Where the OVERUNITY using INDUCTION COILS comes from (eg Joule Thief)

[antijon]

Brad, I know why power increases when a motor is loaded. I also know that efficiency doesn't mean anything with motors under a few horsepower, I think they're all above 85% at least.

You're confusing two very different types of efficiency.

Efficiency 1: high impedance (cemf) results in little power being transferred to a load.
Efficiency 2: the ratio of power supplied and power being consumed by the load.

If you never run a motor at its full load, then I can see why you think cemf equates to efficiency. But in real motors with real loads, if you try to replace a motor with anything under the original motor FLA, it's not going to work.

For instance, two motors with matching RPM and HP:
1075 RPM .5 HP 6 amp FLA
1075 RPM .5 HP 9 amp FLA

The first motor pulls less amps so it seems to be more efficient, but if I'm replacing a motor that pulled 9 amps then I need to go back with 9 amps. I learned from experience. The 6 amp motor actually pulled 10 amps before it overheated.

So I'll say again, the current or power consumed by the motor is directly related to the power transferred to the load.

[tinman]

Revolutions Per Second.


The desired event is to convert electrical potential into mechanical work, it is unfortunate that with the present methods of using the magnetic field interaction there is an opposing voltage created.


https://en.wikipedia.org/wiki/Ampere


There does seem to be a force to amps defined so there is an answer,, 

As i said before--the voltage is not apposing.
The BEMF produced in the self induction of the motor, is off the same polarity as that delivered to the motor.


Brad

[tinman]

Brad, I know why power increases when a motor is loaded. I also know that efficiency doesn't mean anything with motors under a few horsepower, I think they're all above 85% at least.

You're confusing two very different types of efficiency.

Efficiency 1: high impedance (cemf) results in little power being transferred to a load.
Efficiency 2: the ratio of power supplied and power being consumed by the load.

If you never run a motor at its full load, then I can see why you think cemf equates to efficiency. But in real motors with real loads, if you try to replace a motor with anything under the original motor FLA, it's not going to work.

For instance, two motors with matching RPM and HP:
1075 RPM .5 HP 6 amp FLA
1075 RPM .5 HP 9 amp FLA

The first motor pulls less amps so it seems to be more efficient, but if I'm replacing a motor that pulled 9 amps then I need to go back with 9 amps. I learned from experience. The 6 amp motor actually pulled 10 amps before it overheated.

So I'll say again, the current or power consumed by the motor is directly related to the power transferred to the load.

Regardless of the size of the motor,or the load placed on it,an efficient motor has a higher BEMF under any load conditions--it is the quantity of BEMF that makes the motor more efficient.

I asked you if you know why a motor draws more current under load,and you answered -yes,i know why. But it would seem that you dont,or you would know that it is only the value of the BEMF that determines how much current the motor draws-whether it is under load or not.

So,why dose a motor draw more current under load?-->because the value of the BEMF drops.
As this BEMF value drop's,the potential difference between the supply voltage and self induced voltage become's greater,and so it is like i explained in post 120.

high impedance (cemf) results in little power being transferred to a load.

High impedance means that the BEMF/CEMF value is very close to the supply EMF value,and so,very little current flow's.

As i said,an efficient motor maintains the highest value BEMF under any load conditions,and here is what i mean-so as it is clear.

We have an electric motor-lets say a brushed DC motor.
Lets say that unloaded,this motor draws 1 amp at 10 volt's.
We know that when we place a load on this motor,the current draw will go up,and the only !only! reason it go's up,is because the BEMF value go's down(impedance decreases)-->this is why i asked you if you knew why the current draw go's up on a motor when you place a load on it.

Now,what if we could maintain the BEMF(impedance) value as we place a load on that motor?--what would happen to the current draw of that motor?--Thats right,the current would not rise,as the impedance value would remain the same.

So,do you now understand as to why a high BEMF value is important to efficiency?.


Brad

[tinman]

85% for a small motor

More in the range of 50-75%,, some really small little ones as low as 35%

Ideal calculations would be energy in per second to torque out per second, torque in N-m =2piJ per rotation per N-m, 1J per second is 1W
torque=(I*V*60)/(rpm*2pi) for 100% efficiency. or
torque=J/(rps*2pi)

I have used a confusing phrase for some,, voltage drives current,, and as such an opposing voltage will act as a resistance to current flow depending on how much of the source voltage the opposing voltage is,, kind of simple really.
Inside every motor is a generator,, you have a magnetic flux that is changing relative to a coil that is observing it,, and it is this very generator that stops the motor from drawing the same amps and spinning up with the same torque to oblivion, as the speed of the generator increases the voltage it produces goes up and when it reaches the source voltage no current can flow.

If then the CEMF from the induction process inherent inside almost every motor were enhanced,, would that make it a better motor or a better generator?  How about if it were reduced?
Is the same torque at a higher RPM more or less power than at a lower RPM?

I think you can also see that the IR^2 losses might not be such a big thing.

Here is a simple answer for you.
Get your self a unipolar/homopolar motor,that produces no BEMF,and see how much torque you get in relation to the power you supply to it.

You now have your motor that has no BEMF, will spin up to self destruction speed,and draw a sh-t load of current
You will also find it has very little torque-->you have a rotating heater.


Brad

[citfta]

Some people just don't seem to be paying any attention.  So here is how to make you very own motor with no BEMF or CEMF, whatever you want to call it.  Take apart anything you want that has a universal motor in it.  The kind with field coils and brushes.  Now disconnect the field coils and apply whatever value of DC you want to the brushes.  I would suggest you start out with only 12 volts or so.  Now keep increasing the voltage as much as you want.  You will never get any torque from that motor.  The reason is because the same thing that creates the BEMF is the magnetic field of the field coils which are also the reason you get torque.  The same interaction that gives torque also gives you the BEMF.

In large industrial machines a lot of the time they use large DC motors.  These motors have the field coils powered separately from the armature.  Because of the BEMF generated these motors will of course reach a set speed with the field coils fully energized.  Now if there is a reason to want the motor to run at a higher speed than the bases speed then the current going to the field coils is reduced.  This in turns means less BEMF and the motor will speed up until the BEMF again balances out between the load and the applied armature voltage.  But you can only do this if the motor is not heavily loaded.  Otherwise the motor will lose speed because of the lower torque.