The question seems to me "Is there magnetic resistance when you raise the 30 gram weight"? If there is then resetting the machine will take as much force as it took to raise the 60 gram weight, it might be 100% mechanically efficient, but friction would eventually stop it even then. If you get no resistance when raising the 30g weight other than the weight... In short figure away to measure the necessary force to raise the 30g weight, it might not be what you think.
you need to figure out how to over come the flux gate at the point of the magnetic field weight and counter ballance whont work i have all ready tryed for 20 years and belive me its hard its all int he flux gate
the lever arm linkage set is only your control. you can further enhance it to get more shudder(bounce/swing whatever) IN the rotating crank. this what you want. stuff the shudder in the arm. to form a loop you want the longer shudder in the crank going thru a one way gear system which will trigger your control arm and thus a loop.
Could it be that you use to large wieghts in combination with to small magnets. Because the weight is dropping so fast that the stator can't even react that fast.
Would mind elaborating? Do you mean that the speed of the flywheel magnets field entering the field of the stators field determines the force of the reaction with the stators field? Or do you mean something else?
Magnetic fields are parabolic and flexible, so as the flywheel magnet's field enters the field of the stator magnet field they dent slightly being two like fields. Since both magnets are trying to re-exert the natural shape of their fields and are both pushing in the same direction do you get a magnetic bounce effect (?) which drives the stator along the linear bearing with equal or more force?
none of your experiments everwork good try though wait until you see my fuelless heater i ve built defenitly overunity 300watt electric motor for 1500 watss heating soon on youtube not quite finished yet
Edison was once asked how he felt about about failing 1000 times before inventing the light bulb. His reply was that he did not fail a thousand times, but that there were 1000 steps to the answer..
cont... if you take away the magnet you won't get the same effect as long as the height to which the larger weight is raised is not ("approximately" due to the ratio of masses used being approx.1/2)inferior to half the height by which you raise the smaller weight.!
No it won't work at all without the magnets LOL. Its easier to imagine this as a seesaw, 30 grams one side 60 grams the other, since 30 grams is twice as far away from the fulcrum , the moments are the same. The magnet in the middle makes it a bit more interesting
The energy you put in your apparatus goes both to increase the potential energy of the smaller weight and the magnetic energy of the magnet set up(thought it's not clear what this latter set up is). So you are putting in more energy than that needed to raise the larger weight. When you release the smaller weight you also release the magnetic energy stored.
Hi would you mind posting the maths workings? I am curious to see because from your video i have no way of determining the value of h. As an aside if you aline the magents on the flywheel and the stator so you have the magentic fields silgthly off poles do you think you would get more force sliding the stator along? Thanks for this champion effort and allowing us to join you for the journey.
The big weight is only being raised by a small height. Whereas the little weight is droping by a large amount. If the big weight is twice the weight of the little one, i think you need to move the big weight more than half the drop of the little one, for over unity.
looks like pendulum advantage from the speed of the drop. You can get much more work wrapping the magnet in a ferrous material to turn it off then wrap the outside of the shielding with copper wire, takes a very small power to turn off and on the magnet, results will be 4x the electric force used, see parallel path magnet motors - flynn research, they already have overunity (lorentz force free) devices, they got contracts to build generators for the government, the technology = stiffled.
Can't wait to see your results. To me it seems like you're really close to unity. It looks like the top weight is raised half the distance the bottom one dropped (I meant to say half instead of twice earlier).
Using Pe=MGH it appears that this is very nearly 100% efficient. dropping the wieght inputs.06 joules and energy required to raise wieght is .06 joules. The wieght dropped has now been adjusted so that there is no overshoot. So now the standard magnet can be removed, and various other combinations , or arrays of magnets could be used to investigate the steorn effect
i thought about another thing. If the distance the center of mass of the large weight traveled upward is more than twice the distance the smaller weight droped downward, then you would acheive over unity. (Twice because the mass is twice as large). You can convince yourself of that if you apply that the initial ME equals the final ME.
p.s. Make sure you're measuring the distances from the center of mass.
i dont think you're getting over unity because what you think is only 30 grams is actually that plus the mass of the lever arm and so on. Basically everything that moves while you're raising the mass with your hand contributes to the mechanical advantage when you release. i wonder what the effciency would be on that though.
The question seems to me "Is there magnetic resistance when you raise the 30 gram weight"? If there is then resetting the machine will take as much force as it took to raise the 60 gram weight, it might be 100% mechanically efficient, but friction would eventually stop it even then. If you get no resistance when raising the 30g weight other than the weight... In short figure away to measure the necessary force to raise the 30g weight, it might not be what you think.
yojimbo1981 4 years ago
Comment removed
yojimbo1981 4 years ago
you need to figure out how to over come the flux gate at the point of the magnetic field weight and counter ballance whont work i have all ready tryed for 20 years and belive me its hard its all int he flux gate
MYLOW121363 5 years ago
And they say magnets cannot do work....LOL
Eltimple 5 years ago
maybe attach the weight on the axis like a flyweel?
DikkeZwaarte 5 years ago
the lever arm linkage set is only your control. you can further enhance it to get more shudder(bounce/swing whatever) IN the rotating crank. this what you want. stuff the shudder in the arm. to form a loop you want the longer shudder in the crank going thru a one way gear system which will trigger your control arm and thus a loop.
refdachef 5 years ago
Could it be that you use to large wieghts in combination with to small magnets. Because the weight is dropping so fast that the stator can't even react that fast.
somelse 5 years ago
Would mind elaborating? Do you mean that the speed of the flywheel magnets field entering the field of the stators field determines the force of the reaction with the stators field? Or do you mean something else?
siroca 5 years ago
Magnetic fields are parabolic and flexible, so as the flywheel magnet's field enters the field of the stator magnet field they dent slightly being two like fields. Since both magnets are trying to re-exert the natural shape of their fields and are both pushing in the same direction do you get a magnetic bounce effect (?) which drives the stator along the linear bearing with equal or more force?
siroca 5 years ago
none of your experiments everwork good try though wait until you see my fuelless heater i ve built defenitly overunity 300watt electric motor for 1500 watss heating soon on youtube not quite finished yet
theoneagain 5 years ago
Edison was once asked how he felt about about failing 1000 times before inventing the light bulb. His reply was that he did not fail a thousand times, but that there were 1000 steps to the answer..
Eltimple 5 years ago 5
cont... if you take away the magnet you won't get the same effect as long as the height to which the larger weight is raised is not ("approximately" due to the ratio of masses used being approx.1/2)inferior to half the height by which you raise the smaller weight.!
linoatwork 5 years ago
No it won't work at all without the magnets LOL. Its easier to imagine this as a seesaw, 30 grams one side 60 grams the other, since 30 grams is twice as far away from the fulcrum , the moments are the same. The magnet in the middle makes it a bit more interesting
Eltimple 5 years ago
The energy you put in your apparatus goes both to increase the potential energy of the smaller weight and the magnetic energy of the magnet set up(thought it's not clear what this latter set up is). So you are putting in more energy than that needed to raise the larger weight. When you release the smaller weight you also release the magnetic energy stored.
linoatwork 5 years ago
Hi would you mind posting the maths workings? I am curious to see because from your video i have no way of determining the value of h. As an aside if you aline the magents on the flywheel and the stator so you have the magentic fields silgthly off poles do you think you would get more force sliding the stator along? Thanks for this champion effort and allowing us to join you for the journey.
siroca 5 years ago
The big weight is only being raised by a small height. Whereas the little weight is droping by a large amount. If the big weight is twice the weight of the little one, i think you need to move the big weight more than half the drop of the little one, for over unity.
lumsdot 5 years ago
looks like pendulum advantage from the speed of the drop. You can get much more work wrapping the magnet in a ferrous material to turn it off then wrap the outside of the shielding with copper wire, takes a very small power to turn off and on the magnet, results will be 4x the electric force used, see parallel path magnet motors - flynn research, they already have overunity (lorentz force free) devices, they got contracts to build generators for the government, the technology = stiffled.
rhenson32 5 years ago
Can't wait to see your results. To me it seems like you're really close to unity. It looks like the top weight is raised half the distance the bottom one dropped (I meant to say half instead of twice earlier).
ulloa2003 5 years ago
i want one!! yea make that 2!!
hudstone420 5 years ago
Using Pe=MGH it appears that this is very nearly 100% efficient. dropping the wieght inputs.06 joules and energy required to raise wieght is .06 joules. The wieght dropped has now been adjusted so that there is no overshoot. So now the standard magnet can be removed, and various other combinations , or arrays of magnets could be used to investigate the steorn effect
Eltimple 5 years ago
i thought about another thing. If the distance the center of mass of the large weight traveled upward is more than twice the distance the smaller weight droped downward, then you would acheive over unity. (Twice because the mass is twice as large). You can convince yourself of that if you apply that the initial ME equals the final ME.
p.s. Make sure you're measuring the distances from the center of mass.
ulloa2003 5 years ago
i dont think you're getting over unity because what you think is only 30 grams is actually that plus the mass of the lever arm and so on. Basically everything that moves while you're raising the mass with your hand contributes to the mechanical advantage when you release. i wonder what the effciency would be on that though.
ulloa2003 5 years ago
Thanks for this video,hmmm i'll have a think and get back to you.
siroca 5 years ago