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Overunity Machines Forum



Roll on the 20th June

Started by CLaNZeR, April 21, 2008, 11:41:56 AM

Previous topic - Next topic

0 Members and 136 Guests are viewing this topic.

purepower

I understand completely the 9-3 scenario will not work, even before I posted. While the momentum would certainly not allow for the mass to carry from 3 to 9, the imbalance from the other rods could. When I said you would need a "very large amount of rods," mathematically it would need to approach infinity. For every mass center located to the right there would be a mass center located to the right to counter-react in the 9-3 example. The only rod with out a counter-partner would be the rod on the 9-3 location. Once this falls, for motion to continue, another rod would have to follow immediately behind. This is assuming the rod could fall. If there is enough magnetic attraction/repulsion to move the rod, then there would likely be enough magnetic force to hold it there.

The ultimate reason I made the post was to get this thread back on track. I know I came in as a strong opponent to Archer and the forum. Once I felt I successfully debunked his adulterations of physics, I wanted to show my true colors as a free energy advocate.

I really do mean well. I do not want to slow the progress of free energy. I have to pump premium ($4.51 now in Cali!) and every day on my way home from work I dream of driving a car powered by a free energy device on-board or from a FE charge station at home. I share the dream with everyone on this site. However, as a well educated engineer, I will also find faults in current devices and will share my analysis. This is not meant to deter anyone and should rather be used as a tool to overcome the challenges in pursuing overunity.

sm0ky2

@ purepower

first i'll address this::

QuoteSimple proof: check your units. m*(g - M1 - M2)*h == kg*[m/(s^2) - kg*m/(s^2) - kg*m/(s^2)]*m which cannot be simplified to kg*(m^2)/(s^2), or Joules, the SI unit for energy. Where is your proof for the conclusion?

Magnetic fields are not constrained by a time variant, time drops out of the equation and the variable is Distance. The resulting acceleration is directly dependent upon the mass portion of the equation to derrive the time of travel, and thus the final graviational acceleration. 
YES - gravity is still the same Force. But the EFFECT on the moving rod is a combination of the Gravitational acceleration ,and the negative aceleration by BOTH superimposed magnetic fields.
Thus the resulting effect of Gravity, is LESS than what it normally is when those superimposed fields are not present.

Can i calculate how much (input) energy it takes to "allow" a magnet to float upwards in repulsion above another magnet ???  Sure i could, it looks a lot like what we are discussing here.
What good is a calculation without measurement??
Can i measure how much (input) energy it takes? Sure i can,  but the resulting energy would not be what it took to "lift" the magnet through the gravitational field, instead it would be the energy it took to insert the magnet INTO the magnetic field, and hold it there for the test to begin.

So - essentially our measured Input energy is not the PE, but rather the energy required to "break the wall".

Which brings me to my next point:::

QuoteYou also say "The "extension" of the weighted rods - provides leverage with which to overcome the "magnetic wall". This causes the wheel to require far less momentum to overcome that barrier - result being a "breaking of the wall" using a mass small enough for the magnets to lift back to the starting position." Sorry, but no additional leverage is gained by placing the weights out at the end, I have already provided a mathematical proof for this in an earlier proof (will provide diagram tonight if you still dont get it). Place the same weights anywhere on the lever and the torsional effect will be the same

I Challenge this statement. Using an earlier (proven) experiment presented by Archer Quinn.

Wherein a radial shaft was connected with weights Before the magnets, and again tested with the weights Outside of the repelling field. The additional leverage DOES INDEED overcome the magnetic barrier, but when placed at a shorter radii, the same weights COULD NOT.

Need another example??  Get a Walnut. and a Nut cracker - the metal kind, 2 hinged handles.
place the walnut at the very end, where it almost slides out of the nut-cracker.
Then try again with the walnut placed close to the hinge.

Angular momentum DOES factor into this, as the wheel IS spinning when it passes through the magnetic barrier, but as instructed by the wheel's inventor, it should be designed such that gravity alone does this when the rods are fully extended on the operational side of the wheel. (the right side in most of the graphicals)


I was fixing a shower-rod, slipped and hit my head on the sink. When i came to, that's when i had the idea for the "Flux Capacitor", Which makes Perpetual Motion possible.

The Observer

Greetings,

Just hit up Archer's site.

Went through the math thing with the levers and weights.

            http://www.surphzup.com/gpage.html

First there are some agreements to be made ... 1 thru 7 (some I didn't agree with, but anyways)

Then the Proof to the World that Overunity Exists ( which I do know to exist, in the sense that not all energy is properly recognized )
     However, this example may leave a bit to be desired.

In my interpretation of his example, I find that 1 meter of fall suddenly transforms to 2 meters in a 10 to 1 ratio lever system.

There is either a serious error on his part, a hint at something tangible while knowingly putting in an error, or an attempt at slight of words.

Let me explain... 1st there is an example of a 5 to 1 ratio lever system... one end is 1 meter and the other is 5 meters.    --o----------
    and is mentioned that 10 meters of travel on the 5 meter end would produce 2 meters of travel on the 1 meter end.
                   True enough.  It Moves 1/5th the distance of 10 meters on the 1 meter side which is 2 meters.   gotcha.

Then a 10 to 1 ratio is brought up, OooooKK ...  we have to assume one end is 1 meter and the other is 10 meters now...   --o--------------------
    this is not ! mentioned. ( well inferred to later, but anyways)

Apparently this is where the story problem starts.   (I am taking the liberty of making the example clearer than is imparted by it's author.)

                   A 10 to 1 ratio lever is 11 meters long.. with one end at 1 meter in length .. the other at 10 meters in length.
                       There is a 10 kg weight on the short end and a 1.2 kg weight on the long end.

                                                         (10kg)--o--------------------(1.2kg)
 
                                          The 1.2 kg lifts the 10 kg an unspecified distance....
           then the 10kg is tipped off the end, with little energy used, to fall an unspecified distance down.

                The very next piece of information is that the 1.2 kg must be returned to the starting position to conserve energy.
                                         10 X 1.2 is mentioned to get a result of 12 energy units (lets call them)

                                    Here we find the distance not mentioned before...10 meters to return to the top.

Fine... Now we know how far the 10 meter end (with the 1.2 kg weight) dropped, 10 meters...and with a 10 to 1 ratio system .. this means the 10 Kg (on the 1 meter side of the 10 to 1 lever) was lifted 1 meter.

Next this...

     8 ) So back to basics again and the math we agree is correct. 10 kilos falling 2 meters is equal to 1 kilos lifted ummmm????? 20 metres !!!!!! less variables

          YES or NO


Ok... I well agree with this ... but at this point you lost me since it has NOTHING to do with the example you are discussing ! ! ! ! !
         I do REMEMBER a point where you discussed a 5 to 1 ratio lever system that moves 2 meters on one end and 10 meters on the other.
         However your problem apparently uses a 10 to 1 lever system with one end being 1 meter and the other being 10 meters.
         It is highly suspicious that you would refer to the 5 to 1 lever system at this point in your proof !

Since the 10 kg weight fell 1 meter, this means your calculated number is 10 ... not 20.
Any Questions?

     9 ) Do we agree that 20 kilos x 1 meters of lift will cover 12 kilos x  1 meters of lift??

            if you answered yes at this point you have agreed that we have overunity and that it does exist.


Ok, I call Bull Shit. 

You mean (correcting the error) that  10 kilo X 1 meters of lift will cover 12 kilos X 1meters of lift
                 which means you are off by  -2 which means the damned Conservationists of Energy win this Battle !

                       Of course all sorts of things like gradient forces for the angles, since gravity pulls in 1 direction,come into play.
                       And distances may be larger than the system can handle due to the fact that the weight can only fall so far.
                       But I'm not going to analyze those at this point if it's ok with you.

If you are brilliant enough that you want to throw the authorities off when checking you out, the my hat is off to you.

Otherwise, these better not be the calculations you are sending to physics departments, etc or I'd pack my bags
   and find a suitable cave.
   
_________________________________________________________________________________________

To anyone who is interested in a viable, and explainable "overunity": situation, please check out my Thread on

      MAGNETIC PERMEABILITY      http://www.overunity.com/index.php/topic,4831.0/topicseen.html

This is an awesome characteristic of ferromagnetic materials that hasn't been discussed here yet. (well,l 1 day so far)

May you be the light, so others may see,

                                                             The Observer

Slavo

this question was discussed gazillion times here already
Quote
This is where the conundrum of science lies, weight has no correlation to the length or travel of a fulcrum, as the shape and weight distribution of a fulcrum is indeterminate. In short when using a beam to lift a given weight, the pivot point is pushed backward or the beam extended until the length of beam you have can lift the weight. So a fulcrum and pivot that is set will have a set maximum weight to weight ratio, you can set it at 2 to 1 or 10 to 1 or greater. 10 to 1 meaning 1 kilo on the extended end will lift 10 kilos on the short end side of the pivot point or fulcrum

jratcliff

Does anyone else find Archer's behavior a bit bizarre lately?  Especially with his post earlier today?

I posted my thoughts on the topic on my personal blog...

http://jratcliffscarab.blogspot.com/2008/06/archer-quinns-descent-into-madness-is.html