Roll on the 20th June

[Abrillo]

OK, I've been a lurker here for a while, and have followed with interest.

I think  it is much simpler than being considered here:

Think it works like this:

[Abrillo]

In this configuration, the pulse electromagnet repels the disk magnets, and only has to move the weight halfway up the tube to create an imbalance.

[zerotensor]

@ Zero,

i was using 9.8m as my Diameter.

the radius thus would have been 4.9,  in either case, the imbalance of the vertically mounted wheel, does not have the same inertial resistance as a flat wheel, with a weighted string tied to a pulley. One set-up utilizes leverage,  whereas the other utilizes only gravitational force.

you can actually see this exact set-up in a few pendulum-impact-hammers, that use a massive fly-wheel for momentum gain.

The "inertial resistance" of a balanced wheel is constant regardless of its orientation in a uniform gravitational field.  There will be gyroscopic precession if the wheel is not rigidly mounted, and there will be additional stress on the bearings, but in our case we assume perfect bearings and a rigid mount.

There is just as much mass "falling" on one side of the wheel as is being "lifted" on the other.  Net sum gain = null.

It doesn't matter if you use a string-and pulley setup, a lever, a motor, or a ballistic projectile to impart the rotational energy to the wheel.  You put a certain amount of energy into rotating the wheel, and that energy is stored as rotational kinetic energy.  Immersion in a uniform gravitational field makes no difference at all.

[zerotensor]

In this configuration, the pulse electromagnet repels the disk magnets, and only has to move the weight halfway up the tube to create an imbalance.

Yes. This is my interpretation also.  The idea is that by pulsing the electromagnet at the bottom left with a small amount of energy you can get the rod to move up and to the right, where it wants to go anyway due to the attraction of the permanent magnets mounted at the top right, thus completing the loop.  One problem with this is that the faster the wheel spins, the harder it will be to push the rod across the center, owing to the apparent centrifugal force in the rotating frame.  Even assuming that this can be overcome, the magnets moving past each other will eventually degauss.

Here's a thought-experiment:  Imagine that you are inside a ring-shaped space-station which rotates to produce "artificial gravity".  There is a tube that goes from your location to the opposite side of the ring, right through the center.  Now you place a rigid rod inside the tube, with a magnet on the far end.  There is another magnet at the terminus of the tube at the far end of the station, oriented so that it attracts the rod.  By pushing up on the rod with a certain amount of force, you can get the rod's center of mass to get past the center of the spinning ring.  The magnet on the far end makes it easier.  Now, increase the rotational rate of the ring.  The "artificial gravity" gets stronger.  You have to push the rod harder to get it to move.

Not only this, but you have to take the coriolis forces into account.  Have you ever tried rolling a ball on a moving merry-go-round?

Here's another experiment (I just did it):  Take a bunch of long cylindrical magnets like the kind sold in toy stores and attach them end-to-end, forming a "string" of magnets.  Now, hold the string out at arms length and start spinning around.  At some point, the rotation gets too fast and the magnets fly apart.  The constant attractive force of the magnets is overcome by the centrifugal force, which is proportional to the angular velocity.
<edit:  Now I'm dizzy and I lost my magnets :0 >

[sm0ky2]

The "inertial resistance" of a balanced wheel is constant regardless of its orientation in a uniform gravitational field.  There will be gyroscopic precession if the wheel is not rigidly mounted, and there will be additional stress on the bearings, but in our case we assume perfect bearings and a rigid mount.

There is just as much mass "falling" on one side of the wheel as is being "lifted" on the other.  Net sum gain = null.

This is not the case at all.   When you have a single weight, turning the wheel by a pully the force on the wheel
is mass (of the weight) * height * gravity  vs. the inertial resistance of a massive wheel at rest.

The force of a vertical wheel, with a weight on its top edge (12:01),  is mass (of the weight) * D * gravity times a leverage factor, which is proportional to the radius (r).  Because there is no counter balance on the other side of the wheel. This is many times greater than necessary to turn the wheel against the inertial force.


Example::  support a bicycle on its side, and wrap the weighted string  around the axle. and let it spin the wheel

now turn the bicycle upright and attach the weight to the top of the wheel and let it go.

These are two entirely different scenerios.  a kid trying to lift a car with a pulley,
vs.a kid sitting on the end of a long teeter-totter lifting the car with ease.

you actually caused me to run the numbers on this, (and im glad i did....)  there happens to be a horizontal distance which when properly accounted for ammounts to 2*(r), which while it does not consume a substantial ammount of force, does affect the time factor by we actually have  1/3 of 9.8m/s at bottom-dead-center, at a 1m drop would ammount to 3.27 m/s
(in radians its about 9.98somethingRPM?)

thus:   1/2 of 1001kg gives us about 5.3KJ which is still more than the original 96.04J

i had an idea while i was typing this. to place a pendulum at the 'striking' point of equal mass as the weight on the bicycle wheel.  It should be more than enough to swing the pendulum all the way over the top
i'll figure something out to set that up to show that it works (or not), and post a video in either case.