Understanding Roll Momentum

[seafuryfan]

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 (http://bestaviationarticles.com/?cat=6)

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

[Epiphany]

Thanks. I'll try to remember that next time I lift into the hover. :)

[[email protected]]

I don't/can't/won't do maths but the article shows clearly the 3 main scenarios where dynamic rollover can occur. It is, as you say, usually insufficient or incorrect cyclic input that allows the roll to start in the first place and looking ahead at the disc and horizon is the best way to see any tendency to roll developing. Those who like to take-off staring at the '2 o'clock daisy' won't notice the roll and are more likely to incorrectly position the cyclic in the first place.

Once it starts to go, it is the collective that determines the rate of roll, especially in a teetering head helo where you have very little control power - hence the corrective treatment is to lower the lever fully.

[Arm out the window]

Depending on the type, you'll have a certain amount of lateral cyclic range.

Once the roll develops (eg slope takeoff, one skid or wheel snags an obstruction and starts a rolling motion), full opposite cyclic will have an anti-roll effect only up to the point where the angle of bank is still less than the angle through which you can tilt your lift vector the other way with cyclic.

Beyond that, even with full opposite cyclic, any lift from the rotor is pro-roll, which is one reason rollovers can go so wrong so quickly.

Hence the technique of dumping the collective to dump the lift and remove that driving force, although by that stage it's probably too late anyway.

Not sure if that adds any clarity to the discussion!

[delta3]

seafuryfan

The math, nor the scenario are correct.

Many articles that I checked on the reference you quote, are either wrong or incomplete, because the articles don't do the complete math, but assume instead a certain scenario, adress some and not all of the relevant forces.

I leave it there, I suggest you consider carefully all forces and the scenario that should be triggered by pilot controls: will he keep pulling collective and not change cyclic ? I doubt that.

m2c
d3

[[email protected]]

Delta3 - you don't need the maths to understand dynamic rollover and the scenarios in the article are perfect for showing the likely conditions when dynamic rollover might be experienced.

The movement of the cyclic to oppose the roll is only as effective as your control power (how much a cyclic input can affect the fuselage attitude) ie next to zero in a teetering head helo, increasing as the effective hinge offset increases.

In most helicopters, once the roll starts, any movement to oppose that roll with cyclic is going to have a minimal effect compared to raising or lowering the collective. This is the whole reason that dynamic rollover continues to crash helciopters - the correct recovery needs to be taught as it is counter-intuitive - ie most pilots will oppose roll on the ground with cyclic, just like they would in the air - it is a natural reaction.

[fly911]

For the few times it has almost happened it me, I've learned that early recognition is key to prevention. Landing on a slope, in muck, soft dirt, or uneven terrain and I lift off very gingerly, anticipating a rollover situation.

[Darren999]

I unfortunately watched a friend roll his 206 over, on a clear bright sunny day. I spoke with him after the accident. The helicopter was on a slight slope, he was new to the 206, he lifted collective, he felt the left skid lift, assumed, the right skid had come up also. He applied right cyclic, pivoted on the right skid, and over she went.
He said from the time he moved the cyclic to the time he realized he was laying on his side was so fast!
A real shame, he was OK, a little shaken, and now sans helicopter :(

Fly safe...

[seafuryfan]

Thank you for all the replies to my query. Be assured Epiphany, that I won't be doing the calcs in the cabin as collective and cyclic are applied. More like ensuring my own safety in case it all goes wrong.

I couldn't get the link to the AP to work (as usual) but found a hard copy extract. delta 3, yes I have considered carefully all the forces with the help of the AP but I still found the maths helpful in that it factored in the horizontal component of rotor thrust, for a given scenario. If you want to amplify on the maths, please do.

Contributory factors (other than roll momentum) I've come across:

Angle of slope with the ground.
Total mass of the ac.
Distance of ac C of G from the pivot point.
Amount of tail rotor force which adds to the rolling force (if tail rotor is facing the 'wrong' direction into slope).
Offset flapping hinges with disc level, producing a centrifugal loading which tends to roll the ac up slope.

[Peter-RB]

My Lady instructor was paranoid about watching out and feeling the power being fed in, I on the other hand became paranoid about sloping ground, forward or backwards I didnt have a problem but sideways always made a certain part of me tighten up to full Pucker, and still is the main point of control/concentration when in that situation.

Call me "chicken" but I have always tried to get on level ground in 99% of my landings.

Peter R-B

Lancashire

[delta3]

crab


I agree with your practical comments, but:

I still don't like the reference because it should at least show first that the max cyclic counter angle is in all helis (I know of) smaller than the angle between vertical and COG-skid line.

For me it is a geometry question. Force only will explain how fast you roll.


The reference keeps center cyclic...
I would use increasing counter cyclic in the drawings.

d3

[EN48]

Trying to clarify my understanding:

Static rollover: once CG moves "outside" the pivot point, the restoring moment present when the CG is inside the pivot point becomes a roll moment and over it goes.

Dynamic rollover: rate of roll builds in a non linear way with horizontal thrust component increasing rapidly(function of angle and collective and cyclic application); if roll not stopped before momentum carries the CG beyond the pivot point, over it goes.

In looking at the geometry of the B407 I fly (with low skids) it appears that the angle for static rollover would be extreme, at least with full fuel (1000 lbs, which sits low in the airframe). With light fuel load, CG would be higher an the critical angle would be reached sooner in the roll.

Comments?

[delta3]

EN48

You'r correct in stating that the low skid is statically more stable, but this has the reverse effect on dynamic stability:

the counter torque available from the rotor system remains the same, but in case of a stuck skid any vertical lift/motion will make the hull roll faster, making dynamic roll more brutal.

An extreme (hypothetical) case would be that both skids are so high and close together, that the system is statically not very stable but becomes dynamically stable, because the angle between COG and skid becomes smaller then cyclic authority angle.

d3

[[email protected]]

D3 - could you explain and expand on that theory?

A wider and lower skid/wheelbase gives an improved static stability so that you have a restoring moment for longer as the fuselage starts to roll - you have to create a bigger rolling moment with more cyclic/collective to get into dynamic rollover.

With a high and narrow skid you would lose that restoring moment very quickly thus the aircraft would be more prone to dynamic rollover with less cyclic displacement.

[delta3]

Crab

I can't publish a drawing right now, but the point I have trying to make (difficult apparently without a drawing) is that dynamic (un)stability depends mostly on geometry.

Picture a heli on the ground. Look at the lateral angle the rotor can make and draw a line between the hub and the pivotal skid. If the rotor can tilt enough to get an intersection outside the skid, then the cyclic would be able to redress the heli. In the case of commercial heli's due to the static stability requirement this will not be the case (perhaps a sky-crane?)

I think any normal pilot will try to avoid tipping over by applying cyclic, due to surprise perhaps not full cyclic, but at least quite a bit. I would start analyzing at this point.
Keeping the cyclic were it is, allows you to draw the trust vector. The perpendicular to this trust vector from the skid gives the arm (if cyclic is applied this will be less than half width)

Now you will see that exactly for the same reasons width gives static stability, it creates dynamic unstability, because the rotor trust also gets a greater arm to tip the heli. So smaller retaining forces on the skid will be able to make the make heli flip.

A wider heli has however the advantage that it will need to lift higher before tipping over, mathematically this is equivalent to stating that the inertia tensor for roll around the skid will be much larger than the tensor around the COG. This will slow down angular acceleration and give the pilot more time to lower the collective, so at this point I should nuanciate my previous statement.

d3

[EN48]

d3,

Thanks for your reply. Got me to thinking - usually but not always a good thing.

You'r correct in stating that the low skid is statically more stable, but this has the reverse effect on dynamic stability:


It would appear that this is true only if the low skids are wider apart (longer moment arm) than the high skids. On the B407, the standard low skids and factory optional high skids have the same skid to skid dimension according the the Manufactirer's Data section of the RFM. So, if my reasoning is correct, there would be no DR penalty for the low skids. If the low skids had a wider stance then it would seem that the moment arm is longer and the DR characteristics would be more "dynamic." Comments?

An extreme (hypothetical) case would be that both skids are so high and close together, that the system is statically not very stable but becomes dynamically stable, because the angle between COG and skid becomes smaller then cyclic authority angle.


Very useful example - one I havent seen before. And clearly illustrates that the initial rolling moment is produced by the vertical thrust component acting through the skid to mast moment arm.

[[email protected]]

D3 - I see where you are coming from but the ability of the cyclic to affect the attitude of the fuselage is determined by the control power - simply stated as the amount of leverage the blades have on the rotor mast.

Control power on a teetering head is next to zero whilst articulated heads have moderate control power and semi-rigid type heads have lots - it is about the physical distance of the flapping hinge from the rotor head or, in the case of semi rigid heads where there might not be physical hinges, the effective hinge offset.

So just looking at the tip path plane to see how far the cyclic can move the disc is not sufficient - whatever the cyclic displacement, the amount of thrust and roll momentum produced will be determined by the collective position.

It is for these reasons that R22s, 206s and other teetering (or near as dammit) head helicopters are more susceptible to dynamic rollover - moving the cyclic in the opposite direction just doesn't do much especially once the fuselage is already moving in the opposite direction.