OK, I know everyone likes wheels, rotating wheels, so, let's make a wheel!
First I had the idea of the piezoelectric-magnetic hammer:
http://www.overunity.com/index.php?topic=8815.0
Nobody replied me.
Then I watched the video of the electromagnetic launcher made by Gilles:
http://www.youtube.com/watch?v=6nowIn-QdbE
That made me think and try to develop a hypothesis in formal terms:
http://www.overunity.com/index.php?topic=8820.0
Nobody replied me.
So let's try to apply the principle of the launcher, of the hammer and of the hypothesis of a wheel. It's quite simple.
We only would need to attach the aluminum disk of Gilles experiment to a wheel:
Look in the video that the launched disc hit the roof.
How high the disc could go if there were no roof?
How many times the wheel could spin if the disk was linked to it?
It was only 3 volts DC / 480uF. And if the disk was made of a magnetic material stronger than aluminum perhaps the energy required for the electro-magnet was even smaller.
Hmmm... Let's do some calculations...
The formula to calculate the amount of joules in a capacitor, as far as I know, is this:
W = 0.5 x C x U²
Gilles said in the video description that he used 3 volt and 480 micro-farads.
So we have:
W = 0.5 x 0.000480 farads x 3 volt x 3 volt
W = 0.000240 x 9
W = 0.00216 joules.
If my calculations are not wrong, Gilles provided to the coil only 2 milijoules.
2 milijoules = 2 miliwatt/second.
It means that 2 milijoules is the energy that we need provide to a LED to make it have a current of 20 miliamps with 2 volt for just 50 miliseconds.
WHAT ?!?!?
Where did I go wrong in my calculations?
Providing to a coil only the energy required to light a LED for 50 milliseconds with a current of 20 mA and 2 Volt, our friend Gilles made an aluminum disc fly at high speed until it hits the roof??
Please someone tell me I'm wrong and correct my calculations! Otherwise I will go nuts!
Rapa
You feel Gilles has an OU launcher?
Chet
he is using an ammeter. i have worked with electronics for a long time and i can say for sure that his caps are sucking back way more juice than you think. no offence but that launcher would not be remotely feasible in the engineering of an efficient or o/u device
I feel that the "flight" of that aluminium disk can be used to generate enough eletric energy to provide 20 mA and 2 volts to a LED for longer time than 50 milliseconds.
And that scares me! A lot!
Quote from: rice on February 25, 2010, 08:05:13 PM
he is using an ammeter. i have worked with electronics for a long time and i can say for sure that his caps are sucking back way more juice than you think. no offence but that launcher would not be remotely feasible in the engineering of an efficient or o/u device
But the input energy was greater than 2 milijoules? Or not? I just can't believe it was just 2 milijoules!
I'm getting fucking scared!
Someone has to tell me my calculations are wrong and that the input energy was not only 2 milijoules!
I just read on Wikipedia:
One joule in everyday life is approximately:
* the energy required to lift a small apple one meter straight up.
http://en.wikipedia.org/wiki/Joule
WHAT ?!?!?
So with 10 millijoules (five times the input energy of Gilles), we can only lift one hundredth of a small apple one meter straight up?
Anyone else besides me thinks that a small apple does not weigh 100 times more than that aluminium disc?
Anyone else besides me thinks that disk was lifted more than just one meter?
If the input energy was really only 2 milijoules, I think I'm gonna be crazy.
If I had contact with Gilles I would propose to him to repeat the experiment, but attaching the aluminum disk to some kind of vertical rail with the least possible friction.
And on top of the aluminum disc, it could put a quarter of an apple.
If the quarter of an apple goes up more than one meter, we have an output of 250 millijoules. Very good, considering an input of only 2 millijoules.
That would be madness?
I couldn't resist... I just sent an e-mail to him (gilles.charles@univ-orleans.fr)
I wrote:
Dear Monsieur Gilles,
I watched the video in YouTube of your electromagnetic launcher:
http://www.youtube.com/watch?v=6nowIn-QdbE
I was very impressed.
In the video description you said that you used only 3 volt and 480 micro-farads. I did some calculations...
The formula to calculate the amount of joules in a capacitor, as far as I know, is this:
W = 0.5 x C x U²
So we have:
W = 0.5 x 0.000480 farads x 3 volt x 3 volt
W = 0.000240 x 9
W = 0.00216 joules.
If my calculations are not wrong, you provided to the coil only 2 milijoules. 2 milijoules = 2 miliwatt/second. It means that 2 milijoules is the energy that we need provide to a LED to make it have a current of 20 miliamps with 2 volt for just 50 miliseconds.
Then I read in Wikipedia:
"One joule in everyday life is approximately:
* the energy required to lift a small apple one meter straight up."
http://en.wikipedia.org/wiki/Joule
So with 10 millijoules (five times your input energy), we can only lift one hundredth of a small apple one meter straight up... I think that a small apple does not weigh 100 times more than that aluminium disc you used... I think that disk was lifted more than just one meter...
Monsieur Gilles, I would propose to you repeating the experiment, but attaching the aluminum disk to some kind of vertical rail with the least possible friction.
And on top of the aluminum disc, you could put a quarter of an apple. If the quarter of an apple goes up more than one meter, we have an output of 250 millijoules. Very good, considering an input of only 2 millijoules.
Sincerely,
Rapadura
without needing to do the math, i need to tell you that you are way off. he is using somewhere between 50 and 100 joules. no need to get excited
Quote from: rice on February 25, 2010, 10:34:25 PM
without needing to do the math, i need to tell you that you are way off. he is using somewhere between 50 and 100 joules. no need to get excited
W = 0.5 x C x U²
W = 0.5 x 0.000480 farads x 3 volt x 3 volt
W = 0.000240 x 9
W = 0.00216 joules.If not, then at least one "law of physics" must be changed, the law that says the calculation of the amount of joules in a capacitor is possible with the formula "W = 0.5 x C x U²"
Quote from: Rapadura on February 26, 2010, 06:19:35 AM
W = 0.5 x C x U²
W = 0.5 x 0.000480 farads x 3 volt x 3 volt
W = 0.000240 x 9
W = 0.00216 joules.
If not, then at least one "law of physics" must be changed, the law that says the calculation of the amount of joules in a capacitor is possible with the formula "W = 0.5 x C x U²"
There is no evidence that voltage is 3v. I guess it is much more and there is some electronics to elevate the voltage during the capacitors charging process.
Quote from: exnihiloest on February 26, 2010, 09:16:14 AM
There is no evidence that voltage is 3v. I guess it is much more and there is some electronics to elevate the voltage during the capacitors charging process.
he is getting the 3V figure from the 2 C or D cell batteries that were shown in the video which are being used as the circuit supply voltage... yes, the inverter electronics are in the box, along with the capacitor bank. they look like caps from a flash camera circuit. all i see is a gauss or coil gun. there is way more than 2 millijoules input...
Well, the only possibility is that the voltage of the capacitor bank after completing the charge is greater than 3 volt...
Anyway, if the charged capacitor bank has a voltage of 24V, then we have:
W = 0.5 x C x U²
W = 0.5 x 0.000480 farads x 24 volt x 24 volt
W = 0.000240 x 576
W = 0.138 joules.
138 milijoule is not that great...
Does the voltage may have been higher than 24 V ?
Quote from: Rapadura on February 26, 2010, 12:06:42 PM
Well, the only possibility is that the voltage of the capacitor bank after completing the charge is greater than 3 volt...
yes, that is what the inverter circuit does... those caps are 300+ volts and 120uf most likely and he is charging them up to near full capacity.
@Rapadura
From the comments under the video,
http://www.youtube.com/watch?v=6nowIn-QdbE
"gilbondfac" wrote 2 years ago:
"of course, i used a flash circuit photos but with
480 Uf "
And watch the meter upper left hand corner, you can see 310V printed, I assume the final voltage the capacitor bank made from photoflash capacitors were charged up to that voltage level. This voltage then was switched to the toroidal coil via the thyristor, all this built into the plastic box. The meter may have showed the peak current , about 3 Amper, taken from the 3V battery bank, when the circuit charged up the cap bank.
So the stored energy in the cap bank was about 310*310*0.00048/2= 23 Ws, ok?
rgds, Gyula
Hmmm... 23 joules... Well, I don't think that aluminium disc could make a apple be lifted 23 meters!
Perhaps with a neodymium disc! :D