How long have you got if the engine(s) go quiet
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How long have you got if the engine(s) go quiet
Just wondering how quickly you have to react to the engine(s) going unexpectedly quiet and putting down the collective, do larger radius rotors provide a larger time margin. Had a first flight in a Schweisser(?) 300 the other day and the chap reckoned you have at most 2 seconds 
As with everything aviation-ery, it depends on a lot of things:
The inertia in the rotor system (the more inertia, the longer it will take to slow)
The drag on the rotor (the more drag, the quicker the rotor will slow)
The amount of pitch you're pulling (related to the above. The more pitch, the greater the drag)
The phase of flight you're in (related to the above. In descent, you'll be pulling less pitch than in the cruise and in cruise less than in the climb, so your reaction time is less in each case)
Your speed (you can trade speed for rotor RPM)
The rotor RPM at the time of the failure (the higher the revs, the longer they'll take to decay to disastrous levels - though this won't make a huge difference given that rotor RPM should be kept within a fairly tight range anyway)
Larger radius rotors could take longer to slow (greater inertia) but they could be generating more drag so slow quicker.
I've mainly flown the Robinson R22 which has a low inertia rotor system and you only get a couple of seconds between the donkey conking and lowering the collective to catch the revs. I've also flown the Bell 47 and 206. They have rotor systems with far more inertia and (comparatively speaking) you seem to get about half an hour between it going quiet and needing to react.
The inertia in the rotor system (the more inertia, the longer it will take to slow)
The drag on the rotor (the more drag, the quicker the rotor will slow)
The amount of pitch you're pulling (related to the above. The more pitch, the greater the drag)
The phase of flight you're in (related to the above. In descent, you'll be pulling less pitch than in the cruise and in cruise less than in the climb, so your reaction time is less in each case)
Your speed (you can trade speed for rotor RPM)
The rotor RPM at the time of the failure (the higher the revs, the longer they'll take to decay to disastrous levels - though this won't make a huge difference given that rotor RPM should be kept within a fairly tight range anyway)
Larger radius rotors could take longer to slow (greater inertia) but they could be generating more drag so slow quicker.
I've mainly flown the Robinson R22 which has a low inertia rotor system and you only get a couple of seconds between the donkey conking and lowering the collective to catch the revs. I've also flown the Bell 47 and 206. They have rotor systems with far more inertia and (comparatively speaking) you seem to get about half an hour between it going quiet and needing to react.
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How long have you got if the engine(s) go quiet
Do you mean, how long have you got before you hit the ground?
Oh..............., reading on I see that you want to know how quickly you should put the lever down.
Bl##dy quickly if you have a low inertia rotor, and not so quickly, but still quite fast, if you have a high inertia rotor.
It's not so much the diameter of the rotor system, but the blade weight and diameter of the rotor - remember that famous equation E=MC2 (This site doesn't do superscript - that should be E = MC squared)
Oh..............., reading on I see that you want to know how quickly you should put the lever down.
Bl##dy quickly if you have a low inertia rotor, and not so quickly, but still quite fast, if you have a high inertia rotor.
It's not so much the diameter of the rotor system, but the blade weight and diameter of the rotor - remember that famous equation E=MC2 (This site doesn't do superscript - that should be E = MC squared)
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....so in other words, you have to get the collective down at the speed of light?! 
Cheers
Whirls
Cheers
Whirls
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Wigglyamps
Its not as bad as it sounds because you should get some form of indication that the engine may quit, ie a different noise/abnormal vibration etc. You are therefore aware that it may happen and it is not a huge shock when it does and you are kind of prepared for it in a way.
Its not as bad as it sounds because you should get some form of indication that the engine may quit, ie a different noise/abnormal vibration etc. You are therefore aware that it may happen and it is not a huge shock when it does and you are kind of prepared for it in a way.
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Originally Posted by Helipilot1982
you should get some form of indication that the engine may quit, ie a different noise/abnormal vibration etc. You are therefore aware that it may happen and it is not a huge shock when it does and you are kind of prepared for it in a way.
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Could you explain that one?
No I couldn't, but it made you think
You could always try the other famous equation by Newton
E = 1/2 MV 2 or Energy = Mass x (Velocity squared) all divided by 2......
You could always try the other famous equation by Newton
E = 1/2 MV 2 or Energy = Mass x (Velocity squared) all divided by 2......
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E=Mc(squared) is true for the real world; just read it as energy=mass or kinetic energy = half mass times velocity squared.
Cheers
Whirls
Cheers
Whirls
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The point originally being made though was that it wasn't the diameter of the rotor blade disc but that mass that was the factor in how quickly one gets the collective down!
It is easy to make things complicated!
Cheers
Whirls
...another one with a degree in physics!!!!
It is easy to make things complicated!
Cheers
Whirls
...another one with a degree in physics!!!!
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Anyway ...... what was the original question???
PS: Bell 47's come in different shapes and sizes - for example, the G2 has a much lighter rotor system and is therefore low inertia = get lever down very quick.
The G4 model has a similar blade inertia to a 206, so you can take quite a lot more seconds lowering the lever. Some say it is possible to perform an engine off landing with the G4 (and no doubt other types too), lift into the hover (without re-instating the engine), perform a quick spot turn and then land again...... all without any noise from the engine!!
PS: Bell 47's come in different shapes and sizes - for example, the G2 has a much lighter rotor system and is therefore low inertia = get lever down very quick.
The G4 model has a similar blade inertia to a 206, so you can take quite a lot more seconds lowering the lever. Some say it is possible to perform an engine off landing with the G4 (and no doubt other types too), lift into the hover (without re-instating the engine), perform a quick spot turn and then land again...... all without any noise from the engine!!
Purveyor of Egg Liqueur to Lucifer
The formula is actually E = mc²/t
Typically 4x2²/2 = 8 seconds from when the donkey stops to when you lose control!
Typically 4x2²/2 = 8 seconds from when the donkey stops to when you lose control!
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'Ang on Sid, I don't eat Mars bars and never touch coffee which gives me a numerator of zero i.e. no reaction time at all. This concerns me somewhat!
Please can you let me know the additional constant factors to be applied to your formula for a consumption of Jordan's Crunchy Bars (fewer calories and no cocoa) and Tetley Tea (less caffeine).
Cheers
Whirls
Please can you let me know the additional constant factors to be applied to your formula for a consumption of Jordan's Crunchy Bars (fewer calories and no cocoa) and Tetley Tea (less caffeine).
Cheers
Whirls
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The goodness of a rotor design for total power loss is determined fairly easily. It is a measure of the stored energy in the rotor as compared to the power the rotor is consuming at that instant. This term is important for auto entry, and also for the collective pitch pull at the bottom, both situations where the stored rotor energy is your life's blood. The energy ratio-ed to the power produces the time available for a given rotor rpm loss:
[Stored Energy] / [Power] = Time
[1/2 x I rotor x (omega) squared]/horsepower needed = seconds of rotor spin
Where "I rotor" is the polar moment of inertia for the rotor system (a term that is the resistance of the rotor to rotational acceleration, and which is mostly the weight, rotational rpm and radius of the blades.) Light short blades spinning slowly are bad, heavy long blades spinning fast are good.
For a "good" helicopter the calculation is about 2 seconds, for a poor rotor it is less than a second. Most Bell products have excellent numbers, the typical 206 is about 2.5 seconds, most Sikorsky's are moderate, about 1.7 seconds, and some of the very light composite systems are less than that, because the designers were told to make the blades very light, without considering this auto entry/landing as a design condition.
[Stored Energy] / [Power] = Time
[1/2 x I rotor x (omega) squared]/horsepower needed = seconds of rotor spin
Where "I rotor" is the polar moment of inertia for the rotor system (a term that is the resistance of the rotor to rotational acceleration, and which is mostly the weight, rotational rpm and radius of the blades.) Light short blades spinning slowly are bad, heavy long blades spinning fast are good.
For a "good" helicopter the calculation is about 2 seconds, for a poor rotor it is less than a second. Most Bell products have excellent numbers, the typical 206 is about 2.5 seconds, most Sikorsky's are moderate, about 1.7 seconds, and some of the very light composite systems are less than that, because the designers were told to make the blades very light, without considering this auto entry/landing as a design condition.
Remember, also, when practising engine failures, that the fuel scheduling characteristics of both piston and turbine engines are such that simply rolling off the throttle quickly may NOT replicate a true failure! i.e. you are protected from a 'lean cut' or flame out, and so the engine RPM decays more slowly that it may in real life.
I just cannot resist this question.....
Nick,
Some folks suggest a slight application of aft cyclic immediately prior to...or at the same time one dumps the collective into the bargain basement priovides for much less loss of Main Rotor RPM by "loading" the rotor disc.
Have you in your idle time between creating power point presentations for the five sided wind tunnel....ever do any testing of that concept?
One has to assume a forward flight situation for this scenario I reckon.
Nick,
Some folks suggest a slight application of aft cyclic immediately prior to...or at the same time one dumps the collective into the bargain basement priovides for much less loss of Main Rotor RPM by "loading" the rotor disc.
Have you in your idle time between creating power point presentations for the five sided wind tunnel....ever do any testing of that concept?
One has to assume a forward flight situation for this scenario I reckon.
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SASless, I'd say you're "unloading" the rotor, or reducing the power required to maintain rpm. If you then take that into the energy/power formula, with power reduced, time to decay will increase. So the technique does reduce the decay time.
Of course, like anything else it isn't that simple. You also want to control your speed, so once you've pitched back, you may have to pitch forward to regain speed. The pitch forward will have the opposite effect. Not a problem if you have a lot of altitude, but low level this can complicate things.
I'd just use the recommended technique for your type.
Matthew.
Of course, like anything else it isn't that simple. You also want to control your speed, so once you've pitched back, you may have to pitch forward to regain speed. The pitch forward will have the opposite effect. Not a problem if you have a lot of altitude, but low level this can complicate things.
I'd just use the recommended technique for your type.
Matthew.
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SASless,
Having done 1000's of engine cuts in critical flight conditions, I can say with great experience that the behavior with cyclic vastly controls the rotor decay.
A little back stick does help swap airspeed for rotor rpm, as well as help hold the nose up when the collective is rapidly dropped. If the aircraft rolls off to one side during the initial cut, it is far better to let it do so (within limits) than to use strong lateral cyclic, which consumes power and makes the rotor drop even faster.
BTW the back stick loads the rotor (increases thrust and g) but in this case that is ok, because it increases the autorotative flow a bit. The one case where it hurts is at very high speed near or at Vne while heavy and high altitude. If you are near stall, back stick makes the stall factor rise, and the rpm droop can be very big.
Having done 1000's of engine cuts in critical flight conditions, I can say with great experience that the behavior with cyclic vastly controls the rotor decay.
A little back stick does help swap airspeed for rotor rpm, as well as help hold the nose up when the collective is rapidly dropped. If the aircraft rolls off to one side during the initial cut, it is far better to let it do so (within limits) than to use strong lateral cyclic, which consumes power and makes the rotor drop even faster.
BTW the back stick loads the rotor (increases thrust and g) but in this case that is ok, because it increases the autorotative flow a bit. The one case where it hurts is at very high speed near or at Vne while heavy and high altitude. If you are near stall, back stick makes the stall factor rise, and the rpm droop can be very big.