|
Post by cornboy on Feb 13, 2020 14:24:51 GMT -6
OK, understand Guys, thanks.
Regards Cornboy.
|
|
|
Post by Marathonman on Feb 15, 2020 12:17:44 GMT -6
I would like to announce a newest member of the forum Florin has joined the ranks. welcome Florin to the best Figuera site on the net and if you have any questions just ask if you do not understand something. i am so glad you took the first step to energy independence by joining up and together with the rest of the members and builders we will make a difference.
All; remember it is easier to remove winding's on part G then it will be to add them. if you need 5 lbs of pressure on the primaries you should test them first to assure you are in the ball park.
Again welcome, Marathonman
|
|
|
Post by creasysee on Feb 16, 2020 13:36:06 GMT -6
Hi MM, can you explain this, please: if you need 5 lbs of pressure on the primaries you should test them first to assure you are in the ball park. I do not understand this. Always when anyone refers to the force or the pressure must specify a gap. The force's strength of the electromagnet is inversely proportional to the square of the gap. The dependence of the thrust force on the gap is called the traction characteristic. Example of the traction characteristic: So, you cannot say you need 5 lbs. Or you need 18 lbs. You need to specify the gap (distance) between a core of an electromagnet and a metal plate with the digital meter to give you the pull force of the electromagnet. Of course, we understand that the gap depends on the quality of polishing both surfaces: north pole of the electromagnet and one side of the metal plate. This describes the formula ( link): Using this formula (2 Amps, 315 turns, magnetic constant 4xPIx10 -7, 0.001 square meters) I get the following table: Gap, mm | Force, Newton | Force, lbs | 0.05 | 99751 | 22425 | 0.1 | 24937 | 5606 | 0.5 | 997 | 224 | 1 | 249 | 56 | 2 | 62 | 14 | 5 | 9.9 | 2.2 | 10 | 2.5 | 0.6 |
Note: In the table the gap provided in mm, of course in the formula above the same gap used in meters.
Thanks, creasysee
|
|
|
Post by creasysee on Feb 17, 2020 3:20:34 GMT -6
Hi all! I was wrong above in calculations an energy for primaries and accordingly inductance. I've created Excel file that helps to calculate this: and I found that you can use both variants of primary connections: series and parallel. The total current, total impedance, energy and supply voltage aren't changing (almost). The inductance is changing only: big inductance primaries should be connected in parallel, low inductance primaries should be connected in series. Of course, the low inductance should be wound a thick wire, the high inductance - thin wire (but not very - the current is reduced in proportion to the number of triplets). I'll fix the previous messages and pictures in a few days (fixed). Regards, creasysee
|
|
|
Post by creasysee on Feb 17, 2020 10:06:44 GMT -6
Hi Skyrob! that was reason for winding more turns According to the previous post, you cannot get any inductance (not so, of course, you can, but this will not be effective). The inductance (number of turns) should have a value that provides storing energy that will be pushed into primaries from a part G. Regards, creasysee
|
|
|
Post by Marathonman on Feb 17, 2020 12:55:50 GMT -6
QUOTE; "The total current, total impedance, energy and supply voltage aren't changing (almost). The inductance is changing only:"
Your getting closer, the total stored flux does not change much because of the reluctance and free air having to travel through air to get to the opposite pole. it will retain 80 to 90 % of total flux throughout the sweeping action. all we are doing is directing more current to one primary then the other and vice verse which is a shift of two different fluxes that are in two different directions used to compress field lines that occupy the secondary one at a time then removed thus fulfilling the same actions of a standard generator north south/ south north field direction when rotating secondary through them.
Both reducing primaries and half of part G combine to offset the rising side storing into the magnetic field. the secondary feedback is there to replace losses and to give rise to amplification to the rising primaries replacing the reduced primaries reduction in field line pressure.
no negative just the opposite current direction. Regards, Marathonman
|
|
|
Post by creasysee on Feb 18, 2020 1:39:04 GMT -6
Hi all.
I wound 12 mH and I plan to use 7 triplets instead 5. I'm using AWG13 wire (don't have other) and 6 layers look good. The 7th layer I think will be superfluous. So, the primary has 271 turns.
I think so because solenoids work fine when they are long and work worse when they short. I can mistake of course, and my decision is still inconclusive. External layers require a lot of wire and they work less efficiently than internal layers. Let me know if you have any comments.
Regards, creasysee
|
|
|
Post by creasysee on Feb 18, 2020 1:52:09 GMT -6
Note: the first time I will use 1000 min-1 and this gives 16 Hz, so I will have:
This requires 10 Amps @ 176 V from the power supply (on start step) and the thick wire on the primary will work well.
Regards, creasysee
|
|
|
Post by Marathonman on Feb 18, 2020 9:43:26 GMT -6
You are correct on the magnetic field of the short primaries but as your picture from above it is perfect for a secondary because of the bulging magnetic field. primaries on the other hand and in this situation needs to be longer then the diameter to produce a field capable of projecting out farther then the opposite side of the secondary. Quote; "and I found that you can use both variants of primary connections: series and parallel."
EXACTLY as did Figuera.
Regards, Marathonman
|
|
|
Post by Marathonman on Feb 21, 2020 14:42:47 GMT -6
Pretty much every old book i have read on electromagnets or solenoids all say the same thing when building fast acting electromagnets.
"The advantage of parallel grouping is simply that it reduces the time-constant"
just as i have been saying so dividing the winding's of the primaries in groups will reduce the time constant as well as the resistance. reducing the resistance of any such circuit results in a more efficient circuit and use of a magnetic field.
Regards, Marathonman
|
|
|
Post by creasysee on Feb 22, 2020 15:21:56 GMT -6
Sure, absolutely true. But it reduces the inductance and I cannot use it because I don't have thinner wire.
Regards, creasysee
|
|
|
Post by cornboy on Feb 22, 2020 19:36:35 GMT -6
Hi Skyrob and all, i haven't been able to prove it yet, but if G controls the current flow, you will want almost no resistance in your primary electromagnets.
If you need to increase the inductance, without resistance, winding with insulated copper foil will fit the bill. Just saying, have not tried it yet.
Regards Cornboy.
|
|
|
Post by cornboy on Feb 23, 2020 0:09:07 GMT -6
Holy crap Skyrob, that's one heavy duty set up, never thought of doing it that way. Can't wait to see it operational.
Regards Cornboy,
PS. Where did you get those slip rings?.
|
|
|
Post by creasysee on Feb 23, 2020 4:58:49 GMT -6
Hi Skyrob would you help with calculations for "G"? and primary? what i should expect from experiment with inductance of 36mH and 2 ohm or 1.43mH 0.045ohm ? this should drive 16 primaries 8 "N" and 8 "S" im intending to have 4 x triplets only due one triplet will have 2x + 2x parallel each primary is 743mH @ 6.3 ohm (two in paralel for each triplet ) Your primaries have too high inductance for your part G. you will need to use 21 triplets: Or you need to reduce inductance:
Select the current according to the overall power of part G. Regards, creasysee
|
|
|
Post by Marathonman on Feb 23, 2020 13:20:34 GMT -6
Just so you know primaries are not storing vast amounts of energy, they will release as much as it is needed for the increasing side which will off set the storing into the magnetic field drop in potential. the storing potential will be the same as the releasing potential as both sides are mirror images of each other. the secondary feed back is controlled by part G which will take what is needed to replace losses and to give rise to amplification. part G governs the current and potential as it nears saturation which will then limit the current flow. this also includes the starting potential limiting through part G. Can't for the life of me figure out why in the world you need so many slip rings. that is a whole lot of drag on the motor and outside the patent. yes the reducing side of the system includes the reducing primaries, reducing half of part G and the secondary feed back which when combined will raise the overall system voltage forward biasing the rising primaries allowing more current to flow which will replace the reducing primaries loss in field line pressure being reduced to create the electric field the secondary Lenz Law field is push across. so basically part G acts as an amplifier to the original signal giving rise to amplification to the rising primaries. correct, part G controls the current flow NOT the primaries so wind accordingly. This is what i am talking about when referring to siphoning off of the magnetic fields. in the middle the opposite pole will jump to the other core when presented with the opposite pole. then to make matters worse we are compressing the field lines which will make it even worse. this is indicated in the red square. all poles have to be the same or the siphoning will take place with a reduced output. Regards, Marathonman
|
|