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Is a helicopter's flight path (track made good) an arc of constant radius regardless if it's a coordinated or uncoordinated turn? (Assume nil wind conditions and that the pilot maintains the angle of bank, altitude and airspeed constant in both cases.)

I'm confused because texts mention slipping and skidding as sliding downhill and skidding uphill from the turn respectively, which to me reads like the radius of the turn changes.

[EDIT] Updated assumptions to include bank angle and airspeed.

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    $\begingroup$ "Assuming a constant altitude" is a big assumption. In a turn, power must be added to maintain acceleration. This in turn increases torque, and therefore all kinds of side-effects. I've consulted Prouty and Wagtendonk (perhaps the two leading authorities on rotary dynamics) and I can't find an answer to this. My intuition says that yes, the radius will be constant since with no wind, there is nothing to disturb the equilibrium except, perhaps, for the helicopter encountering it's own vortex (which is a good test of skill when making constant attitude coordinated turns) but I really don't know. $\endgroup$
    – Simon
    Feb 1, 2017 at 19:55

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Given your assumptions about bank angle and airspeed and my assumption that coordinated turn is determined by acceleration inside aircraft pointing downward "perpendicular to the floor" (if you mean anything else by coordinated turn, try to be more specific on it), then you have one exact radius for coordinated turn. That one where centrifugal force generated by airspeed and curvature summed with gravity gives vector matching the (fixed) bank angle. If you want local acceleration to point in other direction (making slip or skid), only free variable remaining is turn radius.

So yes, given airspeed and bank angle is constant, turn radius is the parameter determining if turn is coordinated or not.

In a slip the local acceleration in aircraft points toward that side you want to turn (or, in other words, aircraft's floor is banked too much outside the turn). To get this, centrifugal force has to be less compared to coordinated turn (centrifugal force is only quality you allowed to change), so turn radius have to be larger. To get a skid turn radius have to be smaller for the same reason.

Inertial and gravity acceleration as seen and felt inside an aircraft during a turn with bank angle $\alpha$: aircraft in turn where $\vec{g}$ gravity acceleration and $\vec{a_c}$ centrifugal acceleration. Resulting $\vec{a}$ is total acceleration felt by pilot. From basic mechanics $a_c=v^2/r$ where $v$ and $r$ is speed (either TAS or GS, both are the same here) and turn radius respectively. Turn radius for coordinated turn is $\tan\alpha=a_c/g\Rightarrow r_{\rm coord}=v^2/(g\tan\alpha)$. Any smaller radius with the same bank and speed leads to higher $a_c$ and thus skid, larger radius results in slip. Note that this is true for any aircraft, not specific for helicopter.

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  • $\begingroup$ Martin, given that you've stated the turn radius determines whether its a coordinated turn (as we've assumed other parameters remain fixed in this example), would the respective flight paths then look like !this? $\endgroup$
    – jumblie
    Feb 14, 2017 at 3:07
  • $\begingroup$ @jumblie I believe, it is just other way around. Slipping turn gives (with the same bank angle) larger turn radius than coordinated one, skidding turn smaller. If it is not clear from images above, you can try to imagine it like this: would you fly according to the "step on the ball" rule, then in slipping turn the ball is "inside turn" and you have to step on bottom ruder to get coordinated turn. That is you need to "turn more". $\endgroup$
    – Martin
    Feb 14, 2017 at 7:46
  • $\begingroup$ One could tend to believe slip gives "more turning" because we have sideslips, right? But note that sideslip (or forward-slip, it is the same maneuver aerodynamically anyway, only ground track differs) is no turning at all, it is airplane flying straight forward, only difference is the nose points "wrong way". $\endgroup$
    – Martin
    Feb 14, 2017 at 7:49
  • $\begingroup$ I've amended my diagram so that the tighter turn is the skidding turn and the larger turn is the slipping turn, but its still not quite clear in my head. Regardless if the turn is a coordinated/slipping/skidding one, will the aircraft always be tangential to the track? Also, sliding downhill makes it sound as if the radius would get smaller and skidding uphill as if the radius would get larger but its exactly the other way around! $\endgroup$
    – jumblie
    Feb 17, 2017 at 11:08
  • $\begingroup$ @jumblie IMO has any aircraft in skid/slip to fly partially sideways. Approximately like you have sketched it in your drawing. Reason becomes clear when you look on forces from outside. You need, in addition to the lift some aerodynamic force pushing aircraft to the side. Reasoning will be little different for fixed-wing and helicopter (because rotor teeter allowing lift to point somewhat aslant relative to helicopter's fuselage -- someone actually flying helicopter can correct me here if I am wrong), but the result should be same. I'll try to add some explanation into my answer eventually. $\endgroup$
    – Martin
    Feb 19, 2017 at 0:28

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