Mr Desaguliers! please help us...

[path_finder]

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?...

[Gregory]

I think you're wrong, or I just didn't get it.

If it is closer to O than B1 then no matter where pin D is placed. When supported by this pin the red weight on the left will always act as it would be attached there directly on the wheel without pin, rod, ect. As far as I can see this situation is not about Desaguliers balance.

[alexjrgreen]

How far does the weight fall when the wheel turns 5 degrees?

The pinned weight behaves as if it were attached through its centre of gravity.

[Stewart]

You're right Gregory. This is a fundamental concept to understand in OOB wheel construction and one of the first things to learn! It never ceases to amaze me how often we see people who have been designing/building for a while and yet still haven't understood that concept.

Stewart

[Gregory]

You're right Gregory. This is a fundamental concept to understand in OOB wheel construction and one of the first things to learn! It never ceases to amaze me how often we see people who have been designing/building for a while and yet still haven't understood that concept

That means I should make my hands dirty and build some bloody wheel, however I haven't completed my final design yet, nor I got my home made cnc mill to really kick into the construction phase. ;)

[Stewart]

Go for it Gregory! ;-) Looks like you've got good insight into wheel dynamics. Good luck with your designing/building. Stewart

[Trevor Lyn Whatford]

You're right Gregory. This is a fundamental concept to understand in OOB wheel construction and one of the first things to learn! It never ceases to amaze me how often we see people who have been designing/building for a while and yet still haven't understood that concept.

Stewart

Hi Stewart,

It is Bessler's wheel that make you go over old ground in the search for his wheel as you think you may have missed some thing, I think I will call it the Bessler Curse syndrome : )

Regards Trevor

[Gregory]

Thanks Stewart,
I'll post up when my lucky day arrived.

[path_finder]

Implementing the lessons of Mr Desaguliers hereafter is the animation representing the theoretical motion of the above mentioned concept.

The idea is to use some particular levers, some diameters of the wheel, where the pivot and the application point are on one side of the wheel, but the acting force is located at the opposite side. Because the pivot and the application point are fixed on a rotating disk, and because the acting force of the lever is located at the other side of the rotation axis, the applied torque will surprisingly lift-up the pivot, and thus let rotate the wheel like desired.

As you can see, there is no contradiction against the basic laws of the physics, just a particular geometry where the pivot is rotating around the wheel axis and an exchange of pivot at 12:00/6:00 (the 'escamotage' of JC?).

Details of the concept:
1. The concept is based on a single lever of the first kind (the pivot being located between the force and the lifted contact point).
But it can be multiplied like you want: on this animation we have four levers (violet, grey, blue, orange).

2. The pivot is fixed to the main disk of the wheel but eccentered, far from the center.

3. Each lever is made of a rigid rod with a weight at each end (two equal weights per rod).
This is justified by the 180 grades symmetry of the exchange process in the middle vertical plane at 12:00/6:00.

4. The main disk of the wheel has some retractable pins (small red circles) and pivots (small white circles).
As shown in the animation the retractable pins and pivots are absent on the falling half side of the wheel (right part, as the wheel is supposed rotating clockwise), and are extracted on the left side of the wheel.
Theses retractable pins and pivots are synchronous, exiting at 6:00 and disappearing at 12:00.

5. The red pins (fixed on the main disk) purpose is to lock the upper rim of the yellow weights, transferring the upward exerted torque to the main disk.

6. The white pivots are the points where the levers are pushing and keeling (in fact they are rotating around their middle point).

7. The green weights on the opposite end of the lever are the acting force of the lever.
Due to the ratio selected for the two parts of the levers, the force applied on the red pins is four times the value of the weight.

8. There are no physical connection at all between the four levers and the rotation center of the wheel (so far the axis is not free for a pass-through shaft): due to the geometry of this assembly, each rod is centered but not physically linked with the center of the wheel.

Hoping this explanation remains clear (it was not easy), now the question is: what kind of mechanism could be able to manage the pins and the pivots?
IMHO it should be not too much difficult.
Several ways are available for the alternative motion of the pins/pivots.
Another way consists in two thin axles, fixed to the rod, and inserted/extracted in/from some holes of two V shaped planes.
We can also use a grooved path for the weights, modifying the introduction of these axles. Etc, etc...

Waiting on your comments and suggestions.

[broli]

I wish it was that easy.

Edit: To be more constructive and simulate discussion I highlighted the problem in the attached illustration. The real question is how do you analytically determine the magnitude of the force acting on the pivot. And that is where it gets unintuitive and vague.

[path_finder]

Dear broli,
Many thanks for your drawing.
You are right: the exact value of the applied force is very difficult to calculate (may be dependent of the angle and perhaps also from the weight of the whole wheel) because the pivot is mobile and rotates itself around the main axis.
But furthermore, ask yourself the effect of this force wathever its real value: a rotation of the main disk. This is the most important point.

[Trevor Lyn Whatford]

Hi path_finder,

the weight on the pivots is only on the side of the wheel you are trying to lever up if you add the leverage of this weight back to the axle you will see it balance or should I say the negative leverage will counter the positive leverage.

With that said what a fantastic design, wow.

Regards Trevor

[alexjrgreen]

For there to be leverage, the lever has to be able to move. Placing the pivots on a separate inner wheel and limiting the travel of the falling weight might be a start.

[Bill_Mothershead]

I kind of came to the notion that if....

...the rod and its weights are...

...jammed, frozen in place, stuck, unmovable relative to the wheel...

...then all the forces are static.

It does NOT matter, really, where the pivots and stop pins are.

It is as though the weights are super glued to the wheel.

The configuration resolves to simple gravity forces on
weights that act just like they are bolted in place on the wheel.

It should be easy to do some simplified experiment to see
if leverA.gif would actually work. Perhaps even a simulation.

[path_finder]

Dear Trevor Lyn Wathford,

the weight on the pivots is only on the side of the wheel you are trying to lever up

.
I'm not exactly in accordance with your comment, because even if the yellow weight is locked (therefore part of the wheel) and because the lever even stacked remains in rotation, the force exerted by this yellow weight must be taken in account in the global balance calculation, as physical part of the main wheel.

Dear alexjrgreen,

For there to be leverage, the lever has to be able to move

Even stacked a lever remains a lever.
Put your dog on a end of the seesaw, and put a big stone at the other end, despite the lever is stacked, the action remains still active and the dog in the air.
But what is much more important in the suggested design, is the motion of the pivot around the main axis, transforming the application point of the torque.


In view to be more explicit, and in complement of the broli's drawing, I made the drawing hereafter showing the active force of the lever.
Look carefully at the small inserted detail: what will be the consequence of the W force applied to the lever: a rotation of the disk.
If there is no yellow weight at the other side, just a small value for W is sufficient for let rotate the disk.
But due to the presence of the yellow weight the W force must be greater, in a ratio depending from the distance between the pivot and the main axis, and from the distance between the pivot and the yellow weight (also contact point).
In addition if we multiply the number of such as single lever, the ratio should be different (just an intuition).
As Bill_MothersHead said: this concept merits an experiment.