You've all heard the "one inch of fish per gallon" rule of thumb, but every tank is different in terms of gas exchange and volume, and surface area. So why hasn't anyone come up with an accurate stocking method that fits your tank? Well they have. I rented a book out from the library called "The Complete Fishkeeper," by Joseph S. Levin, and it had a stocking method that worked great!!! This book is hard to find and expensive, so i'll tell you how it's done.
If you can add, subtract, multiply, divide, and own a calculator, it should pose no problems.
Multiply the length and width of your aquarium in inches to get your tank's surface area. Let's say that the tanks measurements are 24" by 12", a standard 20 gallon, you'd get 288sq.in. Now before you do anything else, i should say that you should have a list of fish that you want to get, and have their maximum adult lengths at hand. Always start off with a plan, trust me.
The longer the fish is, the thicker the body is, and the heavier it's bio-load in the tank is. So It's not fair to keep 1 20-inch fish in a 20 gallon aquarium, and a 20 gallon stalked with 10 2-inch fish looks pretty empty, so the stocking rule will change for each fish size category. For fish upto 2 inches, 6sq.inches of surface area will be needed for each inch of fish body. For fish between 2-4inches, 9sq.inches of surface area will be needed for each inch of fish body. And finally, for fish between 5" and up, but not too long, 12sq.inches of surface area will be needed for each inch of fish body.
This may seem complicated, but it's not. Let's say that in this 20 gallon tank, you'd like to keep 6 Black Neons. At 1 1/2 inches per fish, it would fall into the "up to 2 inch" category, demanding 6sq.inches of surface area per inch of fish body. Now simply multiply the fishes length by the required surface area per inch of fish, and you get the total surface area that a fish of this size needs, which, in this case comes out to 9sq.inches of surface area per Black Neon (1.5 X 6sq.inches=9sq.inches). I want 6 Black Neons so i multiply 9sq.inches by 6 (the number of fish i want) and get a total or 54sq.inches of surface area for 6 Blach Neons (9sq.inches X 6=54sq.inches). I now subtract this total from the total surface area and find out that i can keep 6 Black Neons in my tank and have 234sq.inches of available surface area in my tank (288sq.in - 54sq.in=234sq.in).
Now lets say that i want 2 Rams in my aquarium. Measuring 3", Rams fall into the "2-4inch" category, they demand 9sq.inches of surface area per inch of fish body. So again, multiply the required surface area per inch of fish body by the length of the fish, and you'll get the total surface area that each of these fish needs. In this case you'd find that each Ram needs 27sq.inches of surface area (3 X 9sq.in=27sq.inches). Since i want 2 Rams, i multiply the the required surface area per fish, by the number of fish i want. In this case, i'd get 54sq.inches of surface area (27sq.inches X 2=54sq.inches). So i'd subrtract 54sq.inches from the 234sq.inches of surface area remaining, and get 180sq.inches of surface area (234sq.in - 54sq.in=180sq.in).
So far in this tank, I have 6 Black Neons, and 2 Rams, with 180sq.inches of available surface area. Just continue this procedure until you can't comfortably fit any more fish, and you'll be set to go.