The summary of my latest studies

[path_finder]

Dear raj,
Please find below the details of my animation, with the torque value for each step.

In your simulations/animations, how have you made sure that the smaller inner wheel turn at twice the speed of the larger wheel?

Look at the spokes of the both wheels.
When the big wheel (red spokes) rotates of 90 grades (by steps of 7,5 grades), the small one (blue spokes) has rotated of 180 grades.

if the computer program that Path_finder is using has been given the instruction to rotate the smaller inner wheel at twice the speed of the larger wheel

I don't use any computer program. The computer is just my brain. Like the middle age monk I draw each image applying some strict geometrical consideration. There is no chance for a crazy calculation.

Now, for the torque calculation you can keep the indicated value for each individual image.
You have just to consider the size of your wheel and the mass of your weights.
The indicated values are based on a big wheel with a diameter of 180 pixels. So far make the correction like this:
Per example, for a wheel with a diameter of 1 meter, and two weights of 500 grams (0.5 Kg), the torque for the first image will be:
L = 1m x (30/180)
W = 2 x 0.5 (The two weights have been exchanged by the COG located at the middle of their distance, and equal to twice the mass)
T = 0,166 Kgm

[raj]

Dear Path_Finder,

I am DELIGHTED with your presentations and explanations.

But I am sorry to say that I am still not convinced with your torques figures.

You have taken the COG of pairs of weight to calculate l and got torque=lxm where m=2xw.
The method of calculation is not in dispute, if the wheels were turning at the SAME speed.

As far as I can see, gravitational forces of weights connected to the smaller wheel are felt on the rim of the larger wheel where the torque is applied.
I believe the gravitational forces of the weights acting on the smaller wheel CW and CCW should be DOUBLED to arrive at the correct torque calculations.

This is JUST my opinion.

I sincerely wish you are RIGHT and I am wrong.
I could still get some credit for part of the design.

Raj

[Tarsier79]

Raj, you are correct. The extra leverage on the inner wheel changes the apparent, but not the actual COG. Dealing with the actual COG, only matters if and how far the COG is actually lifting or falling.

[raj]

Dear Path_Finder and Tarsier79,

You are both RIGHT!

The design provides both the torque requirements and off-set COG requirements:

1. the wheels (large and small), arms, and weights form just ONE combined unit(device) at any instant in time, fixed on the axle.

2. as per your explanatory (torques) drawings above, each drawing, at its instant in time, shows the COG of the combined unit (device) off-set from the axle, slightly towards the descending side of the wheels (assuming that the weights of the arms(elbows) are relative negligible).

Raj

[path_finder]

The completed wheel, including the four elbowed pendula and the stand.
Unfortunately I made two mistakes during the mounting:
- the elbow pinions must be on each end of the same diameter of the doubler, which is not the case.
- the linkage with the main rim of the second pair of elbowed pendula is wrong, and must be in quadrature with the other pair on the first side of the wheel.
Nobody is perfect.
So far I have to dismount the full assembly and take a better care in the new mounting.
Nevertheless even in this state the wheel shows an important torque regarding the modest size.

[path_finder]

I rebuilt everything in view to correct the above mentioned mistakes.
I have taken the opportunity for exchanging the center cam by a more simple, transparent and more accurate part.

The results of my first tests show me some strange effects.
On the mechanical point of view everything is now conform (see the two double elbowed pendula in yellow in the shot below).
But on a dynamic point of view this building don't follow the expected behavior.
After further examination I found what happens: even if the COG of the weights is well excentered, the effective obtained torque acts like a lever on both the doubler and the main wheel.

In my animations the tripod was supposed to remain at the 6:00 o'clock position.
In fact, the system searches the position with the lowest level of energy, and instead to apply the torque only on the rim of the main wheel, it divides the torque between the tripod (which is totally free) and the main wheel.
This explain why the tripod is pushed counterclockwise, until an equilibrium is established between the COG of the weights and the center of the tripod (see the axis of the tripod at 7:30 in rosa).

How to solve this? simple: we need absolutely a clutch (rachet) on the tripod. Once located at 6:00 no backward motion will be possible, and the torque will be applied only in the clockwise direction.

So, next step: implement a clutch on the both tripods (on each side).
Again, no computer emulation program will replace the practical experiments.

[path_finder]

The use of a freewheel at any place don't change anything. The doubler axle is well locked, but this don't forbidden the doubler to overpass the 6:00 vertical limit. We need something new.
Finally I remembered the clue of Bessler: reverse everything.

Trying to center all the axles on the geometrical center of the wheel was not a good idea. In fact the main axis is not where 99% of us are waiting to find it.

Like shown in the first drawing the doubler must roll NOT on the lowest bottom of the rim (like shown in my previous animations) but the flywheel (in red) must roll around the doubler (in fact a reduced disk linked to the main wheel, when the red flywheel rotates at the half speed).

In this configuration the previous assembly is still valid, but becomes a pendulum of the main wheel, hung on the small drum (with a size respecting the famous 1:4 ratio).
The rod (in blue) where the elbowed pendula are linked is just held hung by two rigid rods and is free to swing.

By the way the final primemover shares two third of the internal surface of the wheel.

IMHO this design shall be more efficient when the red drum is full of weights, being in this case a good flywheel.

The rest of the design has not been represented here, but you can yourself imagine the action of the two elbowed pendula (raj Balkee's idea).

edited:
Note : As you can see, the weights are not attached to the main wheel's rim, but at the red drum's rim which is the real primemover.

[Blitzbrain]

Hello Path_Finder,

Without ccommenting your current design I want to tell you that from what I see, you are someone that puts a high amount of effort in the idea of reproducing the Bessler Wheel.

I have not seen anyone here in the Forum, that made so many models, attemps, plans graphics etc.

I just wanted to show respect and acknowledgement for that.

[path_finder]

Dear Blitzbrain,
Many thanks for the encouragement.

The main wheel, in accordance with the remarks above.
(Click on the thumbnail image for an best accurate view)

edited: Note: there are always two weights at the center of the primover.

[johannesbender]

i am as dumb as dirt with the calculations but dear path finder i believe the root of your problem will be that your weights or there points of applications are/results in a geometrical state of balance as any one passes the vertical line through the centre of the wheel (past 6 o clock) gravity will find its division of equality (equilibrium) .

if your wheel comes to a standstill and at that point you cut your wheel and everyting down the centre vertical line ,splitting it into 2 halfs the left will weigh the same as the right (which is what i think may be the problem although like i said i am not all that smart ) i have this notion that gravity seems to be able to divide better than a calculator , if only the result were equall to PI what would happen ??

oh and it is also this division by 2 (half left half right) which is what you should prevent in my oppinion and also pi cannot be reached dividing by 2..

edit : actually may i ask would it not be better if you shifted the whole weight assembly upwards so that your centre weights actually hang in the centre of rotation of the wheel instead of on the bottom ?

[path_finder]

Dear Johannesbender,
Many thanks for your comments.
I'm afraid you have not followed the full script of the story which drove us until my latest proposition. Look at the previous pages of this thread and don't forget the raj's thread 'mayday,mayday' also.

Regarding the question of the permanent unbalance of the primemover, just remember this: http://www.besslerwheel.com/forum/files ... fork4x.gif where there is always a weight delayed and waiting at the center of the wheel, not involved in the COG calculation. But this design cannot work alone, because it needs an extra mechanical assembly furnishing the double speed for the spokes supporting the elbows.

actually may i ask would it not be better if you shifted the whole weight assembly upwards so that your centre weights actually hang in the centre of rotation of the wheel instead of on the bottom ?

This is exactly what has been made before, without success. My previous experiments (see here:http://www.besslerwheel.com/forum/files ... _step8.jpg) showed the validity of all animations ONLY if the center of the rod supporting the both elbows is fixed on the 6:00 vertical line. Any other situation drives to a keeling state, this rod rotating backward a little, removing the excess of torque.
So far the only remaining solution is to leave this rod free to swing meanwhile keeping the design active. This is the reason why we cannot attach the long arm of each elbowed pendulum to the rim of the main wheel, but INSTEAD to attach these arms to the rim of the hung primemover.

The 1:2 rotating speed ratio is obtained by reversing the casting: the main wheel rotates at OMEGA (the 1:4 sized small drum also), and the primemover rotates at OMEGA/2.
This new repartition supposes a lot of weights around the primemover in view to create an efficient flywheel.
This design assumes a permanent rotation of the support rod versus the real speed of the main wheel: any swing of the primemover will be automatically corrected, what is impossible in the previous design.

IMHO one of the reason why we have not been successful until today is coming from our natural tendency to center all the parts (like you suggest).

[johannesbender]

to my understanding you have weights group up in the centre to not affect cog and an arangement to try and offset cog towards the right but with two wheels with a ratio , the smaller wheel being the one who carry's the weight arangment and also helps with the movement of getting the left side weight upward along the vertical 6o clock and into the centre of rotation of that wheel , i find your build interesting for the fact of trying to hide the weights in the centre of the second wheels rotation but what makes it difficult
for me to understand is that the arms would hang down from their connected
positions when the reach the top side
of the small wheel , am i incorrect in assuming that that could be negative towards your goal ?

i ask because i do not know much but you are the first to show an arangement which follows in keeping mostly weight on one side (first i have seen ) which has been difficult to design for myself when i also think about height/raise issues , this is why i thought about raising the assembly you have towards the top to be centred on the big wheel as one single wheel but i see that that wont work as you state .

i like what you do and even if i do not quite understand it all i wish you could get this thing running or as close as possible to understand more .

[daanopperman]

HI path_finder ,

I , like most of the readers of your topic , respect you very much , and have great admiration for your work and builds , but I have to agree that the path you now follow is the broad road .

A weight lying on the ground cannot fall , because it has not risen .
Inside the wheel , we must first raise the weight , see , it must first be placed higher , only then can it fall . It cannot fall if it rotates around the axel and the axel do not fall . It cannot raise another weight by leverage to the same height as it has fallen from . In your drawing you lift the one weight from 6 to the axel , and then the same weight falls down to 6 and lift another same size mass back to the axel , that is impossible , the best you can do is to balance the 2 weights so they will be in equilibrium .

Inside a pendulum clock , where you have the heavy weight to drive the clock , it can power the pendulum for 12 hours to fall 500 mm , 60 sec , X 60 min. X 24 hrs , how many divisions , calculate the mass drop in height , for every sec or one oscillation of the pendulum , that is how much energy we need to keep the pendulum to return to it's original fall height , it is only halve of a fart , and we cannot do that with weights turning in a wheel , I say turning because they have never changed their position in the wheel . If we join the 2 weights with a plank to make it one weight ,
do you still think that you can make the weight turn the wheel . If I add the same weight on the left as on the right side of the axel , I have not changed the balance of the wheel , so to say make the 2 weights into one it might be more easy to see that your build cannot work . I am sure , in a working wheel , there is no weights lifting other weights , the weights must be in one position , doing what it should do , ( i do not know what that will be ) and using that energy to drive the wheel , be it vibration , waves , vacuum . I would think vibrations , or oscillations , it would be the only things that could continue to rotate the wheel so slow that you can hardly see it move if you have one crossbar in your wheel . If I have a bow string plugged , it might continue to vibrate for one halve revolution of the wheel even if so slowly rotating .
So , sit back , think of the single long weight in your wheel .

[path_finder]

Dear johannesbender and daanopperman,
I red carefully your comments, which are for sure pertinent in a situation like the famous Roberval balance. But in this paradox (see below) the length of the both arms is identical, what is not the case in the suggested design.

What will be the Roberval balance behavior if the size of the left arm is the half of the right arm?: it will rotate. Obviously for mechanical reasons the Roberval balance will stop after the collision of the arms with the frame. But lets suppose the sliding of the fulcrum full automatic and restoring the same ratio after a half turn.

Regarding the raising upward of the weights on the left side: due to the difference of rotation speed the rotation torque is obtained on the right side by the long arm, instead the upward raising on the left side is obtained by the short arm , with a more powerfull force.

Another point not taken in account in your explanation is the role of the CF (centrifugal force) in particular just after the push at 12:00. This CF knows an important reduction on the left side also, because the short and then null distance with the axle.

I don't know who is true or wrong. Once again, a building is still the only way to observe the strange reality.

[johannesbender]

i believe you have most of the motions/actions in the design wich would be classicly sought after like the path of those weights travel more towards the right side and the weights hiding in the centre however i also believe that you need to try and stay with those concepts further if this design does not work , the only thing i would have done different is to really try and get those weight on the left to actually
reach the centre faster before they go out that far past the 6o clock ..

but like you said only only one way to test it.