Weight vs ROD in Autorotation
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Weight vs ROD in Autorotation
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?
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?
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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.
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.
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think of it in energy terms
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
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
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.
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.
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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
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
Originally Posted by 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
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
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.
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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.
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.
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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.
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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
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
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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
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
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???
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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.
You will find a difference in ROD and Glide Distance if you change the gross weight and maintain the same NR and same ASPD.
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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.
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.
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.
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.
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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.
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.
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Tourist - So if the Nr is the same, where is the extra lift coming from to balance the extra weight once things settle down?
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.
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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
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