Hello All,

My name is MOA and I am an advanced mathematics and physical science major from southern Idaho. A few years ago I undertook a monumental task--rectifying the errors of many of the common rules on stocking a freshwater aquarium. Surprisingly, I was able to develop a set of equations that proved 100% accurate in laboratory tests. The advantage of using these formulas is that you can predict bioload before the aquarium system is even bought! Unfortunately, this set of equations has only limited field testing and so it should only be considered a theory at this point. Also, it is amazingly complex and therefore not very user-friendly.

That being said, I plan on posting bits and pieces of it over the next couple of weeks so that the useful information of this study will be available to the general freshwater community. Please note that I am not saying that this theory is absolute. In fact, I am hoping for some feedback from whomever reads this so that I can perfect it. I thoroughly welcome conflicting viewpoints so long as you can provide specific examples.

More to come!

"Tears aren't a sign of weakness, they're a sign of poor plumbing."

--Dead Men's Lies

Be warned -- everyone at college has a weird roommate. If you don't have a weird roomate -- then you're the weird roommate.
                                                                                                  Conan O'Brien, Stuyvesant High School.

Hello Again,

It's time for me to make good on my promise and start explaining something I call the MOA, or math-only aquarium. The MOA is, of course, only theoretic in that aquarium dynamics change continuously, but the MOA is a useful way to approximate what goes on in a freshwater aquarium. The MOA accounts for such factors as fish biomass, filtration capacity, surface area, temperature, fish aggression, and other incidental items. Since there is so much to cover, I am only going to post bits and pieces at a time. Following is an excerpt from my thesis on how the MOA accounts for biomass. Please remember that I welcome comments and questions. Therefore, if you got something to say, please say it!

"Before we start creating equations, we have to clear up one major problem: what units do we want our equations to return?. As I said earlier, the MOA is intended to help aquarists stock their aquariums with fish, but how does one define the significance of a fish in terms of units that have a mathematic meaning? If we use the popular stocking rules that we covered in chapter one as a guide, we might decide to use inches of fish length as our base unit. However, just because using fish length as a base is popular does not mean it is accurate. Therefore, we need to ask ourselves another question: how does fish size relate to a fish’s impact on its environment?.

"The answer to this question is that fish have an impact proportional to their biomass. Biomass is essentially a measure of an entity’s mass that produces a biological effect and it follows that the base unit of the MOA must approximate biomass. To do this I have developed a unit called the IFU, which is an acronym (I tend to favor acronyms) for Ideal Fish Unit. The IFU accounts for the errors of using fish length by utilizing a cubic relation. That is, instead of measuring the length of a fish, it approximates the volume of a fish and thus it also approximates the mass of a fish because most fish have similar densities. The justification for the IFU cubic relation is as follows:

"Specifically, an IFU is equal to the estimated biomass of a fish that measures one inch from mouth to the base of its tail. Consequently, the IFU number assigned to any fish is equal to the length of that fish from its mouth to the base of its tail multiplied by itself three times. As an example, a goldfish can often grow to 8 inches long, but only measures about 6.8 inches from mouth to the base of its tail and would subsequently have an IFU number of 314.432 IFUs (6.8”x 6.8”x 6.8”= 314.432 IFUs).

"It may seem odd that the tail is not counted when computing IFUs, but consider this: if a fish has a very long tail in proportion to its body, the tail length will represent significantly less biomass than the actual body length will. Since this condition occurs fairly often in the real world, it is more accurate to measure without the tail fin. Additionally, while we are talking about being accurate, there are fish whose biomass is not adequately approximated by IFUs because they are either too high-bodied, wider than usual, or long and skinny. For these sorts of fish, the IFU number can be made more accurate by multiplying it by a fish-indicative factor. These factors are as follows:

IFUs Computed by Length Alone x 1.2 = Actual IFUs

EXAMPLE: An Angelfish that measures 4 inches from mouth to the base of its tail has a biomass of 76.8 IFUs.

4”x 4”x 4”= 64 (IFUs computed by length alone) Þ 64 x 1.2 (High-Bodied Factor) = 76.8 IFUs

IFUs Computed by Length Alone x 1.4 = Actual IFUs

EXAMPLE: A balloon-bellied molly that measures 1.75 inches from mouth to the base of its tail has a biomass of 7.5031 IFUs.

1.75”x 1.75”x 1.75” = 5.3594 IFUs Þ 5.3594 IFUs x 1.4 = 7.5031 IFUs

IFUs Computed by Length Alone x 0.4 = Actual IFUs

EXAMPLE: A rope fish (type of eel) measuring 18 inches from mouth to the base of its tail has a biomass of 2,332.8 IFUs.

18”x 18”x 18” = 5,832 IFUs Þ 5,832 IFUs x 0.4 = 2,332.8 IFUs"

Well, that covers the basics of how the MOA system accounts for biomass. It might seem a little pointless right now in that biomass alone does not relate to aquarium size, but, as I post more tidbits, it will become much more useful. In the meantime, ask me anything about biomass and tell me if you see any problems with this procedure for computing biomass.

"Tears aren't a sign of weakness, they're a sign of poor plumbing."

--Dead Men's Lies

In my last post I went over how the MOA system approximates biomass, which is well and good, but it only means something if you understand the rudiments of the filtration process. In talking about this I am probably "preaching to the chior" in that I am sure that most of you already have a very accurate and well-justified understanding of what filtration capacity. Nonetheless, as the MOA system defines things just a little differently than most aquarists are used to, I will go over filtration types for the sake of clarity.

By definition, the filtration capacity of an aquarium is a measure of the quantity of biomass it can support safely (no kidding, right). More specifically, filtration capacity can be broken down into main categories: direct removal of wastes and biological treatment of wastes. These categories can be further subdivided as follows:

Direct Removal

Chemical Purification: Chemical purification refers to processing wastes by either absorbing them or making them chemically inert. A classic example is the use of activated carbon (heat-treated charcoal) in a filter reservoir. Because of the nature of activated carbon, it tends to absorb various ionic substances in the water (like dangerous nitrogen compounds and some growth-inhibiting hormones). (In saltwater aquariums, protein skimming is an example of chemical purification.)

Water Changes: Water changes, as most experienced freshwater aquarists know, are the lifeline of a properly maintained aquarium system. They exchange contaminated water with water that is significantly lower in wastes and therefore directly remove harmful substances.

Biological Treatment

Aerobic Bacterial Activity: Aerobic bacteria are those that require dissolved oxygen to carry out their metabolic processes and are also the sort most often utilized by freshwater aquarists to treat fish wastes. Unfortunately, the end result of aerobic biological filtration is nitrate which, though much less toxic than ammonia or nitrite, will accumulate if not removed by some sort of direct removal.

Anaerobic Bacterial Activity: Anaerobic bacteria are not often utilized by freshwater aquarists due to the fact that they can produce toxic compounds if not maintained under ideal conditions. However, they are interesting in that their existence in the absence of dissolved oxygen leads them to denitrify nitrate--or break it up into oxygen and nitrogen. In the natural environment of most fish, the fish to anaerobic bacteria ratio is fairly small. This means that anaerobic bacteria can process a significant portion of nitrogen compound waste in the natural environment of fish. In the home aquarium, though, most people simply keep too many freshwater fish for the natural quantity of anaerobic bacteria to have a substantial impact.

Please not that I did not mention physical filtration at all. The reason for this is that it is nearly impossible to erect a filtration system that accomplishes the above filtration types without physically removing particulate matter from the water flow. Thus, physical filtration is a given in most cases.

"Tears aren't a sign of weakness, they're a sign of poor plumbing."

--Dead Men's Lies

Now that I have discussed what filtration capacity is, I can begin to explain how the MOA quantifies it. Right now, let's consider filtration by aerobic bacteria since it is such an integral part of freshwater aquarium husbandry. Please bear in mind that MOA creates a model that can be given numerical models. Thus, it isn't going to be a completely accurate picture of what is going on; the MOA simplifies things to make the math easier to deal with. As before, if you see a problem with the MOA, don't hesitate to share it.

The MOA Explanation

Aerobic bacteria can only do their job if three conditions are met: 1) they need some sort of medium to grow on, 2) the water must contain dissolved oxygen provided by an air-water interface (surface area of water that touches the air so that gases can be exchanged), and 3) there must be a flow of this oxygenated water through the medium--or a turnover rate (the rate at which the water flows through the water over a given span of time). Thus, the first biomass equation has three factors: medium (M), air-water interface (A), and turnover (T). Also, the product of these factors multiplied by some constant (C) is equal to the relative number of IFUs an aquarium can support by aerobic bacteria. Subsequently, the first biomass equation looks something like this:

IFUs/gal. = C x M x A x T or IFUs/gal. = C(MAT)

Please note that aerobic bacteria leave behind nitrate. Because of this, and other factors, this basic equation will be expanded later on. For now, we will assume that nitrate and other trace compounds are insignificant.

In order to use the above equation you must be able to calculate M,A, and T for a given aquarium system. M is probably the most difficult in that it is the surface area of medium available for the bacteria to grow on. Thus, you need to now how many cubic inches of medium you have and how many square inches are represented by a cubic in of your particular medium. Measureing the total cubic inches of a medium follows a basic format: if your medium is "square" (a rectangular prism of some sort), multiply it length by height by width in inches. As an example, suppose you use an undergravel filter in a standard ten-gallon tank and your gravel bed is about 2 inches in depth. The total volume of such a medium is 400 cubic inches (20" (length of tank) x 10" (width of tank) x 2" (height of gravel) = 400 in.^3). Unfortunately, calculating the volume of a medium that has an awkward shape is much harder. One trick that might be useful is to know that one cup is equal to about 14.4375 cubic inches. Thus, if you use a canister filter with ceramic tubes as a medium, their volume could be approximated by using kitchen measuring cups.

With regard to the number of square inches a cubic inch of medium represents, the following table might be useful:

For Gravel of a Specific Diameter:

3.1416 / average diameter of gravel = conversion factor

For Cermaic Tubes of a Specific Diameter:

6.8 / average diameter of ceramic tubes = conversion factor

For Foam or Sponge Medium

Conversion factor = 450 sq. in. / cu. in.

Bearing in mind that are basic equation is designed to return IFUs/gal., at this point, M will need to be measured on a per gallon basis as well. Using the previous example of the ten-gallon tank with an undergravel filter and theorizing that the average diameter is about a quarter of an inch, the M would be 502.656 square inches per gallon (400 cu. in. x 3.1416 / .25 in. diameter = 5,026.56 square inches for all ten gallons; 5,026.56 / 10 gal. = 502. 656 sq. in. per gallon). Thus the formula for M is as follows:

M = (Cubic Inches of Medium) x (Specific Factor) / (Gallon Size of Aquarium)

While M is a regal pain to calculate, A is much simpler. Since on gallon of water occupies 231 cubic inches, the surface area of air-water interface per gallon is equivalent to 231 divided by the height of the tank in inches, or

A = 231 / (Height of Aquarium)

As an example, the standard ten-gallon aquarium has a height of about 12 inches. Thus, its A is 19.25 square inches per gallon (231 / 12 = 19.25 sq. in.).

The nice thing about this formula is that it works for any aquarium with straight, vertical sides. So if you have a bow-front, hexagonal, or pentagonal aquarium, you can indeed calculate its A without a problem provided you have a piece of paper handy (or a really good math brain).

T is also pretty easy to calculate. To calculate T, take the gph (gallons per hour) rating of your filter pump and divide by the gallon size of your aquarium. In keeping with the ten-gallon aquarium, let's suppose it uses a powerhead rated at 80gph to pump water through its filter bed. If such is the case, its T is 8 cycles per hour, or it can cycle all the aquarium water 8 times in one hour (80 / 10 gallons = 8 cycles per hour).

The product of the M, A, and T yields a relative comparision for the power and filtration capacity of various biological filters. To make this coincide with how the MOA approximates biomass, we can use the relationship I discovered a few years ago regarding tropical freshwater aquariums maintained at 78 degrees Fahrenheit:

IFUs/ gal. = .0005(MAT)

In other words, multiply the M, A, and T of your aquarium and then multiply by .0005 to find out how many IFUs each gallon of your aquarium can support. Using the ten-gallon aquarium again, the procedure would be thus:

M = 502.656 ; A = 19.25; T = 8

MAT = M x A x T = 502.656 x 19.25 x 8 = 77,409.024

IFUs/gal. = .0005(MAT) = .0005 x 77,409.024 = 38.7045 IFUs/ gal.

Therefore, the ten-gallon aquarium could support 38.7045 IFUs per gallon or 387.0451 IFUs total (38.704512 x 10 gal. = 387. 0451 IFUs altogether). Now we have connected the way that the MOA approximates biomass to waste production.

Please bear in mind that this basic formula is only correct if the tank is maintained at 78 degrees, the water is changed appropriately, and there are no mitigating circumstances. Consequently, the MOA does not stop here and I will add on to this equation later on. For now though, play with the numbers and tell me what you think.

"Tears aren't a sign of weakness, they're a sign of poor plumbing."

--Dead Men's Lies