58
• Sorry about the late reply. I don’t have a specific meta discussion relating to this to point you to but in general, the culture of SE as I have observed, is that cross-posting is discouraged in many cases. This question would be a good question for either Physics or Aviation but given the reception at each site, I recommend deleting this one. – dalearn May 15 '20 at 23:40

## 3 Answers

A) Is the magnitude of the wing's lift vector illustrated correctly in each of the three cases? Should it really be the same size in each case?

No (on both counts).

B) What real aerodynamic force generated by the aircraft has been completely omitted from the second two cases, but should be included to make the diagrams more comprehensible? Specifically, to explain why the vector labelled "load" is not the same in all the figures?

Side-slip. This is my attempt at better describing the forces in a slipping, skidding and coordinated turn. These thoughts represent my best guess and should be considered the basis for further discussion rather than fact.

The biggest and somewhat major limitation of my diagram (that I'm aware of), is that it doesn't show the critical role of yaw (due to my limited ability to draw, rather than by design).

Ideally it would show a yaw away from the turn for a slip and a yaw toward the turn for a skid.

This would also serve to clarify that the aircraft depicted is viewed from behind (only relevant to the slip ball displacement shown).

Another limitation of the diagram, due to it's 2D nature, is that it doesn't illustrate conventional drag (along the longitudinal axis) and how it varies with yaw during a slip or skid. ie: the force acting perpendicular to and out of the screen. This is important as the vector diagram doesn't facilitate conceptualising total drag during uncoordinated flight.

The key to understanding slipping, skidding and coordinated turns (as I understand it), is recognising the effect of side-slip on the total aerodynamic force (TAF) which is defined as the net aerodynamic force experienced by the aircraft. As per the previous paragraph, this diagram only considers the TAF along the lateral axis and so doesn't provide the complete picture, though it does serve to fully explain the different turn rates and resulting turning radii.

To be clear, I'm using the term side-slip to refer to the aerodynamic condition where an aircraft has an asymmetrical airflow about it's longitudinal axis resulting in a degree of lateral motion with respect to the relative airflow.

The air impinging on the exposed side of the aircraft effectively creates a lateral force, which is labelled side-slip in the diagrams.

The diagram depicts forces on the basis that, regardless of whether the turn is coordinated, slipping, or skidding - the aircraft is maintaining a constant;

• angle of bank; and
• altitude (thus lift [$$TAF_{vertical}$$] always equals weight); and
• airspeed

This means the main variables are;

• AoA (power) - indirectly represented by the vector labelled TAF
Quick recap on the aerodynamics of a turn:

In any turn (assuming the constants above), the power requirements increase
proportional to the rate of turn.

In vector speak, a turn is achieved by tilting the TAF away from the vertical
in the direction of the turn. The resulting horizontal component provides the
centripetal acceleration required, thus the greater the tilt, the
greater the rate of turn.

The consequence of tilting the TAF away from the vertical is a reduction in
lift. In order to maintain altitude during a turn, the TAF must therefore be
increased in order to keep the vertical component equal to what it was in
level flight. This represents an increased power requirement.


In the diagram, the length of the TAF vector varies in each scenario, therefore each has a different power requirement.

For any given side-slip angle (representing a given amount of drag along the longitudinal axis), a skidding turn requires more power than a slipping turn due to the higher value of $$a_c$$. I am making the assumption that the lateral drag is the same in both cases (but I don't know if this is the case).

I posit that at some angle of side-slip, the power required for a slipping turn will exactly equal that required for a coordinated turn. In this scenario, some of the power used to create the turning force ($$a_c$$) during a coordinated turn is instead used to combat the additional frontal drag associated with a slipping turn, which results in a reduced rate of turn.

It seems a skidding turn does a better job at turning than a coordinated turn. This is true, the rate of turn is higher, so why not use a skidding turn to turn faster than a coordinated turn? Aerodynamic hazards aside, it's simply less efficient.

Key to realise is that the rate of turn achieved per unit of power (ie: efficiency) is less in an uncoordinated turn (slip or skid) as power is wasted in both dragging the aircraft sideways and forwards (additional drag is created in the direction of travel as the aircraft presents a greater cross-section to the relative airflow).

Thus, in order to achieve a given turn radius, it is more efficient (requires less power) to perform a coordinated turn at a higher angle of bank than a skidding turn at a lesser angle of bank or a coordinated turn at a lower angle of bank than a slipping turn at a higher angle of bank.

• turn radius

The turn radius is determined by nothing other than the horizontal component of the TAF, identified as $$a_c$$ in the diagrams.

A higher centripetal acceleration results in a smaller turn radius and a lower centripetal acceleration results in a larger turn radius.

Insights:

• the TAF varies according to how coordinated flight is, in order to keep the value of lift ($$TAF_{vertical}$$) constant (ie: to maintain altitude in the turn)
• apparent gravity / load and how it changes is explained simply as the reaction force equal and opposite to the varying TAF (rather than a result of centripetal force and it's fictitious sibling centrifugal force)
• in a level slipping turn, side-slip opposes the horizontal component of lift ($$a_c$$) increasing the radius of the turn
• in a level skidding turn, side-slip reinforces the horizontal component of lift ($$a_c$$) reducing the radius of the turn

Comments & criticisms welcome.

• I think your diagrams would be more accurate if you deleted the vector labeled "lift", and applied the label "lift" to the unlabeled vector that is aligned with the vertical fin. (In the coordinated turn, this is also the same as the vector labeled total aerodynamic force, but not in the other cases.) But even then I think the diagrams would contain a small error. But they would be much better than the FAA diagrams, because they do include the vectors representing the aerodynamic sideforce generated by the sideslip or skid. – quiet flyer May 21 '20 at 17:03
• There is a paradox that seems to be raised by this answer (which I suggest would not be raised if the diagrams were drawn completely correctly.) The paradox is this: if the lift vector generated by the wing is the same in every case, and the aerodynamic sideforce vector is no larger in the skid than in the slip, then why should the skid require more power than the slip, to maintain power at a given bank angle, as this answer seems to suggest? – quiet flyer May 21 '20 at 17:30
• Another thing to keep in mind is that the drag vector does not appear in the diagram at all, because it acts parallel to the flight path and so would point straight away from the viewer, assuming that the flight path is intended to be aimed straight toward the viewer, i.e. straight out of the page. (Yes, as you point out, ideally the aircraft should be drawn yawed a bit to one side or the other in the slip and the skid.) Presumably power is applied as needed to balance drag, so that airspeed and altitude can both stay constant. Doesn't generating the aerodynamic sideforce incur a drag (ctd) – quiet flyer May 21 '20 at 17:31
• Doesn't generating the aerodynamic sideforce incur a drag penalty? Just because it is not illustrated, doesn't mean it doesn't exist. So the statement that a slipping turn requires less power to maintain altitude than a coordinated turn needs to be reconsidered. Do you mean for a given bank angle, or for a given turn rate (radius)? I would argue that the latter statement would certainly not be true, while the former statement would be true for extreme bank angles but not for shallow bank angles -- the exact bank angle at which it would become true would depend on (ctd) – quiet flyer May 21 '20 at 17:32
• I would argue that the latter statement would certainly not be true, while the former statement would be true for extreme bank angles but not for shallow bank angles -- the exact bank angle at which it would become true would depend on how efficient the aircraft was at generating aerodynamic sideforce, but it would certainly be true for all aircraft by the time the bank angle reached 90 degrees. Even infinite power won't allow a 90-degree bank angle to be sustained in a fully coordinated turn with no loss of altitude. – quiet flyer May 21 '20 at 17:32

Regarding the turn, we still have the same 4 forces .... thrust and drag, lift and gravity. Those are the forces acting on the plane we need to consider.

The lift vector is broken down into vertical and horizontal components. Drag will depend on the relative wind. Ideally, in a coordinated turn, the relative wind is as straight at the nose as possible, which produces the least drag. In a skid it from the fuselage side outside of the turn. In a slip it is from the inside of the turn. The thrust vector also plays a role. In a skid, it is inside the turn, in a slip it is outside the turn.

how do these forces add up to the same force vector opposing gravity and centrifugal force loads for the same turn?

For a given turn of the same radius and airspeed at same altitude, the resultant "G" force will always have the same direction and magnitude. This load or G force consists of gravity, always pointing towards earth and the "centrifugal force" from the acceleration of circular movement.

In all cases, the plane must create forces to match these loads.

In a skid, the rudder pushes the nose inside the turn. The bank angle is lower, but the vertical lift component still equals weight. The side force on the fuselage plus the thrust component inside the turn add to the smaller horizontal lift component to create the necessary lateral force component.

In a slip, just the opposite. Insufficient rudder leaves the nose outside the turn. The bank angle is higher. Vertical lift equals weight. The side force plus the thrust component outside the turn subtract from the larger lift component to create the necessary lateral force component.

In a coordinated turn, the rudder points the nose directly into the line of flight. There for, There are no lateral forces created by thrust offset, slip, or skid. Vertical lift equals weight, and only the more efficient wing creates lateral force.

MUCH LESS DRAG.

what is missing in these diagrams?

1. Proper illustration of the gravity and horizontal acceleration loads, that combined vector is always the same.

2. Proper illustration of how the plane creates matching lift and turning forces, including side forces and thrust vectors, as well as wing horizontal component. It's combined vector is always the same.

3. The aircraft bank angle should be different for skid, coordinated, and slip conditions. One many also present a case of constant bank at different speeds, but this may apply more to race cars,

My compliments to quiet flyer and jumblie keeping this fundamental question alive. Many downvotes, but much learned.

By including the vector labeled "centrifugal force", the illustrators have signalled that they are basing their reference frame on the aircraft itself, not the earth or the airmass.

The reference frame based on the aircraft is not a valid inertial reference frame.

The aircraft cannot accelerate with respect to itself. The net force in the aircraft's own reference frame must be zero. Yet that's not what we see illustrated in the diagrams for the "slipping" and "skidding" cases.

It would better to omit the "lift" vector entirely and just show the load vector, weight vector, and "centrifugal force" vector, than to give the false impression that the actual aerodynamic forces generated by the aircraft are identical in all three cases.

What is missing from the "slipping" and "skidding" diagrams is the aerodynamic sideforce generated by the fuselage as it flies sideways through the air. When this vector is added to the lift vector, we end up with a net aerodynamic force vector that is equal in magnitude and opposite in direction to the vector labelled "load".

To make this work, the lift vector must be reduced in size in the slipping case, and must be increased in size in the skidding case.

Drawn this way, the diagrams would help the reader to understand the real reason why the inclinometer ball is displaced to the side in a slip or a skid. Fundamentally, it is due to the aerodynamic force created by the airflow striking the side of the fuselage. As a result, the net aerodynamic force vector is no longer "straight up" (i.e. parallel to the vertical fin) in the aircraft's own reference frame. So the ball, and the pilot's body, and the other aircraft contents, tend to be displaced toward the low wingtip in a slip, and toward the high wingtip in a skid.

Drawn correctly, the diagrams would teach the reader this concept, regardless of whether or not the choice is made to include the "centrifugal force" vector or not.

Drawn correctly, the diagrams would also teach the reader that the load "felt" by the airplane, slip-skid ball, pilot's body, etc, is the direct result of the aerodynamic forces generated by the aircraft. The "load" vector must always be the mirror image of the net aerodynamic force vector.

Drawn correctly, the diagrams would help the reader understand why increasing or decreasing the lift force by moving the stick or yoke forward or aft during a normal coordinated turn doesn't make the slip-skid ball deflect toward one side or the other, even though the turn rate is altered. As long as the net aerodynamic force vector acts straight "up" in the aircraft's reference frame, the apparent "load" vector must act straight "down" in the aircraft's reference frame, regardless of whether the turn rate is "right" for the bank angle and airspeed, or has been temporarily boosted or decreased via a pitch control input. (Naturally, such variations in the lift force will also cause the flight path to curve skywards or earthwards-- for a given bank angle, there's only one value for the lift vector that permits a stabilized turn at constant airspeed.)

Starting from the "load" vectors that we've been given here, what would the corrected diagrams look like? They would look like the top row of diagrams described in the remainder of this answer.

Diagrams to be added-- for now we'll have to use our imagination.

I'm referencing the second diagram specifically, the one from the 2016 edition of the "Pilot's Handbook of Aeronautical Knowledge", where the aircraft are clearly all drawn at exactly the same bank angle.

Imagine four rows of diagrams, each based on the diagram referenced above, but modified as follows--

First row -- forces in aircraft reference frame (not a valid inertial reference frame)

Weight and "centrifugal force" and "load" as illustrated in original. Note that "load" is the vector sum of weight and "centrifugal force".

The figures will include a net aerodynamic force vector (not shown in original). The net aerodynamic force vector must be exactly equal and opposite to "load" in all three figures.

Lift vector is correct in fig 1 of original (coordinated), and same as net aerodynamic force. So fig 1 of first row of answer is essentially same as original.

Lift vector needs to be shorter in fig 2 (slip), and there needs to be an aerodynamic sideforce vector acting at right angles to the lift vector, pointing to the right side of the page and upwards. That is the force that is missing from the diagram. It is generated by the air hitting the side of the fuselage. The vector sum of lift plus sideforce is the net aerodynamic force vector, and it must be exactly equal and opposite to the "load" vector.

Lift vector needs to be longer in fig 3 (skid), and there needs to be an aerodynamic sideforce vector acting at right angles to the lift vector, pointing to the left side of the page and downwards. That is the force that is missing from the diagram. It is generated by the air hitting the side of the fuselage. The vector sum of lift plus sideforce is the net aerodynamic force vector, and it must be exactly equal and opposite to the "load" vector.

With the figures drawn this way, we are able to see that we actually "load up" the wing in a skid, in the sense that we force it to create more lift than we'd normally need for a given bank angle, assuming that we are not letting the flight path curve earthward. The turn rate is also increased, and the turn radius is decreased.

(On the other hand, if we hold the turn radius fixed and leave the bank angle unconstrained, then a skid actually "unloads" the wing, because it involves a shallower bank angle.)

Second row-- forces acting on aircraft in earth reference frame (or in the reference frame of airmass moving at constant speed)--(these are valid inertial reference frames, at least from an elementary viewpoint that views gravity as a real force.)

Same as above but "centrifugal" force and "load" are omitted. A net force vector may be added which is the vector sum of net aerodynamic force and weight. It is horizontal, and exactly equal and opposite to the (omitted) "centrifugal force" vector. It is the centripetal force vector that causes the turn. For a given bank angle, it is smaller in the slip, and larger in the skid, than in coordinated flight.

Third row-- aerodynamic forces only-- same as above but now weight is also omitted. Now there is less clutter to take our attention away from the net aerodynamic force vector. Note that the net aerodynamic vector is aligned with the aircraft's sense of "up" (i.e. the direction of the vertical fin is pointing) in coordinated flight, but not in the slip or skid. This is arguably the most important row of diagrams. It shows us what the pilot really "feels".

Or if it makes more sense to us-- we can have a fourth row-- "apparent inertial force" "felt" by the pilot-- exactly equal and opposite to the net aerodynamic force vector on row 3. Just this one force vector for each figure, with all aerodynamic forces omitted. It is valid to say that this apparent inertial force is caused purely by the net aerodynamic force vector. It is also valid to note that it is exactly equal to the vector sum of gravity and "centrifugal force", though those won't be included on the fourth row of diagrams. It is also valid to observe that the "apparent inertial force" vector is exactly the same thing as the vector labelled "load" in the top row of diagrams.

Unlike the diagrams in the question, the diagrams in the answer won't show the wing's lift vector decomposed into horizontal and vertical components. No real insight is provided by doing that, especially when we completely omit the aerodynamic sideforce vector generated by the airflow hitting the side of the fuselage.

Unlike the diagrams in the question, the diagrams in the answer won't give the illusion that somehow the net aerodynamic force vector is "balanced with" (equal and opposite) the load vector, or the vector sum of weight and centrifugal force, in the coordinated turn, but not in the slip or the skid. That is simply not true. The load vector is equal and opposite to the net aerodynamic force vector in all three cases. From one point of view, the net aerodynamic force vector is what causes the load vector.

Unlike the diagrams in the question, the diagrams in the answer won't give the illusion that the "centrifugal force" vector is exactly "balanced with" (equal and opposite) the horizontal component of the net aerodynamic force vector in the coordinated turn, but not in the slip or the skid. Again, that is simply not true. The "centrifugal force" vector is equal and opposite to the horizontal component of the net aerodynamic force vector in all three cases. Because fundamentally, in this specific case where the vertical component of acceleration is constrained to be zero, we can observe that the centrifugal force vector is entirely caused by the horizontal component of the net aerodynamic force vector.

Unlike the diagrams in the question, the diagrams in the answer won't give the false impression that some mysterious thing, presumably somehow related to turn rate, yet apparently somehow unrelated to any actual aerodynamic force, is magically affecting the amount of "centrifugal force" the aircraft is generating in the slip or the skid.

If the diagrams in the answer actually include the airplane figure (I'm not much of an artist), it should be drawn yawed to the high side of the turn in the slip and yawed toward the low side of the turn in the skid. The flight path will be drawn coming straight out of the page toward the viewer, so the little arrows suggesting that the plane is sliding down and left in the slip, and up and right in the skid, will be omitted.

One interesting question is whether the aerodynamic sideforce vectors referenced above should be considered to include the sideways thrust vector cause by yawing the thrust line sideways relative to the aircraft. As described above, it seems they should, but interestingly, the sideways component of thrust vector has no tendency to displace the slip-skid ball, because it has no sideways component relative to the slip-skid ball, or relative to the pilot's seat for that matter. If we include the sideways thrust vector as part of the aerodynamic sideforce vector, then we are not really using a reference frame fully aligned with the aircraft itself, but rather a reference frame aligned with the direction of the flight path through the airmass at any given instant. It's interesting to think of other analogs-- for example, perhaps a flat-bottomed sled on an icy lake performing a turn by yawing sideways and then firing a thruster that was aimed purely in the aftwards direction in relation to the driver's seat-- that would be equivalent to a flat skidding turn that was somehow performed using only the sideforce generated by yawing the thrust line to one side, with the aerodynamic sideforce from the air hitting the side of the fuselage somehow playing only a negligible role. One airborne analog would be a perfectly spherical airship with a fixed motor in the back, plus thrusters that could establish any desired yaw angle between the flight path and the heading. Skidding turns in such an aircraft would not disturb the slip-skid ball.

• "The net force in the diagram must be zero", yes, and one can do that by including lateral force vectors from relative wind and thrust. Well done to continue the work here. In the skid (under power), the lateral component consists of the horizontal lift component (smaller) plus the horizontal thrust component plus the horizontal side force from the leeward side (it is skidding). In the slip, the lateral component consists of the horizontal lift component (larger) minus the horizontal thrust component (now outside the turn) minus the windward side force(slipping) – Robert DiGiovanni May 24 '20 at 13:40
• Notice, the vertical lift vector and the net lateral vector are the same for all cases, matching the vertical and horizontal loads. In a coordinated turn (theoretically), the lateral component is all horizontal lift component, the thrust is in the line of flight, and there is neither slip nor skid. The wing is king. – Robert DiGiovanni May 24 '20 at 13:47
• @RobertDiGiovanni -- well, I know exactly how I intend to draw the diagrams-- exactly as described a month ago in this answer-- just need to sit down and do it! – quiet flyer May 24 '20 at 13:59