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jagannath
5th Feb 2014, 05:22
sir
for a given rotorcraft weight, main rotor diameter, chord, airfoil, rotor rpm what may be the 'g' capability the rotor can sustain(steady or transient) during maneuver(pull up/turn/wing over...). Is there any empirical formula to calculate rotor capability in terms of 'g' for a single main rotor and tail rotor helicopter configuration. Suppose if there is no empirical formula as such then for a given weight of 3.2Tonnes, hingeless rotor what may be the rotor capability in terms of 'g' for similar contemporary helicopter weight class. Am talking about aerodynamic 'g' capability of rotor.


Thanks in advance


sincerely
Jagannath

Art E. Fischler-Reisen
5th Feb 2014, 09:02
There is no easy answer because no manufacturer tests G limits in any definitive way. Most Rotorcraft Flight Manuals say "Aerobatic manoeuvres are prohibited" or similar because of this.

If you deliberately pull G you are a test pilot.

skadi
5th Feb 2014, 09:09
If you deliberately pull G you are a test pilot.

so how do you fly f. e. 30° bank without pulling more than 1 G ?????

skadi

Lord Spandex Masher
5th Feb 2014, 09:09
If you deliberately pull G you are a test pilot.

Does one fly in a straight one all day?

HeliComparator
5th Feb 2014, 09:53
Am talking about aerodynamic 'g' capability of rotor.


Such a calculation would be very complicated. It would have to take into account so many variables such as rotor blade section both chord wise and spanwise, blade twisting under load, blade area, rotor speed, air speed, air density and mass.

In practical terms I suspect that unlike a fixed-wing, rather than having a "break" at the stall, attempting to pull increasing amounts of g becomes harder and harder since the rotor system tends to mush, caused I presume by some part of the disc becoming stalled whilst others remain flying, and perhaps an increased amount of blade twist from aerodynamic load. Thus there is a massive increase in drag as the limit of lift is approached.

JohnDixson
5th Feb 2014, 10:46
Do some homework and look for descriptive information on a value termed Equivalent Retreating Indicated Tip Speed or ERITS, for short. It is a short cut way to get an approximation of a rotor's stall boundary. Not perfect, but gets one in the ballpark.

As to the maneuvers to obtain maximum G, look for a copy of the original Mil Standard D 23222. Look in the Structural Demonstration Chapter. Don't do that stuff without adult supervision.

Ascend Charlie
5th Feb 2014, 18:16
Generally you will find it is around 2.7g

You should be able to maintain a level turn at 60 deg bank = 2 g, plus a fudge factor for turbulence and ham-fisted stick manipulators.

BK117 is listed as +3.5 to -1 because of its rigid rotor head and military pedigree via the BO105

ShyTorque
5th Feb 2014, 20:01
I'd hardly call turns of 30 degrees AOB the sort of situation being asked about. Try maintaining the 2G required for a sustained 60 degree AOB whilst maintaining a constant airspeed though. Or pulling to the vertical from close to Vne. Most helicopters will begin to protest.

albatross
6th Feb 2014, 00:16
Old daze for me but I recall in the Astar 350D you had reached it when you felt feadback in the cyclic. In the Twinstar 355 when the Limit Light illuminates.
Yes it is in the RFM.

Shawn Coyle
6th Feb 2014, 01:07
Typically, you can't generate enough G to do any structural damage - any G that you can generate transiently decreases pretty rapidly in both time and amount of G.
Most I've ever seen was on the Flight Data Recorder for a helicopter that had a main rotor servo hardover - I think that was all of 3.2G. Mind you most helicopter pilot seem to black out at about 2G… (he said tongue in cheek)

busdriver02
6th Feb 2014, 01:56
Max G will be bounded by excess power available for a given velocity and onset of blade stall. In other words, if your aircraft can fly strait and level at a given airspeed at 10,000 pounds gross weight and has enough excess power to also sustain level flight at the same airspeed at 20,000 pounds gross weight it should be capable of sustaining a level 60 degree bank turn (2G) at that airspeed, assuming it doesn't run into blade stall. Actual structural limits are a different animal.

AMDEC
7th Feb 2014, 07:42
Try this

n = 0.2 ρ b ℓ R U² / Mg

where
· ρ is the air density in kg/m3 (1.225 in ISA SL conditions)
· b is the number of blades of the main rotor
· ℓ is the average chord in meters
· R is the main rotor radius in meters
· U is the main rotor tip speed (Ω.R, where Ω is the rotor RPM) in m/s
· Mg is the helicopter weight in Newtons

This should be valid in the 80-100kt range where you can reach the maximum load factor. It decreases at highre speeds.

Fareastdriver
7th Feb 2014, 09:56
Why didn't I think of that?

JohnDixson
7th Feb 2014, 12:20
AMDEC posted in part:

"This should be valid in the 80-100kt range where you can reach the maximum load factor. It decreases at highre speeds."

Possibly true for some rotor designs, but newer rotors that I am familiar with produce the best flight test results in the 135-170 kias range. Perhaps you were referring to steady state rather than maneuvering Nz capability? The speed range I cited applied to maneuvering flight, i.e., pull outs of one sort or another. In those maneuvers, there are other factors that enter the picture beyond the Cl max value.