MODIFIED BEDINI SALE ON EBAY

[Omnibus]

From the answers to the questions on the ebay listing:

So let's run the numbers. Take the "starting OUTPUT" and just assume it's constant for the whole 6.5 hours. How much ENERGY has been put through the lighted lamps?

I get this:  0.02 Watts for (6.5)(60)(60)seconds, or 468 Joules.

What do you get?

Now, how much ENERGY is in a typical modern AA battery? Let's just use an ordinary one off the shelf which has 2000 mA-H capacity, and let's use the ENDING voltage of the battery used in the device: 1.25 volts.

So I get this:
The battery contains (1.25)(2000) milliWatt-hours of energy, or
(1.25)(2000)(60)(60) milliWatt-seconds of energy, or
9000 Joules.

So the energy content of the AA battery is about 18 or 20 times greater than the energy output of the device, totalled up over the 6.5 hours of runtime.

It is always possible that I have misplaced a decimal point or divided when I should have multiplied or something, so I encourage everyone to repeat my calculations, just in case this device is 500 percent efficient instead of only 5 percent.

Meanwhile, I am going to have another cup of coffee, with a stout shot of brandy in it, because it's cold in here and my Ainslie heater doesn't seem to be working properly.

Problem is we don't need to know what the energy content of the battery is, as is obviously your impression. What we need to know what the actual amount of energy has been consumed during these 6.5 hours. Do you know it?

[TinselKoala]

He says the battery started at 1.5 volts and the device stopped running at 1.25 volts, 6.5 hours later.
If the battery contained a measly 2000 mA-H to begin with, that means it will supply 2 amps for an hour -- or about 300 mA continuously for 6.5 hours -- BEFORE the voltage drops appreciably, and modern batteries, especially alkalines, have a pretty flat initial discharge curve and a steep dropoff at the end.

So which numbers don't you agree with, the 9000 Joules input (the battery's entire charge) input to the system, or the numbers FROM THE BUILDERS own measurements, of the 468 Joules output of the system?

[Omnibus]

He says the battery started at 1.5 volts and the device stopped running at 1.25 volts, 6.5 hours later.
If the battery contained a measly 2000 mA-H to begin with, that means it will supply 2 amps for an hour -- or about 300 mA continuously for 6.5 hours -- BEFORE the voltage drops appreciably, and modern batteries, especially alkalines, have a pretty flat initial discharge curve and a steep dropoff at the end.

So which numbers don't you agree with, the 9000 Joules input (the battery's entire charge) input to the system, or the numbers FROM THE BUILDERS own measurements, of the 468 Joules output of the system?

I already said what I don't agree with. Read carefully what I write. And ... leave the builder alone. Again, do you know how much energy was consumed? Where are the results of your measurements of the input and the output energy?

[TinselKoala]

The builder's own measurements aren't good enough for you?
The only thing in my numbers that did NOT come from the builder's measurements and statements is the energy content of the battery.

Perhaps he used a nearly-depleted battery of inferior quality that was just about to tumble over its discharge cliff and drop voltage, and only contained 400 Joules to begin with.

So you are absolutely right, Omni. I have no idea, I am assuming that he would use an off the shelf, new battery...like he says he did.

[osiris]

i will say ... its real i have many units that operate identical ..

he is honest .. and he is correct in asking 500 bucks ..
dont like his price .. pay his labour .. or build your own ..

osiris

ill garentee he has spent far more in labour than 500 could buy you in man hours on it ..