Hello. Concerning dynamic roll over, I’ve been trying to understand a little more the build-up of roll momentum. This website
Helicopter | Best Aviation Articles
seemed to explain it in language a bit more friendly than AP3456, and I’ve reproduced the maths below. I’d be grateful if you could have a look and see if this looks right and add comments that would help my understanding.
Given:
Arm distance to the pivot point (skid to the rotor head) = 3 meters
Cyclic design limitation of a particular ac = 10° of roll per second, which may equate to 10000 units of roll momentum. whichever comes first.
On take off, the pilot pulls power set to the equivalent of 1000kgs of total rotor thrust thrust (equalling the ac)
Units of roll momentum =
Horizontal component of rotor thrust
x arm to pivot point distance
x the roll rate²
Initial roll momentum =
100kgs
x 3m
x 1°² of roll per second = 300 (less then 10,000 units, cyclic can counter sufficiently)
Pilot then pulls more power to equivalent of 1100kg of thrust, increasing horizontal component,
200
X 3
X 2°² = 2400 (less than 10,000 units, still within cyclic limits)
The roll momentum has increased by a factor of eight
The pilot feels the roll momentum further developing, pulls the lever to 1200kg thrust….
300
X 3
4°² = 14400 (outside cyclic limits, roll momentum continues until ac rolls over)
Is it fair to say that the situation develops because initally the pilot hasn't applied enough opposite cyclic in the first place, allowed the exponential roll momentum to develop, and he panics in the end by just using more lever? (Not that any of you chaps would panic of course!)
Can anyone also tell me how I describe the roll rate, as I don’t know how to say,
“1°²"
Does it mean ‘one degree roll squared per second’?
Cheers,
SFF