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Post by rmc on Jan 11, 2015 17:38:29 GMT
I've found two answers to the following question, and I am guess that only one answer is really correct: When a thing, (in this case a propeller), has a frictionless pivot-point that is located directly inline with its center of mass, or center of gravity, what is the final resting position of the thing, (assuming it is located at the surface of Earth)?
Answer 1: Like a sea saw, it will ultimately come to rest, level with the surface of the Earth.
Answer 2: Since it is balanced perfectly, and we are assuming frictionless pivot, it will rest in any position we stop it, level or not.
The following two videos demonstrate, regarding balancing a propeller at least, that there seem to be at least these two distinct answers. In the first video, the gentleman declares that the item in question, (a propeller), should come to rest level with the surface of Earth at about 1:17 into the video:
In the next video, another technician declares that a balanced item will rest in any position. I tend to lean this direction myself. But, want your opinions if you have any, or, better, the math that supports one view or the other. He states his position on this at about 2:00 into the video:
The first explanation, (the ending up being level idea) was posed to me at school by the math center's tutor. He said that since potential energy is based on a thing's height, the side highest should swing back down, then back up again, not as high, dampening out, ultimately resting at a totally level orientation.
The opinion I have is that since potential energy is figured on a thing's center of gravity, there is really only one spot worth factoring in with regard to potential energy, and that's the spot right where the thing is spinning. So, the arms of the thing don't really count with regard to potential energy. Only the height of the prop as a whole (measured at its cg) is used to determine potential energy. And, since the thing is balanced perfectly, it will act as though falling in free fall when mounted on its pivot, coming to "rest" at whatever position it has when there is no rotational speed.
Naturally, none of the real things shown or spoken about has a truly frictionless pivot point, but the question is about a thing that somehow has a frictionless pivot. I hope there is a way to parse out the facts for this one.
As an aside, I explain the notion about the sea saw concept as the fact that most true sea saws are actually built with the pivot point slightly off center from the center of mass. For instance, a piece of wood, serving as the sea saw, has some sort of bracket bolted to one side, serving as the pivot. This arrangement causes the wood and the rest of the structure to be above the pivot point, resulting in a leveling-out by way of imbalanced forces, top to bottom, but balanced forces, side to side.
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Post by the light works on Jan 12, 2015 16:42:18 GMT
in theory, a perfectly balanced thing in a frictionless environment, should freely rotate forever. perpetual motion aside - I favor the explanation that a perfectly balanced thing in a low friction environment should come to rest wherever the kinetic energy runs out - by definition, if it has a "resting point" that it returns to, that means it is off balance. factoring theories about diminishing effects of gravity in proportion to distance from the center of the planet, I would expect a long skinny object to be biased towards a vertical orientation. (the lowest point will be more strongly affected by gravity than the highest point, therefore the center of balance is shifted downward in minute proportion to the vertical length of the propeller.
but I still remain with my contention that if it is in balance, the potential energy from the highest point is always equally offset by the kinetic energy required to lift the lowest point. - in agreement with your opinion.
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Post by watcher56 on Jan 13, 2015 1:57:02 GMT
I agree the second is correct. It will stay wherever it is placed. If it has any preferred resting position including horizontal or vertical, it is not balanced.
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Post by OziRiS on Jan 13, 2015 19:12:07 GMT
Eerrmm... What?
Wouldn't that only be possible if the thing we're talking about has equal weight on every possible axis of rotation, i.e. a ball?
As I see it, any other shape than a ball will have a shift in mass if you tilt it. The propeller is a perfect example. Tilt that to one side and you've shifted the center of gravity, making the propeller do one of two things when it stops, depending on how much it's shifted: fall back into the level position or lose its balance and tip over entirely. Sure, it can stay on its pivot point while it's rotating, but that's due to centrifugal energy and nothing else. When it stops, it'll either level out or tip over.
Anything (other than a ball) that doesn't act this way is either not frictionless, has some sort of compensatory system to shift the weight of the object with the tilt to keep in balance on its pivot point, or the bulk of the mass of the object is located near the center of the object.
EDIT: In the case of the second video, I'm willing to venture a quiet bet that it's the last of the three possibilities that makes him able to balance it in any position. We're basically talking a lever here. With any lever, place the bulk of the mass at the pivot point and the mass won't have much effect on where the lever comes to rest. Place the bulk of the mass further out on the lever, then it becomes significant.
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Post by the light works on Jan 13, 2015 20:01:59 GMT
Eerrmm... What? Wouldn't that only be possible if the thing we're talking about has equal weight on every possible axis of rotation, i.e. a ball? As I see it, any other shape than a ball will have a shift in mass if you tilt it. The propeller is a perfect example. Tilt that to one side and you've shifted the center of gravity, making the propeller do one of two things when it stops, depending on how much it's shifted: fall back into the level position or lose its balance and tip over entirely. Sure, it can stay on its pivot point while it's rotating, but that's due to centrifugal energy and nothing else. When it stops, it'll either level out or tip over. Anything (other than a ball) that doesn't act this way is either not frictionless, has some sort of compensatory system to shift the weight of the object with the tilt to keep in balance on its pivot point, or the bulk of the mass of the object is located near the center of the object. EDIT: In the case of the second video, I'm willing to venture a quiet bet that it's the last of the three possibilities that makes him able to balance it in any position. We're basically talking a lever here. With any lever, place the bulk of the mass at the pivot point and the mass won't have much effect on where the lever comes to rest. Place the bulk of the mass further out on the lever, then it becomes significant. I'm not so sure. the center of mass should not be capable of moving around inside the body without the body actually changing shape. therefore if your center of mass is the pivot when the propeller is horizontal, it should still be the pivot when it is at 45 degrees. - the only thing that changes would be if the effects of gravity change with the relative change in elevation between the tips. |
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Post by watcher56 on Jan 13, 2015 23:18:42 GMT
I'm not so sure. the center of mass should not be capable of moving around inside the body without the body actually changing shape. therefore if your center of mass is the pivot when the propeller is horizontal, it should still be the pivot when it is at 45 degrees. - the only thing that changes would be if the effects of gravity change with the relative change in elevation between the tips.[/quote] Bingo. The CM does not change with physical position. It is where it is. |
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Post by silverdragon on Jan 15, 2015 10:00:41 GMT
I look at them both kind of being right.
The idea that the blades should balance horizontally shows the blades are balanced.
The ability to then put the thing in any position at all and it will stay where it is put, shows that the hub is balanced as well.
You need both to be in balance, blades and hub.
However, in the second video, his "It doesnt matter" on sandpapering down the shape of the blade, is wrong.
A LOT of research has gone into the shape of any blade.
As for balancing the rig before you balance the blade, does that matter?
Yes.
If you have a tilt to the axle, as you spin, it may try to answer gravity by moving downwards.
That may cause friction on the rotation of the axle.
It aint much, but it is "Something", and may prevent you getting the right result.
... Or, So I have been told.....
Perhaps that needs to be tested?
I have always seen it done with the axle perfectly balanced before the blade goes anywhere near it.
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Post by watcher56 on Jan 15, 2015 18:10:59 GMT
Beyond this, there is the the fact that static balance and dynamic balance are not the same. The blade can be perfectly balanced statically, but horribly unbalanced when spinning.
A mass concentration has a force directly proportional to it's distance from the center under static conditions, but the force it produced when the body is spinning is proportional to the square of distance from the center.
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Post by rmc on Jan 16, 2015 0:33:58 GMT
With regard to the question of how things behave for a static balanced propeller, (having an ideally frictionless pivot), my math tutor is still clinging tightly to the notion that, when a blade of such a propeller is raised to 45 degrees from level and then released, it will attempt to reach level with the surface of Earth. If there is no air resistance, then the propeller would continue harmonic motion, swinging down to -45 degrees and then returning back to the original 45 degrees, over and over again.
My math tutor, (actually pretty skilled in Newtonian physics, actually) states that since the video (#2) shows a propeller staying stationary when released, it is likely due to air resistance, buoyancy in air, possibly even small electro-static forces, and whatever little friction the bearing has that holds it in place. He again argues that when an object (in this case, one of the propeller blades) is higher, it has a greater potential energy. The two torques created (one for the "left" blade, and another opposite torque for the "right") are not completely equal. And the reduced value for the pull of gravity as we go higher and higher in the sky is not enough to counter the raised potential energy of the higher blade. (or something like that. I'm actually having a bit of trouble following his logic to tell you the truth. So, it's sort of hard to try and argue his line of logic in his place. I am probably a bit stupid or something)
Since, to test this, it requires a perfectly frictionless pivot (impossible), a vacuum, and a perfectly balanced propeller with regard to its static position (perfect balance of any sort is arguably impossible too, of course), this one remains merely a thought problem it seems to me. And, since it depends on what model you use to math it out, I wonder if proving it on paper is feasible at this point too.
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Post by the light works on Jan 16, 2015 1:21:03 GMT
With regard to the question of how things behave for a static balanced propeller, (having an ideally frictionless pivot), my math tutor is still clinging tightly to the notion that, when a blade of such a propeller is raised to 45 degrees from level and then released, it will attempt to reach level with the surface of Earth. If there is no air resistance, then the propeller would continue harmonic motion, swinging down to -45 degrees and then returning back to the original 45 degrees, over and over again. My math tutor, (actually pretty skilled in Newtonian physics, actually) states that since the video (#2) shows a propeller staying stationary when released, it is likely due to air resistance, buoyancy in air, possibly even small electro-static forces, and whatever little friction the bearing has that holds it in place. He again argues that an object (in this case, one of the propeller blades) is higher, it has a greater potential energy. The two torques created (one for the "left" blade, and another opposite torque for the "right") are not completely equal. And the reduced value for the pull of gravity as we go higher and higher in the sky is not enough to counter the raised potential energy of the higher blade. (or something like that. I'm actually having a bit of trouble following his logic to tell you the truth. So, it's sort of hard to try and argue his line of logic in his place. I am probably a bit stupid or something) Since, to test this, it requires a perfectly frictionless pivot (impossible), a vacuum, and a perfectly balanced propeller with regard to its static position (perfect balance of any sort is arguably impossible too, of course), this one remains merely a thought problem it seems to me. And, since it depends on what model you use to math it out, I wonder if proving it on paper is feasible at this point too. how does higher potential energy create higher torque? does he weigh more if his scale is on top of his desk than he weighs if it is on the floor? if I don't have the weight to loosen a bolt, will I do better if I jack the car up? he may be skilled in newtonian physics, but his argument, by my understanding of potential energy, does not follow. |
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Post by rmc on Jan 16, 2015 1:31:30 GMT
Since he won't clarify that part and now only mentions things like electrostatic forces holding it still while telling me to get back to whatever integral I'm working on, I figure he embarrassed himself early on and just wants the topic to go away? Since he won't explain and since video (#1) states basically the same thing he originally said, I was curious if I could find someone with the same opinion who might complete the description.
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Post by the light works on Jan 16, 2015 1:40:57 GMT
Since he won't clarify that part and now only mentions things like electrostatic forces holding it still while telling me to get back to whatever integral I'm working on, I figure he embarrassed himself early on and just wants the topic to go away? Since he won't explain and since video (#1) states basically the same thing he originally said, I was curious if I could find someone with the same opinion who might complete the description. sounds suspiciously so. |
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Post by silverdragon on Jan 16, 2015 7:10:01 GMT
I have showed this thread to my Kid who is doing A-Level maths and OND to HND Physics (College course).
They are trying to understand principals of flight.
Their version, a propeller properly balanced, rested horizontal, should stay horizontal.
If raised at one side, it should return to being horizontal.
The weight of the blades should balance around the pivot.
The hub should not have any overriding effect on that balance.
If the propeller is raided to the vertical, it may balance in the vertical, momentarily, but the slightest rotation of that, it should, like a set of scales, balance in the horizontal.
It is, after all, a balance such as a set of scales, the weight of the blades being more than the weight of the hub.
I think I agree with my Kid...
I was looking at this from what I knew of flight, and as stated a lot, I aint no mechanical genius, I did electronics.
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Post by rmc on Jan 16, 2015 8:53:57 GMT
Okay, Silverdragon! Now, it looks like we share a similar problem, maybe. The individual I know who puts it like that won't fill in a complete explanation. Will your son provide the exact details for us? Is the prop in video (#2) held at 45 degrees, stopped by some sort of resistance, does he figure? If the blade is balanced completely around the center of rotation, what imbalanced forces are at work, leveling it out? (How does one end achieve more torque than the other)?
The concept of a scale with balancing masses may not be a good analog for the propeller, due to the fact that, like most sea-saw arrangements, though they intend to be balanced, they really are only balanced in the side to side orientation, it looks to me. There very likely is some imbalance top to bottom. As in a sea saw, the bracket serving as a balanced pivot is bolted to one side of the board, usually the lower side of the board. So, with this slight off set, we get the balancing scale effect. Balancing scales, the very first anyway, hung the masses beneath a leveling arm, etc.
So, in the typical case of a sea saw, the top-to-bottom off set is caused by setting the board slightly above center of rotation. In a balancing scale, the off set is due to masses hung slightly beneath center of rotation, I think. It is this very slight off set from center of rotation that provides an imbalance of forces, albeit found within a so called 'balancing scale', that sets up the wobble to level scenario. The propeller, before sanded and balanced could behave the same way, should the imbalance be on one side of the hub for instance. But, after removing the slight imbalance, it no longer seeks level, me thinks...
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Post by silverdragon on Jan 16, 2015 9:15:44 GMT
Thats what my Kid is trying to work out... Them at college, so just me at the moment, I will try to get them back to this time permitting, as they have lots of revision going at the moment.
However...
The hub, minus the blades, must have some imbalance, to allow a "Stop anywhere" balance when the blades are attached?...
Is that necessary?...
Is it really necessary that there is a "Four point" balance.
Their initial thought was if you put a half ounce weight on one side of a already balanced bicycle wheel, you need a half ounce weight on the exact other side to keep it in balance.
Because the mass of the wheel exceeds the mass of the weights, that wheel will be in perfect balance.
However, if you put a mass greater than the wheel on one side, it is no longer a wheel, its a balance, and if you match that weight on the other side, its now a see-saw balance, not a wheel.
My thoughts,
So the hub, is it a wheel in its own right.
If the mass of the blades is greater than the hub, then its not a wheel.
Yes you need a sort of equilibrium between the two sides that dont have the weight attached.
So, in conclusion, the Hub weight must exceed the weight of the blades, and in fact what they have done is balance a wheel.
I also have another question, he mentioned a prop spinner.... the bit that is cone shaped that goes over the prop.
Are we to presume that is also balanced in some way?.. it must be, as its now part of that "wheel"....
At this point, I run out of knowledge, as this bit where a prop is a wheel when it comes to balancing it is a tad confusing to my brain...
But it is making sense, if you think of the prop as a wheel with the properties of a wheel, just no outer tyre.
And wheels need balancing on all sides, not just two.
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Post by the light works on Jan 16, 2015 15:13:25 GMT
I agree that a balance beam scale is slightly imbalanced to the bottom to allow it to rise to level.
as for the wheel question: a wheel is a thing designed to reduce the sparks when you drive your car. it doesn't matter what the weight relation is between the hub and the rim.
however, as far as balance is concerned, the relation between static balance and dynamic balance bears investigating, but that opens the question of why triangular propellers remain in balance.
otherwise, as long as the mass is symmetrically arranged around the hub, the propeller should remain wherever you put it. any other result implies imbalance.
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Post by silverdragon on Jan 17, 2015 9:54:42 GMT
I questioned my Kid again...
And got lost.
He started with "Moments", being something to do with balance...
I caught up with him on the following bit, that the art of balancing a pencil on its point.
A Sharpened pencil can, in theory, be balanced on its point, on a stationary object.
You ever tried that?...
But, as he argues, its a very fine balance...
So fine that also in theory the movements of electrons in the sub atomic level of the pencil's mass could be enough to be what eventually makes it fall over....
But he did get back to the prop, with Friction.
Being that you have it finely balanced to be only slightly out, the friction free argument of the shaft its balanced on comes into question, in that its not totally friction free, and the small negligible friction that is in there is probably just enough to stop the prop rotating.
In that its stops anyway if you were to spin it.
The fact that if you spin it, it will eventually stop, proves his point....
So no, its not balanced totally, its just balanced "Enough" that the friction of the axle is enough to stop it turning.
This also brings me on to the eventual mechanical limit of speed on a propeller driven aircraft....
It wasnt the design of the plane that was the problem.
The Spitfire is perfectly capable of doing much more than Mach1.... in theory...
More power?..
No.
What was needed was the ability to balance the engine and other components finely.
The shakes is what stopped it going mach1 (In level flight)
The reason Jet engines were eventually used is that they were by design already built to higher tolerances than piston engines..
Just ONE turning part is probably the key... KISS, keeping it simple.
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Post by rmc on Jan 17, 2015 11:03:36 GMT
A moment of force for our discussion is the product of a force and its distance from the axis of rotation.
The force is weight, in Newtons, figured on the center of gravity (cg), and is the apparent weight or net weight, resulting at the moment of force. The moment itself would be the portion of meters away from the center of rotation (cr) the center of gravity is, multiplied by the force of the apparent weight located there (net weight). But, if the center of gravity is in perfect line with the center of rotation, there would be no moment of force and no rotation, because there would be no distance between cg and cr to provide the "lever arm like" moment of force.
We can further demonstrate this, supporting the frictionless concept, by maximizing masses that are symmetrical and distant from the cg/cr. When everything is in line, symmetrical, and balanced, you will know due to the fact that the object can be placed in any orientation and it stays there.
Take a large, ten-speed bicycle tire, attach two masses at opposite sides from one another, make adjustments until you find that you are able to orient the wheel such that the masses are in line with one another, one rests at 45 degree angle, let's say, while the other is found at 225 degrees. And when everything has been adjusted correctly, letting go of the wheel should cause no motion, no rotation. If you still get some rotation, adjustments to balance are still required.
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Post by silverdragon on Jan 17, 2015 11:19:35 GMT
You sound just like my Kid.... So it must be science then.  But you lost me.... I see the words, I understand the words, where you loose me is their arrangement in a sentence...?...  What you are saying, and thanks for saying that, is making sense, and I am learning something here, but I cant argue with anything just yet until I understand.  Just what is the difference between a balanced beam and a propeller. If a beam is balanced on a point so that it stays in the horizontal, such as a balance beam set of scales, it stays in balance, it wont rotate 45degree and just stay where its put, it will return to horizontal. So how can a two blade prop where the blades are obviously much more mass than the hub they are attached to have any way to overcome that balance beam problem. THATS the bit I can apply any logic that I understand to. Its a two dimensional object. The wheel problem, being that it is a wheel, needs balance in all three dimensions and all points in-between, because its three dimensional. So I am assuming that the prop has crossed the border between a balance beam and a wheel at some point.... Where? Three blade props are three dimensional by nature, its easier to understand. Four blade props must be a breeze..... |
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Post by rmc on Jan 17, 2015 11:31:31 GMT
Balance a pencil on your finger, such that the pivoting allows the eraser-end to dip down while the writing end swings up, let's say. Once you achieve balance on your finger, all you have is the center of gravity directly over your finger, but is also located within the body of the pencil, because, it is after all, the pencil's (cg). The center of rotation lies outside the body of the pencil, along your finger, and the center of gravity is over the (cr), but not in the exact same place. This tiny, tiny distance between cg and cr creates a moment of force, allowing the pencil to be torqued in opposing and symmetric ways that level it out: like the balancing scale and the children's sea saw. They all have a moment of force involved since the cg and cr are not exactly over each other in every way.
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