Tesla's Charging Circuit and it's Application to Pulse Motors

[MileHigh]

Farmhand:

With respect to the motor coils, in theory the magnets on the rotor interacting with the "dead" motor coils will not have any affect on the rotor.  The fly-by is energy neutral, the rotor speed up on approach (assuming the rotor magnet is attracted to a drive coil core) is counterbalanced by the rotor slow-down on departing for a net energy change in the rotor of zero.

However, we know this causes a type of cogging.  The assumption is that the cogging is stressing the rotor bearings and that's adding some friction.  In theory if you had ideal bearings for the rotor there would be no friction due to the cogging.  Also, I am not a bearing expert but for sure there are high-quality bearings out there that will remain silky smooth and barely induce any friction due to the cogging.   In reality, you probably have 100 times more experience in this stuff than I do.

So if you don't feel the presence of the two drive coils is adversely affecting the speed-up of the rotor while the weight falls then don't remove them.  It looks to me that it would be an unreasonable amount of work to remove the drive coils from your setup.  I am really guessing because I haven't watched all of your clips and I am not too familiar the build details of your motor.

Think of the bearings in an in-line skate.  They are designed to withstand thousands of pounds of radial and "roll/pitch/yaw" type stresses and work fine.  To get that extra strength you might have to make a trade-off and live with a bit of friction.  Compare that with the silky-smooth bearing inside a hard drive.  The hard drive bearing is not really designed to handle excessive amounts of stress and it's not designed to handle any type of load itself, just the "load" of the platter.  It's designed for ultra long life and very very low friction.  I am just talking from my common sense here, I haven't changed bearings or ever really read about bearings.

MileHigh

[Farmhand]

Yes I understand that, when I think about it you are correct of course the cogging is drag neutral logically. But it does make the first fraction start slower with the generator coils not there, but it is probably only due to the circumstance of the position of the cog right on where I wanted to start the rotor, I probably should have started it between cogs.

No matter, the result turned out to be very consistent either way and very similar to each other anyway.

Did I make the calculation correctly ? If not just say so and I'll try again.

What do I call that bit the "MOI" in "" Kilograms ?

Cheers

[MileHigh]

Farmhand:

Question what does the number stand for ?

The number is telling you how much resistance to change in speed the rotor has.  So you apply torque to the rotor with the falling weight and put a precise amount of energy into the rotor.  If the rotor has a lot of resistance to a change in speed then it won't be turning too fast after the weight falls.

This is related to but not directly connected to the mass of the rotor.  You can have a 10 Kg rotor made like a long cylinder with a small diameter and you can have a 10 Kg rotor made like a thin flat disk with a very large diameter.  The two rotors will have greatly different moments of inertia.

What you are measuring is the mechanical inductance of the rotor.  I know that I have droned on about this before but it's true.  You will note that electrical inductance is a measure of the resistance of the coil to a change in current flow.  From above, "The number is telling you how much resistance to change in speed the rotor has."

Perhaps this will ring a bell:

The energy stored in an electrical inductor:  1/2 L i-squared

The energy stored in a mechanical inductor:  1/2 I w-squared

Look familiar?

MileHigh

[Farmhand]

Yep I see so it's just the MoI value, similar to an "inductance" value. Makes sense I guess.

Does it seem correct though ? I'll find out later I guess.

Thanks.

[Farmhand]

Farmhand:

Perhaps this will ring a bell:

The energy stored in an electrical inductor:  1/2 L i-squared

The energy stored in a mechanical inductor:  1/2 I w-squared

Look familiar?

MileHigh

Familiar and similar.

Ahah, yes I see now, "the light's just came on, the door flung open and the cool breeze hit me", then I heard the bell ring. 

This takes me back to reply #20 from Tinsel. E = (Iw2)/2 , E= 1/2 I w-squared , E= 1/2 L i-squared, neat.

I'll do some more calculations for the practice now to see how I go then I'll weigh the rotor parts in the dead calm of after midnight.

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