Free ride, one way ticket

[nicbordeaux]

A video of what I call a free ride. There are other systems which do this in a more spectacular manner. If anybody can tell me if I am wrong ?

https://youtu.be/hkHRbPGSx-g

[WaltzCee]

Hello nicbordeaux,

I don't understand why the pendulum wheel has to be loaded/restrained. I think it's an inaccurate reading of the position of the mass at 3 o'clock from where it begins to where it ends.

You did a heck of a job. An improvement would be a steady camera focused on some sight lining up the wheel with the background. Also knowing the mass of the pendulum with the counterbalancing weight would be helpful.

[Tarsier79]

Is the video of you Nic?

There is energy lost. As the pendulum moves to the left, it rotates slightly, and the OB weight drops its absolute height. Gravity only moves anything, if the COM of the entire system drops. The same rule applies for Buoyancy, or any other gravity/leverage system.

[Fletcher]

I agree with what Tarsier just said, in relation to Gravity Only Systems i.e no other force or energy is introduced and applied to the system.

Think of it like this - we have all heard the maxim that 'water finds its own level'.

To do that it must find the position where the CoM has the least GPE.

This applies to leverage principles in simple and complex machines also.

e.g we all know that a 1kg mass can hold up a 4kg mass with a 4:1 reduction pulley system i.e. force x displacements are equal therefore there is equilibrium, and subsequently no movement. Add more mass to one side and we have disequilibrium and the system moves.

But what is really happening is that when the pulley system is in equilibrium then the system CoM cannot fall and lose GPE. Add extra mass to one side and the system CoM can now fall.

So it is the ability for the system CoM to get lower that allows for movement within a system.

So in Nick's case if there is movement of parts (in a Gravity Only situation) then it is a result of a NET loss of GPE - IOWs the relationship of the parts allowed the rearrangement of those parts so that the system CoM could get lower.

If anyone ever finds a situation where the system CoM gets higher then you have a winner.

[nicbordeaux]

Quote, Fletcher: "So in Nick's case if there is movement of parts (in a Gravity Only situation) then it is a result of a NET loss of GPE - IOWs the relationship of the parts allowed the rearrangement of those parts so that the system CoM could get lower."

I'd love to be able to grasp that and find it true. But the parts are locked together in a manner which won't allow a lowering of the whole system CoM except if mechanical loss is stopping full travel which I don't see: the wheel is tethered to the arm with zero stretch dacron. I even used superglue to fix that line.

IMMO locking everything in place excludes any movement of the parts relative to each other. The only thing that changes is the angle of the dangle. The arm is locked whilst the wheel is brought from 3 to 6, and this stops the arm and mass from moving towards where they want to. Which will show up as stress or strain. And that is what causes the whole contraption to move when the arm is released, whilst lifting a small weight. Retain the small gain in height of the weight and dissociate it from the system, and ... well, that is what I believe I am showing.

EDIT: OK, I now understand. Ta.

[nicbordeaux]

Also knowing the mass of the pendulum with the counterbalancing weight would be helpful.

The weight per mm of the 16 mm tubing is wrong, the decimal point needs to be moved one position to the left. There is also some stuff on the diagrams which isn't on the Wheel (linear bearing).

[nicbordeaux]

Well, it can't get any simpler than this. If the CoM is going to drop in this situation when the block is removed and a swing to the left occurs, you guy's are right and I am wrong.

Please note these two darned planks are nailed, glued and screwed together at 90° to each other. Weightless glue and screws and nails, just to avoid nit-picking ;-)

What does a sim say about this ?

[ME]

(nice tuning fork in those pictures)

Duh-challenge

If the CoM is going to drop in this situation when the block is removed and a swing to the left occurs, you guy's are right and I am wrong.

Huh?? I hope you're kidding, right?
(add: please read this as a compliment for your build capacity - But I find this question an odd and unexpected contrast.)

There's no need to 'sim', it should be obvious what would happen here. (hint: why do you need that "breeze block"?)
Before this becomes a yes/no-game, perhaps it's better you just explain to us (or draw) why the CoM will not drop.

[nicbordeaux]

The breeze block is to stop the "L" mass swinging left. It (the double plank) got to an "L" shape by you putting energy into lifting the lower arm or plank to 3 position after having blocked the upper arm with the block.

On the block magically dematerializing, I know, like everyboy else, that the L will swing left or CCW as it finds it's new CoM.

Not being at all versed in physics , I am assuming that this new overall or averaged out position of balance means CoM is displaced laterally, but not vertically.

It is as simple as that.

We all have (except that Leonardo dos Vinki guy) areas where we are good, and others where we fail to see things. In which case only reason a bloke would get even remotely involved in PM stuff would be total ignorance, or some form of mental disorder :-)


You are probably right, the CoM changes in height, but I'd like to see where and by how much. Upon which I will give up.

The tuning fork is not a Bessler clue, but if anybody wants to make it one, please feel free. It has actually been used with three chopped and straightened baritones fixed to a soundboard to see if it is possible to use a tuning fork to demolish buildings and other stuff.

[ME]

We all have (except that Leonardo dos Vinki guy) areas where we are good, and others where we fail to see things. In which case only reason a bloke would get even remotely involved in PM stuff would be total ignorance, or some form of mental disorder :-)

True points there, it just struck me totally off-guard to see your build-skills and use of Latin like some lost Ceasar ordering pizza on a holiday.
I'll try to ease off, my Latin is not that good. But when you put your statement strong, the reply will likely be strong.

You are probably right, the CoM changes in height, but I'd like to see where and by how much. Upon which I will give up.

Please note these two darned planks are nailed, glued and screwed together at 90° to each other.

Lifting (and fixing) the lower arm to the 3-o'clock should have given you a hint (I guessed).

I'll show the boring calculation of the CoM (without the breeze block) where I just assume the large vertical part [nr. 1] is twice (in length and mass) the horizontal part [nr. 2].

Geekly put:
For my convenience I set the Length of plank [no. 2] equal to (6·L), and its mass is (M).

CoMₜₒₜₐₗ = ( CoM₁·M₁ + CoM₂·M₂) / (M₁+M₂) = (CoM₁·2+CoM₂·1) / 3

Determine from the plank-coordinates:
Pivot<x,y> = <0,0>
CoM&#8321;<x,y> = ½( <0, 0> + <0, 12L> ) = < 0, 6L >
CoM&#8322;<x,y> = ½( <0, 12L>+ <6L, 12L>) = < 3L, 12L >
note: With this I just calculate the half-way position

Thus CoM&#8348;&#8338;&#8348;&#8336;&#8343;<x,y> = <L,8L>
And when rewritten as Polar stuff: L&#8730;65 &#8736; &#8776;7&#8539;° <-- This shows the pendulum
The CoM will drop L·(&#8730;65 - &#8730;64) &#8776; 0.06226·L
note: sorry, can't write it easier

To rewrite the CoM for measurement:
The lengths are (Horizontal short Plank&#8322; = 6L) and (Vertical long Plank&#8321; = 12L)
Thus CoM&#8348;&#8338;&#8348;&#8336;&#8343;(ratio short,ratio long) = (L/6L , 8L/12L ) = (1/6 , 2/3 )

Practically put:
1. This setup should be observed to simply swing to the left and thus (when there's not too much friction) almost return to the bricked position.
2. Confirm the CoM (to verify that calculus) by making that hook 'levitate' horizontally by suspending it by a single string: the center of that string (when continued downwards) should pass the hook at a point 2/3rd from the long-end and 5/6th from the short-end.

Humanly put:
It is a pendulum.

To underline an earlier point:

...should somehow have worked by implying some kind of logical analysis at subgoal-level.
http://www.besslerwheel.com/forum/viewt ... 411#150411

You probably assumed the top-lever moves up (correct)
You probably assumed the bottom-lever should do the same (yes, when it would have rotated around a point a bit more to the right)

Not to rub it in (as it might seem), but an opportunity to understand the (unexpected/hidden) complexities of perpetual motion design and then find a work-around.

add: a picture
Blue: the CoM - Red: the pendulum effect.

[nicbordeaux]

OK, I believe you, Me :-) The CoM drops. Whereas I assumed things evened out, the rise of the top arm of the "L" was "equal" in energy terms to the drop of the other arm, meaning no change in total GPE.

So, the first point in explaining this build is invalidated.

Thanks a lot, and most definitely no "rubbing it in" perceived this end. Far from it.

Thank you everybody.

[nicbordeaux]

Edited out of existence.

[nicbordeaux]

This whole "offset weight at right angle to pendulum" thing is marred by one simple omission.

The initial point (which I probably didn't make) is that illustrated in the sketch: locking the pendum arm in position, and maintaining the red OB weight at 3 creates a "preload" .

If you were to lift OB weight (by rotating wheel) to 12, it's precisely twice the input force, but twice the output force minus the force exerted laterally at the 3 position.

How that translates in terms of vertical/downwards force, I haven't a clue. It should be the same (100% at 12, 50% at 3).

[nicbordeaux]

The point being that if the restraints which stop the pendulum arm and the OB wheel from rotating are simultaeously and "instantly" removed , before anything has time to move, the entire mass of arm + wheel in terms of strictly vertical GPE is identical in both cases. And in the case of the wheel with the OB weight at 90° , there is additional force pushing in a horizontal plane. It should follow that there is more force available in the 90° setup ?

If you prefer, that could be measured in "equal and opposing action and reaction forces" being greater in one case that the other for identical input, and identical mass.

[nicbordeaux]

This was an attempt at using that "concept". https://www.youtube.com/watch?v=MwEMpER ... e=youtu.be