by CJMOSS » Wed Jun 28, 2006 2:18 am
Bob, I think I understand. If the negative electrode is 60 sq. in. and the positive electode is 9.8 sq. in. [1.25" dia. = 2.45 sq.in.(considering both sides of the washers), times 4 washers].
Then 60 sq. in. + 10 sq. in.= 70 sq. inches. Next, 70 sq.in. /2 = 35 sq.inches. 35 sq inches x 0.25 Amps/sq.in.= 8.75 amps. Seems to jive with the less than 10 amp reading on the guage. Two (2) cells in series is about 4.4 amps draw.
Although, it nags me the relation is 60 sq in x 10 sq in = X square root or 24.5 sq in.
thanks in advance
"the present day composer refuses to die" FZ