Sjack Abeling Gravity Wheel and the Worlds first Weight Power Plant

[Omnibus]

Actually this is not the case.

Also incorrect.  The first video shows how WM2D can behave incorrectly, just like any simulation package, if the input and parameters given allow for too large of a calculation error.  The video I posted shows how WM2D does in fact correctly predict the action once the input and parameters have been adjusted to eliminate the source of the large calculation errors and with more accuracy.

This is a discussion of something whose workings are unclear to both of us and is just guess work. Therefore, wm2d simulations are to be discarded in this case altogether if we don't want to get into medieval scholastics of counting how many devils reside on a tip of a pin. Instead, we have rigorous scientific tool to rely on -- first and foremost the mass-axle discrepancy as well as the torque calculations. Let's deal with them.

[Omnibus]

@mondrasek,

For example, if you pick up a 10 lb bowling ball with a flat frictionless plate from the bottom, you must hold it with 10 lbs of force.  If you hold it with two plates in a vee, each plate must apply a force.  The total of these forces is not 10 lbs.

That observation is inapplicable in the calculations we’re doing. Like I said, we are only interested in the force resulting from the available 10lb force of the ball. No other forces are of interest. If the point of application of that force, connected rigidly to an axle (to form an arm) is unobstructed by other constraints, the effective force which the ball exerts will be 10lb. Therefore, it is these 10lb that are to be multiplied by the sinus of the angle between the force and the arm to get the force needed to calculate torque. This is the same as your example with the ball sitting on a frictionless plate but stated correctly, that is, focused on the force the ball exerts on the actual point of application and avoiding the forces of reaction. In our calculations no reaction forces take place whatsoever.

So now we have a ball with one constraint â€" the arm.

If there are no other constraints external to the ball-arm system, then the torque will continue to be calculated in the explained way, using the full 10lb value of the force, independent of whether the ball is glued to the arm or there’s a steep surface attached to the arm where the ball is, not allowing the ball to slide out. In other words, the mechanism of holding the ball in place on the arm has no bearing on the calculation of the torque, as long as there are no constraints external to the ball-arm system.

By keeping in mind that it is only the force which the ball exerts on the point of application that participates in our calculation and nothing else, we find out that when there is an external constraint (external to the ball-arm system) the situation changes. Any external constraint (such as the ramp in our case) that forces the ball to go along a trajectory different from the circular one (due to the arm constraint) will cause modification of the initial 10lb force.

Firstly, that modification must be in the direction of decrease of 10lb because 10lb is the initially available value which cannot become more by itself. That is to be understood very well.

Second, the direction of the available force derived from the ball (not from anything else) when the external constraint is present must be different because vertical direction of the force will only cause a particular circular trajectory of displacement with the particular arm constraint.

As is seen, to do the calculation right (as @eisenficker2000 has done), having the arm length fixed in length, we are specifically focusing on what the magnitude and direction of the force derived from the ball is, involving no other forces. Any other forces involved in such calculation (especially forces of reaction by holding mechanisms, part of the ball-arm system) is completely frivolous. If one is to do calculations with such frivolous forces one can prove literally anything.

And, again, the above, proper, calculation, carried out by @eisenficker2000, corroborates with the most important criterion for perpetuum mobile â€" the persistent displacement sideways of the center of mass with respect to the axis of rotation at every position of the wheel. This criterion should be the guiding light for the correctness of any calculation done with the aim to explore as to whether or not the wheel in question is perpetuum mobile. The wheel in question is a perpetuum mobile because it fulfills said criterion and any calculation ostensibly showing otherwise must be in error.

[mondrasek]

By keeping in mind that it is only the force which the ball exerts on the point of application that participates in our calculation and nothing else, we find out that when there is an external constraint (external to the ball-arm system) the situation changes. Any external constraint (such as the ramp in our case) that forces the ball to go along a trajectory different from the circular one (due to the arm constraint) will cause modification of the initial 10lb force.

There are two external constraints, not one.  There are the ramps.  And there are the slots.  The slots change the angle by which the force of the weights act on the wheel similarly to how the force of the weights changes due to their angles with the ramps.  This is easy to see as the weight can be in the long portion of the slots with one angle, or in the short ends of the slots that are at a different angle.

[mondrasek]

Here are the results for the 15 degree rotation.  The equilibrium point due to keeling must be between 15 and 20 degrees (closer to 20).

[rbe]

Hi Guys,

Don't know if anyone posted this before,

http://mooieenergie.nl/index.php/en/home/3-gewichtenergiecentrale/9-nieuwsbrief-mei-2009

Nothing new really, unfortunately!

Cheers
rbe