Mostly Permanent Magnet Motor with minimal Input Power

[Khwartz]

Very thanks Luc.

On my side I've still continued to work on the different factors which increase the efficiency of a (air) coil, in terms of inductance against power consumption, or flux against power consumption, and with 2 different formulas I just find full opposite results, even not a range where both would have coherent results. Then, I really think only own experiments will do...

Best regards,
Didier.

[gotoluc]

Here is the efficiency test video:

https://www.youtube.com/watch?v=-zdfBbDarQw

Here is the Math:

Super Cap start Voltage 2.07197vdc x 650F = 1395.244J

Super Cap end Voltage 2.07110vdc x 650F = 1394.073J

= 1.185J used

then we subtract 0.54J which is the unity amount needed for the 2.35kg coil to travel 23.5mm

=  0.645J 

then we subtract 0.334J collected from the recovery cap bank (12.2mF @ 7.4vdc = 0.334J)

we are left with 0.311J under unity

[TinselKoala]

Here is the efficiency test video:

https://www.youtube.com/watch?v=-zdfBbDarQw

Here is the Math:

Super Cap start Voltage 2.07197vdc x 650F = 1395.244J

Super Cap end Voltage 2.07110vdc x 650F = 1394.073J

= 1.185J used

then we subtract 0.54J which is the unity amount needed for the 2.35kg coil to travel 23.5mm

=  0.645J 

then we subtract 0.334J collected from the recovery cap bank (12.2mF @ 7.4vdc = 0.334J)

we are left with 0.311J under unity

Ignoring the false precision for the moment, energy E in Joules on a capacitor  is
E = 1/2 (CV²)

So your starting energy in Joules is
E = 1/2(650 F x 2.07197 V x 2.07197 V) = 1395.244396292 Joules.
Ending energy in Joules is
E = 1/2(650 F x 2.07110 V x 2.07110 V) = 1394.07294325 Joules.

You got the right answer even though your stated formula is wrong. Therefore you did not use your stated formula, but actually used the correct one.

It is really difficult to check your work if your answers and your formulae do not agree.

Can you really measure voltage on a capacitor to the tens of microvolts precision? I am jealous.

[tinman]

Ignoring the false precision for the moment, energy E in Joules on a capacitor  is
E = 1/2 (CV²)

So your starting energy in Joules is
E = 1/2(650 F x 2.07197 V x 2.07197 V) = 1395.244396292 Joules.
Ending energy in Joules is
E = 1/2(650 F x 2.07110 V x 2.07110 V) = 1394.07294325 Joules.

You got the right answer even though your stated formula is wrong. Therefore you did not use your stated formula, but actually used the correct one.

It is really difficult to check your work if your answers and your formulae do not agree.

Can you really measure voltage on a capacitor to the tens of microvolts precision? I am jealous.

I think Luc was just saying yay amount of volt's in a 650f cap-much like we say 1x bucket of ice cream
Do you not have a DMM that can go to the 5th decimal TK ?.

[tinman]

Here is the efficiency test video:

https://www.youtube.com/watch?v=-zdfBbDarQw

Here is the Math:

Super Cap start Voltage 2.07197vdc x 650F = 1395.244J

Super Cap end Voltage 2.07110vdc x 650F = 1394.073J

= 1.185J used

then we subtract 0.54J which is the unity amount needed for the 2.35kg coil to travel 23.5mm

=  0.645J 

then we subtract 0.334J collected from the recovery cap bank (12.2mF @ 7.4vdc = 0.334J)

we are left with 0.311J under unity

Luc-why is the coil sliding so slowly up the guide's?. It almost looks as though the slides are coated in honey or something . Have you taken into accound the friction on those slides?-there seems to be to much friction there.

P.S-just watched the video again,and even when coming back down it seem'd very slow,regardless of weather or not you where collecting the BEMF. How fast dose it fall with the recapture cap's disconected(free fall)>?