Weight vs ROD in Autorotation
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Tourist - The increase in Nr is due to the higher ROD
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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...
Last edited by Aucky; 10th May 2012 at 23:11. Reason: picture link
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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.
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.
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Prouty has the answer
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.
Originally Posted by 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.
Originally Posted by Tourist
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.
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.
(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.
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!
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!
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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"
"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"
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and therefore cause the pilot to raise the lever more to contain it.
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?
Last edited by topendtorque; 11th May 2012 at 21:54.
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...
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...
Hey, that was my 2c!
Rotorfossil's 2c was quite similar, or perhaps that should have been 2p in his case.
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.
Heavier aircraft, holding some pitch to control Nr in the green, blades are now at a more efficient angle and therefore ROD is lower.
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.
Last edited by Arm out the window; 12th May 2012 at 02:12.
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.
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.
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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.
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.
Last edited by Tourist; 12th May 2012 at 11:17.
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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!!!
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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.
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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.
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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.
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.
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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..
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..
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Hi,
Usually I donīt talk in English about aerodynamic aspects of a helicopter, so iīm sorry when I donīt always use the right terms. I hope it will still be clear what I mean.
Maybe itīs 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.
Itīs just a consideration, so maybe itīs all bull**** i wrote
Bye!
Usually I donīt talk in English about aerodynamic aspects of a helicopter, so iīm sorry when I donīt always use the right terms. I hope it will still be clear what I mean.
Maybe itīs 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.
Itīs just a consideration, so maybe itīs all bull**** i wrote
Bye!
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Energy Terms
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.
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.