What I am saying is(Tommey Reed)

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What I am saying is(Tommey Reed)

Jeff Shanab
Lets please put this to bed. give us the whole picture once or not at
all. Even at the top level, power in = power out.

    Volts*Amps in = [(rpm * torque)/5252] * 746 + magic volts * magic
amps out + waste watts

   If you want to provide the following information we will look at it
two data points of steady state
    Shaft rpm and torque since (rpm * torque )/5252 is hp and 1hp is 745.7W
    Volts and amps driveing the motor
    volts and amps chargeing the 2nd battery.
    Measuring equipment specs.

    At the moment it just looks like a very inefficient generator.
torque is near zero so most the power goes from one battery to the
other. with about 70% loss.
     The rpm climb is probably a change in the inductance is reduceing
losses. The real cost is available torque, torque that would be
developed if there was a load.

Perhaps in some areas(rpm torque,field strength combos) where motors are
less efficient, you can provide a way to recoup some of the loss. But
there generally isn't even 10% to be had.

Take  field weakening in a motor, the rpm's climb at expense of torque
at a different rpm provideing more avail torque at higher rpms.
Take  field weakening in a PM motor, overriddeing the PM field to extend
the rpm range.
I think all this tells us is the motor was running with a fixed field
strength some used to provide torque and some detrimental. Just no way
to tell without the rest of the data...rpm and torque.

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