Fletcher's Wheel - Ingenuity verses Entropy

[triplock]

Fletcher,

A nicely thought out and considered design TBH

I may be stupid, but what is the driver (in an OB sense) ?

Is it the rack and pinion displaced mass or is it the flail impacting on the stop ? Obviously, the former makes it an gravity wheel. The latter a motion wheel ?

Chris

[Fletcher]

Fletcher is the wheel principle the impact of the flail or supposed to be just the transferring of felt mass?

The 'driver' is the rack portion of the rack & pinion mech - the pinion is the blue gear wheel to which is attached the double ended balanced flail which has no torque on its own.

The driver or rack moves right - because it is meshed to the pinion it turns the flail CW - this is a straight gearing ratio so the driver will still have some momentum when the flail connects the rim stop - there will be some momentum & KE transfer to the rim stop as the flail (& rack) is stopped.

But ... this is NOT an "Impact Wheel" though there must be some impact to make connection.

When the flail is resting on the rim stop the rack/driver has ceased to moved right temporarily - but, the distance between the right end of the horizontal rack & pinion mech & the rim stop will increase (the separation grows naturally).

However ... because the rack is not pinned or locked it is free to move further right if the flail loses contact with the rim stop - this is what happens - the gap widens, then flail stays in contact by rotating CW & it does this because the rack moves further to the right to 'take up the slack' so to speak.

So dax to answer your question - it is not a transferring of 'felt mass' but a transferring of 'felt force' - the force is the Centripetal force of the rack/driver which is calculated by Cpf = mv^2/r in Newtons.

The rack & pinion is like a leverage so I expect a reduction in force felt on the rim stop - the upside is that it is felt for a long period of time & over almost 90 degrees (1/4) of wheel rotation.

[triplock]

Fletcher,

If, as you say, the sliding mass in the rack and pinion, creates OB drive through its lateral displacement, then what is the purpose of the flail.

If there is a geared linkage from the ground reference to that sub-system, then the flail brings nothing to the table. ?

Therefore, looking at this trimmed down arrangement, setting aside any CF interaction, do you not just have a height for width scenario - meaning the energy required to radially elevate the driver mass is greater than the torque produced by an opposing mech.

I may be looking at matters in over simplistic terms Fletcher, so I will happily be corrected. I just get alarm bells going off in my head whenever CF / Motion is mentioned.

Chris

[Tarsier79]

Chris
The red mass doesn't create OB. It is pushed outwards by its own inertia, but it sits on a roverbal balance, so its horizontal displacement doesn't effect its "felt mass". The Flail transfers the energy of that movement to the rim. The wheel is "balanced" at all times. The advantage of this means there is no negative torque created by unforseen weight positions. IMO Negative torque is now created from accelerating the driver mass back and forth.

I do however agree with your line of thought. Given the right mechanism, you could design it to operate in a gravity OB drive system.

If you look at the driving device in terms of KE, expecting a gain, the gain is supposed to come from the conversion or rotational energy to the cycling RB mechanism. This part can be simmed and measured, or relatively easily built. Of course you experience energy measurement hurdles in the real world.

[triplock]

Tarsier,

Ok, I am even less convinced now that this design will work.

If the heavy weight just remains on the horizontal plane due to the RB properties, and it displacement is not felt per se, there why would this reciprocation create drive to the wheel via the flail.

Fletcher says its not through an impact action at the stop. Ok.

Under what circumstance does the rotational energy of the flail then get transferred to the rim if not through impact ? Is there a clutched gear train ?

[triplock]

Looking at the rack and pinion / flail set up, would it not be better to use a reciprocating rack and pinion system instead

http://youtu.be/ScB6Wd4r9Jc

Using this will dramatically reduce loses at the point of energy transfer and change in direction of the sliding mass. It is conducive to a much more fluid operation.

Chris

[Fletcher]

Q3; is the flail neutral with both mass' being equal? what causes the flail to rotate CW and not CCW?

The flail has a shaft and end wheel weights (yellow & green) - both end weights are identical with same mass m2 kg [balanced flail] - I coloured them differently so you could see where each was in a cycle & see which one was more effected by Cf's at any particular time e.g like when moving CW to intercept the rim stop at 1.30 o'cl or when the green wheel weight is closest the rim at 7 o'cl & we want to start the reset process to flail vertical orientation.

The flail rotates thru 90 degrees CW on the down going side because the rack/driver is meshed to the pinion gear that the flail is attached to - the rack moves right turning the flail CW - for reset the rack moves left & the flail rotates CCW.

Q4: If gravity neutral why just a drive phase from 130-3?

We need a drive phase to apply force to the wheel i.e. create torque - that torque is created not from gravity imbalance but from redirecting centripetal force from the rack to the rim.

To visualize how the RB system works take a position on the rim - say 12 o'cl - note that the head of the RB (the gear) is directly on the radial from axle stator to 12 o'cl - it is like they are on the same spoke - that never changes - the rack & flail are always orientated horizontally - that is why there is separation between the r&p & flail & the rim stop.

If in doubt research a simple RB system on the net - the Roberval Balance Gearing System (RBGS) that I use is a variation of the simple RB system that uses pantographs - the artificial horizon is the anchor for the system as any RB system needs.

If you take a balanced RB system in a wheel format & apply some of its weight at a greater radius it is no longer balanced - the wheel will rotate - this has been the basis of countless attempts to use RB systems to create OOB systems - mine is different, it uses force transference to create imbalance.

[triplock]

Ever feel like you've walked into a party you've not been invited to, and then to be ignored by the host ? :-D

Anyway, thanks tarsier for your post at least. I'm outta here .

Chris

[Fletcher]

Fletcher,

A nicely thought out and considered design TBH

I may be stupid, but what is the driver (in an OB sense) ?

Is it the rack and pinion displaced mass or is it the flail impacting on the stop ? Obviously, the former makes it an gravity wheel. The latter a motion wheel ?

Chris

It's a 'Motion Wheel' concept Chris, not a gravity unbalanced wheel - the imbalance is caused by force transference to the rim - this is achieved by a rack & pinion system where the flail is attached to the pinion.

P.S. just read your last post Chris - I answer posts in order they occured generally - yours was after dax & dave & I just got to it, so sorry you felt ignored.

[daxwc]

Since the slider weight needs CF to work and thus considerable rpm, is there enough time for the slider to work before being re-pinned to the slider track by the same CF?

Due to the slider weight always being parallel it has very little time in its transfer zone.

[Tarsier79]

If you have WM2D, you can see the movementof the driver mass in this series of posts by helloha: http://www.besslerwheel.com/forum/viewt ... 073#129073

I think Fletcher was just offering a way to use this drive mass to drive a wheel. There will be many different ways to do this, some more efficient.

[triplock]

Right, ignoring my hissy fit above, IMHO the arrangement,as is, is open to question ( just the point of energy transfer to the rim ).

I believe that the following arrangement may be better suited:

The horizontal slider mech on the RB is great.

Instead of the singular rack shown, use the reciprocating one shown in the link in my 2nd to last post.

This means that on spinning the wheel up the flail will not suffer any deceleration at the point of change of direction of the sliding mass.

Now the only point of opportunity here is to utilize that shuttling mass at the end of its travel each end of the shuttle. That rythmic production of G force is a useful source of energy indeed.

Chris

[rlortie]

There was mentioned a page or so back that a catch and release system was required, will this one work?

https://www.youtube.com/watch?v=TpQpAXu4wOg

[Fletcher]

Since the slider weight needs CF to work and thus considerable rpm, is there enough time for the slider to work before being re-pinned to the slider track by the same CF?

Due to the slider weight always being parallel it has very little time in its transfer zone.

The driver/rack doesn't move due to Cf's - they are fictitious forces - it moves because its latch is released - the latch was providing Centripetal forces (mv^2/r N's) - when the Cpf is removed the rack moves because it has inertia (momentum) - this momentum turns the flail etc etc etc - the flail is balanced (like a low mass flywheel) with end wheel weights, so requires very little linear KE to be changed into RKE of the flail - the flail has some small MOI (low mass) which resists the speed of its CW rotation but it is still very fast.

The rack only moves a very short distance back & forth & this happens quickly - in fact, the inertia of the flail dictates how fast that will be, & we don't need much mass or MOI in the flail to do its job, which is to push the rim stop, not to bash it.

[Fletcher]

There was mentioned a page or so back that a catch and release system was required, will this one work?

https://www.youtube.com/watch?v=TpQpAXu4wOg

Thanks Ralph .. it might have a place - I was more referring to a simple latch & release so the rack can be held by Cpf's, do its out & back again routine, & be caught again.