Sjack Abeling Gravity Wheel and the Worlds first Weight Power Plant

[mondrasek]

Indeed.
But, the wheel only slows down if you use the wheel to speed up radially. The ramp is offering a means to convert purely horizontal momentum at 6:00 into vertical. When we visually a radially spoked wheel, the weight will pass spokes "get ahead". If the exit speed at 6:00 is sufficient, it could make it all the way up to 0:00, but would a lacking amount of radial speed, being due right. A second ramp could come into place. Anyways, velocity lost along the rim was traded for getting to 0:00 sooner. But, are these equal?
I keep coming back to a logically occuring stuttering action where exit of the weight would be at top speed, and re-entry at minimum speed, to get the most out of the "launch". What else could be making this work, if anything?

Well, conventional physics says that all the ways of moving a weight in any path where the start and finish are the same results in the same amount of potential energy at the end as at the start.  So no matter how you move it along the path, every gain in energy is balanced by a loss in energy along the circuit.

I have tried every way I can think of to move a weight in a wheel.  So far, with ideal conditions, I always get exactly perfect balance of forces.  That is why I am looking for something that I have not already tried.  And in the Abeling design I have not found anything new as presented.  So I am of the mind that it does not work as presented.  But I'd be thrilled to be proved wrong or to be shown any yet undiscovered phenomenon or undisclosed mechanism!

[Omnibus]

Well, conventional physics says that all the ways of moving a weight in any path where the start and finish are the same results in the same amount of potential energy at the end as at the start.  So no matter how you move it along the path, every gain in energy is balanced by a loss in energy along the circuit.

I have tried every way I can think of to move a weight in a wheel.  So far, with ideal conditions, I always get exactly perfect balance of forces.  That is why I am looking for something that I have not already tried.  And in the Abeling design I have not found anything new as presented.  So I am of the mind that it does not work as presented.  But I'd be thrilled to be proved wrong or to be shown any yet undiscovered phenomenon or undisclosed mechanism!

See, you're making the same mistake over and over again. This is your second misconception (the first one is the implication that guides can somehow create mass). It's not about a given object. The phenomenon we're talking about is sort of a cooperative phenomenon whereby totality of all objects in certain particular configurations display a perpetuum mobile effect. It can never be achieved with just one object or with the combination of objects which are not arranged in a particular favorable way.

[eisenficker2000]

@Stgpcm, some vectors

[stgpcm]

@Omnibus, as promised, a calculation of forces using reactive calculatuions.

As you will see, no mass is created or destroyed.

I hope the sequence of pictures is easy to follow, start at the top left, and work down; when you reach the bottom of the first column, start at thetop of the middle column, and work down; when you reach the bottom of this second column, move to the top of the right column and moev down.

First diagram of first column:
Starting at the intial state, where gravity is is acting the weight to form a force "1". as the guide is vertical, there is no initial interaction.

second diagram of first column
I resolve the vector into its components, as the angle from vertical is 30, the component that generates torque is
1 . sin(30)
= 1/2
The component that acts along the radius between the axle and the centre of mass is
1. cos(30)
= square root of 3/4

third diagram of first column:
I've removed the original weight vector, leaving just the components. if the mass were fixed to the wheel in some way, the radial force component would pull against the wheel axle, without generating torque. and as that is fixed to the earth it would be absorbed (well, it would push the earth a tiny bit left and down).

fourth diagram of first column
at this point I got bored of drawing arrowheads on the vectors, it was much too fiddly, and just made things cluttered
Unfortunately, the weight is free to slide, so the radial component of the original force causes the weight to push against the guide, with force root 3/4. Decomposing that force into the components perpendicular and parallel to the guide, we get a perpendicular force of
sin(angle from parallel). input
sin(30) . root(3/4)
= root(3/4)/2
which is absorbed by the earth, and we can ignore - except for calculating friction (as the god of this mathematical model, I declare the coefficient of friction between the weight and the guides to be zero.)
The vertical component, which the vertical guide cannot act against is
sin(angle from perpendicular) . input
= sin(60) . root(3/4)
= root(3/4) . root(3/4)
= or 3/4

Fifth diagram of first column
I've removed the first calculated radial component, and the horizontal component of that, which has been absorbed by the fixed guide rail. This leaves a vertical vector, which remains to be accounted for of 3/4.

First diagram of second column
Decomposing the unaccounted vector into radial and rotating components, we get a rotating component of
sin(angle from radial) . input
= sin(30) . 3/4
= 1/2 . 3/4
= 1/8
this is added to the first torque producing resultant (1/2)
The radial component, is
sin(angle from rotation) . input
= sin(60) . 3/4
= root(3/4) . 3/4

second diagram of second column
as before, in the third diagram of the first column, I've rmoved the decomposed force, but the weight still isn't attached, so it still attempts to slide down the slot, and still works against the guide.

third diagram of second column
as before I decompose the radial motion into the forces perpendicular and parallel to the guide, as before these are in the ratio 1/2 and root(3/4) of the radial force. As before the guide absorbs the vertical component.

forth diagram of second column
the as yet unaccounted vertical force is
root(3/4) . ( root(3/4) . 3/4 )
= 3/4 . 3/4
= 9/16

fifth diagram of second column
as in the second diagram of the first column, and the first diagram of the second column, we decompose the unaccounted vertical force into radial and rotary components, as before, the rotary component is 1/2 the vertical force, and the radial component is root(3/4) of the vertical force,
rotary = 1/2 . 9/16
= 9/32
radial = root(3/4) . 9/16

Third column
the process is repeated until there is no component of the original force that isn't acounted for, either by being converted to torque, or by being absorbed by the axle, or by being absorbed by the guide.

This takes an infinite number of steps.

the result is the sum of 1/2 + 3/8 + 9/32 + 81/128 + ...
or the sum of (2^n+1) / (2^(2n+1) for all integer values of n between 0 and infinity.

I hope you can see from the diagram, the sum of the rotational vectors forms the hypotenuse of a triangle, the top angel of which is 60 degrees, and the height is the original 1
cos(angle) = adjacent/hypotenuse
cos(60) = 1 / f
f = 1 / cos(60)
f = 1 / (1/2)
f = 2

The distance of the mass from the pivot is 0.9, so the torque is 1.8 Counter clockwise.

You will recall the result I obtained by balance of forces was also 1.8 counter clockwise, and the approximation of the torque by calculating the change in energy as the system rotated through one degree was equivalent to that.

So, my calculation by reactive forces match my calculation by balance of forces, which match my calclation of change of energy.

Unless you can identify an error in ALL THREE, you're making an extraordinary claim, and that requires an extraordinary proof.

[Omnibus]

@stgpcm,

Study carefully what @eisenficker2000 has shown you. Notice also that it is a discussion of your case and isn't the way our guide works.