Bob Boyce 101 efficiency details

[CrazyEwok]

really newbie... so if i had the relevant plate count in a series cell to divide the incoming voltage down to 1.2v per plate pair then that would not produce more gas??? same if not less gas per watt but more gas per applied amp... Volts are 100 percent relevant. Being your Volts control alot of things that happen in a cell. Just of the top of my head your Volts ,by them selves, can help you overcome the resistance of water... oh wait you don't understand how that would possibly work??? Or that your voltage can be increased to overcome larger plate gaps... oh wait forest that wasn't in D14.pdf was it so how can i possibly know that. Or that your pulsing you blessed amps into your cell only increases the efficientcy due to colapsing magnetic fields and the latent lag of them being created. (that one i will give you for free, if you don't believe me that there is an effect on water due to magnetic fields you should look into it). ALSO another monkey wrench that is well known, is that all laws of physics and science have exceptions at which these exceptions are only seen 99.9999% of the time and only in extreme micro scale circumstances. so no you blantant blind following that volts only equals heat and steam is far from accurate...   

[newbie123]

really newbie... so if i had the relevant plate count in a series cell to divide the incoming voltage down to 1.2v per plate pair then that would not produce more gas???

Of course,  In a series cell with "chopped down" voltage  you  will produce much more gas.... But in the context you used, and since you didn't talk about neutral plates...  I thought were implying  6k watts is a lot more energy, therefore it will produce much more gas regardless of the plate configuration...

same if not less gas per watt but more gas per applied amp... Volts are 100 percent relevant. Being your Volts control alot of things that happen in a cell. Just of the top of my head your Volts ,by them selves, can help you overcome the resistance of water... oh wait you don't understand how that would possibly work??? Or that your voltage can be increased to overcome larger plate gaps... oh wait forest that wasn't in D14.pdf was it so how can i possibly know that.

You're trying to change the subject now.......    And everything you just said is irrelevant to Faraday efficiency.  Go read up on it  and explain to me why "his voltage" is significant... Since you think it is.   Amps doing work is significant...  that is it wrt Faraday efficiency!    Of course you need voltage for amps.. But this still has nothing to do with Faraday efficiency.

Or that your pulsing you blessed amps into your cell only increases the efficientcy due to colapsing magnetic fields and the latent lag of them being created. (that one i will give you for free, if you don't believe me that there is an effect on water due to magnetic fields you should look into it)

I already have... And pulsing does nothing for a cells efficiency, but you're right that electric/magnetic fields will affect water molecules (decrease dielectric constant, etc) ..    Read all about it here.  http://www.lsbu.ac.uk/water/magnetic.html

ALSO another monkey wrench that is well known, is that all laws of physics and science have exceptions at which these exceptions are only seen 99.9999% of the time and only in extreme micro scale circumstances

Really?  Do you have a reference?

so no you blantant blind following that volts only equals heat and steam is far from accurate...

Again, did you even read any of my previous posts?

[TheNOP]

Really you can calculate the amps from simply having the watts can you??? i have 4watts of power being produced how many amp am i outputting???
Oh wait you can't tell me the amp because you are missing some information... the voltage produced!!!

in fact, you can convert any units to any other quantifier units, if you have the values you need to do the convertion.

and all conversions of power have relevant losses so no you can't do a simple converstion, because this would involve having to calculate your losses...

can't you make the difference between physicaly converting power using a device and units convertions ?

ex: 1 Farad = 1 Ampere = 1 Coulomb/sec = 1 Farad/1 Volt = 6.241506 × 10¹⁸ electrons , etc...
there are no power losses in those units conversions.

Faraday's law is not dependant on the voltage it is dependent on the current.
Faraday's law formula is using atoms charges to calculate the max effeciency that is theoricaly possible.

can you count the electrons you are using in your cell ?
if not, then trying to further explain Faraday's formula to you is hopeless.

is the current efficiency unit, mmw i think, the one commonly accepted in this forum i mean, use electrons and mol in it formula ?

if you don't know how to get the actual current per cell values, then it is not my problem, it is yours.
my point was that you can use Ohm's law to calculate it.

if you are comparing apples with bananas because you don't know how to convert units then it is not my problem, it is yours.

you can't measure your gases output by mol but...
mol=number of atoms
the number of atoms can be calculated given that you have the gases volume, the temperature and the pressure values of that gas volume.

but i guess peoples are so lasy that those calculations mean nothing to them.
at least that seem to be the excuse they give to hide the fact that they are lasy.
it is easyer to say, some of my cell give me more bubbles, so, it must be OU... 
i am not saying it is not, and i am not saying it is either.
what i said is, regardless of having some cells giving me more output then the others, i never had OU.