Mr Desaguliers! please help us...

[Trevor Lyn Whatford]

Hi path finder,

I say the wheel shown above with the one lever on it will move anti clockwise until it finds the balance point.

Just my thoughts on it, and I hope I am wrong as we need a runner, if it does then it still my be useful with a bit of R&D and fulcrum switching.

Regards Trevor

[DrWhat]

It will seem unbelievable, but this device will still balance.

Mainly because the torque is not being applied at the axis of the entire wheel, but closer to the periphery.

I have done such experiments at a basic level.

[path_finder]

Dear Trevor Lyn Whatford,

(the wheel) will move anti clockwise

On the next drawing is an attempt to better identify the torque.
The green active weight can be replaced by a 2W (twice the single value) downward vertical force located at the middle of the right side arm (the black cross).
The yellow locked weight can be replaced by an W/2 (half the single value) upward vertical force located on the same black cross.
The result shows a clockwise torque. If wrong, where is the error?

[Trevor Lyn Whatford]

Hi path_finder,

Try it and see as you only need a bit of bar or wood, you have the wheel with a set of holes in and you can check out the four settings of the main drawing as well, it will save you time in the long run right or wrong.

Regards Trevor

[Grimer]

I've only just come across this thread and so haven't yet had time to take the argument in.

One thing I will say though is that pinning, escapement, catches - call it what you will - are vitally important since for effectively infinitely stiff weights and pins they do not involve any expenditure of energy, any FORCE times DISTANCE in other words.

This topic is well worth pursuing.

[path_finder]

Unfortunately I'm far from home (Kinshasa DRCongo) and not able for the moment to build any demonstrator before the end of this month.
Nevertheless any comment is wellcome.

[broli]

Dear Trevor Lyn Whatford,

(the wheel) will move anti clockwise

On the next drawing is an attempt to better identify the torque.
The green active weight can be replaced by a 2W (twice the single value) downward vertical force located at the middle of the right side arm (the black cross).
The yellow locked weight can be replaced by an W/2 (half the single value) upward vertical force located on the same black cross.
The result shows a clockwise torque. If wrong, where is the error?

Why didn't you include the force the pivot/fulcrum of the lever is exerting on the wheel?

[Grimer]

Returning back to the theory (I'm far from home and not able to continue my building nor experiments) I was faced with the following problem, illustrated in the first drawing hereafter.

We have a disk, balanced and rotating on an horizontal pivot O.
On this disk we have a lever rotating on the pivot B1, part of the disk.
This lever supports a weight W at one end A1. The other end C1 is free.
If the lever is free, it will rapidly reach its keeling position, like shown in A2/B2/C2.

But if the end C3 is locked by the pin D, where is applied the torque coming from the weight W ?
Will the disk be in equilibrium ?

This question seems to be simple, even ridiculous, but after review of the Desaguliers demonstration it remains not obvious...
(see full text here: http://www.todayinsci.com/Books/MechApp ... /page2.htm)
A simple deduction can be made from this case: by exchanging the D locking pin position, we are able to modify the torque, therefore to build an unbalance .
Another way to declare the story is: we can move a weight without to move it physically, just by introducing a locking pin at the other end of the lever. True?
Could the principle be so simple?...

With respect to your link above, isn't this just an earlier version of the Roberval Balance principle and hasn't this paradox been extensively discussed before?

Or am I missing something?

[Trevor Lyn Whatford]

Hi Path_finder,

If it works or not I still think it is a brilliant peace of ingenuity so thanks for sharing it with us.

Regards Trevor

[fredos]

Hi Path finder,


I have studied your design and i have done calculations about your design.
The results seems interesting but i can be wrong about my calculations done.

However, my calculations give that your wheel must turn anti-clockwise
but my calculations can be false.

I post the document where i have done the maths.

I had an idea close to yours one year ago but i was different because it used the axis of the wheel as the supporting pivot B.

The calculation done about my design have given a total torque equal to zero whereas your design permits to give a torque different to zero.

So, IMO, this design must be verified by prototyping.

By the way, the main problem remaining in the design is the way of managing the pivots inside the wheel in order to shift the pivots when passing the 6 O clock position ( or 12 O clock position) but it doesn't seems to me very difficult to create a device doing that ( i am electronic engineer not mechanical engineer).

Have a nice trip in RDC.

Fredos

[path_finder]

dear Fredos,
Many thanks. I downloaded your document and will examine it.

[fredos]

Hi Path Finder,


I have found a lot of mistakes in my maths.

As a consequence the result is completely different and the torque found is equal to 0.

So , this design seems not to be a winner.

I post the document with the corrections done.


Regards


Fredos

[path_finder]

Dear Fredos,
My preliminary comment:

We have the sum of the forces applied equal to 0

Apparently you start with an assumption wich determines your final conclusion.
This confirms the validity of your calculation: if the summary of the forces is zero, the final torque must be zero also.
I think the calculation should be made on a dynamic point of view (starting from the roatational acceleration at each point).
By the way I think the summary of the forces is not null, this is the only way for a possible rotation.

[path_finder]

Dear Fredos,
(from Goma, the fresh air of the Kivu lake seems to be productive)

Another way to make the calculation is to consider the torques applied on the wheel axis:

1. The A locking point receives a force equal to four times the green weight, minus the yellow weight, therefore equal to 3W upward vertical.
2. Its A end being locked, the B pivot is a lever of the second kind and receives a force equal to four times the green weight: so far 4W.
Now let's calculate the value of each torque:
3. The clockwise torque coming from A is equal to 3W x 2,5e = 7,5 We
4. The anticlockwise torque coming from B is equal to 4W x 1,5e = 6 We
The wheel rotates like under the action of a 3W force applied at the black cross point (3W x 0,5e)
No torque has been taken in account from the wheel itself (supposed to be balanced).

[fredos]

Dear Path Finder,


Normally, the lever+green weight+yellow weight doesn't move in the referential of the wheel so the sum of all the forces applied to the sytem
(i mean system : lever+yellow weight+green weight) must be equal to zero.

Otherwise, the lever will undergo a change of his position inside the referential of the wheel. Or , we don't want that the lever move inside of the wheel itself.

You wrote:
This confirms the validity of your calculation: if the summary of the forces is zero, the final torque must be zero also.

Not necessary, you can have sum of the forces equal to zero but a torque
different from zero.
A basic example of this is the pendulum. The sum of all the forces applied to the pendulum is equal to zero ( the system cord+ weight doesn't move)
but the weight can turn freely around the axis of rotation.


You wrote:
"The A locking point receives a force equal to four times the green weight, minus the yellow weight, therefore equal to 3W upward vertical."

I agree with you

You wrote:
"Its A end being locked, the B pivot is a lever of the second kind and receives a force equal to four times the green weight: so far 4W. "

I disagree.The B pivot receives the force of the yellow weigth+green weigth+the opposite reaction of the contact force applied to A.
So the result gives: W+W+3W=5W

So the anticlockwise torque is: 5W x 1,5e = 7.5 We

So, the sum of both torques is equal to zero.

I know that is a bit ( i mean a lot of ) disappointing but the final torque is null.

A clever way to avoid the pivot B supported by the wheel itself could be to choose the pivot B on the axis supporting rod of the wheel so the torque applied to the pivot B is necessary null in that particular case.

You are better than me in mechanics so you can find a design using this principle to create a working device.

In my opinion, this type of design could be definitively a solution of the bessler problem because it perfectly fit the main requirement of the bessler wheel: its simplicity of principle.

Best regards


Fredos







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