Sjack Abeling Gravity Wheel and the Worlds first Weight Power Plant

[hansvonlieven]

Gains you displacement. It's all about displacement under the action of a force. Nothing more than that.

In a gravity device only upward vertical displacement injects energy. The question is how much does it cost you to do it. Horizontal displacement gets you nothing.

Hans von Lieven

EDIT:

Below there is a way to achieve the same trajectory for the weights in a way that WM2D can cope with. It still does not work though.

[AquariuZ]

AquariuZ, great video links at the bottom of your posts.

It´s up to us.

[AquariuZ]

Below there is a way to achieve the same trajectory for the weights in a way that WM2D can cope with. It still does not work though.

That´s creative, thanks.

[Omnibus]

In a gravity device only upward vertical displacement injects energy.
...
Horizontal displacement gets you nothing.

Exactly. What we’re looking for here is to find a construction which, at the expense of no energy will constantly bring a given generalized weight in a position to be displaced downwards under the force of gravity.

As for the construction you’re offering, it obviously isn’t the one we’re looking for and has long been out of consideration. Neither is it reproducing what’s being discussed. Do it properly, get rid of the driving weight, turn on air resistance (AR), as an already known requirement for a realistic model, set the rigid joints (RJ) to Measurable, as an already mandatory requirement for a proper rendering of a model, and you’ll get what’s trivially expected from weights constrained to a circular pattern (see first attachment). That isn’t the case discussed here.

As was shown, even a construction with somewhat more relaxed rigid constrains than yours, that using @broli’s forced trajectory, doesn’t work (see second attachment).

What seems to work is a model which allows somewhat looser behavior (up to a point) of the weights at certain sections of the trajectory, as in the third attachment.

[Omnibus]

Another quirk of WM2D. The device cannot possibly work as drawn. If you imagine a vertical line through the axle you will see that there is more torque on the right hand side than on the left, yet the device rotates to the left. And that is without taking friction into account.

Hans von Lieven

Let me mention once again. The vertical you've drawn is to the left of the vertical passing through the constantly changing its position center of mass of the system. It is that second vertical that you should use in setting up your inequality.That's a remark in addition to the difference-in-constraints remark and to what Stefan said.