STEORN DEMO LIVE & STREAM in Dublin, December 15th, 10 AM

[k4zep]

Where's everyone. Maybe everyone's building one of these now, including myself.     The only difference is that I'll include a measurement to see if there's bemf >when there is current in the toroid coil!<  To say there's no bemf when there's no current is meaningless, which is what I've been trying to tell everyone far before MH even said anything. Someone in another Steorn thread said that MH is getting the credit, lol.

So IMO the excess energy would be due to magnetic viscosity, so don't think this task will be easy. Congrats to Steorn on an awesome achievement!!! And congrats to Ben & J.L.Naudin on a great first build.

Hi Paul,

Still kicking around here.  Helping wife, and learning a lot from JLN's motor......Wow.  I understand what you are saying about current in the coil or no bemf.  Both in a motor and generator but remember again the old adage about the chicken and the egg.  Remember the old E before I and I before E in an inductor and Cap?  We are all looking at the same motor through our eyes of experience and the final result is the same.  Your right in there paddling with the rest of us.  I apologize that I haven't been over to the other site yet.....Can only keyboard so much.......and I have to build a new motor too.  Got all the info I can out of my tin can motor.

Respectfully
Ben

[hartiberlin]

Hi All,

I must apologize...

I am getting old, it seems:
After 2 hours of calculations and trying to drive my brain mad,
I came to the conclusion, that I was posting something wrong earlier,
when I said, one could extract about 90 % of the inputed coil energy.

Now after recalculation I came to the conclusion that I was wrong and
mixed something up.

The numbers are a bit lower.

If you have a RL series circuit and put an constant voltage pulse in it
with a varying time:

Vsupply x i(t) x t = Input Energy
where
Vsuppy is the constant voltage
i(t) is the current through the RL series circuit
and
t= switch-On time interval
tau= L/R timeconstant

you can calculate the input energy for the time t.

Now input energy for this time interval is defined as:
Vsupply^2 / Rcoil x ( 1 - (e ^-(t/tau)) ) x t

This is the total energy drawn during t from the constant voltage
power suppy.

Now we can compare this with the energy stored inside the coil
which is Wcoil=0.5 x L x i^2

We will find the coil current after the time interval t with the formular:

i(t)= Vsupply / R coil x ( 1 - (e^-(t/tau)) )

So if you calculate i(t) after the time interval and put it into
Wcoil=0.5 x L x i^2

you will know the stored energy inside the coil and then
you can divide this value through the input energy
value:
Vsupply^2 / Rcoil x ( 1 - (e ^-(t/tau)) ) x t

As the results you get:
if you choose:
at
t= 1/10 tau you store 47 % of the energy input into the coil
t= 1/5 tau you store 45 % of the energy input into the coil
t= tau you store 31 % of the energy input into the coil
t= 5 tau you store 15 % of the energy input into the coil


So you see, at 1/10 tau Ontime you can only recover less than
47 % of the input energy back into a capacitor.

So it was not 90 % but only less than 47 %.

But still this is a way to avoid too many losses
to just switch on only  way less than tau and
use a high supply voltage so the saturation current level
is already reached during this short ontime interval.

Regards, Stefan.

[k4zep]

Hi All,

I must apologize...

I am getting old, it seems:
After 2 hours of calculations and trying to drive my brain mad,
I came to the conclusion, that I was posting something wrong earlier,
when I said, one could extract about 90 % of the inputed coil energy.

Now after recalculation I came to the conclusion that I was wrong and
mixed something up.

The numbers are a bit lower.

If you have a RL series circuit and put an constant voltage pulse in it
with a varying time:

Vsupply x i(t) x t = Input Energy
where
Vsuppy is the constant voltage
i(t) is the current through the RL series circuit
and
t= switch-On time interval
tau= L/R timeconstant

you can calculate the input energy for the time t.

Now input energy for this time interval is defined as:
Vsupply^2 / Rcoil x ( 1 - (e ^-(t/tau)) ) x t

This is the total energy drawn during t from the constant voltage
power suppy.

Now we can compare this with the energy stored inside the coil
which is Wcoil=0.5 x L x i^2

We will find the coil current after the time interval t with the formular:

i(t)= Vsupply / R coil x ( 1 - (e^-(t/tau)) )

So if you calculate i(t) after the time interval and put it into
Wcoil=0.5 x L x i^2

you will know the stored energy inside the coil and then
you can divide this value through the input energy
value:
Vsupply^2 / Rcoil x ( 1 - (e ^-(t/tau)) ) x t

As the results you get:
if you choose:
at
t= 1/10 tau you store 47 % of the energy input into the coil
t= 1/5 tau you store 45 % of the energy input into the coil
t= tau you store 31 % of the energy input into the coil
t= 5 tau you store 15 % of the energy input into the coil


So you see, at 1/10 tau Ontime you can only recover less than
47 % of the input energy back into a capacitor.

So it was not 90 % but only less than 47 %.

But still this is a way to avoid too many losses
to just switch on only  way less than tau and
use a high supply voltage so the saturation current level
is already reached during this short ontime interval.

Regards, Stefan.

Nah, when you really want to believe, numbers are always high.  The 47% sounds very good and in the box!  So we need a COP at least 2 to break even and probably 3+++ to overcome charging efficiency.  Damn good work!!!!  Keep an eye on JLN.  When he gets going, he gets going!!!!!

Ben

[PaulLowrance]

Hi Ben & all,

I just built a test unit, nothing as complete as yours Ben coz I don't have the parts here, but so far I can't get the darn FEMM results, lol! When current is applied to the toroid, the secondary AC voltage goes down.

Even stranger yet, I could swear the darn thing is getting cold. Within about 15 minutes I'll post some photos. This setup is only to analyze the effect, not to get it self-running. So the magnets are on a dremel rotary motor. The entire dremel gets warm, which conducts to the magnets, but if I center the magnets it sure seems like they cool down. To give an idea, the ambient temperature from the thermal gun at the time during these measurements was ~ 67.2F, and if the magnets are not centered (or at least they didn't seem centered to me) then I recall them being ~ 71.2F, but after trying to center the magnets I've seen them get as low as 66.6F. It's not always easy to get a good thermal gun reading while one hand is trying to balance a dremel drill and other hand is trying to take a temperature reading, so who knows if there's something to this. Probably not, but conventional physics says the magnets cannot get colder than ambient temperature. Thermal gun measurements need to be consistent since it depends on the material you're pointing at, as some materials have lower thermal emissivity than others.

Has anyone noticed this cooling effect?

[PaulLowrance]

Here's a few snap shots. Now I'll go mount the dremel so I can focus on taking temperature measurements. To be honest, temperature was the last thing on my mind, but I kept noticing how cold the was, so I ran to get my thermal gun.