Mostly Permanent Magnet Motor with minimal Input Power

[Khwartz]

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.

Understand your feelings, Brad, but I was going just a little bit further and more general for the seak of a overunity COP.

But as a matter of fact, I don't know for now any standard operations in physics allowing to get a COP from a none moving pulling force :/

Indeed, if you sustain a mass in the gravity to a steel hook, we don't consider that the hook consumes any energy, so NO consumption power involved.

The only think we can do I think is as we do right now: comparing with other systems; I mean ONLY IF NOT MOVING.

That is very why I didn't care about calculations about the setups you are running presently, Brad, but of the only setups which allow us to have a mechanical POWER [W] (not just a FORCE [Newtons] ), so we are able to get a COP.

Then, it will be only with a TRANSLATING or ALTERNA or ROTATIONNAL setup we will be able to calculate any COP.

Then, it is possible to calculate a COP by just a simple lifting cause we do have A MOVING FORCE in a TIME PERIOD. But you already know the maths as good as me on this one

Still, my maths of my previous post (that I will repost to correct minor errors or clear up few points), could be usable especially if a link-pulley system.

[Khwartz]

I repost here this very very long post of maths about Luc's MMM, to make few minor corrections for seak of more clarity and to add an expression in words of the necessary conditions for a COP in this particular setup but not only:

TO HAVE OVERUNITY COP, THE ELECTROMAGNETIC FLYBACK COEFFICIENT "JUST" NEED TO BE STRICTLY SUPERIOR TO THE INVERT SUM OF ALL COEFFICIENTS OF LOSSES. 

Looks it is "as simple as that". ^_^

---------

@ 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 can see while checking the equations for the system with a spring at bottom, is that if Wm2 would be transferred as per 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 (note that in this case D > 1 while in the previous operations, the electromagnetic flyback coefficient was < 1, for just a question of how it used in the equations).

We see then that the NET ENERGY GAIN for the 2 first periods, for example, are: (we note We1 = We2 = W3 = ... = constant = We).

G1 and G2 are the net energy gain at period 1, "T1", and period 2, "T2":

● T1:
G1 = Wm1 - We

= electrical excitation energy, reduced by electromagnetic and mechanical losses, minus Energy of ElectricalExcitation.

= B × (A × We - We )
= B × (A - 1) × We
-> a loss

● T2:
G2 = Wm2 - 2We

= spring reflected mechanical energy + electromagnetic flyback energy + electrical excitation energy, reduced by electromagnetic and mechanical losses (spring losses and translating losses), minus energy of two electrical excitations.

= 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 and 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.


Well, I have made my 24 hours around the clock now, hope few of you guys will appreciate the work!  ^_^


Regards,
Didier

[gotoluc]

Just a quick update,

both coils are wound, Epoxy and Glass coated.
The next step is making the end glide bushing system.

From today I'll be away for 5 days and on my return I'll start making the glide system. So it may be another 10+ days before I can test the Super build.

Stay tuned

Luc

[Khwartz]

Well received, Luc. Great you've advance on your setup.

Nice 10 days what you need for!

Best regards,
Didier

[gotoluc]

Hi everyone,

I'm now back and will resume the Super build.

I made a video to show the coils finished with the Epoxy and Fiberglass cloth.
I also included a test to demonstrate the opposite pole on the outside of the coil which is not commonly known or used.
Link to new video:
https://www.youtube.com/watch?v=IvfwELDmKIM


The illustration below demonstrates the poles of the electromagnet and how my design uses it.
Below is an old video which demonstrates the double pull torque advantage if you have not seen it yet.
Link to old video:
https://www.youtube.com/watch?v=-eTQ49RcFKM

Luc