Re: [EVGRAY] Re: https://www.youtube.com/watch?v=RMYo1QlvK5g (Newman motor)

Body

Hi James,

That just changes the impedance. It's the total current of all turns that matters. This is ampere x turns. Having 1A x 1000 turns = 1000A x turns or 1mA through 1,000,000 turns = 1000A x turns gives the same total current (1000A) around the core. The difference is that the first coil has 1,000,000 times less inductance than the second one. This makes it react much faster to any change in current than the second coil does. The two coils store the same amount of energy in the examples given.

Regards
Ole
 

---In [email protected], <jglinski01@...> wrote :

 Sorry I mean finer the wire and many many more turns more power out put Newman's motor would be a good platform for it .

 On Saturday, January 12, 2019, James Glinski <jglinski01@... mailto:jglinski01@...> wrote:
 I think the finer the wire the more magnetic force and by shifting the wattage of amps by volts to hi voltages and very little amps you get more power and more back emf .

On Friday, January 11, 2019, onielsen@... mailto:onielsen@... [EVGRAY] <[email protected] mailto:[email protected]> wrote:
   Hi Hector,

That link is the Joseph Newman motor. It seems to be draining the battery of 60 x 9V pretty fast.
https://youtu.be/RMYo1QlvK5g?t =524 https://youtu.be/RMYo1QlvK5g?t=524 8:44 minutes into the video the motor is running.
https://youtu.be/RMYo1QlvK5g?t =1114 https://youtu.be/RMYo1QlvK5g?t=1114 18:34 minutes into the video the motor has slowed down after running for 10 minutes. It looks like the battery is being drained fast.

Looking at the pump it doesn't look like much power being delivered from the motor. The water is flowing at low pressure. Newman ought to do some proper measurements. If reading his book it can be seen that his motor use pulses. His multimeters aren't optimized for measuring pulses which gives false readings. I don't think his motor is more efficient than other high efficient motors like perhaps brushless DC motors or ironless DC motors.

If the pump lifts one m^3 of water (=1000 kg =2205 lb) to a height of 1m it takes 1000kg x 9.807m/s^2 x 1m = 9.807kJ of energy. If that is done in one minute the power required is (9.807kJ / 60s) = 163W.
With 60 batteries of 9V in series the voltage of the battery becomes:
9V x 60 = 540V.

If the power consumption is 163W the current at 540V is then 163W / 540V = 0.3A.
If the battery is rated at 550 mAh (https://wiki2.org/en/Nine-vol t_battery#Technical_specificat ions https://wiki2.org/en/Nine-volt_battery#Technical_specifications) it should deliver that power for almost two hours: 0.55Ah / 0.3A.= 1.8h.

Regards
Ole

 

---In [email protected] mailto:[email protected], <arkresearch@...> wrote :

 https://www.youtube.com/watch? v=RMYo1QlvK5g https://www.youtube.com/watch?v=RMYo1QlvK5g