Aerodynamics ~ Phase Lag

[JoePilot]

Heedm "If this is so, then the disc won't follow, it will stay with the helicopter." - true but the swash plate will help I suppose.

ALSO (various) FORCES don't mean ENERGY is happening. FORCES need to be applied over a DISTANCE eg 1 JOULE if you push a NEWTON for 1 METER. If you do it at a rate eg 1 METER per SECOND then you're talking POWER ... 1WHAT?

Good night (really)

[heedm]

Joe, good point about the flapping velocity changing the angle of attack. It depends on definitions, but I don't think it changes blade pitch. Regardless of what definitions you use, the effect you mentioned is correct.

Beyond my background, but it seems to me that the amount of phase lag change due to flapping velocity, would be dependant on same. For example, small cyclic changes create small flapping amplitudes which means small flapping velocities, which I think will create a small change in phase angle. Large cyclic changes create....fill in the rest. N'est-ce pas?

________________


You're right about energy, but I'm not certain why you brought it up. It does get confusing when you consider a force applied over a distance, even more confusing when you do it in a certain amount of time. What's easier to see is the starting state and the end state of total energy. If it increases then you've added energy. You don't have to measure forces, distances or times.

If a blade flies off, it has velocity relative to the helicopter, so it has kinetic energy (relative to helicopter).

[Dave Jackson]

heedm

>"Dave, sorry this thread didn't stay on topic. Is there any way we can go back to phase lag?"<

It sounds as if the discussion is on 'phase lag' ~ in a roundabout way.

I came across the expression 'aerodynamic precession' in Padfield's (or Leishman's) textbook a couple of days ago. It was used in conjunction with 'gyroscopic precession'. It therefor appears that one of the gurus thinks in terms of a melding of the two precessions.

Now if Lu will only accept that;[list=1][*]there is another precession in addition to gyroscopic; and its 90-degrees.[*]the other precession (blade flying to position) can be less than 90-degrees.[*]the resultant precession (and hence phase angle) is a resultant of these two.[/list=a]
____________

Now when the two precessions are different and fighting each other, what happens

[vorticey]

heedm
if i made up a frame that suspended a blade in the air would this be frame be showing centripital force? if i released the blade from the frame it would fall away from the frame. this to me is almost the same thing as the helicopter blade spining on the hub.
centripital force seems to be a silly way of understanding somthing so simple.
>what force sends the blade flying off on a tangent? inertia i would think.
if the hub was pulling the wheight of the blade in when its spinning, does this mean that if the blades were removed the hub would inplode?
a child on the play ground 'spining thing' holds a chain near the middle, centripital force stops her from flying off? true? now she drops the chain, does this chain pull itself in now the wheight is gone? now the little girl grabes the chain further from the centre. the chian is holding a hevier object now, can the chain pulling the little girl to the middle be actualy pulling, when it doesnt move if the girl is on or off it? why cant you just say the chain is holding her not pulling her inwards.
i think that you can call centrifugal force what you like and i think it is due to the movment and changing of direction which increases the weight of the object. eg a loop
the object doesnt have to be held from the middle, eg a car loop
when an engine is pinging the correct word is pinkin.
so centrifugal force is centripital. the way it works hasnt changed i just think the pulling to the centre bit is wrong.

[heedm]

Vorticey, I don't want to ignore your post, but you've brought this back full circle. Consider what the blade wants to do if it weren't attached. Being attached changes what the blade wants to do. Consider what force is needed to cause that change.

[helmet fire]

Vorticey,

If you go back to my post on page 3 about Lu's incorrect use of the Goliath Theory Of Suddenly Flying Off Rocks, the answer to your quandry lies there. Your quandry is the same as has been raised by just about everyone whom learn't about this (well me anyway.)

Forget giving the force a name for a moment. I will go through the logic steps required to get to the answer. If you do not understand any step in that logic, then you will not understand the end game. Heedm, Nick and the other Gurus will be able to better define this, and please guys, fix flaws in my steps. Simplified, here goes:

1. An object will remain at a steady state of motion (either rest or at a constant speed) unless a force is applied to it.

2. Speed can be thought of as directional. An acceleration is a force.

3. To increase speed in a direction, you must apply an acceleration in THAT SAME direction. Think of a car which accelerates. If the direction of the acceleration was depicted as an arrow, the pointy end of the arrow would be pointing at the front of the car, I.E. in the direction of the speed change.

THE LOGIC THUS FAR: Speed is directional. Speed change requires an acceleration (which is also directional). THUS changing direction requires an acceleration, which is depicted as an arrow pointing toward the direction in which the acceleration force is acting.

5. When a car turns a corner, it changes direction, therefore it MUST have an acceleration (or force) applied to it (remember the speed is directional) in the direction of the corner.

6. Which way do we draw the arrow to depict this force? The arrow MUST be drawn toward the centre of the turn because that is the direction the force is acting, just like we did for the car's straight line acceleration.

THE LOGIC THUS FAR: when turning, you are changing direction, and an acceleration can be seen to be acting toward the centre of the turn.

7. David's sling is wrapped around the rock, and the rock is traveling at speed. but the rock is not travelling in a constant direction - it is traveling in a circle. Thus the rock is constantly changing direction. Thus a constant force MUST be being applied to make it constantly change direction.

8. We can depict the force acting on the rock by drawing an arrow. Just as we did with the car, the pointy end MUST BE pointing in the direction that the force is acting, I.E. toward the centre, I.E. toward David's hand.

THE LOGIC THUS FAR: There is a force that can be seen to be acting toward the centre of the circle.

Physcisists (who can spell rooly gudder than me) decided to give this force a name so that we could refer to it without saying "that force that acts on a mass that is changing it's direction"

They named it centripetal.

9. When the sling releases the rock the force trying to change it's direction is gone, therefore the rock will travel in a straight line again, along a tangent. If the rock were to fly out of the sling in any other direction, an acceleration must be applied (because we know that a change of direction can only be accompanyed by an acceleration).

If this does not help, just indicate which step (by it's number) that you are having difficulty with and I'll see if I can clarify.

Good luck.

Lu,

Please disregard all the above, it will not help you: it refers to a scientifically accepted and standardised theory from a high school text book.

[ 03 December 2001: Message edited by: helmet fire ]

[heedm]

Helmet Fire,

Nice.


Only one point to add. Centripetal means "center seeking". Don't know why. Apparently most of the original physics dudes came from Latin America. Hence everything is in Latin.

[ 03 December 2001: Message edited by: heedm ]

[JoePilot]

Helicopter rotor blades are simple and continue to work very well despite the fact that the words everyone uses are all mixed up... "But what else have the Romans done for us ... eh?"

Is anyone charging Lu for this course?

[vorticey]

heedm & helmit fire
thanks for the good explination helmetfire, i do understand what centripital is.
but i think its wrong to say there's a force pulling the blades to the centre (wich would increase with speed) when we need strong hubs to hold them from flying off(as lu said). if the force increased with rpm then the hub would never break.!
just because scientists couldnt work it in properly with there theory, doesnt meen its not right. the blades are held from flying off by the hub and they change direction because the hub is turning them!
if a tractor is trying to pull a stump out of the ground, you could say there's a force acting on the tractor pulling it towards the stump.
a car turning a corner only feels a force into the corner because the tiers are changing the direction the car is going.
you can say that there is an imaginary force pulling the car to the inside of the corner, but theres not!
i do agree with centripital being an easy way to understand it, but thats all.

[heedm]

vorticey, you said, "the blades are held from flying off by the hub and they change direction because the hub is turning them!"

You're right. That's exactly what happens. Centripetal force is a force that is needed. It must be supplied by something. In this case the centripetal force is supplied by the hub.

Break the hub and the centripetal force is no longer being supplied so the blades quit changing direction and begin to travel straight.

As the speed increases, the centripetal force would need to increase. Since the hub is supplying the centripetal force then the load on the hub increases. Increase the speed too much and the hub breaks.

_________________

"a car turning a corner only feels a force into the corner because the tiers are changing the direction the car is going"

You're right. The tires are supplying that needed centripetal force. Changing the direction of the car requires a force.

[JoePilot]

VOR City: You are definately kidding .... right?

[sling load]

You have to understand something about Lu, he is an unlicensed A&P Mechanic who calls himself an engineer, he is not. He is also not a pilot, this has been Lu's MO for a while, to beleive everything he says is for the foolhardy. If you ever get the chance to one day fly an S76, think about the engineers who designed it, they designed it with centripetal force in mind, and if you want some further proof, if thats not enough, the fella who test flew it is a regular contributor to the rotorheads forum, I suggest you read the posts by Nick Lappos, remember vorticey, Lu is the bloke who also suggested that retreating blade stall does not exist. He also suggested at the bottom of an autorotation, you lower the collective to affect touchdown. BUYER BEWARE!

[ 04 December 2001: Message edited by: sling load ]

[Lu Zuckerman]

To: sling load

Once again I have to correct you. I have been working in the field of product integrity since 1968 and the title of a person working in that field is a Reliability, Maintainability and Systems Safety Engineer. It is not necessary to be a graduate engineer to work in this field but the title is still the same. Again I must also correct you regarding my being an unlicensed A&P mechanic. I got my P license in the 60s and I got my A license in 1976 and it is still valid. Regarding my comment about lowering the collective at the bottom of an touchdown autorotation I was alluding to the fact that if you did not do this there was a possibility of the blades stalling out and hitting the fuselage.

Also please tell me and the others when and where I stated that retreating blade stall does not exist.

ATTENTION: Sling load did find a statement on the Helicopter History site where I stated that retreating blade stall does no exist. Here is my response which also appears in the weather and other nasties thread.

To: sling load:

You can stop gloating. I put that on the web site in conjunction with a thread on Just Helicopters. It is still my contention that retreating blade stall can best be described as a differential of lift across the disc, which results in the disc flapping back. I have even made these comments on these threads. You can use this description and apply aerodynamic precession or gyroscopic precession but in my mind that is what is happening. The blades are not stalling, they are just generating less lift. If you want to call it retreating blade stall that’s OK with me as that is the accepted term but it still ends up being a differential of lift that causes the flapback / blowback.

I’ll use this illustration which I have used on these threads as well as on Just Helicopters.

If you believe that the retreating blade has stalled and further believe that individual blades stall and drop out of orbit and fall due to the stall then try to visualize this. Once the blade has stalled and dropped down over the tail (90-degrees later) it still is attached to a spinning rotorhead and must then immediately get back to the commanded tip path. That means that that blade when it is down over the tail it must fly up until it is now the advancing blade and, will end up being down over the nose, as this is the commanded tip path. If you were to look at the disc from the side of the helicopter the disc would scribe an inverted V or be just the opposite of the cone angle. Can you imagine the vibratory forces that would ensue if the blades had to change their position so radically at anywhere from 250 to 500 times per minute?

Now this may be difficult to comprehend but try to visualize the disc as a single entity. The basic premise in helicopter design is to have an equal distribution of lift across the disc. When the pilot moves his cyclic he alters the lift distribution across the disc and the disc tilts in the direction of cyclic movement. You can visualize this as aerodynamic precession and I think of it as gyroscopic precession. In either case there is a differential of lift across the disc.

In the case of retreating blade stall, the retreating side is generating less lift than the advancing side. This causes either a perturbation of the disc if you accept gyroscopic precession or it is a direct aerodynamic lift that responds 90-degrees later and the result is the disc blows back / flaps back.

I put the comment on the web to find out what other people thought on the subject.

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[ 04 December 2001: Message edited by: Lu Zuckerman ]

[ 04 December 2001: Message edited by: Lu Zuckerman ]

[ 04 December 2001: Message edited by: Lu Zuckerman ]

[ 04 December 2001: Message edited by: Lu Zuckerman ]

[helmet fire]

Lu,
As this thread has turned into a protracted discussion on centripetal & corriolis, I have started a SEPERATE thread for Retreating Blade Stall (again!) and have transferred your comments there, so as not to obscurate the topic at hand.

Can we argue RBS on that thread, and leave this one ???

[helmet fire]

to vorticey:

I see where you are coming from. You are concentrating on the feeling of being thrown outward, as in the car turning, or the girl holding on to the chain on the roundabout. That feeling of being thrown outwards is central to the difficulty in understanding centripetal, and the reason centrifugal is used as an easy substitute to promote understanding.

Perhaps this will help:

Going back to the logic steps I used above to describe centripetal, I used a car accelerating along a straight line as an example of a change of speed. Because speed had direction, and acceleration does too, when we depict the force, we can draw an arrow in the direction the car is traveling. in other words, to where the car is accelerating TO. Thats why centripetal was IN toward the centre of the circle: because it was an acceleration TO that direction.

So why do we feel flung outwards? When the car accelerates in a straight line do we feel a force? You betcha. We sink into the seat. We would get "flung backward" if the seat was not restraining us. In other words - we "feel" the acceleration in the opposite direction to which it is commonly thought of as being applied. In this case, the acceleration is going forwards, but we feel it going backwards.

Thus, when we are turning, we feel the opposite direction to the acceleration. I.E. we feel "centrifugal" because of the application of centripetal. Thus the force is centripetal (same as acceleration forward) and the sensation we feel is the REACTION (Thank you Mr Newton) to that force, but is NOT in itself a force.

Therefore the answer to Butch Grafton's FAA question on the "questions" thread is not amongst the solutions. The actual answer is that the oil goes to the edge due to the equal and opposite reaction of the attempt by the spin to apply centripetal force to it without restraining it. I dont mean to question the FAA, because they have never been inaccurate have they??

Thus, in answer to your statement above that there cannot be a force pulling the car into the corner, you can see that the force is only as imaginary as the force of straight line acceleration. If a force exists there, a force must also exist pulling the car into the turn. And we feel it just like the straight line acceleration - in the opposite direction to the force application - just like Mr Newton said we would.

I guess they gave centrifugal a special name to make it an easy to grasp concept, but at the end of the day, it is the same force we feel as we are resisting the change in speed (acceleration) of a car in a straight line. I dont know why the Latin Americans neglected to name that feel too??

Wait...what about "g forces"?



hope this helps.

Edited to try and address vorticey's specific statements rather than ramble on & on & on...woops, there I go again!

[ 04 December 2001: Message edited by: helmet fire ]

[heedm]

To follow in this terminology, what is wrong with the FAA is not that they didn't supply the right answer. In fact, they asked the wrong question.

Don't ask how the fluid gets thrown off. Ask why the blades don't fly in a straight line and the fluid does. The answer, the blades are accelerated centripetally by the force applied to them by the hub. The oil is not being accelerated.

[helmet fire]

heedm: you are right (again)! Thanks.

[Lu Zuckerman]

To: heedm

It is not oil it is alcohol. The fluid is pumped under pressure and there is a fixed tube just behind the propeller. Mounted to the propeller is what is known as a slinger ring. The fluid is pumped into the slinger ring, which is spinning with the propeller. The fluid spins with the slinger ring due to some level of friction and in a sense becomes a part of the slinger ring. The slinger ring has small tubes that terminate several inches outboard of the hub and each blade has its’ own tube. Centrifugal force is the motivating force that pumps the fluid onto the backside of the blade.
So, the FAA answer to their question is “centrifugal force”.

[helmet fire]

Lu,

You are right about the oil/alcohol detail - but I think we will understand your selection of centrifugal after all of the discussion if you could comment (by the numbers) on the logic progression I provided for vorticey on the previous page. If you address the first step that you cannot follow, then we can begin discussions there and then we can find out if any common ground exists. To do this, I propose that you concentrate soley on the logic steps, we introduce no other examples, and we discuss no other phenomena. Sorry to hijack the phase lag main topic (of which really looks as though it is getting somewhere).

Cheers,

[heedm]

I would have picked centrifugal. Normally multiple choice questions ask you to pick the best answer, not necessarily the right answer.

As far as oil/alcohol, I didn't have a clue what the question was talking about. Thanks.