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In flight Manual of Russian Mi-17 helicopter, it is give that ”max bank permitted in accelerated turns is 45 deg up to 1km altitude”. At the same time it also mention max bank permitted is 30 deg up to 1km altitude. What do we mean by accelerated turns and why difference between max bank permitted values ?

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    $\begingroup$ Can you provide a link to the document that you're quoting from? $\endgroup$
    – Ralph J
    Aug 19, 2023 at 16:07
  • $\begingroup$ Sorry Ralph, I don’t have soft copy with me. Thanks for time n effort. $\endgroup$ Aug 27, 2023 at 5:18

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What do we mean by accelerated turn

If "we" is intended to refer to the aviation community as a whole, rather than to the authors of the Mi-17 flight manual--

The term "accelerated turn" has no generally recognized meaning in aviation, and doesn't really make sense. All turns are a form of acceleration, namely, centripetal acceleration. Meaning that the net force acting on the vehicle cannot be zero in turning flight.1

My interpretation of the content at the top of page 2-5 of the Mil-17 flight manual is that it is trying to convey that there is a certain speed range (120-250 kph) where, if at low altitude and not above normal operating weight, the helicopter may be banked more steeply than otherwise be permissible. The choice to apply the specific label "accelerated turn" to this particular situation may be somewhat arbitrary, and may not reflect a standard terminology in the rotorcraft context.2

Footnotes:

  1. In relation to winged aircraft (rotorcraft dynamics may be different), one could concoct a definition of "accelerated turn" that meant that the wing was producing more lift than the aircraft's weight, so that the G-loading was greater than one. If the aircraft was then brought to the stall angle-of-attack while holding the G-loading constant, a stall entered from this condition would produce an "accelerated stall", where the stall speed was greater than the aircraft's normal straight-and-level stall speed. Conversely, if the wing was producing less lift than the aircraft weight, so that the G-loading was less than one, a stall entered from that condition while holding the G-loading constant would produce a stall at an airspeed lower than the normal 1-G stall speed. It is possible to turn with a G-loading of less than one, but (at least in the context of coordinated flight rather than "knife-edge" maneuvers where the airflow is striking the side of the fuselage and creating a lifting force) the flight path will always be curving earthward due to insufficient lift. And therefore such maneuvers may only be sustained for short time scales. (The "top" part of a steep-banked, exaggerated "wingover" is an example of such a case-- the G-load may drop well below 1 G, and the airspeed may drop below the normal 1-G stall speed, but neither condition can persist for long.) But even in a turn that was "non-accelerated" according to this made-up definition, the net force on the aircraft would be not be zero and the aircraft would be in fact undergoing a centripetal acceleration-- as well as, in most cases, an acceleration component along the instantaneous direction of the flight path. So let's throw this made-up definition into the garbage bin.

  2. Thanks to this related answer for bringing this page in the flight manual to our attention.

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  • $\begingroup$ @RTO-- good point; my footnote was addressing the airplane situation. I don't have much knowledge about helicopters. Will edit to clarify. $\endgroup$ Aug 22, 2023 at 15:49
  • $\begingroup$ Thanks quite flyer $\endgroup$ Aug 27, 2023 at 5:19
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The Mi-17 flight manual shows those limitations on pages 2-4 and 2-5, but under different circumstances.

The normal bank angle limitations have weight and altitude limitations.

The statement of allowed bank angle in accelerated turns states turns up to 45 degrees are allowed in accelerated turns at up to 1000 meters altitude, airspeeds between 120 and 250 km/h and "normal or subnormal gross weight".

Essentially, the flight manual is telling you the limits during normal flight ops, but also what the limits are if you need to engage in extreme maneuvers, like getting out of the way of a missile about to come up your rear, for example.

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I would interpret an accelerated turn in contrast to a steady turn.

A steady turn is one that can be conducted without losing energy -- speed and altitude. It is fundamentally a power limit.

Turns can also be limited by stall or structural limitations.

If (at a certain flight condition) you turn at greater than the max steady turn rate (but still within stall and structural limits), you will have to bleed off speed and/or altitude to continue the turn.

Of course, the other way to have an accelerated turn (i.e. a turn that is not steady) is to turn at a rate that is less than the steady turn limit (while at full power) -- in that situation, the aircraft would be able to climb or increase speed during a turn. I don't think this is what they're talking about here.

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  • $\begingroup$ In other answers on ASE, I'm pretty sure I have seen the word "stationary turn" used to describe what you are calling a "steady turn". So it seems you are suggesting that what might logically be called a "non-stationary turn", is what the manual is calling a "stationary turn". But-- in relation to the loads that the helicopter structure can tolerate-- why would it really matter whether the aircraft is losing energy or not? $\endgroup$ Aug 22, 2023 at 16:09
  • $\begingroup$ I'm largely using fixed-wing language when I say steady turn and then extend to accelerated turn. The structural limit does not care about whether the aircraft is losing energy. I read all this to mean that the energy maintaining turn limit is 30 degrees, but if you're willing to lose energy, you can structurally go up to 45 deg. $\endgroup$ Aug 22, 2023 at 16:12
  • $\begingroup$ Re your last sentence above-- if that's what the manual intends to convey, I'd say that it's poorly written-- $\endgroup$ Aug 22, 2023 at 16:17

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