I'll be setting up an automatic water changer in a few weeks and I've been
having trouble with the math.
The reading I've done suggests that a 20% daily water change is still going
to leave the fish swimming in 82% crap (r = e ^ -.2) as opposed to 50% to
100% crap with large weekly water changes. BUT, the detailed formula I've
seen shows the crap remaining as r=(1-a/n)^n, where a is the percentage of
water changed and n is the number of changes. Now assuming I change 1 gph
and a drop is 1/3600 gallons, then I'm left with 99.99% crap because the
changes are so small. (One drop at a time) Am I missing something here or do
I need to remove some water first to make this work?