Weight vs ROD in Autorotation

[chopabeefer]

I know from experience and countless test pilots telling me so, that the weight vs rate of descent relationship, in autorotation, is counter-intuitive.

I have proven this to myself, and accept it as fact. But why is it so? I have heard a few arguments, all convincing, but 2 of them contradict each other.

I won't mention them as I do not want to muddy the waters - but will if asked. So, here is the statement - I am still looking to find a convincing argument...!


Take 2 IDENTICAL helicopters. They are exactly the same in every respect, except their weight. Both are put into autorotation at the same speed, density alt etc etc- EVERYTHING is identical except their weight - one is heavier.

For a given Nr, the heavy helicopter WILL have a lower RoD than the lighter one. This is a fact, and easy to prove.

My small brain cannot work this out. I know it to be fact, and have read it countless times. But why?

[Aucky]

Interesting - I hadn't heard this before. I'm guessing that if it's true it's due to the fact that at a higher AUM then more collective pitch will need to be pulled in order to maintain the given NR. More pitch, more lift, lower ROD.

It's a funny relationship, the fact that at 90% rpm in auto the ROD is usually lower than at 100%. However given that in the lift equation the V (mostly rpm) is 'squared' one would think that the lift would be noticeably reduced at lower RPM and therefore the ROD higher? However in increasing pitch to reduce the RPM the increase in coefficient of lift seems to outweigh the reduction in Vsquared... the result being a reduction in ROD. Perhaps someone can clarify further as it seems a little counterintuitive.

[dangermouse]

as a first order approximation.

kinetic energy required to keep the rotor going round needs to come from somewhere when the motors stop (this value is pretty fixed regardless of aircraft mass or collective pitch)

KE =1/2xMxVxV where M is the rotor mass & V is rotor speed

an aircraft at height has potential energy

PE=MxGxH where M is aircraft mass, G is a fixed constant, H is height (so a heavy helicopter has at the same height more potential energy than a light one)

therefore converting aircraft potential energy (variable with mass) into rotational kinetic energy (fixed by rotor mass) requires a loss of height.

if the rotational power is fixed (ie rate of energy use) the rate of potential energy change is also fixed, so an aircaft with more mass must lose less height in a fixed time than a light one for the same energy conversion rate, ie a heavy aircraft will lose less height in the same time than a light one when in autorotation as can be demonstrated

This leads to the interesting (but patently nonsense) statement that an infinitley heavy aircraft can hover in autorotation (rate of descent becomes infinitley small at infinite mass)

DM

[Arm out the window]

My gut feeling is that, just as at the top end of the power curve where you can sometimes gain more efficiency by lowering Nr (beeping down) and pulling more pitch to compensate, thereby getting the blades to a more efficient angle of attack and therefore saving fuel, you can do something similar in auto-land.

Light aircraft, bottomed collective to maintain auto Nr, get a certain rate of descent.

Heavier aircraft, holding some pitch to control Nr in the green, blades are now at a more efficient angle and therefore ROD is lower.

No maths there, but it sounds OK to me - critiques welcome.

[TimdeBoer]

My best guess, as a former glider pilot and a few hours in helicopters, that it has to do with the speed.

In gliders there is an optimum glide ratio at a certain angle of attack. Max CL/CD ratio.

The heavier the glider the higher the speed at which this optimum is found.

So a more lightweight equal glider will sink faster than the heavy glider when flying at the speed at which the heavy glider is at max CL/CD ratio.

For helicopter in autorotations there are also speeds for optimum glide ratio and minimum sink rate.

Tim

[lelebebbel]

as a first order approximation.

kinetic energy required to keep the rotor going round needs to come from somewhere when the motors stop (this value is pretty fixed regardless of aircraft mass or collective pitch)

KE =1/2xMxVxV where M is the rotor mass & V is rotor speed

an aircraft at height has potential energy

PE=MxGxH where M is aircraft mass, G is a fixed constant, H is height (so a heavy helicopter has at the same height more potential energy than a light one)

therefore converting aircraft potential energy (variable with mass) into rotational kinetic energy (fixed by rotor mass) requires a loss of height.

if the rotational power is fixed (ie rate of energy use) the rate of potential energy change is also fixed, so an aircaft with more mass must lose less height in a fixed time than a light one for the same energy conversion rate, ie a heavy aircraft will lose less height in the same time than a light one when in autorotation as can be demonstrated

This is the same explanation that I use when someone asks me about this, but it skips one important point (highlighted). Energy use must actually be higher in the heavier machine, because it requires more lift to control its rate of descent - or in other words, more pitch = more drag = more power consumed by the rotor.

For some reason, the additional energy of the extra mass outweighs the higher power consumption of the rotor. I suspect there is no easy way to show why that is the case, and a mathematical explanation might be necessary to really get to the bottom of this.

I also suspect that the relationship between weight and ROD is not consistent, because the efficiency of the rotor system will change at different AOA. Higher weight will only reduce ROD up to a certain point, and there must be a "best weight" for autorotation for any given helicopter.

[mfriskel]

How about this:
If you use a given collective position in a helicopter with all conditions equal except weight, the heavier aircraft will have a higher rotor RPM and a faster rate of descent.

If you use a given rotor RPM in that same helicopter, the higher weight helicopter will have a higher collective angle.

The item that changes will be the autorotative regions of the rotor.

A helicopter with identical conditions except for MR RPM- lower RPM (within limits) will descend slower.

A helicopter with identical conditions except gross weight - heavier acft will glide farther.

If you have been taught to hold the collective against the floor, in all cases during autorotation, you are still at the rote level of learning autorotations.

[rotorfossil]

I suspect that the answer may be that the higher weight requires a higher lever setting and therefore higher angle of attack to maintain to maintain the same Nr. This moves the L/D of the whole rotor disc towards its optimum conditions. The same reason that max range in autorotation is achieved generally at the lower end of the RPM band.

[topendtorque]

Haven't read many maintenance manual but a simple R22 says when adjusting auto RPM, lever must be full down. Their (Beta) difference between min and max AUW is 205 Kg and min and max power off rpm is 102. Variance is thus a straight line affair.

All machines need a design to not exceed either limit at each weight marker with the lever right down. Otherwise how would the ROD be arrested at the bottom if the lever is already half up, godamm?

In the auto there is only one power source

At min AUW, lever down and RPM at the bottom of the green the A of A might be higher than at max AUW because then RPM will be at top of green, even though the pitch setting relative to the rigging is the same.

This is because the blades are travelling at a faster tip speed at the high AUW, so more power is being consumed, thus the ROD should be higher?

But how much different will the A of A be? The relative airflow at a higher ROD will be from a different direction. If uncle Nick was here he'd most likely nail it in a second.

Operationally which is what counts; as far as range is concerned, when helicopter is empty I land way over there, full right up, it's right here now.

The ROD for each weight marker I have experimented with but don't remember just now, I'll check it out in a few days, but at least this is more relaxing than budget debate on telly.

cheers tet

[Tourist]

I think the opening statement is untrue.

A heavier aircraft will not always have a lower ROD, though it may be true in some circumstances.

This can be proved with a simple thought experiment.

If heavier was always lower ROD, then all you have to do is keep adding weight, and eventually it will stop descending.

Since this is obvious bollocks, then the statement is false.

Apologies DM, you got there before me:ok:

[army_av8r]

lets look at this in a different baseline comparison. 1 helicopter at max weight, and that same helicopter at a reduced weight.(say full tank + Pax) and (near empty + no Pax). also must assume that the collective pitch is adjust correctly by maintenance to give you proper autorotational rpm range throughout the expected gross weights. finally we must assume that both are flown at the same conditions(DA, Airspeed, etc...). with all that being said. Both helicopters will have the same ROD if the NR is at 100%. the difference will be... if both pilots put the collective all the way down, the heavier aircraft will see a large overspeed until "some" amount of collective is pulled to arrest the NR back at 100%. once both reach 100% NR the ROD should in theory return to the SAME charted value. but the heavier helicopter will have some collective pitch applied. for my aircraft, we expect 1% NR increase or decrease for each 100 pounds/ 1000 feet DA increase/decrease. if the weight or da goes up, so will autorotational NR with collective full down. if the DA or weight reduces, the full down NR will also decrease. any questions???

[mfriskel]

AAvtr-In practice, you will not find sentence number 6 to be an accurate statement.
You will find a difference in ROD and Glide Distance if you change the gross weight and maintain the same NR and same ASPD.

[Tourist]

Let's think this through.

In steady state autorotation, total rotor lift = weight

A heavier aircraft therefore requires more lift in auto.

More lift requires more blade angle of attack or more rotational airspeed.

Since we have a max Nr in a real helicopter, the only reasonable way of getting more lift is to increase blade angle of attack.

In auto, however we encounter the problem that we can only increase the angle of attack by changing the blade angle with the collective through a reduced range compared to when powered as the arrow must still be positive to maintain autorotation.

That leaves us with ROD as a method of increasing the blade angle of attack.

ie, increase ROD to increase lift till they balance in steady state

Thus I reckon higher weight = higher ROD in auto.



The only occasions where I can think this might be altered is at very low weights where the drag of the gearboxes might dramatically effect the efficiency of the autorotation.

Aside from that I think the original premise is bollocks.

[army_av8r]

as far as i can tell, both on paper, and in practice... if you maintain a certain Nr and fly the minimum ROD airspeed as charted in the helicopter's manual, you will arrive at the charted rate of decent. the one difference being that the heavier aircraft will fall faster and gain excessive rotor until the pilot arrests the NR back to 100%. once they reach steady state. both will have essentially the same ROD if airspeed and rotor are the same. if only my VSI gave me more than a 1000 FPM reading, haha.

[Tourist]

av8r

So if the Nr is the same, where is the extra lift coming from to balance the extra weight once things settle down?

[Fareastdriver]

On a mountain ridge with a steady wind blowing against a smooth escarpment it is possible to soar/autorotate and maintain height backwards and forwards along the ridge. I have done it several times with a light one but never with a heavy helicopter.

[Tourist]

Me too.

One might almost begin to suspect that a heavy helicopter flutters down faster than a light one.

Who'd have thought it?

[chopabeefer]

Tourist. Thanks for your erudite contribution. Unfortunately, you are incorrect. The same is true in airliners - heavy ones take longer to descend.

A call to the ETPS chaps may put your mind at rest. For a given Nr, a heavier helicopter has a lower RoD. It is a fact. Go and give the tefal-heads at Boscombe a call. If they disagree - I'll delete the thread. Some very good theories on this thread by the way - thanks to all the contributors.

Tourist - you are wrong. I just don't know why.

[Aucky]

Tourist - So if the Nr is the same, where is the extra lift coming from to balance the extra weight once things settle down?

It's the other way around - heavier, greater pull due to gravity, a tendency for a higher NR as a result (for a given pitch setting), therefore more collective (lift) required to maintain NR. So, more lift from more pitch...

One might (quite reasonably) think that the increase in lift would at best counter the increase in weight so result in approximately the same ROD, but it seems that the resultant increase in lift actually goes some way further than the increase in weight.

[Tourist]

Erm, no.
A heavier airliner/glider/fast jet etc does not descend more slowly.

You have the wrong end of the stick I'm afraid.

The slightly counterintuitive thing they do do is maintain the same glide slope with change in weight.
ie if you add weight to a glider it does not fly a steeper path to the ground.
It does however fly faster and with a greater rate of descent.

So if you drop it off a mountain in still air it hits the same spot on the ground at the end of its glide no matter what the weight.
The only thing that varies is the time it takes to get there.
(it's slightly more complex than that due to drag effects, but ballpark)

Aucky

Yes more pitch, but only to a point. You still need enough resultant in a pro-rotorspin to keep the rotors turning. The increase in Nr is due to the higher ROD

[Aucky]

Tourist - The increase in Nr is due to the higher ROD

Not sure that's strictly true (although I agree that a higher ROD, and more airflow through the disc would increase rpm), I think it's more the higher disc loading, impact on coning angle, and conservation of angular momentum. In order to contain the higher NR collective pitch is increased. The inner 'blade stall' area will grow outwards, the outer blade 'drag' area will move inwards slightly, the overall autorotative force section will diminish enough that NR can be maintained. Even though the outer 'drag' area of the blade has grown it is still producing more lift too, along with the increased lift in the 'autorotative force' section. So overall result - same RPM, more lift.

[IMG]http://img825.imageshack.us/img825/3230/verticalautorotationpdf.jpg (http://imageshack.us/photo/my-images/825/verticalautorotationpdf.jpg/)


I guess the apparent resultant reduction in ROD airflow further confuses things, moving the autorotative force section more inboard...?

All a stab in the dark, but it makes some sense in correlating with what has apparently been shown. I'm keen to go and try now at different weights...

[Tourist]

The impact on coming angle and conservation of momentum is a transitory effect, rather than a steady state effect, so not relevant in this case.

ie, if you suddenly pull G and effectively increase your weight, the sudden increase in lift required from the blades will cause a coning angle to develop thus increasing Nr due to conservation of momentum.

That is not the case here.
Here we are flying 2 identical helicopters except one weighs more than the other and entering Auto.

The heavier helicopter is already producing more lift to remain airborne and balance it's weight, so already has a greater coning angle therefore no sudden increase in Nr due to conservation of momentum.




A heavier helicopter requires more lift to remain in a steady non accelerative state such as Auto.

The only ways to get more lift from an airfoil are to increase airflow or angle of attack.

In this example we have limited ourselves to a steady Nr, so we are left with angle of attack.

If, we were to believe that an increase in weight leads to a decrease in ROD as per the premise being discussed here, we can see from your diagram that there would be a decrease in rate of descent airflow which would then lead to a decrease in blade angle of attack.

It just doesn't work.


The way to test any theory is to take it to extreme.

Does anybody here really believe that if you added a million tons to a helicopter it would fall more slowly?
Or conversely, if you attached an enormous helium balloon to the rotor head we would plummet in auto?

No, our internal newtonian physics professor can spot a flawed theory quite well.




In reality, as you increase weight, the ROD will increase, and to minimise this you would have to allow the Nr to increase to stop the stalled portion from growing to encompass the entire blade.

There will be an optimum increase in ideal Nr with increase in weight in auto.

[GipsyMagpie]

Firstly its not a constant. There is only some higher weights when you will descend slower. Obviously infinite weight does not equal a hover in auto. Its explained in prouty's books somewhere. I will post his explanation if someone doesn't beat me to it.

[lelebebbel]

I suspect that the answer may be that the higher weight requires a higher lever setting and therefore higher angle of attack to maintain to maintain the same Nr. This moves the L/D of the whole rotor disc towards its optimum conditions. The same reason that max range in autorotation is achieved generally at the lower end of the RPM band.

I suspect you're right. Certainly sounds plausible.


Does anybody here really believe that if you added a million tons to a helicopter it would fall more slowly?
Or conversely, if you attached an enormous helium balloon to the rotor head we would plummet in auto?

No, our internal newtonian physics professor can spot a flawed theory quite well.

A extremely heavy helicopter would fall like a brick with very high RRPM that can't be controlled, because not enough collective is available / the rotor would stall due to excessive AOA.
(However, it is easy to imagine that this heavy helicopter would still be able to autorotate if the rotor was made large enough. A 1:12 scale RC helicopter can autorotate, and so can a CH-53.)

A extremely light helicopter - forget the helium balloon though, that is not the same thing - would fall like a brick because insufficient energy is available to keep the rotor turning.
(However, if the rotor was small enough, it would produce less drag and still be able to autorotate. Think maple seed vs. chinook)


The question is, what happens in between, and why.

[Baldeep Inminj]

Chopa is definately right - I have known this to be true for years, and we demonstrate it to students during the course with us. I also have not definately understood why.

My understanding is at a basic level. I believe that if you were to 'suspend' 2 identical helicopters at a height (but one heavier than the other), then drop them - the following would happen...

They would initially fall at the same rate. As drag increases, the heavier one would be slowed less by it, and develop a HIGHER Rod. This would cause the Nr to rise more than the light heli, and therefore cause the pilot to raise the lever more to contain it.

This means, for the same Nr, the heavy acft has a greater pitch angle. This puts the blades in a more efficient area, and reduces the RoD - so that it is less than the light helicopter.

Is this correct? - I always thought it was. In a Cobra, the RoD difference between min fuel and max fuel, at 70 kts in Auto, and recommended Nr, is almost 500fpm.

Like chopa, I would love to know why (if it isn't why I said)

Tourist - Chopa is right - sorry. I don't know why either!

[Tourist]

There is a very big diference between the premise

"A heavy helicopter always falls faster than an identical, lighter, helicopter"


and


"certain helicopters may under certain circumstances due to optimisation of rotor efficiency fall counterintuitively slower in some areas of their weight range with an increase in weight"

[topendtorque]

and therefore cause the pilot to raise the lever more to contain it.


Aren't you introducing a variable that you should dismiss with. Of course if you want to extend range, slow down ROD, then you will "lift the lever."

Why not carry out the discussion which is the way I read it from the beginning and stay within the premise of the design characteristics of the aircraft being able to maintain the RPM at low weight just in the green and at no more than top of the green at Max AUW with the lever right down. and aren't we talking about one heavier than the other - only regarding range and ROD..

After all where does the engineer start from to set nominal angles? Yep tha's right jacked up with A/C level, collective at the bottom, nominal lengths and angles checked from the bottom all the way up through the control rods etc to the final blade angle.

This will be the pitch setting then with lever down and kept down, in auto. now add weight and what happens? higher ROD and shorter range ain't it?

[Arm out the window]

No, I think discussing it with the same rotor revs is the go, so it's a direct comparison rather than introducing another variable.

[topendtorque]

No, I think discussing it with the same rotor revs is the go,


OK, I'll try that Tuesday.
tet

[n5296s]

I'm not a helicopter aerodynamicist, but since nor are many other people, here are my 2c anyway.

Surely this all has to do with rotor efficiency? I.e. it's the same reason why you get a better glide ratio at 90% Nr than at 100% Nr.

If you just consider the very basic physics of potential vs kinetic energy, then the mass term cancels out. So that obviously has nothing to do with it. So it must be something in the aerodynamics which changes due to the increased vertical force (lift) which is matching the increased weight (mass * g). The heavier heli is producing more lift at the same Nr (by definition of the problem), hence the blades MUST be at a higher angle of attack. And in the 90% Nr auto, the rotor is producing the SAME lift at lower Nr - hence again a higher angle of attack. That is from the basic aerofoil equations.

Now, I have absolutely no idea WHY the rotor is more efficient at higher alpha - that would take someone who can actually plough through the math involved. But it provides a consistent answer to both observed facts.

My 2c, fwiw...

[Arm out the window]

Hey, that was my 2c!

Light aircraft, bottomed collective to maintain auto Nr, get a certain rate of descent.
Heavier aircraft, holding some pitch to control Nr in the green, blades are now at a more efficient angle and therefore ROD is lower.


Rotorfossil's 2c was quite similar, or perhaps that should have been 2p in his case.

I suspect that the answer may be that the higher weight requires a higher lever setting and therefore higher angle of attack to maintain to maintain the same Nr. This moves the L/D of the whole rotor disc towards its optimum conditions. The same reason that max range in autorotation is achieved generally at the lower end of the RPM band.

[n5296s]

Sorry, must have missed those. Anyway, at least we agree.

Tourist seems to have difficulty with non-linear equations, which is a shame because in real life there are few linear ones. Obviously if you try to auto with a couple of elephants lashed to the rotor hub, you will reach the top of the L/D curve, i.e. stall. Using linear extrapolation to the absurd to refute an argument is a risky business.

[Arm out the window]

Not only that, but unless the elephants were perfectly matched in weight distribution and tied down very firmly, you'd have a hell of a vibe.

[Tourist]

n5296s

Nope, not at all, but if you read the op's original statement, there are no bounds or qualifications set, making it a black and white situation so the question is linear in nature.

ie, If a definitive unqualified statement is made that is not true in all circumstances, then it is a false statement.


"Now, I have absolutely no idea WHY the rotor is more efficient at higher alpha "

That is simple. It's just like being "on the buffet" in a fixed wing.
The lift/drag relationship is at its best just before the stall, which increases the efficiency of the autorotative effect etc, just like its the best turn AoA for a fighter.

[stringfellow]

great debate... i was told during some advanced flying in the mountains recently that the rate of descent is no greater with a heavier machine.... but never got an explanation so thank you!!!

[air-bender]

isnt all this a moot point???

The answer I think we're all looking for is: If you have an engine failure with a load under you or external stores then you pickle them as soon as possible...

any advantages of ROD with a heavy load are negated where and when it matters most... The last part of the auto... Higher collective angle at the bottom means less Pitch to arrest ROD and heavier means harder to slow down that momentum so NR Decay will be greater in a heavier ACFT than for a light ACFT.

I'm just curious how many young pilots now think it's better to be heavier now when your donk fails because of this thread.

spell check not used

[chopabeefer]

Air bender,

Good point, and worthy of comment. The weight is not as great a factor at the bottom end of an EOL as you might think (USL notwithstanding - that's different!).

I have flown/taught/sat thru over 800 engine-off landings (yes, really). As long as they are handled properly, and the Nr is conserved throughout the descent and built during the flare, the weight is a relatively minor factor (although still a factor).

This thread however, is a theoretical debate. Most pilots (not all I admit) are not thick, and will hopefully see the difference. I am not, for one second, advocating being light or heavy in auto, and unless you have a load you can pickle it is a moot point anyway - not gonna throw out a passenger when the gearbox lets go...although I did carry Michael Portillo once...

I simply want to know why a heavy heli descends slower than a light one, all else being equal. It does, and it seems wrong. Contributions here (and some people have clearly expended some brainpower) show that others are not sure either.

That said, I take your point, but believe it to be somewhat misleading in that when you have an emergency requiring autorotation, you weigh what you weigh. You enter auto, recover the Nr, turn into wind if you can and select a landing point, diagnose and deal with the emergency, put out a mayday call, secure the engine if there is time, and whilst doing this you manoeuvre the acft to make your chosen landing site. You fly the acft you have, and use the performance it gives you to achieve what you can.

But I still want to know the answer to my question - things are becoming clearer from the replies so far.

[air-bender]

true chop,

watched a co-worker fail a board because he elected to keep the torps and bouys during an auto...

In a low inertia rotor system Nr is life... heaveier ACFT with a low Nr rotorsystem can make for a sporty auto... then again, if you're in a 205 or other Huey variant who cares what where when conditions when the donk fails.

Loved doing Auto's in those things..

[woifal]

Hi,
Usually I don´t talk in English about aerodynamic aspects of a helicopter, so i´m sorry when I don´t always use the right terms. I hope it will still be clear what I mean.
Maybe it´s not all about the angle of attack to the mainrotor and/or the energy, i think we should talk about the forces which decelerate the helicopter.
The aerodynamic drag, parasite drag, and so on stay always the same at a particular airspeed. We all know the formula

F = m * a , F = const. , m = mass, a = acceleration

m1 = light heli
m2 = heavy heli

a1 = F/m1 > a2 = F/m2

As you can see, a1 is bigger than a2, so in case 1 the helicopter will be more decelerated than in case 2. But as we want to have a constant speed in the autorotation, we have to take down the nose of the helicopter so the acceleration by gravity will work against the deceleration produced by the drag.
But if we do so, we will have an higher ROD because of the nose down attitude.
It´s just a consideration, so maybe it´s all bull**** i wrote ;)
Bye!

[6gillshark]

Dangermouse almost had it right.

It does come down to the energy equation. In autorotation the the power needed by the helicopter can only come from rate of change of potential energy i.e. Weight * Descent rate. The descent rate is therefore the power required (rotor profile power, induced power, power to overcome fuselage drag) divided by the aircraft weight. Hence the RoD decreases as weight increases. The helicopter would not hover if infinitely heavy, it just would not descend.

[AnFI]

.... yes 100 ft for a heavy helicopter is more potential energy than 100 ft for a light helicopter

BUT the efficiency of the helicopter generally gets better with bigger pitch angle - and then it gets worse again - so the ROD will increase again when the power required to generate thrust equal to weight starts to increase faster than the energy available as a result of the increased weight...

as for choppabeefer - i beg to disagree - the one place where extra weight does matter is at the bottom where the energy in the rotor either has to hover a light helicopter or a heavy helicopter - it can hover the light helicopter for longer (=easier, the energy lasts more time so cushioning is easier with greater margin, surely? am i missing something?), (as for 800EOL - is that a lot?)

[[email protected]]

Prouty gives some assistance but only highlights that in certain weight bands (quite small) in certain conditions (lower altitude rather than higher) you can theoretically show that the potential energy goes up faster than the power required and gives a small reduction in the RoD as weight increases. He does say that this data has been observed in flight tests in some aircraft.

However, at higher gross weights, because the induced power is proportional to the square of the weight, the RoD increases again.

His findings are based on the reasoning that the power required for autorotation is almost the same as in level flight and the approximate RoD needed to produce this power is found by multiplying the horsepower for level flight by 33,000 and then dividing this by the gross weight.

We all agree that to achieve a steady state auto, the heavier aircraft must fall faster than the light one in order to generate sufficient rotor thrust to balance the weight which will result in a higher Nr. I don't think anyone's flight test schedule will show the the autorevs expected will go down with increased AUM.

So it all comes down to whether or not raising the lever to contain the Nr( to a nominal 100% for the sake of argument) will reduce the RoD back to the same as/more/less than the lighter helo. Mr Prouty doesn't cover that question but the subject of drooping Nr to reduce RoD (and increase range) does involve taking the blades into a slightly more efficient L/D area of operation. You still need the higher Nr for the EOL however.

It appears that there is no absolute answer to this question and different helicopters will provide different results in tests - just as not all helos have a recommendation to droop Nr to increase range - it depends on the type of aerofoil section, what the helo is optimised for, and each helo may well produce different results in differing conditions of DA and AUM.

Would I rather be in a lighter helicopter to do an EOL - yes please. Like others, I have taught and demonstrated many, many EOLs and there is no doubt that a lighter aircraft gives you more margin for error and that what works on a 206 would kill you in a Robinson.

[6gillshark]

That is true, but only for very high weights, density altitudes or configurations in which the induced power (the power required to create thrust) dominates the profile power (power required to rotate the blades) and the peer required to overcome the drag of the fuselage.

Going back to the original issue, the RoD will decrease with increasing AUW in most conditions. As in most cases, there is so simple 'one size fits all' answer because rotorcraft (and their physics) are complicated.

[AnFI]

what works on a 206 would kill you in a Robinson. Crab - I think the R44 has a lot of energy and margin, are u talking about the R22 perhaps?

As for the Horse Power reference you may be interested to know that for each 330lbs you weigh the VSI is in Horse Power