The summary of my latest studies

[daanopperman]

Hi path_finder ,

Your massage said " if the one arm is shorter than the other arm " in a Roberval balance it will turn ,
please take a look at the topic " would it be correct " page 29 drawing Fletcher . gif .

Here is very much what you have posted in your last post , it does not look the same at first glance , but it is the same thing , here the roberval beam was split into 2 pieces , a left and a right side , it is not one piece where the left side intrafere with the action of the right side . Here the left side is the long arms and the right side is the short arms , and will change over automatically after 180 deg of turning as the short arms move to the outside of the wheel rim . If I make the RH side arm so the weight will be over the axel , it will for all purpose disappear from the wheel in this drawing and in this position , ( the arm is longer for weight to reach the center or pivot of the wheel ) of course then the left arm will be the same length as the right side to balance the leavers .
The weight on the center disk is to keep the small wheel in horizontal position to give leverage via the connecting belt , I used belt not gears to link the inside and outside weights , gears make the outside disk loose it's orientation in the wheel so it goes out of horizontal .

[path_finder]

Dear daanopperman,
Sorry but I don't see any connection between the Fletcher drawing with my previous post.
Please look at the red axles in the drawing: the both distances with the center red line are the same.
The question is not to relocate the weights but the center axle.
Relocating the weights (like linked in the drawing) is just another application of the Desaguliers demonstration.

[daanopperman]

Hi path_finder ,
I see no red axels in your previous drawing , only a red base for the construction to stand on , one of us must have the wrong message .

[path_finder]

Dear daanopperman,
I was referring to the red axles in the fletcher's drawing, reproduced below.

[path_finder]

Dear daanopperman,

In the Orffyreus wheel, shifting weights move in towards the axle on the ascending side and out to the rim on the descending side.
The idea is to produce an increased 'leverage' of the weight directed toward the axle on the descending side and a decreased 'leverage' applied to the axle on the ascending side.
Perhaps this acceleration gradient, centrifugally created is used to cause the increases and decreases of leverage applied to the axle. Remember, this wheel and all its internal components rest solely on the axle.

This is an excerpt from the keelynet.com web site: http://www.keelynet.com/energy/orrcent.htm

Dear Grimer,

Centrifuges can be used to produce acceleration gradients.

This comment should be pleasant to you, confirming your jerk concept (third derivative).

[raj]

My dear Path_Finder,

Many, many thanks for the your link above.

I am simply DELIGHTED to read its content.

Raj

[daanopperman]

Hi path_finder ,

I read through the article that you placed a link to , and I don't think it is correctly to apply it in a wheel .

Think of this . Inside a wheel , I raise weight from 6 directly to 12 , trough the center of the wheel vertically , so that it has no weight effect on the wheel , there is no leverage , only it's weight .
A second weight on the rim at 12 fall with the rim as the rim turn till it reaches 6 .
It has traveled a longer path , but it has reached it's destination at the same time as the weight from 6 to 12 , therefor it had to move faster , that was caused by the leverage , but it fell the same distance ( 12 - 6 ) as the weight rised from 6 - 12 . That is directly the same as a avalanche drive , which I think most know cannot work . If 2 same weights travel the same distance in the same time , it is a balance scale , a one to one ratio , no gain .

[raj]

Hello Daano!

Could you kindly answer me the following:

What about the torque provided on the axle of wheel by the two weights in your scenario?

If the net torque is zero, then the wheel won't turn.

Is the torque in your scenario zero?

[daanopperman]

Hi raj ,

There is no torque to turn the wheel , the 2 weights move the same distance , one up and one down . It does not matter if the falling weight makes loop de loop circles , it only falls from 12 - 6 , that is the amount of potential energy that you have available in your wheel .
There will only be torque to turn the wheel if you do not have to raise the bottom weight to 12 o clock .

[path_finder]

Dear daanopperman,
I'm afraid you did not understand the content of the quote above.

First no where is supposed the weight rising from 6:00 until 12:00 on a vertical line, but just reaching the axle by any path reducing the centrifugal force because the reduction of the distance from 6:00.

Then you ground your arguments on the potential energy, what is a typical static approach, as put in front by a lot of members here when the dynamic aspect is occulted.
Nobody will contest the principle of equivalence between the gained potential energy on the downward side and the loss of an equal potential energy during the upward side travel, whatever the wheel is stoped or in motion.
The subject of the latest topics was about the dynamical energy, in particular the variation of the acceleration under the centrifugal force.

The torque is the product of the distance to the axis by the orthogonal component of the force. If the distance to the axis increases, and in addition if the force (centrifugal) increases more stronger, the torque obviously increases. This is totally independent of the potential energy assessment.

I do not pretend (for the moment) the success of the suggested design.
But this design assumes the requested path in the exact accordance with the quote above.

[raj]

Hello again Daano!

Please allow me to show you how I think a design could work.

Just as in your scenario above, with two weigths, one moving from 6 o'clock to the 12 o'clock through the centre and the other moving with the rim of the wheel from the 12 0'clock to the 6 o'clock, I have drawn two weights moving, one up and one down.

If the weight on the ascending side is made to follow FREELY the curve path nearer the axle as in the drawings, the weight on the descending side WILL provide torque for the wheel to turn, even though, as you said, no work has been done and no gain.

Raj

[daanopperman]

Hi raj ,

I have put some dimensions to your drawing ,
in nr. 2 the red is the amount of movement of the weights , up and down , see the top weight has gone down 1/2 as much as the bottom weight was lifted so your leverage was nullified .
The blue is the amount of leverage , see the top lever is 2 X the length of the bottom lever , that is just enough to make the wheel balance .
In nr. 4 , the falling weight has only 1.5 units to fall till 6 , but the ascending weigtht has more than 2.5 units to rise , so you will need to increase velocity at the rising weight , this eats up you leverage .
If you do not have to pick the bottom weight up the same height as the top weight falls , you will have rotation of the wheel .
If you want the wheel to turn with leverage and weights , you have to lift the weight up at 6 AND lift the weight up at 12 .

[raj]

You have twisted the arguments.

I am talking of TORQUE by the two weights in each separate drawing.

Are the torque on the left and the torque on the right EQUAL?

The answer is: NO, the torque on one side by the weight is more than the other, because the distance of the weights from the axle are not EQUAL.

By simple seesaw logic, the weight further from the axle will move downwards, turning on the axle.

A wheel with weights on the ascending side moving INWARDS and UPWARDS simultaneously nearer to the axle FREELY from the 6 o'clock to the 12 o'clock positions and the weights on the descending side moving on the rim of the wheel from the 12 o'clock to the 6 o'clock position WILL BE A RUNNER.

Work done and energy gain or lost have no relevance in this scenario,
as long as the net torque requirements are there.

[raj]

The most difficult part in this type of design, is how to make the ascending weights move INWARDS and UPWARDS at the same time.

[daanopperman]

Hi raj ,

In the attached drawing I have a bar pivoted in the center , with 2 sliding same size weights on the bar , can I lift the bottom weight from 6 to 12 with leverage of the top weight . For the wheel to keep on turning , I have to have a weight at 12 always so it can fall to 6 to replace ( lift ) the other weight continuously , if I do not lift the next weight to 12 the wheel will keel .
I am not twisting the argument .
If you say you can do that with a bar and 2 weights , I will believe you , and your wheel will surely rotate by itself . If you cannot do that with a bar and2 weights , put a rim on the bar and test it , if you can now do it , then we know it is the rim that makes all the extra energy to power the wheel .