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Author Topic: Mostly Permanent Magnet Motor with minimal Input Power  (Read 252470 times)

Khwartz

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Re: Mostly Permanent Magnet Motor with minimal Input Power
« Reply #240 on: August 29, 2014, 01:55:53 PM »
Hi Brad. Nice we agree :)

For your vid on UFO torque motor measurement, Very Nice that you've noticed the error of adding dynamometers reading instead of subtract them, and Very Kind i.m.o. you took time to demonstrate it :)

For the calculations using only the metric system, I find same result as you at 2%: 257 [W], 0.345 [HP] :)

But is that my bad English but I couldn't find the way you have run your calculations until many comparisons between your calclations and mine ^_^.

The reason why is that looks to me you didn't describe a step while you have used the value of the diameter and not the radius, then dived by 2 and then multiplied by Pi. I couldn't find from where were coming the 7.17 inches! Lol but now it's ok:

58 [mm] = 5.8 [cm] = 5.8 [cm] × 0.3937 [inches/cm] = 2.283 [inches] of diameter;

2.283 [inches] × Pi = 7.17 [inches].

This step was missing or you said it verbally and I didn't get ^_^ but thanks for the vid, it Very Shows imho how you care about truth in the Free Energy Quest. Thanks for helping here, Brad :)

Regards,
Didier






gotoluc

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Re: Mostly Permanent Magnet Motor with minimal Input Power
« Reply #241 on: August 30, 2014, 05:49:07 AM »
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

tinman

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Re: Mostly Permanent Magnet Motor with minimal Input Power
« Reply #242 on: August 30, 2014, 06:51:15 AM »
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

Every morning and afternoon i check this thread,waiting for the MMM's first sign of life-first run. After seeing the results of your first test Luc,and seeing how much extra force/pull your setup had from that of what we have today,makes this such an interesting project. It has been a long time since something has sparked my interest this strongly. I have been thinking of a way to use your effect much more efficiently-but so far,nothing has come to mind.

A truly excellent project and design.

gotoluc

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Re: Mostly Permanent Magnet Motor with minimal Input Power
« Reply #243 on: August 30, 2014, 05:33:48 PM »
Thanks Brad for posting your thoughts towards this design.

It's encouraging to know you "one of the few real builders" is so interested to see the results of what a super build could do.

It may take another week before I have it ready for testing as the side glide and guide system will have to be accurate and very solid before testing.
Once I magnetically close the outer steel plates together, it will be very difficult to separate them. I think it's over 2,000 pounds (per side) of holding force what 16 of N52 1" cubes and 4 N52 2" x 2" square will hold. So I've got to get the coil supports right the first time :P

I know, it's not easy to think of an improved design, like one that could rotate!...  I challenge all great minds out there to submit your ideas on a rotary version of this design.

Luc

synchro1

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Re: Mostly Permanent Magnet Motor with minimal Input Power
« Reply #244 on: August 31, 2014, 01:08:56 AM »
735.5 watts to lift 180 pounds equals a horsepower. 1 watt can lift 4 pounds. Anything better then that is overunity!

gotoluc

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Re: Mostly Permanent Magnet Motor with minimal Input Power
« Reply #245 on: August 31, 2014, 01:57:37 AM »
There must be a certain amount of distance the weight needs to be lifted in a certain time frame no?

Luc

tinman

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Re: Mostly Permanent Magnet Motor with minimal Input Power
« Reply #246 on: August 31, 2014, 03:37:19 AM »
There must be a certain amount of distance the weight needs to be lifted in a certain time frame no?

Luc
@ Luc
Syncro is using metric horse power there.If you wish to use that,then the calculations are:

1 metric horse power =735.498 watts
To have one metric horse power,you must lift 75kg's 1 meter in 1 second.

Power= work/time
Lifting 1kg per second, converts 10 joules of energy to 10 watts of power.
It takes 10 joules of energy to lift 1kg 1 meter.
So to lift 1kg up 100mm,it would take 1 joule of energy.
 As an example-10v in a 20000uf cap is 1 joule of energy.

What we need is a constant p/in(watt hours) to suspend say a 1kg mass. I think it can be done using the acelleration of G in this case. I will look into it.

Khwartz

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Re: Mostly Permanent Magnet Motor with minimal Input Power
« Reply #247 on: August 31, 2014, 10:52:38 AM »
@tinman

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 :)

tinman

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Re: Mostly Permanent Magnet Motor with minimal Input Power
« Reply #248 on: August 31, 2014, 11:22:38 AM »
@tinman

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.
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?.

gotoluc

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Re: Mostly Permanent Magnet Motor with minimal Input Power
« Reply #249 on: August 31, 2014, 06:15:34 PM »
Hi everyone,

I got my 1" cube mostly magnet motor back and have put together a new test which I feel should be a fairer comparison.
In this test I use a large 20VDC 400 RPM Permanent Magnet Motor since it's torque should be very higher and a better comparison to the torque of my 3xM design.

My reasoning for this is the 3xM design is a back and forth solenoid action which has a limit of how many times it can efficiently oscillate per second. At this time it seems to do well with 6 to 10Hz. So if we added a crank shaft to convert it to rotation it should turn in the 360 to 600 RPM range.
So using the above motor seems to be a more ideal comparison. However, keep in mind that the 3xM is only 1" wide compared to the DC motor rotor which is about 6" wide. So obviously surface area is an important factor for high torque hence my new 8" wide super build.

Link to video:  https://www.youtube.com/watch?v=cS5pmKBruuU

Luc

DreamThinkBuild

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Re: Mostly Permanent Magnet Motor with minimal Input Power
« Reply #250 on: August 31, 2014, 07:37:34 PM »
Hi Luc,

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.

gotoluc

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Re: Mostly Permanent Magnet Motor with minimal Input Power
« Reply #251 on: August 31, 2014, 08:44:05 PM »
Hi DreamThinkBuild,

Thank you for your post and sharing your ideas and illustration work.

The way I have come to understand the fields around an electromagnet coil is, it is not just North South on each end of the coils.  Actually half way through the thickness of the coil there is a blotch wall and the polarity shifts.
It would look more like you illustration below where I have identified the poles on one side of the coil.
This is why in my design the magnets poles are not on the vertical plain like your illustration.  They are on the horizontal plain with the blotch wall half way trough the coil windings.
Using a coil of the same thickness as the magnet will gives most torque output.

Look at the below video and take close attention to the magnet poles. Here I demonstrate you can double the torque when you use the coils outer magnetic field which is in opposite poles of the coils inner field.
If this wasn't the case we would have a cancellation of torque since the magnets on the outside cores are in opposite field of the inner cores.

Link to the video: https://www.youtube.com/watch?v=-eTQ49RcFKM

Let me know what you think and understand

Luc

Khwartz

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Re: Mostly Permanent Magnet Motor with minimal Input Power
« Reply #252 on: September 02, 2014, 08:31:24 AM »
@ All: Please, skip the algebraic developments of this VERY-VERY LONG spot, if you're if algebra is not really your friend ;)

yes-not sure why i put watt hour's,should just be watts.
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?.
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

Khwartz

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Re: Mostly Permanent Magnet Motor with minimal Input Power
« Reply #253 on: September 02, 2014, 08:50:32 AM »
Hi Luc,

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.
Very Well Done for the design and ideas!  :)

tinman

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Re: Mostly Permanent Magnet Motor with minimal Input Power
« Reply #254 on: September 02, 2014, 12:37:44 PM »
@ 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
???

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.