Imhotep's Lab Interactive FAQ

Hello,

I really interested in seeing how dead CFL can made work by adding relay and car coil. But I am wondering if utilizing wall electricity directly is possible.

Instead of using 12 Volt lead acid battery, using wall electricity is a lot cheaper in our country. People would not consider power saving if they have to buy 2 batteries and one car coil for it. Their price is enough to covers years of electricity cost in here, which offset the power saving. Same problem if we use DC power supply.

It would be great if there are ways to utilize wall electricity directly. Would attaching self oscillating relay to our wall electricity be safe? I imagine it would light out the CFL even if we don't use coils.

this is something i have done here.

I used an Imhoteb/Bedini fan with an ignition coil connected in parallel with the output coil of the fan. the HT lead of the ignition coil was powering the CF bulb and the output of the fan is charging ordinary zinc carbon batteries. I use a universal charger (one of those which has four different output plugs and switchable voltage) that is rated at 12v 400 milliamps to power it. 12 x0.4 = 4.8 watts

The power consumption is 12v at 280 milliamps. 12 x 0.28 = 3.6 watts.

The Bulb is 220v 12w compact fluorescent with the electronics removed. it is not as bright as when it was used with 220v but gives about as much light as a 3w compact fluorescent.

The motor was rewound bifilar style as shown on Imhoteps web site because it would only oscillate using the standard windings. it gives about half the flow of air that an unmodified fan would give. the fan used is a 5" 12v 560 mA unit similar to a computer fan but modified of course. The trigger coil is 27# wire the power coil is 30# wire. The power consumption is 12v at 280 milliamps. 12 x 0.28 = 3.6 watts.

The ignition coil is a standard Hitachi 12v coil.

I charge either AA or 9v zinc carbon batteries in about 6 to 10 hours. a lead acid 12v 6ah takes about 2 to 3 days in this setup

Total input
3.6 watts x transformer efficiency (assume 80%) = 4.5 watts

total output
fan = 3.6 watts
+
Bulb = 3 watts
+
battery charge 12 x 6 divided by 52 hours = 1.38 watts
=
7.98 watts

this is very approximate but it is clear to see a cop of 1+

Mick

Can you post calculation in watt-hour?

Watt-hour for charging the battery
Watt-hour for powering the lamp and fan

So if 12V6A battery is used:
Watt hour = charging time (Hours) x Power required
= 52 x 4.5

Watt hour = powering time (Hours) x Power required
= ? x 6.6


How long can 12V6A battery powering the lamp and fan? If COP is greater than 1 than it should be able to power for more than 36 hours.

If i run the setup from a 12v 6ah battery assuming the capacity is what it says it is, 6ah. Lead acid batteries that have been charged several times using a radiant charge often increase in capacity.

12 x 6 = 72 watts

power consumption of fan = 3.6 watts. therefore

72 / 3.6 = 20 hours to flatten the battery


Both the CF bulb and the charging battery are being powered by the radiant energy. we are not able to measure amps for this energy so power consumption is unknown, but we can see the resultant light and measure the capacity of charge by draining the charging battery. because of this we are making assumptions.

the apparent light from a 12 watt CF bulb is around that of a 3 watt softlight CF bulb, powered normally, as the light is less intense. If i use a 3w bulb the output is visibly lower than that of the 11w in the same setup, although more intense, plus there is more charge going to the battery. I do not have the means to measure light output.

Apparent light output for the 12v6ah battery

3w x 20h = 60 watts

The fan, we can measure the input power and therefore its consumption is known. I do not have the means to measure its torque or speed so the mechanical output is not. It will not be 100% efficient because of heat in the coils and friction but a reasonable assumption is 80% efficient. We know that the sum of the mechanical and the losses (friction and heat) will be 100%. I made a mistake in my first post, the trigger coil is 32# and the power coil is 30#.

3.6w x 20 hours = 72 watts

The ignition coils power consumption is included in the fan calculation. It must be said that most of the radiant is produced in this coil.

A lead acid 12v 6ah takes about 2 to 3 days in this setup so we know it will not be charged fully. A battery that has been charged several times using the radiant energy, accepts a charge faster than one that hasn't.

Assuming the charge is linear.

1.38 x 20 = 27.6 watts a little over 1/3rd charged

So using a 12v6ah battery to power a fan, light and charge a lead acid battery we end up with an output of (and this is approximate):-

fan and losses 72 watts
CF bulb output 60 watts
Battery charge 27.6 watts
Total output 159.6 watts

apparent free power 87.6 watt hours

Remember we do not have the losses of the transformer in this setup

From ac power to fully charge the battery

4.5 x 52 = 234 watt hours consumption.

loss in transformer assuming 20% losses

234 x 0.2 = 46.8 watts

total output
fan 3.6 watts x 52 = 187.2 watts
+
Bulb 3 watts x 52 = 156 watts
+
battery charge 1.38 watts x 52 = 71.76 watts
=
414.96 watts

Apparent free energy = 414.96 - 234 = 180.96 watts

The gain is huge but unfortunately it is not simple to put all the gain in one form of energy. if you don't use the CF bulb, that power will go to the battery and the charge rate will be close to 100% efficient, but you still have the fan power. if you don't use the fan and just pulse a coil, you are wasting some of the magnetic energy.

In an Ideal setup you will convert the magnetic energy into electrical power too and this is the main aim of most people.

My first post gave figures per hour and this one gives figures per discharge and per charge of a battery. sorry its so long winded. I hope i got everything right this time hehe

Does this answer your question?

TAKING a totally skeptical viewpoint, (I'm actually a believer)
one can eliminate (or question) your CFL bulb output, so let's
eliminate it entirely...
7.98W - 3W = 4.98W output

MINUS FULL RATED INPUT of your charger
4.98 - 4.8 = .18 watts over unity wattage!!!

This assumes that ZPE or Cold Electricity is waaaay more efficient
than electron flow in powering fluorescent bulbs (doubtful)
so in theory, another factor totally IN YOUR FAVOR
is that the power supply was rated at full load (but measured under it)
and was not compensated lower for pulsed operation
(like the skeptics like to point out that the output is pulsed output
and therefore lower) so IMO it is a clear-cut case of OVERUNITY!!!

...But I'd like to replicate (just in case!!!)



PS: Assuming 80% transformer eff. is good, but some cheapo ones might be slightly lower...
I'll check a cheap 10:1 X-Former I have for an
old fiber optic X-mas tree.
But if memory serves, it was 80%!

I don't like to use the term "Overunity" as it implies more out than in which is not the case. The energy is coming from somewhere and i think Tom Beardon best explains this.

As i said, i do not have the ability to measure the input or output of the CF bulb but i say apparent output. I agree this is very approximate but it is giving usable light similar to that of a 3watt softlight CF Bulb so that is why i put 3 watts in the calculation. A normal CF bulb does have electronics that have to be powered and there are losses, i ignored these losses so i agree 3 watts is probably more than it is using; however there is still the light output and that cannot be ignored.

I am no Bedini, and i cannot back this up with any evidence but i believe that ZP energy is equal and opposite to electron flow less losses. This translation does seem to be very efficient, almost 1 to 1. So in any well designed system, not using any type of negative resistance, a COP = to or <2 is possible. All the experiments we are doing seem to work this way.

Fan =3.6w = mechanical energy + losses

But now we are tapping the ZP so we have to add this.

Fan =3.6w + (3.6w ZP) + Mechanical energy + losses.

Negative resistance is a proven thing so im not going to try to explain it because ill get it wrong, but when we are charging Lead acid batteries with this ZP energy we do get a negative resistance effect so the error in my calculations is more likely to be in the battery calculation because of this.

Fan =3.6w + (3.6w ZP) + Negative resistance + Mechanical energy + losses.

Mind bending inst it.

when you replicate what i did, connect one end of the CF tube to the HT and then (carefully because it bites) connect the other to things to get the most light. I use a metal trolley like an antenna for best results. I also get better light just before electrical storms and sometimes the CF gives almost as much light as it would in normal configuration.

Thanks for the answer. Even if it long, I gladly read it all.

I currently trying to charge battery using original Imhotep Lite design. It seems to charge at 0.2V per hour rate for almost dead 12V battery (7 Volts left). Currently draw 9mA, required to oscillate the relay. So charging takes 0.108Watt. I'll see how many hours it takes to full charge this 12V7Ah battery.

that depends upon the condition of the battery, the level of sulphation and the power of the spikes, so too many variables to give an answer.

when it is desulphating a battery it will be very slow to charge. It may take days or even weeks. just keep charging it until the voltage reaches 14.4 volts for a normal lead acid battery or 14.8 for a lead calcium battery (sealed type)

If the voltage does not rise for 24 hours, discharge 5% of the voltage slowly then start to charge again. if after several cycles the battery will not rise, then maybe the battery is too damaged to recover.

Thanks for that . I don't know that charging can take weeks .

I currently increasing the amp of my unit by adding coil in parallel. It is able to do low amp elelctrolizing (50mA worth of released bubble) too while charging battery.

Not shown charged battery is between the HHO cell and collapse point.