Basic Free Energy Device

[sm0ky2]

is the motor a load? yes
is the generator an input? yes
^^^^ those two balance each other.
the motor is not using any significant energy when you are turning around and generating it back
is the light a load? if you don't count the motor, then yes. otherwise you're double counting the same energy....
and forgetting to subtract the input from the generator.
it doesn't matter what the motor and generator are doing. you don't even need to count them.

you can observe the system strictly from the perspective of batteries and light bulb
energy in, energy out.
when the energy in the initial batteries, is less than the energy used by the light bulb + the energy in the re-charging batteries
then you have something.
until then, you are fooling yourself, or trying to fool others,.

" you get longer run times"
longer run times than what? running an unloaded motor?
running the motor under the generator load, without re-charging the batteries ?

[citfta]

If you have two 12 volt U1 batteries connected to the original 3BGS that Dave worked with and they power an inverter for 5 hours that is lighting a 120 volt 100 watt light bulb and the batteries don't show any loss of voltage does that mean anything at all to you guys? The batteries were also powering a motor that had no load on it.  The U1 batteries are the size used in lawn and garden tractors.   They only have a CCA of 200 to 300 depending on the brand.  I don't have any info on the AH rating but size wise it is probably in the 35 to 45 range.  I only saw this one time with my own testing.  Dave said from the very beginning his system was unstable.  So he has been working for all these years trying to find a way to stabilize the system to be able to repeat the power gains and make it a usable system.  He now believes he has found that. 

By the way I am not some young kid at this.  I am 69 years old and have worked in electronics since I was 14.  I KNOW how to read a meter.  I KNOW that battery voltage alone is not a good indicator of power used.  I KNOW those batteries should have shown a drop in voltage but they didn't.

Carroll

PS: A side benefit of the original 3BGS is that is was a very good way to recondition batteries.  It cleaned the sulphation from the old batteries fast than any other method I have used.

[TinselKoala]

Well, let's see. First of all how many Joules of energy are contained in 2 12 volt, 45 A-H batteries fully charged? The nominal 12 volts should actually produce an unloaded terminal voltage of something over 13 volts, but let's disregard that for the moment.

12 v x 45 A-H = 540 Watt-hours per battery.
540 Watt-hours x 60 minutes/hour x 60 seconds/minute = 1,944,000 Watt-seconds or Joules per battery, so the two together will have 3,888,000 Joules of stored energy.

Now let's see how much energy it takes to run an inverter powering a 100 watt light bulb for 5 hours. 100 Watts is of course 100 Joules per second.

100 Joules/second x 60 seconds/minute x 60 minutes/hour x 5 hours = 1,800,000 Joules... less than half of the energy stored in the 2 batteries in the first place.

If an OTS inverter is only 65 percent efficient, that means that 1,800,000 Joules/0.65 = 2,769,231 Joules would be required to provide that output. Still much less than the 3.9 megaJoules stored in the batteries.

The current required to run the lamp at a true 100 Watts of output power will be 100W/120V = 0.83 Amps output. With a 65 percent efficient inverter this would require around 154 Watts input to the inverter, and at 12 volts this means 154W/12V= 12.8 amps.


But we have a pair of 45 A-H batteries in parallel, so we have 90 A-H at 12 V available. 90 AmpHours/12.8 amps = a little over 7 hours. It is entirely possible that the batteries would still read an opencircuit terminal voltage of 12 volts or more after only  5 hours running at that load.


Please check my work, I'm not that good at doing math over the internet.


But the batteries are also running an unspecified "unloaded" motor! How convenient.

[citfta]

Hi TK,

You obviously have not even looked at the circuit of the original 3BGS.  The batteries are in series with each other and in series with the motor and in series with the inverter.  There also a "dead" battery in parallel with the inverter.  The "dead" battery is connected in reverse polarity to the other batteries.  In other words the negative of that battery is connected to the negative of the first of the two that are in series and the positive is connected to the other side of the motor.   The motor was an unmodified scooter motor.  The unloaded current though it is about 1 amp depending on how heavily the inverter is loaded.  You also over looked the part where I said the battery voltage at the end of the run was the same as at the beginning of the run.  They didn't start out at 13 volts and end at 12 volts.  They started at about 12.8 if I recall correctly and ended at 12.8.  I am sure they ended at the same voltage as they started.  I know the math doesn't add up to the batteries maintaining the same voltage for the whole run but they DID.  If you guys aren't interested that is fine but don't try to tell me I don't know how to judge what I have seen.

Carroll

[MarkE]

Hi TK,

You obviously have not even looked at the circuit of the original 3BGS.  The batteries are in series with each other and in series with the motor and in series with the inverter.  There also a "dead" battery in parallel with the inverter.  The "dead" battery is connected in reverse polarity to the other batteries.  In other words the negative of that battery is connected to the negative of the first of the two that are in series and the positive is connected to the other side of the motor.   The motor was an unmodified scooter motor.  The unloaded current though it is about 1 amp depending on how heavily the inverter is loaded.  You also over looked the part where I said the battery voltage at the end of the run was the same as at the beginning of the run.  They didn't start out at 13 volts and end at 12 volts.  They started at about 12.8 if I recall correctly and ended at 12.8.  I am sure they ended at the same voltage as they started.  I know the math doesn't add up to the batteries maintaining the same voltage for the whole run but they DID.  If you guys aren't interested that is fine but don't try to tell me I don't know how to judge what I have seen.

Carroll

Not to be patronizing, but if anyone draws conclusions without performing appropriate measurements, then frankly they are in no position to attempt to judge what they have seen.  Batteries mystify many people and voltage based measurements have allowed some such as Bedini to collect significant sums of money for more or less worthless junk for decades.  I know that appropriate testing protocols were suggested to David years ago and he did not see fit to employ them.

Many batteries, particularly lead acid batteries cannot be reliably be evaluated for capacity be terminal voltage. This is particularly true where voltage spikes occur.  Capacity can be evaluated by running them down into any:  a resistive, constant current, or constant power load.  Ditto if one wants to measure the charging energy: integrate the applied voltage and current.  Perform any experiment you like, but measure power to reasonable accuracy, say: +/-3%, for each phase, including the initial charging of the batteries from complete discharge, operation of the device under test, and then depletion of the batteries back to complete discharge, and then apply the book keeping.  What you will find no matter what you do is that you will be hard pressed to get even 2/3s the energy back out of the batteries that you put into them.  Why?  Because lead acid batteries have a 60% - 65% round trip energy efficiency.  Anything that you add on top such as 12V - 120VAC inverters will only serve to reduce the total efficiency.