Author Topic: Rotational PMM by wizkycho (Read 68316 times)

wizkycho · « **on:** June 05, 2005, 08:32:26 PM »

http://starglider.netfirms.com/RPMMonly.html

both linear and rotational motors works
read carefully.

it is logical to reason (only first look) that two magnets will attract shield two times as hard as one magnet. even in that situation
2* Emag(left side) - 1* Emag(right side)= 1* Emag AT THE SHAFT (INPUT ENERGY).
This means i need exactly ammount of energy to overcome ONE magnet at the input.
As a result of invested energy (shielding removed) I get 2 * Repulsion Emag at the OUTPUT (piston).
therefore my device operates at COP=2 - some friction => COP=1.8 which is still FE

You all (i'm little dissapointed) overlooked one property of permeable materials it is satturation (of shielding)

If shield is satturated to maximum it will "REPEL" !?! (bounce) from magnet to the place on magnet where is less satturated and field density is lower. You don't believe this see this very simple experiment I assure You that this is not happening cause of induced current in permeable object.
the experiment: http://www.geocities.com/wizkycho/index_files/magexp.avi
back to linear:
So if one magnet is sufficient to satturate shield (with respect to its thickness) to it's satturation knee there is no place for additional magnetic field from the other magnet niether to attract nor to repel that shielding.

Callibration doesn't have to be that precise and can be done by 0.5mm gradient (for thickness of shield).

in worst case (unsuitably thick steel) you'll get? 2:1 for OUTPUT Yeah !
if you experiment with different thickness of shield you are in 7:1 range

I think with this we have depleted all open questions for linear motor

What Ya think 'bout Rotational.... cmon I dare you . I think it is very very powerfull
there is an error in COP estimation. It is well beyond 14.

Cmon Brainstormers

m668004 · « **Reply #1 on:** June 05, 2005, 09:08:11 PM »

@wizkycho

I think I understood. The permeable rotor is already saturated by the statormagnets before to rotor with the magnets is inserted. So when the permeable steel is moved the rotormagnets move not because they are attracted to the steel (they can't, its saturated) but they moved only because they were repelled by the outer magnets when they are not longer covered by the steel.

Maybe the rotormagnets will begin to swing between the two repelling zones at the ends of the steel covered zone if the repellation is to strong. What do you think?

M.

wizkycho · « **Reply #2 on:** June 05, 2005, 09:55:33 PM »

Quote from: m668004 on June 05, 2005, 09:08:11 PM

@wizkycho

I think I understood. The permeable rotor is already saturated by the statormagnets before to rotor with the magnets is inserted. So when the permeable steel is moved the rotormagnets move not because they are attracted to the steel (they can't, its saturated) but they moved only because they were repelled by the outer magnets when they are not longer covered by the steel.

Maybe the rotormagnets will begin to swing between the two repelling zones at the ends of the steel covered zone if the repellation is to strong. What do you think?

M.

if you are refering to rotational design:
why would it swing, does it have AC current applied somewhere or the shaft is frictionless!?!

1.there is shielding and no matter how inprecise you make it will represent "walley" of mag field strenght so rotor will rest where this "walley"(lowest repel) is.

2.You can make better (thicker shielding) that just very slightly attracts Output rotor magnets, so the walley becomes
abbys and Output rotor shurely must go and stay where that shielding is.

3. Please observe that this shielding is U (fliped hor.) shaped so both NORTH vectors (from Output rotor magnets and ring mags.) goes to attractional SOUTH (on ring mags.) through that U (fliped hor.) shield

SO THE SHIELD CAN BE VERY THIN AND STILL SECURING ABBYS FOR OUTPUT ROTOR MAGNET TO REST THERE.(if shielding is resting)

No swinging sorry !
THIS STILL WORKS VERY VERY GOOD, COME AND STOP IT.
Next...

ooandioo · « **Reply #3 on:** June 05, 2005, 10:28:19 PM »

Here is whats going to happen.
The inner rotor (magnet) will follow the outer rotor (permeable steel) with exactly motion.

The inner rotor is repelled on both sides with same F - so it will do nothing. After mounting the shield it will head it, because 1. its magnets are attracted by the steel or 2. the steel provides the only "magnetic free area" in this circle.

As the shield is moved the magnets will follow it.

-- Andi.

m668004 · « **Reply #4 on:** June 05, 2005, 10:54:07 PM »

Perhaps we can generate electricity directly with the movable shielding, when we replace the inner magnetic rotor by a coil, a toroidal coil in this case. The magnets and the coil are static, only the shielding moves, so a periodicly changing magnetic field is created and a current in the coil is induced - without moving the magnets or the coil! The big question is now: will the rotation of the shielding slowed down by lenz-law?

M.

wizkycho · « **Reply #5 on:** June 05, 2005, 11:14:25 PM »

Quote from: ooandioo on June 05, 2005, 10:28:19 PM

Here is whats going to happen.
The inner rotor (magnet) will follow the outer rotor (permeable steel) with exactly motion.

The inner rotor is repelled on both sides with same F - so it will do nothing. After mounting the shield it will head it, because 1. its magnets are attracted by the steel or 2. the steel provides the only "magnetic free area" in this circle.

As the shield is moved the magnets will follow it.

-- Andi.

I won't answer to this but I can say (scince you start GUESSING) that You are very very far or to fast in undrstanding
magnet phisycs.
Please Correct Yourself.
Maybe You want slower animation !?!

Once Again
Green(with brown) - input rotator (little enrgy only for friction)
Yellow - output rotator capable of Full repel action (if under max load - why wouldn't it be)

Magnet at outer rotator just MUST (high energy output) follow shield (little energy input)
is it so hard to see. Cmon people there are only TWO ROTORS.

Look again

wizkycho · « **Reply #6 on:** June 05, 2005, 11:26:27 PM »

Quote from: m668004 on June 05, 2005, 10:54:07 PM

Perhaps we can generate electricity directly with the movable shielding, when we replace the inner magnetic rotor by a coil, a toroidal coil in this case. The magnets and the coil are static, only the shielding moves, so a periodicly changing magnetic field is created and a current in the coil is induced - without moving the magnets or the coil! The big question is now: will the rotation of the shielding slowed down by lenz-law?

M.

scince there is now relevant change in field of the shield you induce very little or nothing.
sorry !
Am I right ?

m668004 · « **Reply #7 on:** June 05, 2005, 11:30:05 PM »

@wizkycho
@ooandioo

I think its important for wizkychos device, that the magnetic rotor is under load, so it resists an easy moving. If under these circumstances the shielding moves, than the magnetic rotor will and must follow with the same speed and more energy output than input is arrived.

M.

ooandioo · « **Reply #8 on:** June 05, 2005, 11:34:42 PM »

Quote from: wizkycho on June 05, 2005, 11:14:25 PM

Quote from: ooandioo on June 05, 2005, 10:28:19 PM
Here is whats going to happen.
The inner rotor (magnet) will follow the outer rotor (permeable steel) with exactly motion.

The inner rotor is repelled on both sides with same F - so it will do nothing. After mounting the shield it will head it, because 1. its magnets are attracted by the steel or 2. the steel provides the only "magnetic free area" in this circle.

As the shield is moved the magnets will follow it.

-- Andi.

I won't answer to this but I can say (scince you start GUESSING) that You are very very far or to fast in undrstanding
magnet phisycs.
Please Correct Yourself.
Maybe You want slower animation !?!

Once Again
Green(with brown) - input rotator (little enrgy only for friction)
Yellow - output rotator capable of Full repel action (if under max load - why wouldn't it be)

Magnet at outer rotator just MUST (high energy output) follow shield (little energy input)
is it so hard to see. Cmon people there are only TWO ROTORS.

Look again

Sure - i understand the system. I only explained it in my words to help understand...

In my opinion, thats an idea worth spending time for. Any more details for building it? Are you going to build it?

-- Andi

wizkycho · « **Reply #9 on:** June 05, 2005, 11:46:46 PM »

Quote from: ooandioo on June 05, 2005, 11:34:42 PM

Quote from: wizkycho on June 05, 2005, 11:14:25 PM
Quote from: ooandioo on June 05, 2005, 10:28:19 PM
Here is whats going to happen.
The inner rotor (magnet) will follow the outer rotor (permeable steel) with exactly motion.

The inner rotor is repelled on both sides with same F - so it will do nothing. After mounting the shield it will head it, because 1. its magnets are attracted by the steel or 2. the steel provides the only "magnetic free area" in this circle.

As the shield is moved the magnets will follow it.

-- Andi.

I won't answer to this but I can say (scince you start GUESSING) that You are very very far or to fast in undrstanding
magnet phisycs.
Please Correct Yourself.
Maybe You want slower animation !?!

Once Again
Green(with brown) - input rotator (little enrgy only for friction)
Yellow - output rotator capable of Full repel action (if under max load - why wouldn't it be)

Magnet at outer rotator just MUST (high energy output) follow shield (little energy input)
is it so hard to see. Cmon people there are only TWO ROTORS.

Look again

Sure - i understand the system. I only explained it in my words to help understand...

In my opinion, thats an idea worth spending time for. Any more details for building it? Are you going to build it?

-- Andi

well I didn't mean to be rude. but I got impression that you thought about this design with only 1% of your barin.

You all must Know that there has not been more powerfull OU device based on permanent magnets that surely works
than this one..

hartiberlin · « **Reply #10 on:** June 05, 2005, 11:54:10 PM »

Quote from: wizkycho on June 05, 2005, 08:32:26 PM

http://starglider.netfirms.com/RPMMonly.html

@wizkycho

When you move the shield you wil have to use more force, cause you want to move the shield
also out of the attraction of the rotor magnet.
So EnergyIN is higher than you think !

So I don?t know, if this is really better than the linear version.

Regards, Stefan.

ooandioo · « **Reply #11 on:** June 05, 2005, 11:56:32 PM »

The different forces that attract here are hard to imagine...

Don't you think that there also is an attraction between the shield and the inner magnets, when inner magnets are forced to follow the shield cycling?
This force would count towards the F needed to rotate the shield...

Mh, need a practical sample.

-- Andi

ooandioo · « **Reply #12 on:** June 05, 2005, 11:57:15 PM »

Quote from: hartiberlin on June 05, 2005, 11:54:10 PM

Quote from: wizkycho on June 05, 2005, 08:32:26 PM
http://starglider.netfirms.com/RPMMonly.html

Right what i say...

-- Andi

@wizkycho

When you move the shield you wil have to use more force, cause you want to move the shield
also out of the attraction of the rotor magnet.
So EnergyIN is higher than you think !

So I don?t know, if this is really better than the linear version.

Regards, Stefan.

m668004 · « **Reply #13 on:** June 06, 2005, 12:07:56 AM »

I think the attraction between the shielding and the inner magnets are not the problem, because the inner magnets move always parallel to the shielding and not perpendicular. Look at Butch Lafontes Device. A problem would appear if the inner magnets were moved totally out of the shielding but that can't happen because of the repelling forces of the outer magnets.

M.

Light · « **Reply #14 on:** June 06, 2005, 12:09:30 AM »

In God's sake, Wizk, WHY the rotor's magnet will follow the "screen"? System is closed, they (rotor and screen) just attract each other and all system will stop.

"THIS STILL WORKS VERY VERY GOOD, COME AND STOP IT".
Even if you buy a ticket for me, I won't go to waste time, couse it will not (and can not) ?work this way.

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