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Autorotation Aerodynamics


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Sorry to bother y'all, but I'm having a heck of a time tracking down a real incisive explanation of the aerodynamics of autorotation. What perplexes me the most is that if vertually all helicopter blades have a twist from root to tip, when the collective pitch is set to its almost lowest setting during an autorotation, what section of the blade is in negative pitch (if any)? And further to this, most explanations of said aerodynamics that I have found to date claim that the middle 45 or 50% of the blade length begins to act as a wind turbine during autorotation, and the resultant forward component of the lift vector on that section of blade drives the outer 25% of the blade to continue to act as a propeller (which provides some lift). What confounds me is how the outer 25% of the blade can provide lift if it has a more negative pitch than the middle 45 or 50% of the blade; which during autorotation must be set at close to zero pitch.

 

I will gladly explain why I want to know this if anyone wants to sit still for about fifteen minutes while I do so.

 

Thanks lots.

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Sorry to bother y'all, but I'm having a heck of a time tracking down a real incisive explanation of the aerodynamics of autorotation. What perplexes me the most is that if vertually all helicopter blades have a twist from root to tip, when the collective pitch is set to its almost lowest setting during an autorotation, what section of the blade is in negative pitch (if any)? And further to this, most explanations of said aerodynamics that I have found to date claim that the middle 45 or 50% of the blade length begins to act as a wind turbine during autorotation, and the resultant forward component of the lift vector on that section of blade drives the outer 25% of the blade to continue to act as a propeller (which provides some lift). What confounds me is how the outer 25% of the blade can provide lift if it has a more negative pitch than the middle 45 or 50% of the blade; which during autorotation must be set at close to zero pitch.

 

I will gladly explain why I want to know this if anyone wants to sit still for about fifteen minutes while I do so.

 

Thanks lots.

 

Interesting question.

 

As far as I am concerned this question is irrelevant because I am a pilot and so therefore I m only interested in the following.

 

1. If the engine quits what happens if I don't reduce collective. ( oh crap!!)

 

2. If the engine quits what happens if I do reduce collective. ( better )

 

Other than that I really don't give a crap.

 

The people you want to talk to are engineers. They know all this stuff plus a whole lot more.

 

:P

 

 

 

 

 

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I'm no expert on this stuff......but here's my guess.

The blades are moving at a relatively faster speed out at the tips. That extra speed and the low angle of attack out there creates the small amount of lift needed to descend at a manageable rate.

 

Note that the "lift" that we are talking about here produces a rate of descent of about 1300 to 2000 feet per minute.......so that ain't much "lift".....(but hopefully it is just enough!!)

 

Also, the air is flowing up through the rotor system in autorotation, so it changes how we have to think about 'negative' pitch. And remember that the blades aren't at negative pitch....in fact the inner region must have some pitch applied for it to provide the driving force. This pitch is built into the blades, and is still slightly positive, even when the collective is fully down.

 

Here is a manual (see pages 3-9 through 3-12) that may explain it a little more clearly.

http://www.faa.gov/library/manuals/aircraf...a-h-8083-21.pdf

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Hope this helps without sending you to sleep (I'd love to know why you want to know!)

 

In autorotation, the rotor drag that is normally dealt with by engine power is now handled by the air coming up through the disc. Induced flow now comes from below, so the angle of attack is now the combination of pitch angle and induced angle, so it is rather large.

 

Juan de la Cierva, of autogyro fame, found that the wing of an aeroplane, when gliding, not only provides lift, but a forward pull as well, and this quality is made use of in autorotation to keep the blades rotating, since, if you tie one end of a blade to the hub, the remainder will want to turn in a circle.

 

The inner 25% of each blade is stalled, and the outer 30% is providing a small drag force (in other words, it is being driven). The middle part of each blade is therefore the only part providing lift. The only negative lift is required by the tail rotor because the fuselage is now trying to go in the same way as the rotor blades, due to friction.

 

As you increase speed in autorotation, the driving region moves towards the retreating blade side (together with the stalled region) until it touches the edge, where the angle of attack is larger, which is your power-off VNE, because once the driving region goes beyond the edge of the disc, the surface area of the driving region is reduced, resulting in rotor decay. At this point, the advancing blade contains a higher proportion of the driven region and the retreating blade contains more of the stalled region, with a reduction in the size of its driven region.

 

Phil

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Interesting question.

 

As far as I am concerned this question is irrelevant because I am a pilot and so therefore I m only interested in the following.

 

1. If the engine quits what happens if I don't reduce collective. ( oh crap!!)

 

2. If the engine quits what happens if I do reduce collective. ( better )

 

Other than that I really don't give a crap.

 

The people you want to talk to are engineers. They know all this stuff plus a whole lot more.

 

:P

 

 

Deuce ',

 

I agree with you that it isn't essential to understand the aerodynamics of autorotation for a pilot to safely land a helicopter in that mode, but some people's curiousity drives them to a more thorough understanding of issues than is absolutely essential.

 

I HAVE in fact consulted an engineer on this matter, and his understanding seemed to be not significantly deeper than yours. I did track down a VERY high powered aerodynamics consultant who could undoubtedly explain it to me. He wants to be paid 600$/hour ; which, in the end might prove cheap.

 

Thanks for your response.

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Deuce ',

 

I agree with you that it isn't essential to understand the aerodynamics of autorotation for a pilot to safely land a helicopter in that mode, but some people's curiousity drives them to a more thorough understanding of issues than is absolutely essential.

 

I HAVE in fact consulted an engineer on this matter, and his understanding seemed to be not significantly deeper than yours. I did track down a VERY high powered aerodynamics consultant who could undoubtedly explain it to me. He wants to be paid 600$/hour ; which, in the end might prove cheap.

 

Thanks for your response.

 

 

I was being factitious which is a failing of mine after the third martini.

 

I apologize if my reply was offensive.

 

There is so much information to hold in my poor little head in order to do this job that I tend to stick to the things I know I may need to survive and leave the in depth analysis to my more intelligent brethren. :rolleyes:

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I was being factitious which is a failing of mine after the third martini.

 

I apologize if my reply was offensive.

 

There is so much information to hold in my poor little head in order to do this job that I tend to stick to the things I know I may need to survive and leave the in depth analysis to my more intelligent brethren. :rolleyes:

 

Deuce',

 

No offense taken. I kinda thought that you might be being facetious, but wasn't sure.

 

I operate on the assumption that for every piece of new information that I can shoehorn into my mind, another piece escapes out the other end.

 

Inventor'

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I'm no expert on this stuff......but here's my guess.

The blades are moving at a relatively faster speed out at the tips. That extra speed and the low angle of attack out there creates the small amount of lift needed to descend at a manageable rate.

 

Note that the "lift" that we are talking about here produces a rate of descent of about 1300 to 2000 feet per minute.......so that ain't much "lift".....(but hopefully it is just enough!!)

 

Also, the air is flowing up through the rotor system in autorotation, so it changes how we have to think about 'negative' pitch. And remember that the blades aren't at negative pitch....in fact the inner region must have some pitch applied for it to provide the driving force. This pitch is built into the blades, and is still slightly positive, even when the collective is fully down.

 

Here is a manual (see pages 3-9 through 3-12) that may explain it a little more clearly.

http://www.faa.gov/library/manuals/aircraf...a-h-8083-21.pdf

 

Hey Over-talk,

 

For my purposes, I might have to re-name you UNDER-talk.

 

With respect to the 'lift' produced, as long as all parties to the discussion understand what the term 'lift' means in the context of autorotation, a useful exchange can ensue.

 

When speaking of the pitch of the blade, I am making the assumption that positive blade pitch means positive in the context of the blade being driven by the motor. One factor that complicates this discussion of blade pitch IS the issue just mentioned of whether one is discussing pitch with respect to air flowing down through the rotor disc (as when the blade is driven by the motor) or when the air flow is up througth the rotor disc (as during an autorotation).

 

But there is another issue that complicates the discussion of blade pitch far more. And that is the fact that most helicopter blades have a twist to them along their length. Said twist represents a change in pitch from root to tip at any blade pitch setting. Sooo, when one discusses what a specific blade pitch setting is, one must also state which section of the blade is under discussion.

 

But the REAL connundrum for me is "In what direction is the blade twisted?" I read a patent recently that depicted a graph whereon were plotted the blade twists of the blades of several models of Blackhawk helicopters. From that graph is seemed that the twist of the blades were all counterclockwise from root to tip for a blade that rotated counter-clockwise when view from above. Is this correct?

 

If this IS correct, then when the middle section of a blade is set to a slightly positive pitch during autorotation, the tip of the blade must be negative (because of the twist). And if the pitch of the blade at the tip is negative, and the apparent wind vector is at its LEAST angle to the plane of the rotor at the tip because the tip of the blade is moving the fastest, how can the tip of the blade under these conditions provide lift??

 

I DID try to download the FAA handbook for rotorcraft, but because I have a dialup Internet connection, it is a rather slow process to begin with, and the download stalled twice. I will try again however.

 

Thanks for this.

 

 

 

 

 

 

 

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Hope this helps without sending you to sleep (I'd love to know why you want to know!)

 

In autorotation, the rotor drag that is normally dealt with by engine power is now handled by the air coming up through the disc. Induced flow now comes from below, so the angle of attack is now the combination of pitch angle and induced angle, so it is rather large.

 

Juan de la Cierva, of autogyro fame, found that the wing of an aeroplane, when gliding, not only provides lift, but a forward pull as well, and this quality is made use of in autorotation to keep the blades rotating, since, if you tie one end of a blade to the hub, the remainder will want to turn in a circle.

 

The inner 25% of each blade is stalled, and the outer 30% is providing a small drag force (in other words, it is being driven). The middle part of each blade is therefore the only part providing lift. The only negative lift is required by the tail rotor because the fuselage is now trying to go in the same way as the rotor blades, due to friction.

 

As you increase speed in autorotation, the driving region moves towards the retreating blade side (together with the stalled region) until it touches the edge, where the angle of attack is larger, which is your power-off VNE, because once the driving region goes beyond the edge of the disc, the surface area of the driving region is reduced, resulting in rotor decay. At this point, the advancing blade contains a higher proportion of the driven region and the retreating blade contains more of the stalled region, with a reduction in the size of its driven region.

 

Phil

 

 

Thanks for this Phil.

 

Juan De La Cierva's observation makes sense to me because the angle of attack of the apparent wind upon an airplane wing in a glide condition is rather steep (one assumes that the 'pitch' of the airplane wing is at zero with respect to horizontal). And in that condition, the air pressure extant on the bottom of the wing is higher than the air pressure on the front of the wing. And that is because the front of most airplane wings is generally round in cross section. so, even at a high angle of attack of the apparent wind, the wind that is routed around the front of the wing must be accelerated (and therefore its static pressure is reduced) while the air impinging directly on the bottom of the wing imparts some of it's momentum to that surface (INcreasing the air pressure on that surface). And the airplane wing is pushed from the high pressure side to the lower pressure side (I.E.-forward).

 

This observation however does not shed any light directly on the heart of my confusion.

 

Most of your further explanation is consistent with several other explanations that I have discovered online, but for one point. When you state that, "the middle part of each blade is therefore the only part providing lift.", this is concradictory to the other explanations that I have found. The most incisive explation that I have discovered so far states that the middle (driving) portion of the blade has a lift force vector that is tilted somewhat forward of the axis of rotation; while the lift vector generated by the outer 25% of the blade is tilted behind the axis of rotation. By adjusting the collective pitch, a pilot can balance these lift vectors to maintain the angular velocity (RPM) of the rotor at an appropriate level.

 

But my confusion still remains as to how the outer 25% of the blade can provide any lift at all if it is in a more negative pitch (with respect to the plane of the rotor disc) than the middle 45% of the blade; unless, during autorotation, the angle of attack of the apparent wind on the outer 25% of the blade (even on the advancing side) is still significantly below the negative pitch on that section of blade.

 

I suppose, given a rate of descent of about 25 feet/second (as stated by Over-talk), and an autorotation RPM of 360, and a blade diameter of thirty feet, the ratio of apparent wind caused by blade tip advance to the apparent wind caused by upflow through the rotor disc is about 24 to 1. That would translate into an angle of attack of the apparent wind below the plane of the rotor disc of about 2.5 degrees. Perhaps the twist on the blade along its length is not severe nough to result in the tip assuming a negative pitch as severe as 2.5 degrees when the collective pitch is set as low as it will go.

 

My confusion MAY arise out of a conceptual misunderstanding as to the direction of the twist on a typical helicopter blade, but I doubt it. It would seem to make intuitive sense that the inboard sections of the blade have a steeper pitch than the tip in the context of the blade being driven by the motor because the inboard sections of the blade move slower than the tip.

 

Hmmmmmm.

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