OT: Anyone Know Where to get 30-40% H2O2 in Ontario, Canada?

Shouldn't you?

Post the final answer.

Hi Jerry,

You use so little silver that a set of electrodes may cost $10 and last for years. I get my silver from monsterslayer. They have no minimum order, but the shipping cost to Canada is outrageous:

With the simple 3 nines cs generator commonly used, you can calculate how much silver you drank. The rods wear down to points after long use and take a uniform cone shape.

Consider the volume of a cylinder is

Cyl = pi * r1 * r1 * h

and the volume of a cone

Cone = 1/3 * pi * h * (r1^2 + r1 * r2 + r2^2)

where

h = height of cone p1 = 3.14159... r1 = first radius r2 = second radius

The amount of silver used is Cyl - Cone.

The diameter of 12 ga wire is 0.0808 inch and the tips were worn down to 0.024 inch dia over a distance of 4 inches. There are two identical electrodes. Silver has a density of 10.5 grams per cubic centimeter.

Normally, calculations like this are highly prone to error due to the numerous metric conversions needed. However, a program called "Mercury" written by Roger Schafly does the necessary calculations for you. It only needs are the conversion factors and the input parameters and it will figure out what you want and solve the equations for you.

In my example, the amount of silver drank was 0.361 cubic centimeters, which weighs 3.79 grams. (Calculations are below.)

Consider the published data indicates the amount of silver needed to produce argyria is 2 grams. I have ingested nearly twice that amount and am still as rosy-cheeked as the day I was born, 63 years ago.

Well, almost, anyway:)

Mike Monett

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Mercury Calculations ~~~~~~~~~~~~~~~~~~~~ ; Volume of Cylinder Minus Truncated Cone

; Roger Schafly's Mercury is available at ;

; 1 cubic inch = 16.387064 millilitres ; 1 millilitre = 1 cubic centimeters ; silver weighs 10.5 grams per cubic centimeter

Unit Conversions ~~~~~~~~~~~~~~~~ Ag = 10.5 * cc cc = 16.387064 * CuIn * 2 Cone = 1/3 * pi * h * (r1^2 + r1 * r2 + r2^2) CuIn = Cyl - Cone Cyl = pi * r1 * r1 * h

r1 = d1 / 2 ; radius of first diameter r2 = d2 / 2 ; radius of second diameter

Input Parameters ~~~~~~~~~~~~~~~~ d1 = 0.0808 ; original diameter d2 = 0.024 ; tip diameter after years of use h = 4 ; height of cone

Solution ~~~~~~~~ Ag = 3.7990 cc = 0.3618 Cone = 0.0094 CuIn = 0.0110 Cyl = 0.0205 d1 = 0.0808 d2 = 0.0240 h = 4.0000 r1 = 0.0404 r2 = 0.0120