Fletcher's Wheel - Ingenuity verses Entropy

[Fletcher]

Fletcher this sim had me bamboozled for a while till I realized the rim is not free of the wheel and that is why the RBGS has rotated downward.

If you look at the diagram of the sim of a normal weight force displacement RB then you can see a circle orientation line (the black radius line from COR to 3 o'cl) for the background wheel - to that background is pinned the gears of the RBGS, as are the rim stop boards - they are always located in the same relative position on the background wheel - one can't move without the other moving also - this makes it a lever on a fulcrum.

If you have a lever with equal weight at 1 meter each side of the fulcrum then it is balanced - if you take some weight off the rhs & put it at 2 meters the lever is no longer balanced & the lever rotates CW - this is what this system does in this diagram - we shift weight force to a greater lever arm distance.

If the rim and stop was independent of the wheel like your wheel proposal then the RBGS would have back torqued due to the counterweight.

Yes, if the rim stop board were just a shelf fixed in space to put the hanging weight on then there would be CCW torque turning the RBGS anticlockwise - IOW's some weight has been removed from the rhs making the lhs heavier - just like a lever.

But my proposal has the RBGS & background wheel (with rim stops) always locked together, they are not independent.

The RBGS will back torque till the string is tight again. Yes, then it will coast because it is weight balanced again.

The back torque after the load is dropped is why your wheel needs a feedback system to turn the RBGS. Enough CF might provide enough power to do both.

[Fletcher]

... could you explain why it is RB doesn't carry "moment of inertia penalty" other mechanisms do. I am having trouble seeing how flail weight, RB/ rack/ pinion mechanism has same MOI from wheel's frame of reference.

Before I decide what examples to give you or how complex to pitch the answer please tell me your understanding of MOI in a rotating environment ?

You might save me a lot of time.


http://en.wikipedia.org/wiki/Moment_of_inertia

The natural frequency of oscillation of a compound pendulum is obtained from the ratio of the torque imposed by gravity on the mass of the pendulum to the resistance to acceleration defined by the moment of inertia.

[triplock]

Fletcher,

I need your intellect on this one as I don't understand something.

What is the relationship between radial mass position at x R and Centripetal Force felt.

Graphically , starting at dead centre, and with the mass displacing outwards , is CF change linear or expotential ?

The wheel has a fixed RPM.

How is CF affected by the changing circular distance travelled by the mass over a fixed time as it moves outwards on a fixed RPM wheel ?

Chris

[Fletcher]

Here is a basic study of MOI (Rotational Inertia).

Two pendulums.

The first (lhs) pendulum has a drive mass of 1kg at 1meter arm & two masses of 2kg each at 1meter arms.

The second (rhs) has a drive mass of 1kg at 1meter arm & two masses of 1kg each at 2meter arms.

If we were to make a lever & fulcrum then 2kg at 1m would balance 1kg at 2m arm i.e. it would be weight balanced with no torque.

IOW's ... m1d2 = m2d1 (Archimedes Law of Levers).

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But ... the inertia of the system is different depending on where the masses are placed on the lever - so 2kg at 1m is not the same as 1kg at 2m in terms of how hard it is to rotate the system.

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This is why for simple point mass systems I = mr^2 and not I = mr (law of Levers).

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We know this by some simple arithmetic - the drive mass will lose height & PE - as it does this it must spin up the masses on the end of arms - the combined KE's of all 3 masses can not be above the PE lost by the drive mass - the difference in the drive masses KE & the balance of KE in the others must be due to Rotational Inertia.

As a further check, if m1d2 = m2d1 then the drive masses would have lost the same PE & have the same KE's & be at the same height loss in a given time - if the one with the long arms were then the sum of the KE's would be greater than the PE lost by the drive mass, which cannot be, & isn't.

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In the RBGS it does not matter where the mass on the horizontal beam is placed i.e. the distance to the COR can be variable - firstly, a RBGS is a torque equalizing mechanism - secondly, because of the nature of the RBGS it always loses the same PE & has the same period so MOI is the same for all arrangements of beam mass position.

In any other pendulum like device the mass distance for COR effects the MOI of the device.

[daxwc]

Changed my wee mind.

[triplock]

Daxwc

Now that is a diagram I understand !

I hope Fletcher isn't trying to lift himself up by pulling on his own shoe laces as it will killed the concept.

Chris

[triplock]

Bloody hell Daxwc, that wS a quick edit !! Now I look a complete twat :D

Chris

[Fletcher]

Fletcher,

The wheel has a fixed RPM.

What is the relationship between radial mass position at x R and Centripetal Force felt ?

Cpf = mv^2/r in Newtons

If RPM is fixed at say 60 rpm & the mass is at 1r then we can calculate the masses velocity & Cpf.

Circumference of a circle is 2Pi.r = 2 x 3.14159 x 1 = 6.28 meters
If rpm = 60 then the mass travels 1 revolution per sec, or, 6.28m/s – so its velocity/speed is 6.28m/s.

If the mass is 2kg then the Cpf = mv^2/r = 2 x 6.28 x 6.28 / 1 = 78.88 N’s.

Graphically , starting at dead centre, and with the mass displacing outwards , is CF change linear or expotential ?

If rpm is constant then doubling the radius (2r) will double the speed of the mass – that’s because it travels twice as far in the same amount of time.

Linear for the same rpm (see below) – that’s because it is proportional to speed & divided by radius so it moves it from a quadratic to a linear relationship.

How is CF affected by the changing circular distance travelled by the mass over a fixed time as it moves outwards on a fixed RPM wheel ?

Cpf = mv^2/r = 2 x 12.56 x 12.56 /2 = 157.75 N’s which is twice the Cpf’s at 1r for the same rpm.

If rpm is constant then increasing mass radius increases Cpf.

[triplock]

Fletcher,

Now that mathematical explanation should be held up as an example of how to perfectly illustrate a point. Clean, concise and understandable use of formulae.

Well done ;-)

Chris

[Fletcher]

Changed my wee mind.

Glad you removed that drawing dax - you can't bootstrap.

What you are possibly forgetting is that you need a gradient or differential to release Potential Energy - your lever diagram didn't have a pathway - the RBGS does because the distance between the horizontal beam & the rim stop grows as they move by each other - this provides the pathway or gradient.

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However, I want to go back to basics - we know we could build a RBGS with a rack & pinion, & a flail type mechanism to push on the rim stop - I have no doubt that can be done.

I am confident that we could harness the inertia (momentum, or energy if you prefer) of the sliding rack/driver as a force & apply that force to the rim stop thru the flail action.

We know that Newtons Laws says that for every action there is an equal & opposite reaction - so even if we take a 10th of the racks inertial force & apply it at the rim it will be applied for a long time (about 1/4 of a circle) - this is a long push time - since there must be an opposite force felt by the wheel then an equal but opposite force is felt by the RBGS (the back torque).

But ... unlike the regular standard RB where mass is lifted onto the rim stop which causes rotation (N.B. we can't reset) here force only is applied at the greater arm of the lever that the RBGS & background wheel form - so we have imbalance causing rotation, using Law of Levers.

IOW's we take momentum of the rack & convert it into momentum at the rim, not unlike taking KE of a mass & converting it to PE (height).

Before you ask, ...

In the regular RBGS that tries to create rotation by lifting a mass & placing it on the rim stop the in-contact distance is dictated by the string length as you pointed out (a short arc of distance) - in my force transferrence version that in-contact distance is increased considerably.

More time & distance pushing means higher rpm !

[oldNick]

Hi Fletcher,

I could show you how to build a wheel with no stator, and how to bootstrap (make a weight lift itself ) But you would have to assess one of my wheels. Interested.?

No disrespect intended with this off-topic subject.

Nick

[daxwc]

Fletcher:

Glad you removed that drawing dax - you can't bootstrap.

What you are possibly forgetting is that you need a gradient or differential to release Potential Energy - your lever diagram didn't have a pathway - the RBGS does because the distance between the horizontal beam & the rim stop grows as they move by each other - this provides the pathway or gradient.

No I realise that RBGS has movement on the outer gear and why I took the picture off, but obviously have concerns it binds into a lever. I was going to draw what I thought was the correct picture because there is some movement. In truth today I am unsure of what really will happen as there are too many forces at play on pivots then you add in momentum.

Hard to believe the Sim doesn’t have a linear to rotation option of the rack and pinion. I know you said you tried to sim the slider and the lever and it doesn’t work, but that even makes me more skeptical that the rack and pinion will not work either. So exactly why will the slider and lever in the picture not work?

[Fletcher]

Hi Fletcher,

I could show you how to build a wheel with no stator, and how to bootstrap (make a weight lift itself ) But you would have to assess one of my wheels. Interested.?

No disrespect intended with this off-topic subject.

Nick

That's very kind oldNick .. I'd love to know these things first hand, because they are 'challenges' to say the least (obviously most OOB wheels don't need a stator) - bootstrapping is also a sort of holy grail for some PMM researchers.

If you'd like to start a new thread or resurrect one of your others I'd be happy to assess one of your wheels, if I am able to do so.

[Fletcher]

No, I realise that RBGS has movement on the outer gear and why I took the picture off, but obviously have concerns it binds into a lever. I was going to draw what I thought was the correct picture because there is some movement. In truth today I am unsure of what really will happen as there are too many forces at play on pivots then you add in momentum.

There are many things in play at the same time - I can see the movements in my minds eye quite clearly - as I said earlier I am trying to explain something that looks very promising to me but is impossible to quantify in known physics terms & math expressions - that is because a PMM is impossible according to the physics, so no one is able to explain it in those terms, or with the math, no matter how hard you try - you can only get some sort of approximation - that's because the secret of nature isn't known or quantified yet, if it exists.

Hard to believe the Sim doesn’t have a linear to rotation option of the rack and pinion. I know you said you tried to sim the slider and the lever and it doesn’t work, but that even makes me more skeptical that the rack and pinion will not work either. So exactly why will the slider and lever in the picture not work?

It does have a linear to rotation option - a while ago Ed & I had a discussion about it - you just unclick the rod length box in the gear - as Ed explained it is quite unreliable in terms of accuracy - I tried it myself & it wasn't close to a physical model in performance.

The slider & lever with rod connections will work to a degree - I was previously saying that a Kinetic Impact transference doesn't work that well because you get incomplete transfer & mass bounce back etc, which all wastes energy or is not complete energy transfer - then you have to read & consider my above comments - you can't expect to show a PMM if the secret of nature is not known by your program, in this context, IMO.

P.S. maybe I'll have a go at building some sort of push rod connection & see how that looks in the next few days.

[daxwc]

The slider & lever with rod connections will work to a degree - I was previously saying that a Kinetic Impact transference doesn't work that well because you get incomplete transfer & mass bounce back etc, which all wastes energy or is not complete energy transfer

Agreed, but did the Sim's and wheel's dynamics play out as the one in your minds eye? Did anything unexpected happen? How large is the bounce back and counter torque?