Hi Luc
Wow I've never seen anything like that before,fantastic :o I would like one for Christmas ;D
I found it very hypnotic, also reminded me of one of those executive gadgets.
Did you mount the coil on square box section to stop it rotating ?
I would have thought you got less friction with a round tube.
cat
No problem Luc, I will see who is in ;)Dear Cat and Luc,
cat
Dear Cat and Luc,
Interesting. We worked with:
1. One coil in the middle repelling two magnets to soft and hard surfaces to achieve unbalanced force in one direction.
2. Two coils with a moving magnet in the middle to achieve unbalanced force in one direction
http://www.youtube.com/watch?v=FafYVGbgULw
Noiw you have the arrangement of:
A Moving Coil in the middle of two magnets.
We shall be working on:
A non-moving coil in the middle of two magnets with a changing magnetic flux to draw out energy.
We already have a prototype demonstrated to over 25,000 visitors in the two Open Shows in Hong Kong. See the Lee-Tseung Lead-Out Energy theory thread in this forum.
Welcome to the team. You will have fun and many unexpected experimental results. Overunity and unbalanced force are two examples.
@gotoluc
First thanks to @powercat for letting me know about this thread.
I saw your video, and son of a gun, I had been looking for a suitable linear motion method for an idea I had been mulling over 2 years now. At the time I made some drawings and an animation that is now located here;
http://purco.qc.ca/ftp/Wattsups%27%20stuff/linear-wheel/
The gif file is the animation.
The idea is not to use a crank type motion of the linear movement but to simply use the linear movement as a wheel balancing offset.
My only question is, can your idea work standing up. Even as a catapult, can it have enough force to push up that coil. If it can, then there may be a way for it to work, by mounting it a ring that has a center axle that is aligned to the center of your iron bar, and then controlling the pulse to the coil to always keep it on the right side then use gravity as the main force.
Also, think about it. That coil is moving. Put some other coils around it and it will become a generator. Kind of reminds me of @handyguy1's Thingamajigger. Or you can hold the coil steady and let the iron bar with the magnets slide.
Very nice work. Geez now you are on motor type devices. Good progression.
I wonder if I recognise your coil.
Is it the low current coil on the 8:1 mains power transformer of a microwave oven?
If so, they are available to us all for peanuts - a visit to a recycling centre, and
an amusing evening in front of the TV cutting through the transformer with an angle
grinder (hack saw will do), and soaking the result overnight in water to soften up
the paper/card packing around the coils in order to get them off.
@powercat
thanks for your PM on this thread, in the meantime I also noticed it at the energetic forum where I answered in the meantime too.
@Luc
regarding the vertical arrangement wattsup mentioned, I recall an old patent where a permanent magnet was placed under the core of an electromagnet and this extra flux helped more than double the lifting height against gravity with the same input current to the electromagnet, see this link and the patent drawing: http://www.overunity.com/index.php/topic,1621.msg16347.html#msg16347
rgds, Gyula
I wonder if I recognise your coil.
Is it the low current coil on the 8:1 mains power transformer of a microwave oven?
If so, they are available to us all for peanuts - a visit to a recycling centre, and
an amusing evening in front of the TV cutting through the transformer with an angle
grinder (hack saw will do), and soaking the result overnight in water to soften up
the paper/card packing around the coils in order to get them off.
Hi Gyula,
yes this works! but how do you get the magnet (weight) back up once it has used gravity?
Luc
Hi Luc,
Well, sorry I did not mean using it in your present horizontal setup but in wattsup wheel setup he has showed where he wishes to offset the wheel balance by pushing objects linearly up whenever the vertical course for them just passed the 12 o'clock clockwise. I thought here to help the upward linear move with that 'under'magnet, then, as the wheel rotates clockwise, these objects will return by sliding back again due to gravity. I only meant here using this trick, you get more upwards movement for the same input to the electromagnet coil.
rgds, Gyula
Okay I see,
it would work but one thing to consider in wattsup's design is maybe the quick launch (at desired time) may cause a braking of the wheel at that timing point???
Luc
Cat,
I am affraid thalin wishes to see the video user magnetman linked to in his Reply#17 here in this thread, previous page.
Now I also get 'video removed by user', two days ago or so it was still there.
Hi, I removed all my videos on the U tube except three of them. My last video is a summation of all my work so far and by far is the best motor device I ever constructed. It works on a reciprocating basis also. Here is the link to my SWING MOTOR 360. I also have this link posted on the first page of magnet motors in this forum.
I am 64298 on U tube.
http://www.youtube.com/watch?v=4L_UG0sezNU&feature=player_embedded
Tom magnetman12003
Gyula
Gotoluc,
Hi my friend, I did not notice your design two years ago, but it just caught my attention today. After looking at your YT clips, my conclusion is : very interesting.
Your conception indeed does have validity in utilizing the magnet power and the flux concentrating characteristic of the steel( or ferrite) bars to multiply the force exerted by the moving coil.
I think using the motor to drive the device is a bad idea. This is because you are wasting the input power to overcome the magnets, rather, may I suggest that the flywheel be attached to a DC generator or AC alternator, using the multiplied torque to spin it. I think you will find the output be almost 4 times more than the input( as a minimum rule of thumb).
This is a very efficient piston motor, and you can stack several magnetic "pistons" to the flywheel like a modern Internal combustion engine does, and drive a big conventional generator with the crankshaft.
The last thing needed to be optimized will be to reduce heat loss due to eddy currents in the steel laminate bars when power goes up. But if speed is maintained in low level, this issue can be ignored. Once again, congratulations, Gotoluc, for disclosing this invention publicly. Now it belongs to the whole Humanity and not to the power that be.
Gotoluc,
I like what you are doing, I love pinch fields, that is what I call it when you have the same poles facing each other like that.
I hate crankshafts, they are not efficient and so I made for myself a mechanical rectifier that is much more efficient and works wonderful with a solenoid input.
There are bidirectional solenoids that work in a similar, but not the same fashion, they have a core and a magnet out at the end of the coil, supply the power one way and it pulls the core in, the other way and it pushes it out, similar but not the same as yours.
Tom Webb
3 one way needle rollers, 3 gears, 3 shafts, 2 connecting arms and one of those is also the input arm.
Hi Luc,
Very inventive design, thanks for sharing. The weight of the coil moving back and forth at low speed would be good for a lever/pendulum setup.
This design I used a small motor with weight.
http://www.youtube.com/watch?v=hr0KyMqMZtc (http://www.youtube.com/watch?v=hr0KyMqMZtc)
But if you place your motor centered on top it can use the displacement of the coil instead of a spinning weight. That would eliminate some side to side frame wobble and the powered side dead zones. Then lever through lenz with a big block N52 on the bottom over some coils... :)
Hi Luc, are you sure you explained the scope channels right ?
Isn´t in the first scopeshot the green trace not the coil voltage or is it really the coil current ?
IMHO only green can be the coil voltage, otherwise I am pretty puzzled...
Hmmm...
Also how long do you energize the coil ?
Is it about these 16 milliseconds so about 4 x DIVs ?
Also it would be nice if we would have the same RPM in both scopeshots
to compare the waveforms... So I don´t know, where the zero point crossing voltage induction is
located regarding the coil location inside the motor ?
At which location doesn´t the coil generate any voltage when it is moved ?
Is it at the center location of the motor ?
Well the only time I know that there is no counter induction when a magnet passes
a coil or a coil passes a magnet is in the ORBO device,
where 2 ferrite toroid coils are 180 degrees out of phase, so the induction is canceled out,
but the toroids attract a magnet and when energized the magnet can pass
all happening WITHOUT any induction inside the toroid coils... So there LENZ law is fully
violated.
But here in your case I am not sure, it seems that Lenz is there, if the green trace in the first
scopeshot is the voltage across the coil.
Maybe you can try to use a charged capacitor to pulse the coil and see, how far it lifts
up the coil.
If you put additional weight on it or addtional magnets and if you can raise the coil higher than
the stored energy in the capacitor then you have already shown overunity.
The stored energy in the cap is 0.5 x C x Voltage^2
and the raised weight energy = mass in Kg x g( earth acceleratiion 9.81) x height-difference in meters
So if you can show with a single shot from a charged capacitor, that you can raise the
coil weight higher than the stored cap energy you would have already shown overunity.
So lets make a calculation example:
If you use 1 Farad as the Cap and have it charged up to 10 Volts you have an energy stored
as 0.5 x 1 x 100= 50 Wattseconds.
Now, if you put additional weight onto the coil so it is about 10 Kg
you should be able to lift the 10 Kg weight:
height difference=50 / ( 10 kg x 9.81 ) = 0.509 Meters
So if you could lift it vertically against gravity higher than 0.509 Meters you have Overunity.
This would be probably work best with bigger coils with just very fine wire and many turns and
much higher cap voltages.
Okay, so please let us know the fire timing of your circuit and the colors of your scopeshot
of what is what and I can have another pondering...
Many thanks.
Best regards, Stefan
Maybe you can try to use a charged capacitor to pulse the coil and see, how far it lifts
up the coil.
If you put additional weight on it or addtional magnets and if you can raise the coil higher than
the stored energy in the capacitor then you have already shown overunity.
The stored energy in the cap is 0.5 x C x Voltage^2
and the raised weight energy = mass in Kg x g( earth acceleratiion 9.81) x height-difference in meters
So if you can show with a single shot from a charged capacitor, that you can raise the
coil weight higher than the stored cap energy you would have already shown overunity.
So lets make a calculation example:
If you use 1 Farad as the Cap and have it charged up to 10 Volts you have an energy stored
as 0.5 x 1 x 100= 50 Wattseconds.
Now, if you put additional weight onto the coil so it is about 10 Kg
you should be able to lift the 10 Kg weight:
height difference=50 / ( 10 kg x 9.81 ) = 0.509 Meters
So if you could lift it vertically against gravity higher than 0.509 Meters you have Overunity.
This would be probably work best with bigger coils with just very fine wire and many turns and
much higher cap voltages.
Many thanks.
Best regards, Stefan
Hi Luc,
okay, now I had another closer look.
As I don´t know, how your H-bridge circuit looks alike and how the resistances of the transistor
or MOSFETs change over time there, the yellow trace= coil voltage is a bit strange, but okay....
In the green trace you can see exactly that there is the LENZ law occuring, as in the motor operation you
see a quick accelerating current pulse and then the counter induction from the coil kicks in,
so the input current strongly gets reduced and as the coil gets slower again at the other turn
point the input current rises again as the counter EMF of the coil is reduced due to lower speed and
thus the input current rises again...
But the coil voltage should be different then also again, but maybe you H-Bridge is mostly powered
by cap discharges as the batteries might have a high internal resistance, so the coil voltage
behaves this strange...
Well, it surely also depends how long the coil is.
If you would have a short coil, that is only 0.5 cm long and the core would be 10 cm long
it would behave quite differently as it is now, where the core is maybe 10 cm long and the coil
is at least about 4 cm long ?
Also it could help to place in the center of the outer core some additional
iron core pieces so the return flux is concentrated in the center....then the counter EMF
would be also different and it could help to reduze Lenz law effect...
P.S. Yes, a device that can show more vertical lift height in ONE shot , that
does not comply to the energy conservation formular:
m x g x h = 0.5 x C x Voltage^2
will show overunity.
So if you can lift a weight vertically up into the air higher than the stored cap
voltage will tell you, you have a winner....!
Regards, Stefan.
Hi Stefan
As per your request I have made a video demo to prove Overunity. Does this mean I can apply for the OU prize ;D
Link to video: http://youtu.be/OxuotFUWVGQ (http://youtu.be/OxuotFUWVGQ)
Let me know what you think
Luc
Luc,
was the bigger coil placed as a weight ontop shortcircuited or open circuit ?
This could affect very much your magnetic repulsion !
Hi folks, awesome work luc, thanks for sharing again the idea.Hi Gotoluc, I tend to side with Tyson that a prony brake method to check the output of the DC generator will be a better proof. Still, this contraption is a VERY efficient piston, granted.
Hi harti, 3.7% efficiency does not sound right, his coil was hanging there for a moment also.
Math aside, lucs design is somewhat like garry stanleys dual rotor air core pulse motor, in that you can use many more magnets and get ever greater shaft work for the same input, I think it's possible that the math is not correlating between units properly to account for this.
Prony brake would be a better method, no cross unit errors.
Though in lucs design, to add more magnets and keep it from becoming an air core essentially, we'd have to prevent saturation, more ferro material in center core.
peace love light
tyson
Just some casual thoughts from an innocent bystander.....
I'd probably try this: a ferromagnetic or soft ferrite coil core that is a hollow cylinder, and a non-magnetic axis for the core to slide on, like an aluminum or plastic rod. I think a lot of flux is being wasted or channelled into parts where it's not doing useful work.
Also.....I think I'd like to achieve "critical damping" in the discharge waveform. That is, one doesn't want ringing, one wants all the cap's energy dissipated into the coil in the first cycle of the "ring". This requires tuning inductance and capacitance together. Diodes will stop ringing but at the cost of dissipating energy where it can't do useful work. You get the most "bang" for your buck if you can dump all the cap energy into the load in the first cycle without any coming back or voltage reversal: the critically damped waveform.
Stefan's math up above looks more complicated than it is. Don't forget "significant digits".... you can safely ignore all but the three leftmost digits in all of those numerical results, because they represent false precision and are certainly wrong, anywhere except inside the calculator. If you measure 28.7 volts on a 0.3 microFarad capacitor.... which probably has a 20 percent tolerance in its value anyway...... your answer is NOT going to have six or seven meaningful digits, so there is really no point at all in regurgitating them from a calculator display... because they are just wrong (remembering the tolerance in the component's value).
So the energy on a capacitor in Joules is one-half of the Capacitance in Farads, times the square of the Voltage in Volts (this is why higher voltages are better than higher capacitances, usually). The energy in Joules required to lift a mass (in kilograms) to a certain height (in meters) is just the mass times the height times the local gravity acceleration. If you avoid false precision and just use the digits you are sure of, it is easy to compare the results of the two calculations.
Ok, I'll go back to lurking quietly now.
Go for it, Gotoluc !
Thanks for your post TinselKoala.
Math is not my thing!... I have other talents.
I have a new test video uploading now but I can give you the results. So please help me and calculate how this test scores.
The Inductor weighs 115 to 120 grams. I can now lift it 1mm or more using a 0.272uf capacitor charged at 325vdc
How does this now score?
It could get better with stronger magnets but at this time this is all I have
I will post the new video demo as soon as it is ready for viewing
Thanks for your time and help
Luc
Thanks Tom,
I now see how your system works.
So you think all these bearings and gears are more efficient then a crankshaft!... I would not of think so but since you built it I'll take your word on it.
Very interesting and thanks for sharing
Luc
Hi Luc,
Here is another clever crankshaftless design
Ron
Hello Luc i did some experiments along these lines in the late 2005.
It can be as easy as sticking magnets to a modified transformer see below.
These are used is SMPS for noise suppression.
The coil form usually has 4 segments so i cut it into half leaving two segments only, and then it can move over the core due to the created free space.
The coil form you see in the picture is specifically wound the first segment is wound clockwise while the second segment is wound counter clockwise, and i think you can guess what the field looks like when it is energized on a DC current :)
According to my calculation you gained 0.17J in potential energy and used 1.185J electrically so 6.9% efficiency in that case. But this test alone is not really good on its own due to joule heating and wasted inductive energy.
Interesting Stuff Luc, did you make those laminations yourself?
Hi Luc.
I've stuck the maths in a spreadsheet for you.
Here is the Microsoft Excel version :
http://internut.webspace.virginmedia.com/Lucmotor%20Efficiency%20Test%20-%20Capacitor%20Version.xls (http://internut.webspace.virginmedia.com/Lucmotor%20Efficiency%20Test%20-%20Capacitor%20Version.xls)
Here is the Open Office version :
http://internut.webspace.virginmedia.com/Lucmotor%20Efficiency%20Test%20-%20Capacitor%20Version.ods (http://internut.webspace.virginmedia.com/Lucmotor%20Efficiency%20Test%20-%20Capacitor%20Version.ods)
All the best,
DC.
Im not sure I agree with the first statement above. The magnets applied to the top and bottom of the core are same poles facing the core. If you know how typical speaker technology, the magnets fields are not inline with the voice coil. They are "directed" to be from the inner side of the coil to the outer of the coils circumference.
So Lucs inner core with N facing in on top and on the bottom, those are repelling fields.
Those fields N will be directed outward from the central core. All along that center core are north fields outward, perpendicular to the core, looping back around to the south poles at the top and bottom.
Then you add the outer cores amd the path for thos outward N fields become more concentrated because the loop cannot expand outwards to get to the south. The outer cores help contain and direct the fields, very similar to a speaker voice coil and magnet structure.
So basically Luc, If you want to enhance your knowledge greatly with your project, study speaker design. Or more to the point, high power sub woofer design.
There are special speaker designs that will interest you also is a planar speaker design with a spiral coil on a diaphragm between 2 flat disc magnets with lots of tiny holes for the air pressure(sound wave) to pass through the front and the rear. Pioneer made a very nice set of headphones using the planar method years ago. And the disk magnets were same pole facing in, like your setup. The bass reproduction from those headphones was UnReal.
So dont think about the coil being attracted and repelled by the magnets at the top and bottom, but how the coil reacts to the fields escaping outward all along the central core.
The coils powered interaction to those fields lines perpendicular to the core is what is really going on. If you remove the core, or replace it with a non magnetic one, your output will not be better. It will be an inefficient speaker in comparison.
And, I dont believe there is any ring happening when Luc discharges the cap into the coil. It actually is taking a good amount of time for it to totally discharge from high to low then nadda. Electrolytic caps dont provide a good oscillating environment anyways.
Try it, its not very good at all.
Luc, look into the speaker and high power subwoofer design that can be found. There is not as much as you might think. Not for people like us anyways. There is a program out there that allows you to design speakers in a cad environment. You can design the cores, magnets and the rest any way you want. I have to look the name up as it was years ago that I have looked into the subject. Google books has some good old stuff also. ;]
here is a sub woofer design I had made near 15 years ago. The key features were shallow depth with longer throw, elimination of the lower spider suspension by utilizing a second inverse rubber surround of which helps to keep the cone and coil on axis.
Earthquake audio company got it out before I did with the dual rubber surround suspension eliminating the spider and reducing depth.
The magnets in the magnet housing are N facing inwards to the center and S outward to the outer core. Alumapro conquered that part of it soon after.
I just couldnt get the funding to break it out at the time. too little too late
You're welcome, and you know I'm not knocking your efforts.
Here's what I get, running the energy in the cap against the potential energy of the lift.
You've lifted 120 grams by 1 millimeter, using a cap of 0.272 microFarad charged to 325 Volts.
Putting everything in mks units we have
0.120 kg lifted 0.001 meter using a cap of 0.000000272 Farad at 325 Volts.
The energy on the cap is
(CV^2)/2 == (1/2) x (0.000000272) x (325) x (325) == just under 0.015 Joule.
The energy of the lifted mass (gravitational potential energy, the energy it takes to lift the mass against gravity) is
(mgh) == (0.120) x (9. 8) x (0.001) == just under 0.0012 Joule.
(I usually just use 10 m/s^2 for g, the local acceleration due to gravity, but 9.8 is more correct if more difficult to calculate with. I've always believed that Earth's gravity was a bit light, anyway.... it should be one Standard G of 10 meters per second per second exactly.)
Now, you are almost certainly also fighting against friction and other drag forces like eddy current drag so it will actually take somewhat more energy to lift your mass a given height, but unfortunately these will also work against you in the other direction as well and so represent (probably unrecoverable) losses to the system. So you can say that it takes _at least_ 0.0012 Joule for your system to raise your mass, possibly much more. Ten times more, due to losses? So if your system actually expends ten times the GPE, or 0.012 Joule, to lift the mass .... that is still less than the 0.015 Joule that you started with in the capacitor.
I'm bad about decimal points, though. I've checked this a couple times and I still wind up with a dismal efficiency of around 8 percent, from cap energy to mass lift.
(ETA: I think the first way I'd try, to improve the energy transfer, would be to tune for the "critically damped" condition. )
Oh and i can already hear your next question:
What if we change the magnets to North-South-South-North?
Well here it is see the vid.
Luc,
I had to do a quick test on your neat idea.
The coil is 27mm wide with about 730 turns of #25 wire for 12 ohms. Running on 12 volts is about 12 watts.
The magnets are 1/2 x 1 1/2 inches (12 x 38mm) the rod is 3 inches of 1 1/2 inch CR steel (76 x 35mm)
For the pull test I am starting with the coil to the left and pulling to the right.
With single magnets the initial pull is .75 KG's finishing at 1.02 kG's
With two magnets each end the pull is 1.2 kG's and 1.4 kG's
This is quite a contrast with a solenoid where the pull is very slight at the start and ramps up to maximum only at the end of the pull in.
On the use of a crankshaft: I would suggest timing the pulse to only cover say 80% of the stroke and so when the crank pin is immediately in front of or 180 at the other end of the stroke there is no pulse input. A small flywheel would carry the machine through the dead band.
Ron
Hi Luc,
well done new video.
In your video 5 about it,
you would need to lift this coil up 1.2 cm to get to 100 % efficiency.
All over 1.2 cm lift would be overunity at this cap size and charged voltage.
But you only lifted it maybe 1 mm ? So you need to get it lifted 10 times better,
which could be probably done by just using many more and much stronger magnets...
Magnet motors and things have to be build at least 10 times more bigger than this to get into the
overunity mode. If you build it too small, you will never have the chance to get it to overunity !
Here always the truth it: Size matters and bigger is better !
Regards, Stefan,
Hello Luc i did some experiments along these lines in the late 2005.
It can be as easy as sticking magnets to a modified transformer see below.
These are used is SMPS for noise suppression.
The coil form usually has 4 segments so i cut it into half leaving two segments only, and then it can move over the core due to the created free space.
The coil form you see in the picture is specifically wound the first segment is wound clockwise while the second segment is wound counter clockwise, and i think you can guess what the field looks like when it is energized on a DC current :)
@ gotoluc
I just recently got to watch a good chunk of your videos. I'm incredibly late, having avoided research for a few years; but Great job with all the work and videos!
That's pretty much it ATM, lol. Just wanted to say good work!
Thanks
Hi Luc,
you need at least to get 10 times better with your lift to get into the
unity range.
To get conclusive overunity you need to get about15 to 20 times better so you could also close the loop.
So you really need to get much bigger magnets and setups and also much bigger cores.
Size really matters here and this is the case with ALL magnetic setups to get good efficiency and
reduce losses...
Regards, Stefan.
Luc asked me to update the spreadsheets to use grammes instead of kg, millimetres instead of metres etc.
New versions are same names so same download links used.
Here is the Microsoft Excel version :
http://internut.webspace.virginmedia.com/Lucmotor%20Efficiency%20Test%20-%20Capacitor%20Version.xls (http://internut.webspace.virginmedia.com/Lucmotor%20Efficiency%20Test%20-%20Capacitor%20Version.xls)
Here is the Open Office version :
http://internut.webspace.virginmedia.com/Lucmotor%20Efficiency%20Test%20-%20Capacitor%20Version.ods (http://internut.webspace.virginmedia.com/Lucmotor%20Efficiency%20Test%20-%20Capacitor%20Version.ods)
*** edit add ***
I think they may have downloaded as read-only before but this should be fixed now.
Stupid UNIX file permissions ;+}
*************
All the best,
DC.
@gotoluc thanks mate luvyawork. Just had a play with a Rodin coil and your idea.
http://www.youtube.com/watch?v=Q3Q_1c11wRQ (http://www.youtube.com/watch?v=Q3Q_1c11wRQ)
Edit: link fixed
Thanks for your interest and trying it out on your Rodin Coil mate ;)No worries mate here's the latest http://www.youtube.com/watch?v=f2yhfyjLqBs wondering if the way it sails past the middle mags is significant
She's flying high now
Luc
No worries mate here's the latest http://www.youtube.com/watch?v=f2yhfyjLqBs (http://www.youtube.com/watch?v=f2yhfyjLqBs) wondering if the way it sails past the middle mags is significant
Hi Jimboot,
well done.
What is the weight of your coil and the power input ( voltage x current)
Many thanks.
P.S. Maybe you can also try it on a charged capacitor ?
Do you still have more magnets that you can use to put on the stator side ?
@gotoluc:
Yes, I meat 12 mm = 1.2 cm
so you need to do at least 10 times better than now.
Regards, Stefan.
I just told you in my previous post even showed it in two video's so which part did you not get?
When the magnets are placed NORTH-SOUTH NORTH SOUTH there is no induction because the core is exactly in the middle which is the DEAD ZONE.
This translates into NORTH-CORE-SOUTH where core is in the dead zone.
When the magnets are placed NORTH SOUTH SOUTH NORTH there is induction because there are many field lines cutting the coil and that's what it is all about here the number of field lines cutting the coil.
No worries mate here's the latest http://www.youtube.com/watch?v=f2yhfyjLqBs (http://www.youtube.com/watch?v=f2yhfyjLqBs) wondering if the way it sails past the middle mags is significant
Luc asked me to update the spreadsheets to use grammes instead of kg, millimetres instead of metres etc.
New versions are same names so same download links used.
Here is the Microsoft Excel version :
http://internut.webspace.virginmedia.com/Lucmotor%20Efficiency%20Test%20-%20Capacitor%20Version.xls (http://internut.webspace.virginmedia.com/Lucmotor%20Efficiency%20Test%20-%20Capacitor%20Version.xls)
Here is the Open Office version :
http://internut.webspace.virginmedia.com/Lucmotor%20Efficiency%20Test%20-%20Capacitor%20Version.ods (http://internut.webspace.virginmedia.com/Lucmotor%20Efficiency%20Test%20-%20Capacitor%20Version.ods)
*** edit add ***
I think they may have downloaded as read-only before but this should be fixed now.
Stupid UNIX file permissions ;+}
*************
All the best,
DC.
Excuse me but looks to me that to know the energy consumed need to have 2 voltage while a capacitor: the one before the test and the one after; am I right?
Yes, you are correct! but in this test the capacitor was connected long enough to discharge completely.lol, ok V2 = 0 Thanks for having confirmed. Cheer. :)
Thanks for your interest
Luc
if you used two cube magnets on top of each other to double their length this way you would also achieve a double distance between the coil and the side bars may show the effect if any.
I assume that the improvement in coil inductance (when you use a side bar to bridge the outer like poles, say South poles) comes from removing the "fringe" South pole flux from the coil area and direct them into the outside bar. I say this because seemingly you indeed close the magnetic circuit by using an iron side-bar but the magnets to be bridged being like poles they cannot close to any other pole in the bar because at the other end of the bar the same like pole enters the bar from the other end magnet.
So I assume that the increase in pulling force does come from the increased coil inductance and this inductance increase comes about by applying more flux from the permanent magnets, shifting the core's working point towards a higher permeability area of the B/H curve. Of course as you also mentioned in the video, there is an upper limit in increasing the flux by using stronger and stronger magnets: this upper limit is set by core saturation.
The possible ultimate core and magnet shape for your setup would be to use a cylinder core (say a ferrit rod) in the middle (on which also a cylinder coil would slide) and use a ring magnet at both core ends with its ID matched to core's OD and then use a soft iron tube (with certain wall thickness) outside instead of the bars to connect the ring magnets' outer like poles. No stray flux would escape from such setup.
Hi gotoluc!
Thanks for sharing your very valuable experiments.
I see very brillant your use of bacefm as an alternative drive synchronized with the moving of the coil.
You can use imho this as a very interesting piston motor; a "motor" indeed. But true you can drive a generator and try to find the limit of efficiency of the system.
But before to do so, could you scientifically and rigorously measuring the power output of it? Could you add a pouly with a baril, a wire, a mass, and measure the time your motor will pull the mass up to 1 meter? This giving us the effective mechanical output power to compare to the electrical power. A graphic of the evolution of the mechanical power in function of the electrical power input would be a very must; imho.
Cheer, Khwartz.
If you increase the air gap (distance) between the outer surface of the coil and outer core, the result will be a drop in pull force, even though you double the magnets. As the outer cores and coil can do as much work as the inner core and coil. So the closer to zero air gap will give the best results.
I don't know if you follow me. I've been trying to tell about this before but people are not picking up on it.
I don't know about that. How can a cylinder work better the what I just demonstrated?... also, doesn't a ferrite loose inductance when you apply magnet flux?
Hi Luc,
Okay, I accept that and although it is irrevelant now, I think you have tested how the coil inductance changes when you increase the air gap and found the inductance increases in a lesser extent compared to a previously smaller air gap? I mean you measure 867 mH for the coil with the single magnets at the ends (when there is no bridging bar between the outer like poles) and then you apply the bridge and get 942 mH.
Then you double the air gap for the bridge by using double magnets (i.e. doubling their thickness) and you measure also an increase in inductance with respect to the no bridge 867 mH value but now this measured value would not reach the 942 mH value received for the previous smaller air gap case, so this is what I figured.
Because you measured the pulling forces for the different cases, obviously the size of the air gap has a much stronger influence on the force than the increase in inductance does.
Well I follow you but this time I did not revise your years ago tests on this setup, sorry. Yes, a coil will also have a Bloch wall in its middle part just like a permanent magnet does, no doubt on that.
Well, a cylinder coil in itself would not work better than your rectangular coil (assuming equal inductance and number of turns) but I meant also using ring magnets instead of the cube magnets. I simply thought of occupying all the space around a cylinder coil by having covered it all around by a ferromagnetic tube to replace the bars (in fact you would have an infinite number of "bars" from the generatrix of the cylinder tube), this way the opposite poles could be utilized the most from the coil.
Now I looked for radially magnetized ring magnets (i.e. their ID is say North all inside and their OD is South all around outside) which would be needed for my suggestion but unfortunately such is not manufactured... can only assembled from arc magnets like is shown here: http://www.supermagnetman.net/product_info.php?products_id=380 (http://www.supermagnetman.net/product_info.php?products_id=380) or also could be assembled from say smaller cube magnets also wrapped up in an enclosure. Understand now what "cylinder" setup I mean? Of course such setup would cost more to assemble than your present one and I also prefer initial testings with a cheap solution, I just indicated an 'ultimate' setup in this respect.
Regarding a ferrite loosing inductance when influenced by (strong) outside magnets, yes it normally does, mainly when cross section area is small versus the flux strength, I mentioned ferrite as a first straigthforward thought for cylinder shaped cores...
rgds, Gyula
Thanks for your positive comment.You're wellcome! Luc.
The coil weight is 100 grams. I can place it vertically (against gravity), power it and calculate the amount of mm it moved up in x amount of time using x amount of watt.
Would that work?
Or much easier for me is a capacitive (joule) discharge and tell you how many mm it moved up.
Luc
You're wellcome! Luc.
I have thought this possibility to put vertical the axis and work against grativity but if it very easy to realize, it is not so easy to have an enough accurate time measurement on so short period of time.
The formula any way is: P[W] = M[kg] * g[m/s2] * L[m] / T.
Cheer.
So then, how about a capacitive discharge against gravity and just measure how many mm the 100 grams traveled up.So far, at least mathematically it is in both cases energy measurement. So it could worth the try in a first stage; i.m.o.
Luc
...
1) You're limited by the strength of the PMags. The best neodymium is less than half as strong as saturated electric steel (.5T vs 1.6T). It looks like an optimised plain steel solenoid would outperform your design, when running at full power. (Whereas your design is better at low power - below saturation). Also, I think I'm right in saying that you can't saturate iron with any permanent magnet - none are strong enough.
2) Your piston is effectively 'pushing' against the H-field of the core - which is extremely weak compared to the B-field of iron. There are ways round this - but you'd have to change the basic design.
...
Your piston is effectively 'pushing' against the H-field of the core - which is extremely weak compared to the B-field of iron. There are ways round this - but you'd have to change the basic design.
Hope this helps.
Tim
Hi gyulasun - you're right, neos are according to Wikipedia >1 Tesla. I've been using the magnet calculator here: http://www.kjmagnetics.com/calculator.asp (http://www.kjmagnetics.com/calculator.asp) - which says an N52 has a field of 6619 Gauss - i.e. about .66T. I'm not sure how to reconcile that difference TBH so I was working on the assumption that the folks who were making the magnets were probably right - and to assume the lower field.
Electric Steel - http://en.wikipedia.org/wiki/Electrical_steel (http://en.wikipedia.org/wiki/Electrical_steel)
I've calculated power at 1000RPM for each of the above. Obviously my calcs could be totally wrong, but I'd be happy to go thru them with you guys... One problem is that I don't believe how little power it takes to fully saturate steel:
http://en.wikipedia.org/wiki/Saturation_(magnetic (http://en.wikipedia.org/wiki/Saturation_(magnetic))
...says 100 Amp Turns per inch is more than enough (and diameter doesnt seem to matter!). So for my 8" coil below - I should only need 800 Amp Turns, not 8,000. In which case the coil can be made much thinner... I've been trying to err on the side of caution.
Hi Luc,
was very excited looking at your videos - great work!
Just want to point out that in order to generate 1 w of energy you should be able to lift weight of 10 kg by 1 centimeter at 1 second,
or 1 kg by 1 cm 10 times per second, or 250 grams at 1 cm 40 times per second!
But generally speaking, in order to achieve this you need a) - much bigger magnets, b) - much smaller travel of the coil.
According to my calculations the magnet ( the source of the energy) should have the pull of about 400 kg, and the travel about
.5 cm. With your setup this doesn't appear to be practical since magnets will be fighting each other (IMHO).
Hello gotoluc, thanks for continuing your experiments :D
Very interesting to know it is not linear :)
When you say: "pulls 520g for 1.2 watts input. However, if I drop the current it can pull 200g with .3 watts input", it is just ratio of force against electrical power; could tel us in how many th of second it lifts them in each case?
Cheer.
Hi Luc,
this is a big step in the right direction.
If it takes 1 second, than the mechanical equivalent of .3 w X sec (.3 joyles) equals
3 kG x cm, and the efficiency is still below 10 %.
Perhaps you may add few more magnets, or increase number of strokes per second or both?
Hi telecom,Great,
I think it would be more like 1/4 of a second to travel 1 cm in the modified version.
All I did to double the pull force (from last video) is, if you look towards the end of my video I added 2 small pieces of magnets on the ends of the center core and it helped a little, so basically I used this same model but added 1 inch N52 cube magnets on those ends and got 520g of pull instead of the 250g. So as you can see it's all about the strength of the magnets that will make a stronger pull force. I'm sure a design with more coil and core surface area will also boost things.
However, one important thing is the generator effect. The stronger the magnets the stronger the fight with the generator effect is (Lenz Law) This is what I'm more interested in finding, is a way we can use the Lenz effect to our advantage.
Luc
That's the problem! I don't have an accurate way of calculating such a fast and small amount of time for a movement.What about a gearing counter or what ever kind, and you let it run for 1 minute? then you could make the ratio and have the average duration with more precision; what do you think?
All I can do is do that ca be accurate is a test of Joule energy (capacitive discharge) and measure how high it travels.
Luc
Looks good ideas, Luc :) and nice project, your home boat all seasons ;) you just have made a little confusion in your initial post, but significant while been an electrician mysel: it suppose it was not "48 volts 100 amp/hours lithium ion battery" but "100 amp*hours" ;)
Cheer..
???
One amp-hour is not one amp per hour! So you cannot write A/h for amp-hour.
???Yes, but that is not a "rate". It is a quantity. The Ampere is a rate, so if you go at a rate for a time, you have what? A quantity. Go 1 km per hour for one hour. How far have you gone? One km. 1 km/hr x 1 hr = 1 km.
One amp hour is a steady current of one ampere flowing for one hour,-is it not?
No-all clear
Now are we completely confused?
Here is the second part of the demo video with the Signal Generator attached to it.
https://www.youtube.com/watch?v=oDXq8wkhV2c (https://www.youtube.com/watch?v=oDXq8wkhV2c)
Luc
Yes, but that is not a "rate". It is a quantity. The Ampere is a rate, so if you go at a rate for a time, you have what? A quantity. Go 1 km per hour for one hour. How far have you gone? One km. 1 km/hr x 1 hr = 1 km.So we agree :) just Luc was speaking of the quantity of electrical charges provided by the battery then I only correct him on this point; great you went further :)
Just as "per" denotes division, "for" denotes multiplication.
Consider what we mean by "three amps per hour". This is written 3A/H. How I got that number is I started with zero amps and I turned my variac up slowly and at the end of an hour I was up to three amps, and at the end of two hours I was up to six amps, and at the end of three hours I was up to nine amps. I turned the current up at a RATE of 3 A/H, or three amps per hour. If you want to know my current at, say, 90 minutes, you multiply the rate by the time: 3 a/h x 1.5 h = 4.5 A. This use is in effect the "rate of change of the rate" or the first derivative, the slope, of the graph of the current vs. time.
One Amp-Hour is a _quantity_. To get there mathematically you _multiply_ the number of amps by the number of hours. So if I have a system that produces one amp-hour at 12 volts and I run it for one hour, I have sent (one amp-hour) x (12 volts) or 12 Watt-hours past my measuring point, for that hour. A quantity, distributed over the hour time. And for that one hour, then, I am dissipating the energy at a _rate_ of 12 Watt-hours per hour. 12 W-h / 1 h = 12 Watts... a _rate_, the power dissipation, the rate of energy dissipation. Power in Watts = Joules of energy per second -- a rate.
Now are we completely confused?
Hi Luc! :) How is going your reactive genertor ? :)
Hi Khwartz,Lol, no problem Luc, just do like you feel better to do :) and sorry if you perceived me too much academic ^_^
I have next to no schooling. The things I can do are mostly self taught. The experiments I do are my way of trying to learn more in hopes that one day I'll know enough to build a device to be energy independent.
I don't know why but learning all those math equations does not interest me.
I've always been hands on and not into reading patents and theorizing about this or that.
If I find something interesting from the stuff I build that's not commonly done, I share it and hopefully the more knowledgeable ones will find a use for it, suggest different tests or explain in a laymen way why it's not useful.
Anything thrown at me from an academic level is like water on a ducks back.
Luc
Hi gotoluc,
Your last videos outperforming a speaker by almost 10 times is great!!
In this other video at instant 14:19 sec. you state that the current generated while moving (concentrating in its generator mode) has the same polarity as the current that you send to excite the coil.
https://www.youtube.com/watch?v=GYoXmDvFqQs#t=14m19s (https://www.youtube.com/watch?v=GYoXmDvFqQs#t=14m19s)
In case so, this motor would not have back emf, and the generator mode of this device will create a current which will reinforce you input current. Could you elaborate a bit deeperow you tested this feature? I am really interested because of the big consecuences of this fact !!!
I think that the simplest way is to testing it adding a second small coil outside the first coil. Then with a scope you can compare the applied voltage in the first coil and the generated voltage in the second coil. How did you test it?
Thanks and congratulation for your great work!!!
To everyone,Hi Luc
since this topic has been re-opened I may as well share a better performing version of my Mostly Magnet Motor which I compared its pulling force to an off the shelf speaker.
So here is the link to the unlisted video if anyone is interested to see it.
https://www.youtube.com/watch?v=qa1dO8qWPQU
Luc
Hi Luc
very excited at your progress.
Perhaps you can make a jig where you can lift up a certain weight so many times per second
= you may be are getting very close to OU, if not there already!
For example, 1 W = .1 kg X m/sec
In your case, for 0.4 W, you will need to lift 500 g by .08 m (8 cm) once a second,
or 500g by 2 cm 4 times per second...
...
I can confirm the current created by the coils generator effect goes in the same direction as the input current used to move the coil. How I tested this is simply connect a DC volt meter the the coil and move it by hand in one direction and note the coils polarities. Then, connect the coil to a DC voltage source with the same polarities to see if the coil moves in the same direction as moving it by hand.
It does go in the same direction.
Somehow I don't think this means it has no back emf since I did the same test with a DC permanent magnet motor and the results were the same.
Where this design may have an advantage compared to a standard motor is when the coil reaches the end of the stroke and is switched off. The coils Flyback goes in the same direction as the next input pulse would needs to go to move the coil in the opposite direction. So it seems the flyback is working in a complementary way compared to what may be happening in a DC motor.
....
I just have to build the thing. However, the main problem is I don't have enough steel lamination core material to build it. So I was thinking to use solid steel just for testing and proof of concepts and wondering if the eddy currents using solid steel would make such a big difference if the coil is only on for a second and traveling less the 10 cm?
....
Hi telecom,
last night I was actually thinking that maybe there could be a possibility of OU since I have a newer tandem design that should double what I already have done. So 1 Kg pull with less then 0.5 W should be possible.
I just have to build the thing. However, the main problem is I don't have enough steel lamination core material to build it. So I was thinking to use solid steel just for testing and proof of concepts and wondering if the eddy currents using solid steel would make such a big difference if the coil is only on for a second and traveling less the 10 cm?
Anybody care to comment on that?
Luc
Hi hanon,Luc
back emf could become quite a topic of conversation.
I can confirm the current created by the coils generator effect goes in the same direction as the input current used to move the coil.
How I tested this is simply connect a DC volt meter the the coil and move it by hand in one direction and note the coils polarities. Then, connect the coil to a DC voltage source with the same polarities to see if the coil moves in the same direction as moving it by hand.
It does go in the same direction.
Somehow I don't think this means it has no back emf since I did the same test with a DC permanent magnet motor and the results were the same.
Where this design may have an advantage compared to a standard motor is when the coil reaches the end of the stroke and is switched off. The coils Flyback goes in the same direction as the next input pulse would needs to go to move the coil in the opposite direction. So it seems the flyback is working in a complementary way compared to what may be happening in a DC motor.
This is my own observation and would need to be confirmed by the more qualified individuals.
Hope this helps to answer your questions.
If anyone else would like to add or correct anything please feel free to do so, as this would only help many to better understand the working of a motor or an Inductor in motion.
Thanks for your interest and participation
Luc
Luc
The current flow from the flyback from an inductor(coil) remains in the same direction,but the voltage polarity invert's. The answer as to how to capture the flyback and use it again is right there in Bendini's SSG pulse motor circuit.Just swap the charge battery out for a good size cap,and you have your power for your reverse pulse right there in the charge cap.You could actually use a double acting relay to achieve the switching-infact i have the perfect device to achieve this switching and collecting of power for your setup.I used it in the little moter bellow.You will also notice i use the term backEMF where i should have said flyback collection-the days befor i knew the difference between the two. So insted of using the flyback to run the LED,you can collect that in a cap,and use it to pulse your reversing coil.
https://www.youtube.com/watch?v=Z4VJG8-9izQ&list=UUsLiBC2cL5GsZGLcj2rm-4w (https://www.youtube.com/watch?v=Z4VJG8-9izQ&list=UUsLiBC2cL5GsZGLcj2rm-4w)
Hi Brad,Hi Gyulasun
If I got it right, Luc drives his moving coil from the output of a full H bridge (to have current hence pole polarity change in the right moments as the coil reaches the ends of its course) so this involves ground independent coil ends in his present setup. So to catch the flyback pulse across such coil you have to use a circuit to handle that ground-independently.
To solve this, perhaps the simplest thing would be to use a bifilar coil to get the exact "image" of the spike in the coupled coil. I mean one coil in the bifilar is the normal working coil and the other (identical) coil in the bifilar will be a ground independent output for the flyback pulse. This solution may also help to add this captured energy to the input voltage source perhaps more easily than with other circuit variants where there is no such output coupling coil.
Gyula
PS Sorry I have had no time to watch your above video so if you have a double coil in it, then ... I :'(
I would guess you would be referring to the one I connected to my H-Bridge?... if so, sorry but no.What i will do first Luc,is make a standard solenoid setup,using a PM insted of a steel plunger(pistion). We can then compare results of lift/distance,and power consumption between the two setup's.This will let us know if your on the right track,and how much better your setup is than a standard setup.
How about just testing the basic design to see if you feel it's worth more investment of time. I'm sure when the time comes Gyula can help with the schematics for more advanced switching.
Luc
Hi Luc,This is an interesting test,but what are the ruel's based around the test.?
I keep thinking about this aspect of the testing:
For example, if you do the lifting test once per second,it
consists of on/of duty cycle, which means that you apply power only
half of the time of the cycle, which will be 0.2 W, instead of .4 W per second, to perform the same work.
Which means that for the present setup of 500g pull, you only need to lift
it twice per second by 2 cm to reach OU. If you do it more often - you are in a surplus territory.
This is an interesting test,but what are the ruel's based around the test.?Hi Tinman,
Would this be allowed-we have a spring sitting under the 500g weight.If Luc can get the weight with spring into resonance,and the 500g weight is lifted that 2cm's twice a minute,dose that still count?-are we aloud to use the spring. If a spring is not an energy source,then i dont see why not. this would then become like a kid on a trampoline.
@ telecom, when I came up with this design 5 years ago I ordered many large 2" square by 1" thick N52 magnets to build a super powerful version of it.
I never did build it because most said it wouldn't give OU and then the topic died.
I still have at least 10 or more of these super magnets in the original delivery box in storage . Today they are worth a fortune compared to when I purchased them.
I guess it's time to open the box and build the thing to see the real results!
I was thinking to build one that's at least 8 inches wide by using 4 of these 2" wide by 1 " thick magnets next to each other. That way most of the coils outer magnetic field is used. Only 1" on each sides where I attache the guides would be lost compared to my latest design where only half of the outer field is used.
Luc
Hi Tinman,If a spring is not an energy source,but only an energy storage device,how will it make the output higher?
I really don't know the answer:
On one hand, the spring will distort the actual output of the device, will make it higher;
If a spring is not an energy source,but only an energy storage device,how will it make the output higher?When weight is being lifted, we increase its potential energy.
Think of the weight as an inductor,and the spring as a capacitor. All you need to do then is hit the right frequency of the tank circuit to gain resonance. Once resonance is achieved,we will get a large amplitude-in this case,the 1/2kg weight being lifted that 2cm twice per minute
To be within test guidlines,the 2cm lift would have to start from the point at which the spring is not compressed-just off the surface from which it will contact as the weight drops.If the spring is to compress 1cm,then the total travel of Luc's slide(coils travel) on his device would need to be 3cm.
Why the spring would allow the weight to be lifted more efficiently than if it wasnt there?.
The answer is simple. If there were no spring,and the weight was just allowed to hit the surface as it fell to a resting state,then energy is disipated from the system as noise/vibration. The spring removes this loss,and stores that energy that would have been disipated as noise/vibration,and returns it back to the system. So the spring wont actually distort the output,but it will make it higher due to less loss in the system.
When weight is being lifted, we increase its potential energy.No,the kinetic energy isnt being absorbed by the string,it is being transformed into vibrational energy via the string through the framework of the device,and then to the bench the device is mounted on,and finally to ground(earth). The spring stores the kinetic energy that would be normally transformed into vibrations/vibrational energy/sound,and returns it back to the system apon spring decompression. Energy is never absorbed,it's stored or transformed.
When it drops, the potential energy becomes kinetic, which is being absorbed
at the end by the holding string(w/o the spring).
When the spring is underneath, it absorbs the kinetic energy into the potential energy of the spring when it compresses, and releases it on the way up, which helps to pull
the weight up. This will distort the actual performance of the device.
No,the kinetic energy isnt being absorbed by the string,it is being transformed into vibrational energy via the string through the framework of the device,and then to the bench the device is mounted on,and finally to ground(earth). The spring stores the kinetic energy that would be normally transformed into vibrations/vibrational energy/sound,and returns it back to the system apon spring decompression. Energy is never absorbed,it's stored or transformed.
If you have a brick and drop it on top of the spring, the spring will first compressThat is correct-the spring stores the energy,and then returns it back to the system.This is in relation to this thread in that a system be designed and understood,so as Luc can achieve maximum efficiency from his DUT.A correct understanding as to how your system opperates, where losses may occur, what those losses are,and how to remove those losses,is the best way to achieve the results you are after,and make your DUT the most efficient it can be. Although the spring itself will have losses (vibrational/noise),it will increase the efficiency of the DUT,as those losses in the spring are not as much as they would be without it.
to a certain length, after that it will straighten up and propel the brick to a height
which is smaller than the initial height due to the losses...
What this discussion has to do with the topic of the thread?
That is correct-the spring stores the energy,and then returns it back to the system.This is in relation to this thread in that a system be designed and understood,so as Luc can achieve maximum efficiency from his DUT.A correct understanding as to how your system opperates, where losses may occur, what those losses are,and how to remove those losses,is the best way to achieve the results you are after,and make your DUT the most efficient it can be. Although the spring itself will have losses (vibrational/noise),it will increase the efficiency of the DUT,as those losses in the spring are not as much as they would be without it.It may increase the efficiency of this particular setup, but may be irrelevant
All in all,we agree that the spring will increase the efficiency of the DUT.
It may increase the efficiency of this particular setup, but may be irrelevantIndeed that is correct. But think of this-if Luc can raise that 1/2kg weight 2cm 2 times per second,he has a unity device.Thats a 100% efficient motor,and thats something that dosnt exist today. Every device i have tested so far,has no where that pull force that Luc shows with his device. I would hope that the high end builders here would try and build something that shows the efficiency of Luc's DUT-in fact,i make it a challenge to all here.I would spend more time on it,but im working on my Inertia drive at the same time,which i want to get finished.
to a real life application of the device.
In real life it will be connected to a generator with a resistance in both parts of a stroke - forward and return. There will be no free fall like in a setup.
Here is the pull force(or torque) from a standard small dc motor.
https://www.youtube.com/watch?v=pDLLphaRC-k&list=UUsLiBC2cL5GsZGLcj2rm-4w (https://www.youtube.com/watch?v=pDLLphaRC-k&list=UUsLiBC2cL5GsZGLcj2rm-4w)
Hi Brad,That sounds about right Luc. The less current your coil draw's,the higher the backEMF voltage will be in your coil-and of course,the higher the backEMF voltage,the lower the current draw.This is the very same in my little DC motor,and in most electric motors i know of.The more load you place on the motor,the less the BackEMF voltage in the inductors will be,so up go's the current to try and maintain the forward voltage within the inductor.
I was giving your test more thought and remembered that the Watts consumed for a specific amount of grams pulled is not linear, well not in my device anyways.
Let me give you an example, if I can pull 500g with 0.43W and I reduce the input to pull 250g the watts are not divided in half, it's much less than that!... more like 1/3 or less. Unfortunately I don't have my latest test device with me but if I pulled 130g like you did it would probably use around 0.08W to do it.
If you can re-test your motor and raise the voltage till you achieve 500g of pull and calculate the watts used you may see what I mean.
Luc
So here is that second test Luc,and yes-more than double the power to achieve twice the pull force.
250g's-.888 volts@2.02 amps= 1.793 watts
500g's- 1.45 volts@3.18 amps= 4.8 watts
https://www.youtube.com/watch?v=VSvp5PtffUo&list=UUsLiBC2cL5GsZGLcj2rm-4w (https://www.youtube.com/watch?v=VSvp5PtffUo&list=UUsLiBC2cL5GsZGLcj2rm-4w)
If a spring is not an energy source,but only an energy storage device,how will it make the output higher?Very Good and Accurate analysis! :)
Think of the weight as an inductor,and the spring as a capacitor. All you need to do then is hit the right frequency of the tank circuit to gain resonance. Once resonance is achieved,we will get a large amplitude-in this case,the 1/2kg weight being lifted that 2cm twice per minute
To be within test guidlines,the 2cm lift would have to start from the point at which the spring is not compressed-just off the surface from which it will contact as the weight drops.If the spring is to compress 1cm,then the total travel of Luc's slide(coils travel) on his device would need to be 3cm.
Why the spring would allow the weight to be lifted more efficiently than if it wasnt there?.
The answer is simple. If there were no spring,and the weight was just allowed to hit the surface as it fell to a resting state,then energy is disipated from the system as noise/vibration. The spring removes this loss,and stores that energy that would have been disipated as noise/vibration,and returns it back to the system. So the spring wont actually distort the output,but it will make it higher due to less loss in the system.
That is correct-the spring stores the energy,and then returns it back to the system.This is in relation to this thread in that a system be designed and understood,so as Luc can achieve maximum efficiency from his DUT.A correct understanding as to how your system opperates, where losses may occur, what those losses are,and how to remove those losses,is the best way to achieve the results you are after,and make your DUT the most efficient it can be. Although the spring itself will have losses (vibrational/noise),it will increase the efficiency of the DUT,as those losses in the spring are not as much as they would be without it.Very Correct!
All in all,we agree that the spring will increase the efficiency of the DUT.
That sounds about right Luc. The less current your coil draw's,the higher the backEMF voltage will be in your coil-and of course,the higher the backEMF voltage,the lower the current draw.This is the very same in my little DC motor,and in most electric motors i know of.The more load you place on the motor,the less the BackEMF voltage in the inductors will be,so up go's the current to try and maintain the forward voltage within the inductor.Please, dear Tin, don't use "torque" when you speak about "tangential force". The first is in N.m ("Newton.meter" or "kilogram.force.meter" if you want) and the second in N only ("Newton" or "kilogram.force"). This is confusing and may lead to erroneously concepts and ways in experiments and calculations them :/
I dont know if that little motor would have 500g's of torque in it,but we can make some smoke and try lol.
Will go do it now.
Would be good to have some test perameters right about now.Dear TinMan, I think you're on the Very Right path in term of Correct Methodology for o.u. checking :D Caps are Very Great to provide accurate maths on energy delivering or input :)
So we could do it like this.
It takes 1 joule of energy to lift 1kg of mass 10cm high.
If we are setting the weight to 1/2kg(500g's),with a lift hight of 2cm,then we need 100millijoules.
A cap that has say 5000uf would then need 6.325volts in it to give us our 100mJ.Or if we have a 10000 uf cap,we would need only 4.47 volts in it to give us our 100mJ.
So Luc,if you can lift that 500g weight by 2cm using a 10000uf cap with only 4.47 volts in it-you have hit unity. If you need only 4.46 volts in the cap to do it,you are OU.
Please, dear Tin, don't use "torque" when you speak about "tangential force". The first is in N.m ("Newton.meter" or "kilogram.force.meter" if you want) and the second in N only ("Newton" or "kilogram.force"). This is confusing and may lead to erroneously concepts and ways in experiments and calculations them :/The torque of an electric motor is measured exactly as i did it.We are measuring static rotor torque.Torque is a measure of rotational or "twisting" force. I can dig up my torque dial if you like,and show you that the results will be exactly the same for the P/in we used.
Would be good to have some test perameters right about now.
So we could do it like this.
It takes 1 joule of energy to lift 1kg of mass 10cm high.
If we are setting the weight to 1/2kg(500g's),with a lift hight of 2cm,then we need 100millijoules.
A cap that has say 5000uf would then need 6.325volts in it to give us our 100mJ.Or if we have a 10000 uf cap,we would need only 4.47 volts in it to give us our 100mJ.
So Luc,if you can lift that 500g weight by 2cm using a 10000uf cap with only 4.47 volts in it-you have hit unity. If you need only 4.46 volts in the cap to do it,you are OU.
Dear TinMan, I think you're on the Very Right path in term of Correct Methodology for o.u. checking :D Caps are Very Great to provide accurate maths on energy delivering or input :)How can be they so great if voltage varies and there is always a small charge remaining?
How can be they so great if voltage varies and there is always a small charge remaining?A cap with a set value(uf) and set voltage,will have X amount of energy(joules).After running the test,IF there is any voltage left in the cap,you simply calculate the joules of energy left in the cap,and subtract that from the amount you started with in the cap befor running the experiment.This is a very accurate way of seeing how much energy you used in the test.
Hmmm.... what if I take a one nF cap and charge it to 15 kV and discharge that into a water arc chamber, full of water, with the bottom of the 500g "piston" in contact with the water? Or rather, since the arc will cut off before the cap is completely discharged, let's say I start with 17 kV and end with 2 kV still on the cap. If I can get 2 cm rise (or 2.2 cm) out of the 500g piston, will I have OU?There is no reason i can see that this wouldnt be a valid test TK. Why dose this example you give,give reason for caution? Do you believe that your 15kv in your 1nF cap is enough to cause that greater explosion,that it will raise that 500g piston 2.2cm?. i doubt that very much. Infact,it is very hard to get an ark in water,and i dont think 15kv at 1nF would do anything at all.I have tried to get a spark plug to work under water,and even with that small spark gap and 25kv ,i couldnt get it to spark.
You have to be very careful with this kind of reasoning. It's almost the same thing that caused Peter Graneau to waste years of his life and millions of dollars of grant funding, chasing an imaginary unicorn through a dense forest of experimental data.
A cap with a set value(uf) and set voltage,will have X amount of energy(joules).After running the test,IF there is any voltage left in the cap,you simply calculate the joules of energy left in the cap,and subtract that from the amount you started with in the cap befor running the experiment.This is a very accurate way of seeing how much energy you used in the test.
Hi everyone,
I'm working on building a Super Tandem version of the Mostly Magnet Motor
Here is a video update of the building process
Link to video: https://www.youtube.com/watch?v=XhDnqw_le88
Luc
Where will you get a cap with an exact value - they all within at least 10 %Only as per the spec's on the cap. A decent DMM will give you the correct value,and this is the value you use. An AC cap can be used in a DC situation aswell,and this way the flyback could also be collected.
Hi everyone,This is a very large setup you have got going here Luc.
I'm working on building a Super Tandem version of the Mostly Magnet Motor
Here is a video update of the building process
Link to video: https://www.youtube.com/watch?v=XhDnqw_le88
Luc
Here is the video on winding the first coil and what it looks like when finished.Great craftsmanship!
I checked the DC resistance of each bifilar stands and they are 1.3 Ohm each. The coil winding program had calculated 2.7 Ohms as a single coil, so we are right on track. I don't have my Inductance meter with me but will have it on Thursday.
Link to video: https://www.youtube.com/watch?v=ULds78-fDoQ
Luc
..... The CA is just like resin, its not real strong unless its with fiber, what ever it may be.
Mags
The torque of an electric motor is measured exactly as i did it.We are measuring static rotor torque.Torque is a measure of rotational or "twisting" force. I can dig up my torque dial if you like,and show you that the results will be exactly the same for the P/in we used.Dear Brad, there are 3 distincts concepts here:
How can be they so great if voltage varies and there is always a small charge remaining?Hi telecom, needn't we just the difference of voltage between the start of the experiment and the end? as per:
A cap with a set value(uf) and set voltage,will have X amount of energy(joules).After running the test,IF there is any voltage left in the cap,you simply calculate the joules of energy left in the cap,and subtract that from the amount you started with in the cap befor running the experiment.This is a very accurate way of seeing how much energy you used in the test.;)
Where will you get a cap with an exact value - they all within at least 10 %I understand you now: as you had the false datum that caps can be only 10 % maximum of precision, sure if the overunity gain is below 1.1 we won't see it. But what if the COP is already > 1.1 ? If you obtain 1.3 for example, haven't you proved your o.u.? ;)
Hi everyone,Very Clear explanations, Luc, thanks for the video :)
I'm working on building a Super Tandem version of the Mostly Magnet Motor
Here is a video update of the building process
Link to video: https://www.youtube.com/watch?v=XhDnqw_le88
Luc
Here is the video on winding the first coil and what it looks like when finished.Agree with telecom: Great craftsmanship! Luc :)
I checked the DC resistance of each bifilar stands and they are 1.3 Ohm each. The coil winding program had calculated 2.7 Ohms as a single coil, so we are right on track. I don't have my Inductance meter with me but will have it on Thursday.
Link to video: https://www.youtube.com/watch?v=ULds78-fDoQ
Luc
Dear Brad, there are 3 distincts concepts here:Didier
1. "Shaft torque",
2. "Rotor torque",
3. "Moment", or just "torque",
and as more we differenciate the concepts more we allow us to be accurate and so not leading to misinterpretations, so wastes of time, good to clear up the differences in between. So,
• "Shaft torque" is about twisting the axis,
• "Rotor torque" is about the reaction of the fluid against the rotation of the rotor (like for helicopters rotors),
• "Moment" is "the rotational force times meter" reduced to the center of rotation when there is rotation of course (which is on the axis necessarily). This moment corresponds to "torque" when nothing is specified.
So yes, indeed, your way to measure the "torque" (definition 3) is absolutely correct but not your way to interpret it. YOU DO NOT MEASURE "TORQUE" BY YOUR READING ON YOUR PONY-BRAKE, YOU ONLY MEASURE THE "ROTATIONAL FORCE" (or "TANGENTIAL FORCE") AT A SPECIFIC RADIUS.
To get the "torque", or "moment", YOU DO NEED THEN TO MULTIPLY IT BY THE LENGH IN METERS. (To have directly the reading on the pony-brake you would need to set a 1 meter radius disk, not just been 1:1 with the radius of the rotor ^_^).
For more details about these concepts see for example:
http://books.google.fr/books?id=FiEapaNgjLcC&pg=PA423&lpg=PA423&dq=%22rotor+torque%22+definition&source=bl&ots=NIXpY_U_yw&sig=JH7FFM3jvQLjFDj6Z2aMOnjEfjY&hl=fr&sa=X&ei=wmj9U5z5LoHtaL7fgtAO&ved=0CCsQ6AEwBzgK (http://books.google.fr/books?id=FiEapaNgjLcC&pg=PA423&lpg=PA423&dq=%22rotor+torque%22+definition&source=bl&ots=NIXpY_U_yw&sig=JH7FFM3jvQLjFDj6Z2aMOnjEfjY&hl=fr&sa=X&ei=wmj9U5z5LoHtaL7fgtAO&ved=0CCsQ6AEwBzgK)
http://www.physicsforums.com/showthread.php?t=571073 (http://www.physicsforums.com/showthread.php?t=571073)
Otherwise, still your idea of torque measurements and comparisons looks to me most relevant in the case of the Luc's MMM :)
Regards,
Didier.
Here is the video on winding the first coil and what it looks like when finished.Fantastic job on the coil Luc,and i can wait to see what kind of pull force this will give. Those are some nice magnets you have there as well.
I checked the DC resistance of each bifilar stands and they are 1.3 Ohm each. The coil winding program had calculated 2.7 Ohms as a single coil, so we are right on track. I don't have my Inductance meter with me but will have it on Thursday.
Link to video: https://www.youtube.com/watch?v=ULds78-fDoQ
Luc
@gotoluc
Great job - why no center taps?
wattsup
DidierYes, sorry, I am not English tongue, it is indeed of DYNAMOMETER meter I was talking about, and DYNAMOMETER is all about your PRONY BRAKE!
It's a prony brake(dynamometer),not pony brake,and i never used one in my video. A prony brake is used to messure power and torque to get the horse power. Like i said,i was getting the static torque force of the rotor in the motor,as Luc was getting from his MMM.
Yes, sorry, I am not English tongue, it is indeed of DYNAMOMETER meter I was talking about, and DYNAMOMETER is all about your PRONY BRAKE!Hi Didier
Your PRONY BRAKE is indeed a DYNAMOMETER which measures the TANGENTIAL FORCE at the SPECIFIC RADIUS you set the hook and screw on the disc.
Then, all what I have noticed remains: YOU STILL HAVE TO TAKE CARE OF THE RADIUS LENGH, to multiply the TANGENTIAL FORCE you read by it.
Torque [N.m] = F [N] × L [m].
I indeed didn't know about the prony brake system (even if it was a French citizen mate who invented it ^_^) but I pretty know now about this physics. Just go in the article of Wikipedia about prony brake and you will read this, brad:
"The difference between the two readings multiplied by the radius of the driven drum is equal to the torque."
http://en.wikipedia.org/wiki/De_Prony_brake (http://en.wikipedia.org/wiki/De_Prony_brake)
So, unless you have a system in your electronic PRONY BRAKE which permits to enterTHE VALUE OF THE RADIUS you set, it won't give you any TORQUE, even "STATIC", but only the FORCE [N]. But yes, if you care about the RADIUS LENGH and multiply it by the FORCE, you'll get indeed the "STATIC TORQUE".
As for the 1:1 comparison, would only work if you were taking the exact same radius than luc had used while testing on rotor disc, but surly not for example in his test of lienar force (without rotor system); but you probably know this already. :)
Regards.
I now have my Inductance meter and the below are the measurements of the new coil.
Each bifilar coil strands measure 31 mH and when connected in series they measure 88 mH.
Measurements are Air core. So it looks like my coil software did not calculate the Inductance value very well probably because the actual shape of the coil center opening is not a circle. It got all the rest of the measurements right. Just checked the weight and it's 2,500 grams including the center core board which may be about 300g. So the program got the weight right as well.
Once the coil is between the cores the Inductance should be much higher. It all looks good so far.
Luc
There must be a certain amount of distance the weight needs to be lifted in a certain time frame no?@ Luc
Luc
@tinmanyes-not sure why i put watt hour's,should just be watts.
Your perfectly right in your calculations, Brad.
Just at the end , if you mean by "watt hours": "watts times hours", I remember you it is no more a power but an energy ;)
Good idea to use gravitation but then only 1 phase of the cycle will be needed as excitation, right? And we would need to take in account that only half the time period the consumption will occur but that gravity will work the second part of the cycle; isn't it?
Theoretically, the pulling work made to lift the crank would be equal to the work made by the gravity in the second phase of the cycle. I am not sure that the system would be better if only excited on 1 half of the cycle, but experiments should tell.
@gotoluc
Please, be careful with your fingers and skin hands when you'll handle these so powerful neodymium magnets ^_^
Yes, if we can get o.u. in the straight design, a rotary system would be A Must! :)
Thanks for keeping going and sharing, Luc :)
yes-not sure why i put watt hour's,should just be watts.No problem Brad, for the adding hours; aren't we here to correct, give ideas and share experiments to each other, in what we like the most to do? :)
So would we use 1/2 G in this case? 4.45m*/s
So as 1 watt /4.9x1kg=204mW.
A pull force of 1kg @ 204mW of power?.
Hi Luc,Very Well Done for the design and ideas! :)
Great work on the build.
I had two untested ideas for your design one is a linear generator.
Instead of having continuous bars along the edge replace them with steel pipes. A split is in the middle so that when the coil is powered it makes the top of the pipe south while the bottom is north. The output coil will see this change as it collapses toward the north pole. I'm not sure on how this will be effected by the opposing field though. Also a split may be needed along the side of the pipe so that eddy currents are minimized. It might be something to test.
The rotary idea is just a thought in progress. Magnets can be used to polarize the metal as all north. There is a falloff toward the center but it is still north facing, this difference is needed otherwise we end up with a static system. The gates would have the strongest field to push from while being pulled toward the next gate.
The polarizing is based off of this observation:
http://www.overunity.com/14843/are-partial-monopoles-possible/
This current design has issues but maybe it will inspire a better idea or design.
@ All: Please, skip the algebraic developments of this VERY-VERY LONG spot, if you're if algebra is not really your friend ;)???
No problem Brad, for the adding hours; aren't we here to correct, give ideas and share experiments to each other, in what we like the most to do? :)
For your question: no, cause we don't even have to care about G! Lol, cause it is NEUTRAL in the cycle, as I see it and as per the following algebraic operations:
Say:
• Wmt, the total effective mechanical energy of the whole cycle
• T1, the time period of lifting
• T2, the time period of droping
• Wm1, the effective mechanical energy of the lifting
• Wm2, the effective mechanical energy of the droping
• We1, the electrical energy consumption of the lifting
• We2, the electrical energy of the electromagnetic flyback at the droping
• Wg, the gravitational energy as per: Wg = F × g = (M × L) × g, with:
• F, weight [Newtons]
• M, mass [kg]
• L, crank or lifting or droping length [meter]
• g, gravitational acceleration, or "gravitational intensity field"
• a, the ratio of electrical energy losses respect the electrical energy consumption
• We': effective electrical energy, as per We' = We - a × We = (1 - a) × We
• A, the conjugated quantity of "a", "1 - a", which is the corresponding coefficient which allows to obtain the effective electrical energy directly from the electrical energy consumption: A × We = (1 - a) We = "effective electrical energy" = We'
• b, the ratio of mechanical energy losses respect the difference (electrical energy consumption - gravitational energy)
• Wm': effective mechanical energy, as per Wm' = We' - b × We' = (1 - b) × We'
• B, the conjugated quantity of "b", "1 - b", which is the corresponding coefficient which allows to obtain the effective mechanical energy directly from the effective electrical energy minus the gravity energy "Wg" when lifting, but plus the gravity energy "Wg" when droping: B × (We' +/- Wg) = (1 - b) × (We' +/- Wg) = "effective electrical energy" = Wm'
• c, the ratio of electrical energy flyback respect the electrical energy consumption of the lifting,)
• Wm": effective mechanical energy returned, as per Wm" = Wm' + c × We1
• C, the conjugated quantity of "c", "1 + c", which is the corresponding coefficient.
● Wmt = Wm1 + Wm2
● Wm1 = [We1 - a × We1 - Wg] - [b × (We1 - a × We - Wg) ]
○ Wm1 = (1 - b) × (We1 - a × We1 - Wg)
○ Wm1 = B × ( (1 - a) × We1 - Wg)
○ Wm1 = B × (A × We1 - Wg)
● Wm2 = [We2 - a × We2 + Wg] - [b × (We2 - a × We2 + Wg) ]
○ Wm2 = (1 - b) × (We2 - a × We2 + Wg)
○ Wm2 = B × ( (1 - a) × We2 + Wg)
○ Wm2 = B × (A × We2 + Wg)
○ Wmt = [ B × ( A × We1 - Wg) ] + [ B × ( A × We2 + Wg) ]
○ Wmt = B × { [ A × We1 - Wg ] + [ A × We2 + Wg ] }
○ Wmt = B × [ A × We1 - Wg + A × We2 + Wg ]
○ Wmt = B × [ A × We1 + A × We2 + Wg - Wg ]
○ Wmt = B × (A × We1 + A × We2 ); QED.
Note 1: In fact I didn't care of the time periods cause the time frame is fixed (even if not necessarily equal).
Note 2: I didn't care too of a spring. I considered that at the end of the first period, the lifting one, the mass is pulling back down by the gravity itself, so then no need of a spring (little more complex equations if we care! ^_^ ). And I have considered that at the end of the second period, it would have no returning mecanical energy system, so all the droping energy lost (which is placing us far below in efficiency than with a spring to reflect the energy or if a mechanical flywheel with a link).
Note 3: The electrical supplier voltage is only applied at the beginning of the lifting period.
Note 4: I used to use "expletive parentheses", they are to help the reading and understanding of the equations, even if it is not "mathematically necessary": "(...)", "[...]" and "{...}" is a personal use of parentheses to signify the different level of intrication, like: { [ ( ...) ] }. ;)
THEN, If I haven't messed up in my operations, it comes an interesting equation, imho, Brad: it the one of the conditions of o.u. for the device in the conditions I've just described:
As fixed time frame,
COP of mechanical output under electrical consumption
= mechanical output power / electrical consumption power
= mechanical output energy / electrical consumption energy
= Wmt / We1
● COP = Wmt / We1
○ Wmt = B × (A × We1 + A × We2 )
○ Wmt = B × A × (We1 + We2 )
○ Wmt = B × A × (We1 + c × We1)
○ Wmt = B × A × (1 + c) × We1
○ Wmt = B × A × C × We1
○ Wmt / We1 = (B × A × C × We1) / We1
○ Wmt / We1 = B × A × C = COP.
So, as we have to have:
● COP > 1
so, we have to have too:
● Wmt / We1 > 1
○ B × A × C > 1
Then, we have our basic condition as a low estimate (remember, we didn't care of any spring reflection or mechanical flywheel):
○ C > 1 / (B × A)
☆☆☆☆☆☆☆☆ C > 1 / (B × A) ☆☆☆☆☆☆☆☆
Example:
If,
• a, the ratio of electrical energy losses respect the electrical energy
= 0.10, or 10%, so A = 0.9,
• b, the ratio of mechanical energy losses respect the difference (electrical energy consumption - gravitational energy)
= 0.10, or 10%, so B = 0.9,
We get:
● C > 1 / (B × A)
○ C > 1 / (0.9 × 0.9)
○ C > ~1.235
So, to have o.u. we should have
• c, the ratio of electrical energy flyback respect the electrical energy consumption of the lifting
> 24%.
○ For a = b = 20% ; c > ~57%
○ For a = 5% and b = 20%, or for a = 20% and b = 5%, c > ~32%.
~○~
The problem we hmcan see while checking the equations for the system with a spring at bottom, is that if Wm2 would be transferred as per "d" spring coefficient losses "d", to the third time, BUT AT THE END OF THE THIRD TIME, ALL THE MECHANICAL ENERGY TRANSFERRED IS CONSUMED BY THE GRAVITY, so Wm4 = Wm2 :/
Then, we realise that luc's arrangement with 2 springs is decisive!
Indeed, it allows us the transfer each mechanical energy of each time to the next, as per the spring losses coefficient "d", and not consuming it just because of this gravity.
~°~
TWO SPRINGS SYSTEM WITHOUT GRAVITY INVOLVED:
After 34 pages of algebra and calculations, soon 24 hours non-spot of checking and rechecking ^_^, I got these following equations which state the conditions for overunity COP with my simplified operations.
The simplifications are:
▪ considering losses or gains always linear, proportional, when it not necessarily, and
▪ equalising the losses or gain coefficients as A = B = C = D = K, with:
□ A, coefficient of electromagnetic losses
□ B, coefficient of mecanical losses
□ C, coefficient of reflection losses (due to the springs)
□ D, coefficient of electromagnetic flyback gain
We see then that the NET ENERGY GAIN for the 2 first periods, for example, are: (we note We1 = We2 = constant = We)
● T1:
G1 = Wm1 - We
= A × We - We
= (A - 1) × We
-> a loss
● T2:
G2 = Wm2 - 2We
= (W,mech + W,ElectromagneticFlyback + W,Electrical Excitation) - 2We
= C × [ B × (A × We) ]
+ D × (A × We)
+ (A × We)
- 2We
= (A × B × C) × We
+ (A × D) × We
+ A × We
- 2We
= [ (A × B × C) + (A × D) + A - 2 ] × We
Then, to have COP overunity, we need:
● G2 > 0
So:
● [ (A × B × C) + (A × D) + A - 2 ] × We > 0
○ (A × B × C) + (A × D) + A - 2 > 0
○ (A × B × C) + (A × D) + A > 2
○ A [ (B × C) + D + 1 ] > 2
Setting A = B = C = D = K,
○ K [ (K × K) + K + 1 ] > 2
○ K^3 + K^2 + K > 2
○ [ (1 - K^4) / (1 - K) ] - 1 > 2
○ (1 - K^4) / (1 - K) > 3
Numerical examples:
• For K = 0.5 = 1/2, (1 - K^4) / (1 - K) = 30 / 16 = ~1.8 and 1.8 < 3; no overunity.
• For K = 0.75 = 3/4, (1 - K^4) / (1 - K) = 175 / 64 = ~2.7 and 2.7 < 3; no overunity.
• For K = 0.8 = 4/5, (1 - K^4) / (1 - K) = 369 / 125 = ~2.95 and 2.95 < 3; no overunity.
• For K = 0.9 = 9/10, (1 - K^4) / (1 - K) = 3,439 / 1,000 = ~3.4 and 3.4 < 3; WE HAVE OVERUNITY.
This means that with 10 % losses for each kind of losses encountered here and for 90 % of electromagnetic flyback recycled, the numbers say that we would be able to have overunity with these first calculations. :)
But now, here is "The Big One",
THE OVERUNITY EQUATION ON LONGUE PERIOD OF TIME:
□□□□□ COP = { [1 / (1 - K) ] - 1 } / K □□□□□
And the solution for COP > 1 is:
□□□□□□□□□□ K > 0.618 □□□□□□□□□□
Means that theoretically, if my calculations abd reasoning are correct (and you may correct me at anytime):
WE SHOULD BE ABLE TO GET AN OVERUNITY EVEN WITH 38.2% OF LOSSES AND ONLY 61.8% OF ELECTROMAGNETIC FLYBACK. :P
Well, I have made my 24 hours around the clock now, hope few of you guys will appreciate the work! ^_^
Regards,
Didier
???Understand your feelings, Brad, but I was going just a little bit further and more general for the seak of a overunity COP.
Ok-something went wrong in translation there.
I am trying to convert a constant pull force x watts-no moving parts.If we are to lift a weight of x amount by a distance of y amount in a fixed amount of time,then that is much easer. But to convert a non moving pull force x continuous watts input is a little harder.
What we want is to be able to hook up the scales to the moving coil,and get a pull force. Then we also have a stedy whatt input aswell-nothing moves. So we want to go say 500 grams pull x 1.2 watts input,and then be able to calculate efficiency from that with the correct math.
OK-some one please explain to me as to why the magnetic field is opposite on the inside of the coil,as aposed to the outside of the coil,when all wire is wound in the same direction,and current flow is uniform.\
Hi Ron,
Many thanks for doing these tests.
Would like to ponder on the resulting poles when an open soft iron core is used in or on the close outside of such oval shaped coil. Cores collect and guide magnetic flux so I assume the unlike poles that are near to each other tend to close in the core surface so not likely to exit the open core. Obviously there will be a resultant flux, the result should come from all the flux collected and not closing in the core. Surely there are plenty of flux remaining that would not close in the core to neutralize themselves but do work when current appears in the coil.
For a normal cylinder shaped coil and core, the strongest flux manifests at the core ends of course. If the cylinder coil is also a multiturn, multilayer type, then one has to think it over how the outside flux from the cylinder coil could be utilized for increasing the ends flux (if the goal is to maximize the force from such electromagnet). Probably a similarly closed magnetic path with the inside and outside cores applied (as Luc showed in his earlier MMM force test videos) should also be used.
Thanks,
Gyula
The shield material properties and its structure are important to the coilgun's performance. In present article, a two-dimensional (2-D) transient finite element model was established based on the structural characteristics of the coilgun and launching process. Effects of variable shield materials with different dimensions on the launching performance have been investigated. The magnetic field distributions at different times have been presented. The results show that, the eddy currents induced in the conductive shield will reduce the projectile's muzzle velocity, and the electromagnetic shielding effect of the conductor is not ideal. Strengthening the main magnetic flux, permeable non-conductive material shield increases the muzzle velocity of the projectile, and the electromagnetic shielding effect is good. Increasing the thickness of the shield, the magnetic materials enhance the magnetic flux better, projectile muzzle velocity increases, and the electromagnetic shielding effects are better. In order to achieve higher projectile muzzle velocity and better electromagnetic shielding effects, high permeability silicon steel sheets can be used as the shield. The distance between coils and shield should be reduced. The thickness of silicon steel sheets should be as small as possible to reduce eddy current effects.
because the "sense" of direction of rotation inside the coil is, lets just say, up and on the outside it is down and it is the "sense" of direction of rotation that is polarity.Hi! webby1. Thanks for your input but I am not sure to get the concept :-\ (which directions and sens you speak). Could you please join drawing with arrows?
Thumb! Ron. Thanks for having taken time to do this experiment and sharing it to us :)
Hi tin man,
First, great work on the inertia drive! most interesting.
Luc does come up with some interesting things, right? Gyula and I were talking about this so did a quick and nasty vid for him...the polarity gets even more interesting with a gauss probe
http://youtu.be/GXkVpZYT76I (http://youtu.be/GXkVpZYT76I)
This is unlisted so refrain from passing it around please
The gauss meter is this one
http://www.coolmagnetman.com/magmetr1.htm (http://www.coolmagnetman.com/magmetr1.htm)
Ron
Thanks for your post ShyloI am amazed again by the Power of you Intuition, Luc, and your obvious to me capability to look accurately at things :)
I have no schooling, so I experiment to understand how things work. I found this about 7 years ago when I built a motor out of 2 MOT secondaries with a ceramic permanent magnet rotor. I still have that build if anyone wants to see it.
Ever since then I assumed it was common knowledge and would mention it here and there but now I'm finding out it's not common knowledge at all. I've just been told it's not in the textbooks.
If experimenters would like to test it, I would suggest a MOT secondary since it's more difficult to detect the poles with a round coil. You need a long strait side of a coil which separates the poles and makes it easily detectable. Also, best if the coil winding width and height are the same, so a square Brooks winding is best.
Please note as I have mentioned in my video that the center opening pole will be stronger because they are double together and in close proximity as opose to the outside poles away from each other but if you combine the two outside opposite poles (as I do in my MMM design) they are as strong as the inner pole.
So now you can double the torque of a coil if you understand this and know how to use it.
Luc
just check this link out guys at:I see you are here just to peddle your religion. Please tell us all what this has to do with Luc's project?--> maybe take your rubbish else where.
http://www.youtube.com/watch?v=6XsZG-6Lp1o
oh wow really check this out all
Tom
OK-some one please explain to me as to why the magnetic field is opposite on the inside of the coil,as aposed to the outside of the coil,when all wire is wound in the same direction,and current flow is uniform.\
https://www.youtube.com/watch?v=IvfwELDmKIM (https://www.youtube.com/watch?v=IvfwELDmKIM)
Hi poyntThere are no neutral zones between the poles. "Poles" is actually somewhat poor terminology. Poles really only tells you what direction (by standard convention) the main central flux is going. Also causing confusion is the way magnets are illustrated with one half being blue and one half being red. This leads us to believe that there is no field (or that there is a "Bloch wall") right in the middle where the color changes. This is simply incorrect. Magnets really should be illustrated with one solid color and a single arrow from one end to the other.
thanks for your post.
The term Bloch wall, I thought it meant the Neutral point between two magnetic poles and that's what I was using the word as.
If that is not correct, then I have no problem calling it the Poles Neutral Zone if that's a better way to describe it.
In the illustration of the solenoid magnetic field you posted, we can see -> arrows going in different directions. Do these arrows represent a pole direction?All magnets and solenoids have a magnetized direction (permanent magnets by how they were magnetized, and solenoids by the direction of current). In the illustration, that magnetized direction is in the center (centre if you are Canadian) of the solenoid, so bottom to top. Magnets and solenoids have "curl". The flux lines curl back to the "x" of the arrow and exit out the "." of the arrow. Notice that there is no neutral zone in the middle of the arrows?
I assume the the fine black rectangle line is the outside of the solenoid coil?... is that correct?Correct.
Luc
Thanks for the reply and explanation poynt
Looking at your new illustration above, would it be correct to say the point of the arrow is in the direction of North and South would follow behind?
Luc
Hi, point of the arrow is North. Travels North to South.
In the illustration, that magnetized direction is in the center (centre if you are Canadian) of the solenoid, so bottom to top.
Okay
if we look at your first illustration (attached below) of an energized solenoid and we follow the arrows, how can we tell which end will be North or South?
Luc
Hi poynt
thanks for your post.
The term Bloch wall, I thought it meant the Neutral point between two magnetic poles and that's what I was using the word as.
If that is not correct, then I have no problem calling it the Poles Neutral Zone if that's a better way to describe it.
In the illustration of the solenoid magnetic field you posted, we can see -> arrows going in different directions. Do these arrows represent a pole direction?
I assume the the fine black rectangle line is the outside of the solenoid coil?... is that correct?
Luc
Hi all,Hi Hanon, thanks for your input but how it would be possible the 2 sides, top and bottom, having exactly the same behaviour (flux going outside), would be 2 different poles? :/
Maybe all that you are discussing is what Howard Johnson demostrated by experimental tests in his book "The Secret World of Magnets". He states, and measured with a Hall probe, that the lines of force do not move from one pole to the other but they move from one pole to the center point, forming a double vortex.
Please see the attached picture from his book.
Regards
Luc indicated 4 block walls, my sketch indicates there are actually 8
The core makes no difference
The following photos show exactly what my sketch indicated
Ron
Magnets are strongest at their ends,
not the middle
In the mean time this is not an engineers forum,
rather a hobbiest/hands on experimenter group where the transition between two opposite poles is
commonly referred to as the bloch wall... lets keep it simple and on track.
If you don't want to continue to discuss your way to interpret and use the probe when you get your read it is your choice and you're very free to not participate to the discussion.
LOL
It is pointless to encourage this ridicules line of endeavour any longer. Lets return to the TOPIC,
Luc's graphic. Luc raised a very valid point. I added to it. If anyone does the experiment with different
results, I would be pleased to correct my material. In the mean time this is not an engineers forum,
rather a hobbiest/hands on experimenter group where the transition between two opposite poles is
commonly referred to as the bloch wall... lets keep it simple and on track.
Ron
Thanks for the comeback. Not too clear, would need a diagram.
However, confirmation of the poles shown in the video is if you place a magnet over the energized coil.
In this case I have drawn it inside but can be on any side of the coil... see the N pole locates adjacent
to the N pole of the coil. Now everyone knows like poles repel, right?
So the S pole that the hall sensor indicates on the top inside is present.
OK, I have redone the 'sketch' to better indicate all the block walls present in an air cored
rectangular coil. The external magnet shows the natural attachment direction.
poyntie's cored solenoid coil is invalid in this discussion as my video is a representation of the polarity
present around a cross section of an air cored coil, which enhances Luc's original diagram, not the whole external field of a solenoid coil, which is a diversion.
Ron
Ron, please do keep up your experiment and share your findings as it is worth more then words.Yep! Except that if "by second" it means you ask for Watts and not Joules ^_^ but I will give you both:
Khwartz, can you please calculate how much Joule energy it takes at Unity to lift 2.35Kg. 1mm, 2mm and 3mm in 1 second.
Thanks for your time and help
Luc
Yep! Except that if "by second" it means you ask for Watts and not Joules ^_^ but I will give you both:
● For 1 mm height:
W [J] = 2.35 [kg] × ~10 [m.s^-2] × 1/1000 [m]
= 23.5/1000 [kg.m^2.s^-2] = 0.0235 [J]
If not clear enough, please just tell me :)
Don't worry Luc,
I'll not "intrude" with my "words" any longer in any of your threads.
I'll have no difficulty putting that time to good use. :)
Cheers!
Ron, please do keep up your experiment and share your findings as it is worth more then words.
Luc
Here we have the sensor in the orientation that I suggested earlier.
Notice how the lines of force with the sensor in the middle, the sensor should read more than to the left or right, sorta opposite of the way you measured it in the vid. Not that you did anything wrong. Im just giving a view of looking at the fields in a second dimension to show a more complete view of the fields. A third dimension of the field would be to face the sensor with the length of the wire/leg of the coil.
Mags
Thanks for your post Ron
How about a new test. Find 2 separate cores and inset one in the center of your coil and one on the outside of the coil and probe to see how the core fields compare to the coil fields.
The core may need to be higher then the coil so it's away enough that your probe won't picking up coil fields.
Thanks for your help
Luc
I wonder if this denial of the Bloch wall is a NWO dumb down tactic?
Thanks Ron for the extra tests and Bloch wall Images.
So, what's your take on why an energized coil has all those pole domains compared to a magnetized bar or a permanent magnet?
You may be interested in viewing this video: https://www.youtube.com/watch?v=T1R44WjyppQ (https://www.youtube.com/watch?v=T1R44WjyppQ)
Thanks for your time
Luc
Nobody is disputing the existence of Bloch walls. The dispute is about talking about fake nonsensical allegations of the existence of Bloch walls.
Again, you are welcome.
My take? just if we want to know how things work this is a valid experiment. Please see my reply to synchro.
I had seen a bit of that video but didn't find it to definitive.
Ron
So, what's your take on why an energized coil has all those pole domains compared to a magnetized bar or a permanent magnet?
Luc
Ron:
I annotated your drawing and attached it here. In position 1 the magnetic field comes in at the "top" of the hall sensor (where the black band is). So in your interpretation you are calling that "South" because your sensor is telling you flux is coming in the top and going out the bottom of the sensor. You say that flux going in an "up to down" direction is "South" so you label that part of the coil "South."
Likewise in position 2 you see flux entering by the "bottom" of the hall sensor so you call that "North."
The reality is what you see in the annotated sketch. There is NO "North" or "South," there is just the convention that we have adopted that says "flux direction coming towards you" is "North" and "flux direction going away from you" is "South" when you position yourself above or below the core in your sketch and you look towards the core.
I don't know if you can understand that but your drawing of the cross section of the coil core with the polarities at each corner of the cross section is completely wrong. The reality is the orange lines of flux with the directional arrows as I have drawn them out. There is NO true "North" or "South" there is just flux direction as per the convention that we have defined. All that "North" and "South" really mean on a bar magnet is that flux is "exiting" from the "North" end of the magnet and "entering" at the "South" end of the magnet. It's a just a naming convention. In reality the flux lines are closed loops with no "start" or "end" or "North" or "South."
The only thing the hall sensor is showing you is the amount of flux and the direction of the flux. It's up to you to make sense of that information and right now you clearly are not making sense of it and you are drawing conclusions and making statements that are wrong.
A stack of cylindrical magnets will line itself up with the field created by the coil. You are stating "how can the North of the stack of magnets be attracted to the "North" part of the coil?" The answer is that there IS NO "NORTH" PART OF THE COIL. Repeat, the cylindrical magnets simply line themselves up with the field created by the coil.
This madness, this endless spinning and grinding and arguing and getting nowhere, could be remedied if you locked yourself away in a room for one week with a determination to understand magnetic fields and to understand how magnetic fields interact with ferromagnetic materials. Then you will be able to look at a relatively simple setup and nearly instantly be able to visualize and understand what is going on.
Hi everyone,
here is a quick 2nd test of the Triple M Super build.
Link to video: https://www.youtube.com/watch?v=f6pc-XNS9uo (https://www.youtube.com/watch?v=f6pc-XNS9uo)
The results are for one coil and two wire strand connected in series
The simple 1 inch wide Triple M Pulled 500g for 0.43 Watts
The 8 inch Super build single coil one stand gave a result of 2Kg Pull for 0.43 Watts
and now the 8 inch Super build single coil 2 series stands gave a result of 2.5Kg Pull for 0.43 Watts
So by using only half of the 8 inch Super Build we are pulling 5 times more then the one inch build
Stay tuned for the full results of the Tandem (double) Triple M Super build
Place your bets on what you think it will pull for 0.43 Watts
Luc
Luc, Aren't you just getting closer and closer to 100 % efficiency ? And when you get the best possible efficiency you can get won't
that be the closest to 100 % efficient you can get ? Is there an efficiency rating for those lifts and what an OU lift would figure as ?
Please forgive my "Apparent skepticism" and feel free to ignore my questions if you wish.
@Gotoluc,
Are "2 series strands" the same as "Series bifilar"? I forecast a very exciting historical precedent on the approaching horizon!
Hi everyone,
here is a quick 2nd test of the Triple M Super build.
Luc
Quote from Gotoluc:
"The 8 inch Super build single coil one stand gave a result of 2Kg Pull for 0.43 Watts, and now the 8 inch Super build single coil 2 series stands gave a result of 2.5Kg Pull for 0.43 Watts"!
Luc's test results prove conclusively that the value of Tesla's "Coil for Electromagnets" is genuine and not imaginary as falsely claimed by some.
I'll post some questions and concerns to the collective group.
For starters, the classic error is being made by discussing "efficiency" without defining what it is. Take the example of pulse motors. You read comments all the time that say, "Wow, your pulse motor is really efficient." When you are watching a pulse motor spinning in a YouTube clip and you don't define your parameters for "efficiency" that is a totally meaningless statement. So if you are going to discuss "efficiency" then you have to define what it means and then show measurements to demonstrate the efficiency.
The major problem is with the variables that you are tracking as you do these experiments because in the strictest terms they don't actually make sense. This is a place to discuss energy and related matters and you live and die by the data that you generate. So this should be resolved before you go further. Right now as it stands the data generated would have to be reverse-engineered to turn it into proper data. This is research and you should be presenting proper data on paper (like in this thread and not on a YouTube clip) that has true meaning. If you were missing the information that you would need to reverse-engineer the data as being presented right now, then the data would be meaningless.
So the choice for the group is to ignore these comments and sleepwalk through the rest of the experiments, or discuss the issues and brainstorm and figure it out for yourselves and then take the proper corrective measures so as you go forward you generate valid data.
Spoon-feeding you all the answers is not the route I will take for this one. I suggest that you discuss these issues in this thread amongst yourselves and resolve them.
That's nothing but a STEAMING HOT CROCK OF UNADULTERATED HORSE SHIT!
Hi Mags,
do you know how the coil is wound?
Luc
Hi Mags,
okay I see
The problem with Zeropoint132 test is similar to what I was trying to tell synchro some posts back.
If you wind a coil with many wires together, 4 in this case, the coil ends up being 4 times the size. So now if you energized only one wire all the other stands cause 3 times more space between the energized wire turns and weaken its magnetic field since the turns are not concentrated and as close together as possible.
I'm quite sure if you took the same length of wire and wound it alone in a nice Brooks coil the magnetic field would be stronger then it being spread out like that.
So to me his test is not valid unless he had made a single coil to disprove this possible problem. Better yet, he should of had 4 individual single filar coils and place them one after the other to prove there's a benefit.
Come to think of it, my first Super build test 1 uses one of two strand is possibly suffering of the same problem ::) ... I'll need to recheck that!
Hope you understand?
Let me know if this makes sense
Luc
Hi Mags,
okay I see
The problem with Zeropoint132 test is similar to what I was trying to tell synchro some posts back.
If you wind a coil with many wires together, 4 in this case, the coil ends up being 4 times the size. So now if you energized only one wire all the other stands cause 3 times more space between the energized wire turns and weaken its magnetic field since the turns are not concentrated and as close together as possible.
I'm quite sure if you took the same length of wire and wound it alone in a nice Brooks coil the magnetic field would be stronger then it being spread out like that.
So to me his test is not valid unless he had made a single coil to disprove this possible problem. Better yet, he should of had 4 individual single filar coils and place them one after the other to prove there's a benefit.
Come to think of it, my first Super build test 1 uses one of two strand is possibly suffering of the same problem ::) ... I'll need to recheck that!
Hope you understand?
Let me know if this makes sense
Luc
Hi Mags and all,
I can now confirm what I described above is exactly the problem.
I retested my super build coil by connecting both wires in parallel and the pull force is 2.56Kg. so 60g. better then with the two leads connected in series using the same 0.43 Watt
Luc
Hey Luc
So you reduced the input voltage for the parallel test as compared to the voltage for the series test to come to a common watt figure?
Mags
Hey Luc
So you reduced the input voltage for the parallel test as compared to the voltage for the series test to come to a common watt figure?
Mags
I'll add that the test above was with the coil connected in bifilar Series since when it was connected in Parallel it needed a little more Joule energy.
So to raise the 2.325kg coil up 10mm
Coil in Series with 7170uf cap @ 9.3vdc = 310.07mj
Coil in Parallel with 7170uf cap @ 9.9vdc = 351.37mj
Luc
So this is quite different in terms of input/output as compared to the previous tests, correct? More power needed for parallel than for series bifi?
Mags
Yes must be because the input is a capacitive discharge compared to the previous test was continuous DC pull force test
I've also noticed the coil holds up a little longer when connected in series. Maybe it's the small coil capacitance? or maybe it's just the coil has more inductance so now the generator effect is more as the coil drops down with gravity and tries to charge the cap bank in opposite polarity?
Luc
Yes must be because the input is a capacitive discharge compared to the previous test was continuous DC pull force test
I've also noticed the coil holds up a little longer when connected in series. Maybe it's the small coil capacitance? or maybe it's just the coil has more inductance so now the generator effect is more as the coil drops down with gravity and tries to charge the cap bank in opposite polarity?
Luc
If you suspect the increased generator effect of the series coil configuration is also to blame for the little longer holding time then consider to include a diode in series with the capacitor with forward direction to the discharge process i.e. lifting, so it will block any reverse voltage hence current in the falling phase of the coils.
Gyula
Hey Luc
I agree with Gyula that most likely the resistance change would have a difference in drain time of the cap. I would recommend a constant dc input, using caps to stiffen the supply, trying to maintain solid input to either coil for these tests.
The effect of capacitance of the bifi increases with voltage increase. The higher the voltage, the acceptance of that voltage by the bifi gets better, overcoming the bifi inductance, as suggested in Teslas pat for electromagnets. Would be interesting to see 100v - 500v cap discharge into each version of the coil. ;) Just for the heck of it.
Gota do laundry. ;D
Mags
Hi Mags
I think I understand what the benefit of "overcoming the bifi inductance" would be but if you don't mind I would like to get your version of the benefit... once you have time between loads that is
Luc
Something that just came to mind. Thane wound his coils in a random fashion from what I was told. This would put a lot of empty space in the windings. So would he have been better off winding straight and tight? How much better? Better output than Z shows with increased resistance of adding all the existing 4 wires in series? There may be more to this, considering the level of output change. ;D
Mags
Hey Mags, I forgot to answer this one.
As you may know Thane lives in my city and we've known each other before he was doing the ReGenx thing.
I do some work for him here and there and I can tell you he no longer winds his coils in random. He now takes much care to wind coils tight and keeping strands side by side.
I think you mentioned litz wire, as far as I know that's not a good choice for bifilar coils as the strands are usually twisted.
I think if you don't keep your stands next to each other as you wind and you allow then to cross back and forth you no longer have coil capacitance.
I'm not the only one who thinks along these lines. This was posted by "Farmhand" in a topic at EF
"Twisting the conductors in multifilar coils then series connecting them
will increase the inductance when the windings are connected together
as does any coils inductance increase with more turns and wire, however
the "capacity" Tesla speaks of is related to the potential difference
between adjacent turns, which twisting randomizes, a twisted coil is not
making a coil as Tesla describes in the patent and will not secure the results
the patent describes. The "self capacitance" is not the "capacity" of the coil.
The "capacity" of the coil is how much energy it can store."
Just thought I would share that
Luc
Yes must be because the input is a capacitive discharge compared to the previous test was continuous DC pull force test
I've also noticed the coil holds up a little longer when connected in series. Maybe it's the small coil capacitance? or maybe it's just the coil has more inductance so now the generator effect is more as the coil drops down with gravity and tries to charge the cap bank in opposite polarity?
Luc
Hi Gyula,
I did place a diode between my 7170uf electrolytic capacitor and the coil but all that does is charge the cap to an negative value when the coil starts to fall back down since the current is going in the opposite direction when it fall.
Cap starts at +10vdc and coil travels up 12.3mm then the cap goes Negative as the coil starts falling back down and cap charges to minus -6.72vdc
The cap being electrolytic don't hold the negative charge too well but only loose about 1vdc.
Make a recovery system though!
Maybe if I use AC caps it would be more efficient?
Luc
Ok, here is my idea behind using litz...
Say we have 20awg mag wire and some 6 strand litz that is equal to 2 of those 20awg wires.
So we wind 2 20 awg wires together as a bifi. If we were to look at a cross section of those 2 wires, we can see that only a very small amount of each wires outer surfaces are in physical contact, and the rest of those surfaces are further away from each other.
lets call 1 of the 2 20 awg strands A and the other B.
Now the 6 strand litz, equal to 2 strands of 20awg, say equal resistance by length. If we look at the cross section of the 6 strand litz, where half of the strands, are connected at the end of the windings to form 1 of the 20awg wires, and the other 3 strands are the other, sorting the strands so they alternate A and B strands rather than to have an A strand next to an A strand only.
Within that litz bundle, the first thing we should notice is that the surface area of the A and B strands in the litz is greater than the surface area of the 2 20 awg wires. The second is there are more contact areas within the litz of A and B strands than the 2 20 awg wires.
So there is more possible capacitance between A and B in the litz than there is with the 2 20 awg wires.
We are talking about capacitance between A and adjacent B strands, let alone contact and proximity with other adjacent windings.
Litz doesnt only come in weave form. Examples below.
Just something I had been thinking of for some time.
Mags
Luc:
Did you consider doing tests where the coils were first connected bifilar, and then you redo the tests with essentially the same coil configuration, but monofilar? Same number of turns, same gage of wire, etc. From a cursory observation of what you are doing, there is no logical reason whatsoever for a bifilar coil configuration to be different from a regular coil configuration for this setup. You basically have a large electromagnet with a large inductance picking up a large weight, and whether the setup be bifilar or monofilar, the coil capacitance will be minuscule and totally insignificant and have absolutely no affects whatsoever on your tests.
The important principle to understand here is to take stock of what actually affects your experiment. It'd just like the thread about myths and misconceptions for magnets. There may be a specific narrow set of cases where a bifilar coil will make a difference, but for a brute force electromagnet the expectation would be that a few extra nano or microfarads of capacitance will make absolutely no difference in the performance of the electromagnet. If you don't agree or believe me then simply run the tests for yourself if you are so inclined. It's simply not right to see people make clips where they make their coils bifilar when there is no logical reason to do so - it's just another myth. The inductance of the electromagnet is the elephant in the room. The capacitance of the electromagnet proportionally might be the size of a fly on the elephant.
With respect to the efficiency of doing the lift, I asked you where all the lost energy went. Gyula mentioned the resistive losses in the wire which I agree with. Here is another thing to think about: As the capacitor discharges it is building up the magnetic field in the coil. So for a certain number of milliseconds as the capacitor discharges, there are i-squared-R losses in the wire, and energy is going into the coil to build up the magnetic field. Before the weight even moves, the electromagnet is building up in strength. So that means you have resistive losses in the wire, and there is energy put into the coil, before the weight even moves. Both of these components will factor into the losses.
Sorry Synchro1 but that is just another nonsensical fantasy posting. I don't want to disrupt the thread, but that is the honest rebuttal to your posting. You have made other fantasy postings in this thread that disrupt what is going on. I will ignore them from now on.
Facepalm.
What MileHigh fails to understand is that the series bifilar coil has no magnetic field inside the coil. The coil vectors a monopolar magnet wave like a smoke ring, with the positive pole nested in the center. This field forms outside the coil when pulsed.
Facepalm.
Well, not much luck finding anything with the coil connected in bifilar series and using high voltage discharges. Voltage and Current seem to be in phase.
The current sensing resistor is a carbon 1 ohm 1% resistor. Voltage probe is no.1 (green) and current probe is no. 2 (yellow)
The Scope shots below range from 50vdc to 1000vdc and cap value from 16.66uf from 50 to 250vdc, 10.16fuf rom 300 to 800vdc and 3.44uf for 1000vdc
Each pic title has the details. The recovery is from the diode as the coil falls back.
I did not bother with the coil height measurement for each test as they were all higher Joule energy for the height reached then when using low voltage. Basically, the higher the voltage the more Joule energy it take for the coil to travel up. The coil wants to go up faster as the voltage increase but so does the generator effect increase at the same time. So more and more energy is wasted as voltage goes up.
I'm not going beyond 1kv as the coils wire insulation could be compromised.
Luc
In other words, its possible that the parallel might outperform distance wise than series if the resistor is not there at all.
Night
Mags
Hi Mags,Hi Luc! Nice to see you are now fully self-related with these kind of calculations; and nice you now use and share your results on this base! Cheers! :)
the results are very bad with high voltage :P
With coil connected in series and no resistor the 2.35kg weight raised by 1mm using a 3.44uf ac cap charged to 1000 vdc = 1.722 Joules
With coil connected in parallel using same cap and charge, the rise was too low to measure.
The low voltage wins hands down with a 23mm rise with 1 Joule input
Luc
Hi MH,indeed, as inductance expresses the ration of magnetic flux, in Webers (Wb) against the current flow in Ampers (A). So this is a kind of efficiency ratio. Higher the inductance is, higher is the magnetic flux will be for the same current. (But you probably know that :) ).
Today I found there is a difference between the coil connected in parallel vs series.
In today's tests the best efficiency results for the coil to move up against gravity was with it connected in series. However, I think the gain could be mostly from the boost in Inductance. Gyula suggested a way to measure the Inductance and the results are 8mH vs 30mH in series. So I think this would need to be considered.
Today's testsso, this looks to verify my result number 1. :)
All tests done with low voltage.
To maintain a stable low voltage through the coils power stroke, I used a 650F Super Cap fully charged and stable at 2.7vdc
By using this low voltage the coil moves up slowly and resulted in less losses as these were the most efficient results to date.
Using the coils maximum upward travel stroke of 23.5mm it used 1.2J connected in parallel and 0.94J in series.Very Great you had, Luc, the idea to make this calculations 8)
If we use the 0.94J and we subtract 0.54J which would be the unity amount needed for the coil to travel 23.5mm (if my calculations are correct?) then we are left with 0.40J under unity.
However, by having the coil travel to the maximum height I was able to collect (in another cap bank) 0.32J as the coil fell back down with most speed.
So if we deduct that we are basically 0.08J under unity.
So it looks like at best around 90% efficiency.No so bad :)
Now that it's confirmed the device is Under Unity, can you please help me to rate it as it's ideal work solution, a Solenoid.I would need some of you guys, make systematic tests of different kind of coil so I could analyse it and produce a spreadsheet able to give the optimum coil in any case against any factor.
There must be an established protocol to test Push or Pull force vs power
and stroke to rate the efficiency of a Solenoid as I've found charts in some solenoids pdf data (see below)The protocol is just what you do: measuring pulling force against electrical input power, and for torque: not forgetting to divide the pulling force by the radius of the application point of the dynamometer.
This would be of most help
Thanks
Luc
Thanks Khwartz for your interest and posting your ideas.You're welcome, Luc, it was "for the cause" ;)
Looks like no one knows how solenoids pull charts are done or they are not interested.
If OU is not a possibility or claimed, people are just not interested, even those who believe OU is not possible.
Luc
the total inductance (http://en.wikipedia.org/wiki/Inductance) of non-coupled inductors in parallel is equal to the reciprocal of the sum of the reciprocals of their individual inductancesIOW inductors in parallel add up in the same way as resistors in parallel.
Thanks for your post TKLuc, could you indicate the frequency of the run up and down while pulling 2.5 kg 1 time (lifting) on 2 (lifting and dropping)? (I suppose you run it against the gravity, to talk about "kg", so mass, otherwise only about "pulling force" and so "newtons"...).
I don't see or maybe don't understand the point of making a curve chart when my coils push or pull force is equal (does not change in any position) throughout its 1 inch of travel?
This is where my design blows away a standard solenoid.
As you know, solenoid are not capable of pulling over 2.5kg at the beginning of their stroke and maintain the same pull force for 23mm of travel and only consume 0.43 Watt continuous 24/7 without the coil even getting warm.
Luc
Way to go Luc.
With out a doubt one of the best achievements I have seen on any OU forum-weather it's OU or not.
Brad
Would be great, guys that if you speak about "force" you speak about "gram.force" (gf), or "kilogram.force" (kgf) but not just "gram" or "kg", otherwise it is confusing.
See on the charts you provide, Luc, it is indeed "gf" for the "force" scale. Because if you just speak about "2.5 kg" this then just a mass and this can lead to wrong interpretation of the results. Now that I have seen your vid on the very large one you've just provided, I see it had nothing to do with lifting a mass.
But you're right, even if not OU it may have a industrial interest.
BTW, I would be very happy if "just for the pleasure" you could make a video, Luc, of the energy consumption against de lifting of a mass by cap discharge, like you have done apparently for yourself already: having the numbers is quiet good but seeing the thing been done is great too ;)
Here is the efficiency test video:
https://www.youtube.com/watch?v=-zdfBbDarQw (https://www.youtube.com/watch?v=-zdfBbDarQw)
Here is the Math:
Super Cap start Voltage 2.07197vdc x 650F = 1395.244J
Super Cap end Voltage 2.07110vdc x 650F = 1394.073J
= 1.185J used
then we subtract 0.54J which is the unity amount needed for the 2.35kg coil to travel 23.5mm
= 0.645J
then we subtract 0.334J collected from the recovery cap bank (12.2mF @ 7.4vdc = 0.334J)
we are left with 0.311J under unity
Ignoring the false precision for the moment, energy E in Joules on a capacitor isI think Luc was just saying yay amount of volt's in a 650f cap-much like we say 1x bucket of ice cream :D
E = 1/2 (CV2)
So your starting energy in Joules is
E = 1/2(650 F x 2.07197 V x 2.07197 V) = 1395.244396292 Joules.
Ending energy in Joules is
E = 1/2(650 F x 2.07110 V x 2.07110 V) = 1394.07294325 Joules.
You got the right answer even though your stated formula is wrong. Therefore you did not use your stated formula, but actually used the correct one.
It is really difficult to check your work if your answers and your formulae do not agree.
Can you really measure voltage on a capacitor to the tens of microvolts precision? I am jealous.
Here is the efficiency test video:Luc-why is the coil sliding so slowly up the guide's?. It almost looks as though the slides are coated in honey or something ???. Have you taken into accound the friction on those slides?-there seems to be to much friction there.
https://www.youtube.com/watch?v=-zdfBbDarQw (https://www.youtube.com/watch?v=-zdfBbDarQw)
Here is the Math:
Super Cap start Voltage 2.07197vdc x 650F = 1395.244J
Super Cap end Voltage 2.07110vdc x 650F = 1394.073J
= 1.185J used
then we subtract 0.54J which is the unity amount needed for the 2.35kg coil to travel 23.5mm
= 0.645J
then we subtract 0.334J collected from the recovery cap bank (12.2mF @ 7.4vdc = 0.334J)
we are left with 0.311J under unity
your formulae do not agree.You're not jealous!... you're sarcastic ;)
Can you really measure voltage on a capacitor to the tens of microvolts precision? I am jealous.
Luc-why is the coil sliding so slowly up the guide's?. It almost looks as though the slides are coated in honey or something ??? . Have you taken into accound the friction on those slides?-there seems to be to much friction there.
P.S-just watched the video again,and even when coming back down it seem'd very slow,regardless of weather or not you where collecting the BEMF. How fast dose it fall with the recapture cap's disconected(free fall)>?
Hey Brad,Mmm-ok,so this is working much like the maglev train dose.\
what you're observing is the generator effect that I've been talking about. When I connect the coil lead to the super cap, as the coil moves up current is also produced in the coil while moving in that powerful magnetic field, so it's also trying to charge the supper cap as it's being fed by the super cap, so it's a slow climb uphill ;D
So now you're observing a visual of this effect and why this cannot go to OU.
If only there was a way to have separate power source leads (unlink) to each coil lead, one for positive and one for negative, then the current the coil produces as it moves would have no where to go and the coil would move freely. But how could current flow if the positive and negative arn't linked?
Maybe EV Gray found a way to do that and the link is using the environment to complete the circuit by a super abrupt discharge of some kind?
When the coil comes back down the same thing is going on. It's charging a 12,200uF cap bank from zero to 7.5vdc, so it's fall is being dampened by this load. Without the cap it crashes back down like a falling 2.35kg weight would. The guides are working just fine ;)
I think GM Corvette had shock absorbers working under this principal?... ReGen shock absorbers would be a good option if you have an off road electric racing car.
Hope this clears up the visual effect?
Luc
Take a well wrapped bifilar solenoid coil connected serially with a welding rod core. Hang a steel carving knife by a string. Aim the center of the coil at the dangling knife and pulse it like you would a Leedskalnin device shorting the coil across a 12 volt battery. Watch what happens to the knife. Next, move the coil further away and try it again. This laser dimension magnet wave was broadcasted and received by Tesla at a distance of 40 miles from his downtown laboratory to West Point on line of sight. This was the first wireless transmission in History!. The broadcast and receiver coils were identical and both were grounded. The wave carried power that Tesla believed traveled through the ground!Here's a picture of a magnetic ray:
I re-discovered this effect by accident as I've recounted in the past: My first shop wound bifilar coil, 350 turns of 22 gauge, with welding rod core slid around ten feet to collide with a cutlery box that was drawn an equal distance along my kitchen counter, from one direct short pulse. The experience was traumatic! The magnetic force produced this way had nothing whatsoever to do with the customary D.C. Joule to coil flux ratios. Try it!
Take a well wrapped bifilar solenoid coil connected serially with a welding rod core. Hang a steel carving knife by a string. Aim the center of the coil at the dangling knife and pulse it like you would a Leedskalnin device shorting the coil across a 12 volt battery. Watch what happens to the knife. Next, move the coil further away and try it again. This laser dimension magnet wave was broadcasted and received by Tesla at a distance of 40 miles from his downtown laboratory to West Point on line of sight. This was the first wireless transmission in History!. The broadcast and receiver coils were identical and both were grounded. The wave carried power that Tesla believed traveled through the ground!Hi synchro1. Thanks for your input :)
I re-discovered this effect by accident as I've recounted in the past: My first shop wound bifilar coil, 350 turns of 22 gauge, with welding rod core slid around ten feet to collide with a cutlery box that was drawn an equal distance along my kitchen counter, from one direct short pulse. The experience was traumatic! The magnetic force produced this way had nothing whatsoever to do with the customary D.C. Joule to coil flux ratios. Try it!
Hi synchro1. Thanks for your input :)@Khwartz,
I am not English tongue and have hard time to well understand your experiment, may you publish a schematic of with the according values of the different characteristics?
Ignoring the false precision for the moment, energy E in Joules on a capacitor ishehe, you're right, TK, Luc didn't put the formulas right but he got the right results cause he simplified in his optmised mind its expression :)
E = 1/2 (CV2)
So your starting energy in Joules is
E = 1/2(650 F x 2.07197 V x 2.07197 V) = 1395.244396292 Joules.
Ending energy in Joules is
E = 1/2(650 F x 2.07110 V x 2.07110 V) = 1394.07294325 Joules.
You got the right answer even though your stated formula is wrong. Therefore you did not use your stated formula, but actually used the correct one.
It is really difficult to check your work if your answers and your formulae do not agree.
Can you really measure voltage on a capacitor to the tens of microvolts precision? I am jealous.
Mmm-ok,so this is working much like the maglev train dose.\Very Great imput: Cery Relevant imho, Brad! :D
Quote: So now you're observing a visual of this effect and why this cannot go to OU.
If only there was a way to have separate power source lead's.
Wait just a minute there-lets not give up yet,as i believe there is a way to fix this problem. You have done before what is needed here Luc,and now you must do it again.It's time to put the effects of two of your projects together. You need to offset the voltage and current by 180* during the P/in cycle to your coil,so your going to need a pulsed input at the right frequency for your setup. As you know,when you switch of an inductor,the current will keep flowing in the same direction,BUT the voltage polarity will reverse-->If only there was a way to have separate power source lead's.
You dont need seperate lead's,as one set will do both job's,but you have to get the offset and frequency right for that coil.Normally with an air core inductor(coil) you would need a high frequency,but with that strong magnetic field i think the frequency needed would be quite low.
First to nut out a circuit to do the job,then find the right frequency to offset the current and voltage.To do this you need to fully understand as to what happens when an inductors P/in is suddenly cut off when running with a 180* current/voltage offset in the strong magnetic field it is in.This is something i dont remember anyone here ever looking into,or try doing. But i have actually done this myself,although the setup was a little different. I believe i called that particular project the magneformer-not quite sure,as it was some time ago. I remember showing TK my result's at being able to get a 180* offset of current over voltage,and if i remember rightly,he said i was creating a standing wave within the unit.This way i was able to maintain a continuous current flow within the inductor,but charge a cap with an opposite potential than that of my P/in-useing the same two wires. The circuit wasnt all to different to that of the SSG,which basically dose the exact same thing if you look at how the circuit work's-the negative of the charge battery is hooked to the positive of the run battery.
I guess the thing you will need to make this work is a signal generator-do you have one of these?.
Sorry Synchro1 but that is just another nonsensical fantasy posting. I don't want to disrupt the thread, but that is the honest rebuttal to your posting. You have made other fantasy postings in this thread that disrupt what is going on. I will ignore them from now on.
Hi Luc,
The concept about bucking coils is very interesting to me, and it seem to have been used in many overunity devices: http://www.hyiq.org/Downloads/Guidelines%20to%20Bucking%20Coils.pdf (http://www.hyiq.org/Downloads/Guidelines%20to%20Bucking%20Coils.pdf)
Hi Luc,
This post is about your video: https://www.youtube.com/watch?v=PTykNjDD0CM (https://www.youtube.com/watch?v=PTykNjDD0CM)
It is just to reference you, for your info, this other video about this scheme , published in hyiq.org about Bucking Coils: https://www.youtube.com/watch?v=Z-V1z2TdQJA (https://www.youtube.com/watch?v=Z-V1z2TdQJA)
I suppose that if in your video you would have used an intermediate tap between the CW coil and the CCW coil to extract the induced voltage you would have got current without the Lenz effect...
The concept about bucking coils is very interesting to me, and it seem to have been used in many overunity devices: http://www.hyiq.org/Downloads/Guidelines%20to%20Bucking%20Coils.pdf (http://www.hyiq.org/Downloads/Guidelines%20to%20Bucking%20Coils.pdf)
Hanon,
I agree one can find the bucking coil in many concepts. Yet it bends my mind... Each coil would generate a CEMF, but as the currents collide at the center tap, no current may flow at all ..? And of cource, no CEMF.
Have you successfully obtained any current flow from a bucking coil?
So far most of what I have seen here uses grade 8 ceramic mags. What would the torque be if using neo's? The fact that this doesn't draw more amps under load is a plus. Although it does slow down under load. This is a possible candidate for a gen head wired for a certain rpm so as to run in a 80% bell curve. As it is loaded, it will slow some and back down first into the high point of the curve and then back down to the 80% level.
thay