The Ossie motor

[gyulasun]

I think the resistance is magnetic The other difficulty is that the terminals are on the rotor of course! I have to spin the stator. mmmm

Yes I thought of that is magnetic.

I think the best construction for this hub would be to fix its axle (i.e. its rotor) vertically into a wooden board and fasten a non magnetic wheel to its outside stator.  And this wheel would be your rotor for the magnets that pass in front of your stationary (air core?) coils. This wheel ought to be at least 25-30cm of diameter at minimum.

Gyula

[Augen]

Hello All!

Jimboot! I'm new here, but I've been following your work for a long time. Nice work! 

I'm doing some research myself in that topic, and I have a few thoughts about them.

I'm sure you are familiar with the following formula: B=N*I/l. Consider this. If you use your wire from your 1.8mH coil to make a new coil, but you lower the length of the coil, it will be much stronger then the original coil. For example your coil is 1" long and it has B1. Make a new coil with 0.5" length. This means "N" and "I" is still the same, but "l2" is half of the original "l1", so B2 = 2*B1. So the repulsive force between the magnet and the coil will be twice as much as it was before. But I don't know how much can you lower the length of the coil. I mean if your coil is 2 mm long, the upper formula is still good? Now I'm using 1 coil. I attached it below. What do you think? Later I will show some photos about my work.

Regards, Augen

[Magluvin]

Hey Jim
If its the type of gen or dynamo Im thinking of, when you turn it, it has many resting points right? Well you will need to add weight to the rotor to get it to run through those sticky spots. For example, if you added some flywheel to the gen as it sits, you will see that it  will spin longer, and the more weight, the smoother it will spin.
Your motor will spin a heavy rotor, it will just take time to get it to top speed. So just give it a spinup then connect the power so the reed wont take a slow beating.

Mags

[gyulasun]

...
I'm sure you are familiar with the following formula: B=N*I/l. Consider this. If you use your wire from your 1.8mH coil to make a new coil, but you lower the length of the coil, it will be much stronger then the original coil. For example your coil is 1" long and it has B1. Make a new coil with 0.5" length. This means "N" and "I" is still the same, but "l2" is half of the original "l1", so B2 = 2*B1. So the repulsive force between the magnet and the coil will be twice as much as it was before. But I don't know how much can you lower the length of the coil. I mean if your coil is 2 mm long, the upper formula is still good?
...

Hi Augen,

The formula you wrote above is not familiar in that form because you used B instead of H.  H is the H field in A/m, H*l gives the magnetomotive force, while B is the magnetic field in Tesla.   So the correct formula is H=I*N/l
If you wish to use B here, you can because B=u*H where u is the permeability.  So H=B/u  and the formula becomes B=u*I*N/l

What you mean with the formula sounds ok to me and the physical explanation for the stronger force is that for a coil of short length the magnetic field has to stay and cover a shorter path in the air than in case of a longer coil and as is known the air's magnetic conductance is poor. Possibly the 'pancake' coil form is the shortest construction possible because its 'length' is just the wire diameter.
But you have to find a tradeoff between the coil length and diameter because the shorter you choose the coil the higher its diameter becomes, assuming the same Amperturns, I*N is to be maintained.  And higher diameter coils involve higher diameter permanent magnets.

Thanks, Gyula

PS: here is a link to the coil calculator for those wishing to use it:
                   http://www.pronine.ca/multind.htm

[Augen]

Hi Gyula,

You are right. B=u*I*N/l. That is the formula I'm using, but I forgot about the 'u' part in my previous post. Thanks for it.

My problem is the following. I did some tests, but the result wasn't what I expected. A made 2 coils. One of them is on the  picture I previously uploaded. The other one has the same outer and inner diameter, but half the length (2,5 mm). I connected them in series (so they've got the same current). Using the formula you mentioned, both coils should have the same strength. In my test, the thicker coil was a bit stronger. And I don't know why. Today I'm going to do some other test. I'm gonna connect my coils (I have 4 coils) in series and in parallel, and check the results. I hope I will get some answers. But I'm almost sure, the coil Jimboot is using could be much stronger with a shorter length (using the same current).

Thanks, Augen