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An aircraft in a 45 degree banked turn has a horizontal lift force vector equal to the weight of the aircraft. Although the Coefficient of Drag is much larger for the aircraft profile, it seems there must be a significant side drag force on the aircraft as a result of the lateral velocity created.

Using the ROT formula: 1091 × Tan bank angle/V knots

a 45 degree turn at 100 knots gives an ROT of around 11 degrees/sec

Sin 11 degrees x 100 is around 19 knots lateral velocity.

Is this an accurate number? Must we consider this side force even if the plane is "coordinated" from the pilots perspective?

Because the nose is also rotating into the turn, this number may be less. Is it more than 0?

Are these the vectors describing a coordinated turn?

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    $\begingroup$ “Side drag”? What’s missing from established explanations about aircraft turning that prompts such new terminology? (Or dare I ask…) $\endgroup$
    – Michael Hall
    Feb 21, 2022 at 16:15
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    $\begingroup$ This is Physics 101. The velocity vector is "straight ahead" and tangent to the circle. Period dot end. True on a 2D chalkboard, and still true in a 3D atmosphere with bank angle and lift components and all the rest. $\endgroup$
    – Ralph J ♦
    Feb 21, 2022 at 21:09
  • $\begingroup$ If velocity is not changing (in your defined Frame of Reference), then side force is zero. Problem is, What frame of reference are you measuring things in? The airmass? The Earth?, the Aircraft? In each of these, the measured velocity will be different, and to get the equations to balance you may need to add one or more fictitious forces, (or a "SideForce") to make it work out. Without specifying (and talking about) the frame of reference, everything else is ambiguous and undefined. $\endgroup$
    – Charles Bretana
    Feb 21, 2022 at 21:43
  • $\begingroup$ @RalphJ so what you're showing on your blackboard is sideways movement, driven by the wing. Extra lift (therefor extra drag) is borne by the wing. Agree the velocity vector is straight ahead (projecting from the board) and the fuse as aligned with the relative wind as much as possible. Again thanks, before moving on to turning techniques specific for aircraft such as hang gliders. Though there is a lot of sideways acceleration, constant re-alignment of the nose to the turn minimizes side velocity, like catching a 10 m/s$^2$ accelerating stone after a 2 inch drop. $\endgroup$
    – Robert DiGiovanni
    Feb 21, 2022 at 22:21

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An aircraft in a 45 degree banked turn has a horizontal lift force vector equal to the weight of the aircraft. Although the Coefficient of Drag is much larger for the aircraft profile, it seems there must be a significant side drag force on the aircraft as a result of the lateral velocity created.

A sideways force component does not imply a sideways velocity component. That is an Aristotelian way of thinking. Nowadays, we understand that a force component drives an acceleration component, not a velocity component. And turning flight is a form of acceleration, even though the magnitude of the total velocity vector may remain constant. The aircraft is constantly accelerating toward the center of the turn. But this does not imply that there is any velocity component towards the center of the turn.

How to calculate lateral velocity of a turning aircraft

In a "coordinated" turn1, the lateral velocity toward the center of the turn, in the airmass reference frame, is always zero, because the aircraft is always pointing in the same direction as it is moving through the air.2

The critique of the actual formulae presented in the question, will be left to other answers.

Footnotes:

  1. It's important to note that turns tend not to be completely coordinated unless the pilot (or the autopilot) is making a rudder input to accomplish this. An aircraft has some tendency to sideslip-- the nose tends to point slightly toward the "outside" or "high side" of the turn. This tendency is typically most pronounced as an aircraft is entering the turn, but is typically present to some degree even in a constant-banked, steady-state turn, especially at lower airspeeds, or more precisely, at lower "scale speed", which is inversely related to the time to cover one fuselage-length. The reasons for this involve the difference in airspeed (and therefore drag) between the inboard and outboard wingtip, the effect of the curving flight path and relative wind on the vertical fin, etc, and are best addressed in detail elsewhere, but this effect does have a very significant effect on lateral stability. It is the reason that geometries like dihedral, sweep, and the multiple different effects related to high-wing or low-wing geometry, do affect an aircraft's lateral stability, including an aircraft's rolling-in or rolling-out tendencies in a stabilized constant-banked turn, in a situation where the pilot is making roll inputs as needed with the ailerons, but is not using the rudder to ensure that the turn is completely coordinated. Calculating this sideslip angle in any given situation is far beyond the scope of this answer!

  2. It's pretty obvious that the question was intended to be about velocities in the airmass reference frame, or in no-wind situations. Computing the lateral component of the groundspeed at any given instant is a solvable problem, but best addressed elsewhere.

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  • $\begingroup$ Just want to be sure. What you say makes sense, and this is why PK concluded has "super high wing" would do fine in a turn. $\endgroup$
    – Robert DiGiovanni
    Feb 21, 2022 at 12:54
  • $\begingroup$ @RobertDiGiovanni -- added content in what is now the first footnote, may be helpful-- $\endgroup$
    – quiet flyer
    Feb 21, 2022 at 13:07
  • $\begingroup$ Multiple points made in this answer could be associated with links to other ASE answers where the point is developed in more detail-- don't have time to do a full search right now, stand by-- $\endgroup$
    – quiet flyer
    Feb 21, 2022 at 13:13
  • $\begingroup$ No velocity component towards the turn, got that, but in the context of an orbit the plane moves a certain distance forward and a certain distance to the side (while it rotates to maintain low drag). It does not have a velocity component towards the center because the forward motion and inward acceleration keeps it at the same distance. This is for academic pursuit, we all know coordinated turns are best, but what does a straight yaw string on a banked aircraft tell us? $\endgroup$
    – Robert DiGiovanni
    Feb 21, 2022 at 15:48
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    $\begingroup$ Again, before talking about or performing an analysis of the forces, velocities, and accelerations involved in any scenario, it should be specified and defined as to what frame of reference the comments, discussion, analysis is going to be done in. Indeed, the choice as to what frame of reference we will do the analysis in is often the most important decision affecting the ease of the analysis, and the clarity of the discussion around it. $\endgroup$
    – Charles Bretana
    Feb 21, 2022 at 21:50
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Easy calculation: lateral VELOCITY is ZERO in a coordinated level turn, using the air-mass as the frame of reference.

Lateral ACCELERATION is non-zero, and is moving the velocity vector at the same rate that the heading is changing. The velocity vector is, in a coordinated turn, aligned with the nose of the aircraft, so lateral velocity can only be zero.

The uninteresting exception to this occurs when using a ground-based frame of reference and wind exists. At that point, you would have lateral velocity to the extent that the wind gives you some crosswind component. But that's the case also in straight & level flight, and is unrelated to all dynamics of the aircraft & its turn.

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    $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – Farhan ♦
    Feb 23, 2022 at 19:32

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