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Overunity Machines Forum



hho wirecell concept largest surface area question

Started by jikwan, February 22, 2008, 01:31:23 PM

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0 Members and 2 Guests are viewing this topic.

jikwan

RR2
got to thank you for imput
i did view your calculations before delete
i felt i was defeated
they were pretty logical.........

MRE
my intuition tells me very well theres more surface area
if i get math backup i got something more concrete

yes understood thankyou
zen is the art of seeing everything and noticing nothing

ResinRat2

Sorry about the previous deletion, I found an error in my calculation. Fixed now.

Here's how I see it:

Formulas obtained from : http://www.math.com/tables/geometry/surfareas.htm



Surface Area of a rectangle: 2ab + 2bc + 2ac

If a and b are 0.5mm let?s assume the rectangle is 1 mm long.

Total surface area is 2(0.5)(0.5) + 2(0.5)(1) + 2(0.5)(1) = 2.5 mm3.

------------------------------------------------------------------------



Surface Area of a cylinder: 2Ïâ,¬r2  + 2Ïâ,¬rh

If r is 0.25mm (0.5 mm diameter) and h is assumed to be 1mm long then:

Total Surface Area is 2Ïâ,¬(0.25)2 + 2Ïâ,¬(0.25)(1) = 1.96 mm3.

--------------------------------------------------

At this point it looks like the plate has the advantage, but now lets line up three plates together so they are touching and make one plate. This leaves us with a rectangle with the following dimentions:

0.5 mm x 1.5 mm x 1 mm. This gives a total surface area of:

2(0.5)(1.5) + 2(1.5)(1) + 2(1)(0.5) = 5.5 mm3.

Notice that four areas would be covered over where the plates are now touching each other. This area is now lost once melded into the shape of a single plate.



What happens when we line up three cylinders side by side?

Since only a small amount of surface area would be lost at the points where the three cylinders are touching each other, we could almost say a negligible amount of surface area would be lost. So we would have the previous area we calculated for one cylinder, multiplied by three, or simply:

1.95mm3 x 3 = 5.88 mm3  with only a small amount of area lost.
This would indicate that the greater the length of the plate, the more of an advantage in surface area would be gained using the ?wires? or cylinder shape.

So your idea appears to be correct, the use of the wires would give a greater surface area if the wires are joined together, side by side, in a continuous line to make a ?sheet? made of cylinders.
Research is the only place in a company where you can continually have failures and still keep your job.

I knew immediately that was where I belonged.

ResinRat2

I guess a picture shows it better.

Which would have the greater exposure of surface area? The plate or the cylinders (wires)?

The calculations in the previous post indicate the advantage goes to the configuration using wires.
Research is the only place in a company where you can continually have failures and still keep your job.

I knew immediately that was where I belonged.

Farrah Day

Farrah Day

"It's what you learn after you know it all that counts"

jikwan

RR2
thankyou for your time and effort on this
your 2nd drawing   i calculate aprox
tubes--70mm
thick plate--24mm    surface area

wire, then has max surface area
tubes and lastly plate

so then the most efficient use of the material being used is wire

my intuition, at this point tells me theres a lot more advantages in using wire

high output for low amperage

im about to order a few hundred metres of the stuff and make a 7 "plate" cell

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                                 I                    I
                                 I                    I
                                 I                    I
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some kind of plastic frame like this
wound horizontaly and vertically     makes one "plate"
pos-neu-neu-neg-neu-neu-pos
aprox 200metres should be enough

the two end plates can be solid plates
jhon aarons points out that the negative charged electrode produces hydrogen
                               
zen is the art of seeing everything and noticing nothing