Hi Luc and All,
I try to help with the calculations, if somebody does not agree with my numbers, please tell.
Assuming your 1 Ohm resistor is not inductive so we can use the simple formula of the voltage squared and devided by the resistance. (I did not consider here your capacitors in parallel with the this resistor either.)
Also, at your Ryobi motor we do not know the phase angle the motor makes but the relative current increase in the loaded cases are what counts so the voltage times the relative current increase will be considered as input power.
TEST #3
Case 1, coil is facing the magnets
Input power increase at the Ryobi when you attached the 1 Ohm load is: 67.1V*0.042A=2.818W (this can be considered as your input power in this loaded case, I considered the current increase only, the 42mA)
Power dissipated in the 1 Ohm is 1.492V*1.492V/1Ohm= 2.226W (here I use your multidigit RMS meter values, not the scope)
Efficiency, output/input power: 2.226/2.818=0.7899 i.e. 78.99% or a COP of 0.7899
Case 2, coil is sidewise to the magnets in video#3
Input power increase at Ryobi when you attached the 1 Ohm load is: 67.1V*0.005A=0.335W
Power dissipated in the 1 Ohm is 0.687*0.687/1=0.471W
Efficiency: 0.471/0.335=1.4 i.e. 140% or a COP of 1.4 Well I used the same 67.1V input voltage to the Ryobi because you mentioned in the video it was the same like in case 1, see my comments on this below.
TEST #4
Case #1 coil pair is side wise to the magnets
Input power increase at Ryobi when you attached the 1 Ohm load is: 67.1V*0.008A=0.536W
Power dissipated in the 1 Ohm is 0.609*0.609/1=0.37W
Efficiency: 0.37/0.536=0.69 i.e. 69% or a COP of 0.69
Case #2 coil pair is facing the magnets
Input power increase at Ryobi when you attached the 1 Ohm load is 67.1V*0.007A=0.469W
Power dissipated in the 1 Ohm is 0.629*0.629/1=0.395W
Efficiency: 0.395/0.469=0.8422 i.e. 84.22% or a COP of 0.8422
I think the result of Case 2 in test 3, the COP of 1.4 is doubtful because how can the Ryobi consume 0.572A (unloaded case) from the 67.1V variac voltage if in Case 1 it consumed 0.564A (also unloaded)?
The variac voltage (or the input mains to the variac) must have been increased from 67.1V to a bit higher value, to get the higher current consumption, right? but I used the same 76.1V in the calculation. So I do not think this is indeed overunity, would you mind repeating if you can?
MY questions:
If I got it right from test#4, the coils were in 'normal' parallel and not in flipped parallel, right?
The 1 Ohm load was inserted into the paralleled coils and not across the paralleled coils, right? (on the word 'across' I mean the single coil case as a comparison where you placed the 1 Ohm in parallel with the single coil)
Regarding your suggestion on using a second magnet on the other side of the coil when in the sidewise position will double the output: well I do not know, all I can say is it needs to be tested.
I tend to agree with David's comments on why you may not get similar results now like you had with your hand tests. What do you object the (perhaps less) output or the bigger drag or both?
Thanks for doing these tests.
Gyula
EDIT: I like your figuring out the possible best 'operation point' of the Ryobi motor by picking the voltage value to which the lowest current belongs because this should be the most sensitive point for testing any current increase due to the smallest load imposed to the shaft. However you have to find why the voltage or the current of the motor changes, probably the variac is not so stable, provided the mains 120V input is stable? I mean here also your test #4 where the current of the Ryobi was 0.585A unloaded in case 1 and 0.558A in case 2, also unloaded, this is too big difference when you think the input is still at 67.1V to the motor... which cannot be the case?
I try to help with the calculations, if somebody does not agree with my numbers, please tell.
Assuming your 1 Ohm resistor is not inductive so we can use the simple formula of the voltage squared and devided by the resistance. (I did not consider here your capacitors in parallel with the this resistor either.)
Also, at your Ryobi motor we do not know the phase angle the motor makes but the relative current increase in the loaded cases are what counts so the voltage times the relative current increase will be considered as input power.
TEST #3
Case 1, coil is facing the magnets
Input power increase at the Ryobi when you attached the 1 Ohm load is: 67.1V*0.042A=2.818W (this can be considered as your input power in this loaded case, I considered the current increase only, the 42mA)
Power dissipated in the 1 Ohm is 1.492V*1.492V/1Ohm= 2.226W (here I use your multidigit RMS meter values, not the scope)
Efficiency, output/input power: 2.226/2.818=0.7899 i.e. 78.99% or a COP of 0.7899
Case 2, coil is sidewise to the magnets in video#3
Input power increase at Ryobi when you attached the 1 Ohm load is: 67.1V*0.005A=0.335W
Power dissipated in the 1 Ohm is 0.687*0.687/1=0.471W
Efficiency: 0.471/0.335=1.4 i.e. 140% or a COP of 1.4 Well I used the same 67.1V input voltage to the Ryobi because you mentioned in the video it was the same like in case 1, see my comments on this below.
TEST #4
Case #1 coil pair is side wise to the magnets
Input power increase at Ryobi when you attached the 1 Ohm load is: 67.1V*0.008A=0.536W
Power dissipated in the 1 Ohm is 0.609*0.609/1=0.37W
Efficiency: 0.37/0.536=0.69 i.e. 69% or a COP of 0.69
Case #2 coil pair is facing the magnets
Input power increase at Ryobi when you attached the 1 Ohm load is 67.1V*0.007A=0.469W
Power dissipated in the 1 Ohm is 0.629*0.629/1=0.395W
Efficiency: 0.395/0.469=0.8422 i.e. 84.22% or a COP of 0.8422
I think the result of Case 2 in test 3, the COP of 1.4 is doubtful because how can the Ryobi consume 0.572A (unloaded case) from the 67.1V variac voltage if in Case 1 it consumed 0.564A (also unloaded)?
The variac voltage (or the input mains to the variac) must have been increased from 67.1V to a bit higher value, to get the higher current consumption, right? but I used the same 76.1V in the calculation. So I do not think this is indeed overunity, would you mind repeating if you can?
MY questions:
If I got it right from test#4, the coils were in 'normal' parallel and not in flipped parallel, right?
The 1 Ohm load was inserted into the paralleled coils and not across the paralleled coils, right? (on the word 'across' I mean the single coil case as a comparison where you placed the 1 Ohm in parallel with the single coil)
Regarding your suggestion on using a second magnet on the other side of the coil when in the sidewise position will double the output: well I do not know, all I can say is it needs to be tested.
I tend to agree with David's comments on why you may not get similar results now like you had with your hand tests. What do you object the (perhaps less) output or the bigger drag or both?
Thanks for doing these tests.
Gyula
EDIT: I like your figuring out the possible best 'operation point' of the Ryobi motor by picking the voltage value to which the lowest current belongs because this should be the most sensitive point for testing any current increase due to the smallest load imposed to the shaft. However you have to find why the voltage or the current of the motor changes, probably the variac is not so stable, provided the mains 120V input is stable? I mean here also your test #4 where the current of the Ryobi was 0.585A unloaded in case 1 and 0.558A in case 2, also unloaded, this is too big difference when you think the input is still at 67.1V to the motor... which cannot be the case?
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