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**Sephiroth** · 12-25-2008, 02:45 PM

Hi Mart,

In step two do you mean discharging battery two into the energiser to charge battery one up again?

There are complications of this method since there is some debate about wether a radiantly charged battery is better suited for running inductive or resistive loads... but you know that already

Why not do a simple load test? That way you will know the precise COP of the energiser and batteries.

**theremart** · 12-25-2008, 03:01 PM

Thinking about the testing of COP > 1

Originally posted by Sephiroth View Post

Hi Mart,

In step two do you mean discharging battery two into the energizer to charge battery one up again?

There are complications of this method since there is some debate about wether a radiantly charged battery is better suited for running inductive or resistive loads... but you know that already

Why not do a simple load test? That way you will know the precise COP of the energiser and batteries.

Yes use battery 2 to charge up battery 1 again.

The reason for this method is to make the math very simple. If Battery 1 ends up with more than initial charge then cop is present. But I see your point of "Negative Radiant Energy" as a roadblock. This could be overcome by using an energizer with a cap, as I understand it the cap discharge method gets around this "Negative Radiant" energy problem.

Let me post the SSG groups load test method, and I will think this out some more. I am wanting a self running system and I want to do the math right to get there.

**theremart** · 12-25-2008, 03:24 PM

From the Yahoo SSG group load test

( Note there is more info, but I cut out everything but the load test )

STEP BY STEP INSTRUCTIONS:

1. Take a small, 1 coil, smooth running, SSG. No capacitor dump [setup], just take the diode output of the coil and put
it to a secondary battery positive [that is the SSG setup with the diode and not the SCR]. The secondary negative terminal should be hooked to the primary positive. Set it to where the mechanical efficiency of the motor peaks out at 75ma input current or below. [If you cannot build that small of a machine, then report your results with higher draws.] The way I tune this is to measure the ratio of rpm and convert to magnets/minute then divide by the ma. You'll have to have a tachometer, or count scope traces to get your mpms. [Or use a meter that measures Hertz] Measure your input current with both your analog meter and your digital meter, and note any major discrepancies. If there is, report the measurement from your analog meter.

Step 2
Now that we have optimized motor output, the next step is to measure the current going into the output battery. At this point we will not be concerned with the current being used by the input battery. This will come into play later; so for now, as hard as it is, just forget it. Now everyone who has played with this circuit knows that measuring the pulsed dc current is a little trickier than normal. I try to measure it with at least two different methods until I can get the two to agree with one another. Then I know that I am close. Here are the two easiest ways I have found to measure this energy.

#1) The easiest. Get yourself an ANALOG current meter and put it in series with the 2ndary battery. Measure the voltage of the secondary battery. Multiply the two together to get the watts.

#2) Get 2 identical small flashlight bulbs (12v 100ma for example). Replace the secondary battery with one of the light bulbs. Run the motor. Now take the other light bulb and put it in series with a potentiometer and a current meter. Hook it to the same battery you are running the motor on. Adjust the potentiometer until both light bulbs glow at the same brilliance. Then read the ampmeter. Put an voltmeter across the lightbulb. Multiply the volts times amps to get the watts.

Step #3) Stop the motor. Take your secondary battery out of the circuit and put a known load across it (I like to use a 10watt 10 ohm, for example [that is too high for really small batteries which could use a higher ohm resistor like 50 ohm more or less). Get a stopwatch and measure, as precisely as you can, the time you are discharging the battery. Put a digital volt meter across the resistor and note the voltage levels during the discharge time. You will be needing to get an approximate average voltage across the load resistor during this time. You want to take the voltage down to a predetermined voltage level which you will be watching precisely say 12.40V for example [Lower levels like 10.50V need to be tried after]. When it gets close to this level (12.41 in this case) disconnect the load from the battery and stop the stopwatch as soon as the meter first flashes 12.40. We are going to call this, for future simplicity's sake, the load termination point. Now measure the power in joules (or watt-sec) that you took out of the secondary battery. Multiply the duration of the load from the stopwatch (in seconds) * AVERAGE voltage * AVERAGE voltage / the resistance of the load (in ohms). This will give you the output of the system in joules. Write this down. Now, take the battery and put it back in the charge circuit in the SSG and run the motor for an hour. Set a timer and run the motor charging the battery back for precisely an hour (3600 sec). (If the battery tops out before this just shorten the duration but make it constant.) Multiply your input energy figured in step 2 (watts) times 3600 (or your arbitrarily decided charge time) and you will have your input power in joules. Write this down. Repeat this experiment at least 5 times in a row or until you can get consistent input and output measurements. The amount of power you can take out will reduce significantly the first few times until it stabilizes. When you do the experiment twice in a row and get the same results you will know that your particular system has stabilized and you can get an exact figure of the cop on the back end.

**theremart** · 12-25-2008, 04:12 PM

Ok, now to comment on the group instructions.

STEP BY STEP INSTRUCTIONS:

1. Take a small, 1 coil, smooth running, SSG. No capacitor dump [setup], just take the diode output of the coil and put
it to a secondary battery positive [that is the SSG setup with the diode and not the SCR]. The secondary negative terminal should be hooked to the primary positive. Set it to where the mechanical efficiency of the motor peaks out at 75ma input current or below. [If you cannot build that small of a machine, then report your results with higher draws.] The way I tune this is to measure the ratio of rpm and convert to magnets/minute then divide by the ma. You'll have to have a tachometer, or count scope traces to get your mpms. [Or use a meter that measures Hertz] Measure your input current with both your analog meter and your digital meter, and note any major discrepancies. If there is, report the measurement from your analog meter.

I fully agree that digital meters are not fast enough to get the voltage. Now after seeing this on a scope I now see that the voltage fluctuates in great degree for the digital meter to get a true fix on it.

Step 2
Now that we have optimized motor output, the next step is to measure the current going into the output battery. At this point we will not be concerned with the current being used by the input battery. This will come into play later; so for now, as hard as it is, just forget it. Now everyone who has played with this circuit knows that measuring the pulsed dc current is a little trickier than normal. I try to measure it with at least two different methods until I can get the two to agree with one another. Then I know that I am close. Here are the two easiest ways I have found to measure this energy.

#1) The easiest. Get yourself an ANALOG current meter and put it in series with the 2ndary battery. Measure the voltage of the secondary battery. Multiply the two together to get the watts.

Ok getting the amount of Watts going to the target battery. Again using an Analog meter is a very good idea.

#2) Get 2 identical small flashlight bulbs (12v 100ma for example). Replace the secondary battery with one of the light bulbs. Run the motor. Now take the other light bulb and put it in series with a potentiometer and a current meter. Hook it to the same battery you are running the motor on. Adjust the potentiometer until both light bulbs glow at the same brilliance. Then read the ampmeter. Put an voltmeter across the lightbulb. Multiply the volts times amps to get the watts.

Ok, now I am trying to understand why you would do this... I guess the light bulbs are a "dead" load. The batteries store energy, but the light bulbs just would consume it. Hmm I am not sure what they are doing here..

1. put a light bulb in place of charging battery ( ok so far understand )
2. put a light bulb in series with the Pot??? ( I guess on the trigger coil ?? )
I am not sure why this is added in the testing...

Step #3)

Stop the motor.

Take your secondary battery out of the circuit and put a known load across it (I like to use a 10watt 10 ohm, for example [that is too high for really small batteries which could use a higher ohm resistor like 50 ohm more or less).

Get a stopwatch and measure, as precisely as you can, the time you are discharging the battery. Put a digital volt meter across the resistor and note the voltage levels during the discharge time. You will be needing to get an approximate average voltage across the load resistor during this time. You want to take the voltage down to a predetermined voltage level which you will be watching precisely say 12.40V for example [Lower levels like 10.50V need to be tried after]. When it gets close to this level (12.41 in this case) disconnect the load from the battery and stop the stopwatch as soon as the meter first flashes 12.40. We are going to call this, for future simplicity's sake, the load termination point.

It seems the first test found the charging power of the energizer. This test seems to be finding out what energy is stored in the battery. This is were my computer battery tester would come in.

Now measure the power in joules (or watt-sec) that you took out of the secondary battery. Multiply the duration of the load from the stopwatch (in seconds) * AVERAGE voltage * AVERAGE voltage / the resistance of the load (in ohms). This will give you the output of the system in joules. Write this down.

Now, take the battery and put it back in the charge circuit in the SSG and run the motor for an hour. Set a timer and run the motor charging the battery back for precisely an hour (3600 sec). (If the battery tops out before this just shorten the duration but make it constant.) Multiply your input energy figured in step 2 (watts) times 3600 (or your arbitrarily decided charge time) and you will have your input power in joules. Write this down. Repeat this experiment at least 5 times in a row or until you can get consistent input and output measurements. The amount of power you can take out will reduce significantly the first few times until it stabilizes. When you do the experiment twice in a row and get the same results you will know that your particular system has stabilized and you can get an exact figure of the cop on the back end.

Ok, now this is where the excel spread sheet comes in handy. I am wondering about the math of "joules"..... Is this the standard way of measuring energy for load on a battery? ( I will google around and see if I can see where others have measured energy from a battery and see if it matches )

Peace

**theremart** · 12-25-2008, 04:31 PM

Found a good web site...

Seems to have all the the math spelled out here nicely...

How to calculate battery run-time when design equipment using batteries; Battery Technical Resources for Design Engineers from PowerStream

**Sephiroth** · 12-26-2008, 12:05 AM

#2) Get 2 identical small flashlight bulbs (12v 100ma for example). Replace the secondary battery with one of the light bulbs. Run the motor. Now take the other light bulb and put it in series with a potentiometer and a current meter. Hook it to the same battery you are running the motor on. Adjust the potentiometer until both light bulbs glow at the same brilliance. Then read the ampmeter. Put an voltmeter across the lightbulb. Multiply the volts times amps to get the watts.

Ok, now I am trying to understand why you would do this... I guess the light bulbs are a "dead" load. The batteries store energy, but the light bulbs just would consume it. Hmm I am not sure what they are doing here..

1. put a light bulb in place of charging battery ( ok so far understand )
2. put a light bulb in series with the Pot??? ( I guess on the trigger coil ?? )
I am not sure why this is added in the testing...

they mean a seperate pot for the other bulb...

calculatewatts.jpg

When both the bulbs are glowing at the same brilliance, then (in theory) the same amount of watts is being used by both of them and you can then accurately measure how many watts is being used in the second bulb since it is steady DC. That figure will be the watts being output by the SSG as well.

I don't really like this method since the SSG reacts to different loads on the charing side... if you tune your SSG so that it is at its "sweet spot" for charging a 12v battery then replace the charging battery with a different load (a 12v bulb or 24v battery for example) then the SSG will react to that by changing RPM, pulse width and duty cycle therfore also changing the amp draw and amp output.

The best methods of measuring amp input (in my opinion) is either with a good analogue meter, a true rms meter or by measuring the voltage across a 1 ohm resistor in parellel with a capacitor like this:

calculateamps.jpg

**dambit** · 12-26-2008, 12:21 AM

I agree Seph, they should have simply said "get an anologue current meter and put it in series with the battery being charged." So much easier than faffing about with lightbulbs etc.

**theremart** · 12-26-2008, 02:44 AM

Ok....

and this was for beginners with an SSG whew.....
LOL

I found a better site that shows the calculation of Joules from a battery discharge.

Battery energy storage in various battery types

Joules are units of energy or work

The Joule is the International Standard unit of energy defined as one watt-second. One watt-second of mechanical work is the work done by a force of one Newton (or 0.2247 pound) pushing through a one-meter distance. 3600 Joules are contained in one watt-hour, since an hour contains 3600 seconds,. Batteries are often rated in milliampere-hours instead of watt-hours. This battery rating can be converted to energy if the average voltage of the battery during discharge is known. For instance, a 3.6-volt Lithium-ion battery rated at 850 mAh will maintain a voltage of 3.6 volts with little variation during discharge. Multiply the voltage of 3.6 volts times 850 mAh to yield 3060 mA-volt-hours, or 3060 milliwatt-hours. 3.06 watt-hours equal 11016 watt-seconds or Joules. Compare this value to those found in the tables below.

Joules may be converted to other familiar units using the numerical factors given below. Divide the number of Joules by 3.6 million to obtain kilowatt-hours. Divide the number of Joules by 1.356 to obtain the number of foot-pounds, a popular unit of work in the English system. Divide by 1055 to obtain the equivalent number of BTU (British Thermal Units). Divide by 4184 to obtain the number of food Calories! Yes, food Calories are energy, of course. This comparison does not put batteries in a good light compared to peanut butter. Two tablespoons of smooth peanut butter contain 191 Calories, or almost 800,000 Joules! It takes a huge battery to contain this much energy.

**ren** · 12-26-2008, 03:12 AM

Originally posted by dambit View Post

I agree Seph, they should have simply said "get an anologue current meter and put it in series with the battery being charged." So much easier than faffing about with lightbulbs etc.

I believe this step was added to the procedures for those that didnt have access to an analogue gauge. It popped up quite a bit in the forums there so they devised another method, hence the above. Anyone who has a gauge would skip this.

Joules in vs joules out is definately the best method Mart. The problem being you cant pop the back battery onto the front of the energizer, so you need another load.

There is another PDF for the monopole3 group that may be a little clearer.

John Koorn posted it in the files section here http://tech.groups.yahoo.com/group/B...nopole3/files/ under Bedini Monopole 3 Group Experiment V1.00.pdf

**dambit** · 12-26-2008, 10:34 AM

Originally posted by ren View Post

The problem being you cant pop the back battery onto the front of the energizer, so you need another load.

Hi Ren, JB said in that last DVD that you can run it through an inverter and there won't be this problem. I have just started doing this with mine. I have an inverter coming off of the primary battery, then a 15V DC power supply from the inverter to the energizer. (I use a 15V 10amp supply only because the shop ran out of 24V ones.

It still runs very well.)

I have not been doing this long enough to be able to make a comparison, but so far it seems to be working well.

Mart, that's amazing about the peanut butter. Makes one realise how inefficient our power systems are compared to the human body.

Cheers,

Steve.

**theremart** · 12-26-2008, 01:53 PM

RE: test

Originally posted by ren View Post

Joules in vs joules out is definately the best method Mart. The problem being you cant pop the back battery onto the front of the energizer, so you need another load.

There is another PDF for the monopole3 group that may be a little clearer.

John Koorn posted it in the files section here http://tech.groups.yahoo.com/group/B...nopole3/files/ under Bedini Monopole 3 Group Experiment V1.00.pdf

Can you post that or email it to me, I am no longer part of that group.

Thanks for you input Ren!

**theremart** · 12-26-2008, 03:04 PM

Understanding Joules

Ok, starting to get a grasp of this now on calculating joules... I just wanted to get and understanding of this math, now I can see how the spreadsheets I have been given do the math. I guess to get a good average voltage you need multiple readings to get the numbers right for said duration.

Joule's laws - Wikipedia, the free encyclopedia

Q = I * I * R * T

Q = Joules

I = Current or Amps

R = Resistance or ohms

T = Time in seconds

**chemelec** · 12-26-2008, 03:15 PM

Originally posted by Sephiroth View Post

they mean a seperate pot for the other bulb...

[ATTACH]1735[/ATTACH]

When both the bulbs are glowing at the same brilliance, then (in theory) the same amount of watts is being used by both of them and you can then accurately measure how many watts is being used in the second bulb since it is steady DC. That figure will be the watts being output by the SSG as well.

I don't really like this method since the SSG reacts to different loads on the charing side... if you tune your SSG so that it is at its "sweet spot" for charging a 12v battery then replace the charging battery with a different load (a 12v bulb or 24v battery for example) then the SSG will react to that by changing RPM, pulse width and duty cycle therfore also changing the amp draw and amp output.

The best methods of measuring amp input (in my opinion) is either with a good analogue meter, a true rms meter or by measuring the voltage across a 1 ohm resistor in parellel with a capacitor like this:

[ATTACH]1736[/ATTACH]

An Analogue Current Meter will NOT Accurately Measure the current of Very Short Spikes.

True RMS meters, Also have Many Limitations on many short waveforms. Also a Limitation on Higher Frequencies.
So they can also give false readings.
** You need to look at the Specs and Understand them, on your True RMS Meter.

Using a Diode to drive the Light Bulb also has Losses.
First, It MUST be a HIGH SPEED DIODE, to actually rectify the output efficiently. Even Than there will besome Losses.

But Remember, (On a Sine Wave) the Output voltage is typically 1.4 Times the RMS voltage and the Output Current is only .707 of the input current.
** For Other Waveforms, such like this device will give, Heaven on knows what this Relationship will be?

The Resistor with that Parallel Capacitor and meter, is Even Worse.

The best option would be a 1 ohm resistor, and using a VERY High Speed Op-Amp to make a Full Wave Peak Detector.
Than Average it to get an Actual Current flow.

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A proposed test to find out if energizer is giving greater than 1 COP

A proposed test to find out if energizer is giving greater than 1 COP

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