Hi Tomte

You asked some figures to clarify the calculations. Below I add an example and also suggest another method.

These calculations are examples only and are not based on any actual measurements.

Shaft diameter = 2 mm (radius 1 mm)
Rpm = 1500 which means 1500 * 2 * PI / 60 rad / sec = 157
Mass = 10 grams

Then torque = (0.001 m x 0.01 kg x 9.81 kg m /s²) * 1000 mm/m = 0.0981 Nmm
And power = 0.0981 Nmm x 157 rad/sec = 15.4 mWatt

Disadvantages of this method:
· requires rpm measurement
· torque arm varies when the mass is hoisted because the thread is wound around the shaft (shaft diameter increases).

I’m afraid the the “work correction” I suggested in my previous post is not correct. However, based upon measurement of work I think the following method is much easier to estimate output power.

Power = work / time
To hoist up the test mass, work is done so if you measure in how much time the mass is hoisted up over a certain distance, you also know the average absorbed power.

Example:
Mass = 10 grams
Hoisting distance = 1 meter (height difference)
Time = 8 seconds

The power = mass x 9.81 x height / time = 0.01 x 9.81 x 1 / 8 = 0.0125 Watt or say 12,5 mWatt

I did some test to compare both methods and the results are very well in line. The conclusion till now is indeed that my motor required more input power than the output power it delivers.

If I understand right, your motor does not give more output power than the input power? Please can you add some figures to your calculations?

Hi,

After some thinking hope I found a way to estimate the motor’s output power.

Method:
Use the motor as a winch: attach one end of a thread to the motor shaft and the other end to a weight. When the weight is hoisted up, it will exert a torque to the motor shaft and as such act a a load.

Formulas:
Mechanical output power:
o Torque = force x arm = mass x 9.81 x arm
o Power = Torque x speed (speed in rad / sec).
o Since in this case, the weight is lifted, work is done and this work has to be taken into account. Required power = force x speed = mass x 9.81 x lifting speed.

Electrical power input:
o Power = current x voltage
o Battery only supplies power when the switch is closed (my motor: +/- 20% of the time).
o Voltage = battery voltage
o Current can be visualised with an oscilloscope and an average value can be determined.

Motor efficiency = (mechanical power / electrical power) x 100%

The graph below shows my first test results. W = weight of the load.
Intermediate conclusion:
o efficiency of my motor is well below 100% (even if the work correction as described above is included).
o the higher the load, the better the efficiency
o effiency drops when motor speed increases.

Of course, there will be quite a lot of error on these measured values I used but my intermediate conclusions are probably correct.
If you see any errors in my reasonning or method, please let me know.

Has anybody purchased the motor kit or the manual and used it to create a working Keppe motor ?

Would like to see the step by step and know "in a nutshell" what this motor is really doing.

Is it merely using back EMF, and what similarities/differences to Bedini's work.

Hi,

When reviewing the test data I posted earlier, I found out that the picture as such was OK but had been made upside down. The correct '' picture should be like this:



As I understand it, the sinuslike voltage pattern in the coil is caused by the induction of the magnet (generating an AC voltage). When the switch closes, the coil is put at the battery voltage and current can start to flow through the coil.
When the switch is opened again, circuit is no longer a closed loop and current drops back to zero. When it opens, high voltage peaks occur and for a very brief period, voltage oscillates till it has dampened out (my motor: total cycle time of 42 msec, switch closed during some 7 msec and oscillations during 0.4 to 0.6 msec. Peak voltages up to 200 Volts detected, compared to a battery of 9 Volts).
Hope this is better data.

j greef wrote:Hi all,

I would like suggest to post the test results of your motors under this topic so that we can learn from each other.

As an example, enclosed I post an oscilloscope picture of my motor's coil current and voltage readings.
* The current is always zero, except when the reed swich closes.
* The voltage varies according a sinus function and drops almost to zero when the switch closes.
* Timescale: 10 msec per division (so one cycle = +/- 42 msec which means motor rpm = +/-1430).
Note: this motor deviates from the Keppe manual so your own motor readings certainly will be different.

Hope to see soon some of your data.

Hi all,

I would like suggest to post the test results of your motors under this topic so that we can learn from each other.

As an example, enclosed I post an oscilloscope picture of my motor's coil current and voltage readings.
* The current is always zero, except when the reed swich closes.
* The voltage varies according a sinus function and drops almost to zero when the switch closes.
* Timescale: 10 msec per division (so one cycle = +/- 42 msec which means motor rpm = +/-1430).
Note: this motor deviates from the Keppe manual so your own motor readings certainly will be different.

Hope to see soon some of your data.