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RE: MOA's: Advanced Mathematics for Aquarists

math-only aquarium — Tue, 01 Jul 2008 19:46:15 GMT

Hello,

I just wanted to tell anyone who might read this that I've changed some of my theories and disproved some of what is posted on this thread. The general concepts remain the same, but the aplication is very different.

MOA

RE: MOA's: Advanced Mathematics for Aquarists

math-only aquarium — Thu, 20 Mar 2008 10:05:23 GMT

New spreadsheets in the FAMA section!

RE: MOA's: Advanced Mathematics for Aquarists

math-only aquarium — Sat, 16 Feb 2008 16:10:44 GMT

Hmm...it's been a while since anyone posted in this thread.

First, tigerfish, I like the idea of keeping aquarium dynamics simple, but the bottom line is that anything that involves living things is going to involve a fair amount of science and therefore some math as well. I do admit that using the level of math that I use is definitely extreme, but the scientific nature of aquariums can not be avoided. Simplifying is OK, but, tigerfish, be careful not to oversimplify. The problem is that oversimplification results in, as you admitted in your own example, exceptional cases. For instance, you mentioned that, "55gal+30 fish = bad," but when on to mention the exceptional case that very small fish negate this guideline. Because of this fact (that simplification leads to exceptional cases), I prefer to present the complexity and then move on to the generality.

Also, reefman13, check out my spreadsheets in the FAMA section. They are in the first post of the thread labeled "Freshwater Stocking Guide: A Reference for Beginners." You will find that they agree, in principle, with what you are saying.

MOA

RE: MOA's: Advanced Mathematics for Aquarists

tigerfish — Fri, 15 Feb 2008 21:19:20 GMT

its easy . feed your 2 or 3 times a week at the most. dont let any food fall to the bottom if any gets there u feed to much . they smaller amounts twice that day. use floating pellets. the more filters the better. big ones on low flow work better than littleones on high. over feeding is bad, overstocking is bad to u dont need math, u just need to know what your doing.55gal+30 fish = bad. les u like guppys and such

RE: MOA's: Advanced Mathematics for Aquarists

reefman13 — Fri, 15 Feb 2008 20:11:25 GMT

Very interesting thread. I find it very interesting.

I do have to say however, that a thread like this can be very misleading. I think that one very often over looked thing is the active characteristics of fish.

The size does play play a part in choosing the fish for the right aquarium, but one of the major thing is how active you are. I would say that you could have around 15 neon tetras in a 10g and they will be absolutely fine.

The thing about it is that the fish (when they are collected) are collected in VERY large schools in about 200, and they are in small areas.

Just something to take into account.

RE: MOA's: Advanced Mathematics for Aquarists

premnas — Thu, 14 Feb 2008 12:36:26 GMT

If you like math with your fish try a book by Escobal called Aquatic Systems Engineering.

RE: MOA's: Advanced Mathematics for Aquarists

math-only aquarium — Fri, 25 Jan 2008 19:39:32 GMT

Hmm.

RE: MOA's: Advanced Mathematics for Aquarists

math-only aquarium — Sun, 13 Jan 2008 18:28:30 GMT

Spreadsheet has been moved to FAMA section now and has over 2,400 views! Check it out!

RE: MOA's: Advanced Mathematics for Aquarists

math-only aquarium — Wed, 02 Jan 2008 13:10:14 GMT

There is an updated spreadsheet in the Beginners' Section now that tells you how large a tank each specific fish needs, how many of particular types of fish you should get to keep them happy, which types of fish are not suitable for a community aquarium, etc. It is in the thread labeled "Freshwater Stocking Guide: A Reference for Beginners."

RE: MOA's: Advanced Mathematics for Aquarists

math-only aquarium — Fri, 28 Dec 2007 19:52:02 GMT

Hmm,

RE: MOA's: Advanced Mathematics for Aquarists

math-only aquarium — Wed, 19 Dec 2007 19:44:56 GMT

A Spreadsheet for you.

RE: MOA's: Advanced Mathematics for Aquarists

starlette11 — Tue, 18 Dec 2007 11:10:51 GMT

wow...im gonna have to read all of this later tonight when my babies are in bed.. i cant wait

RE: MOA's: Advanced Mathematics for Aquarists

math-only aquarium — Tue, 18 Dec 2007 10:59:44 GMT

Hello Again,

Some readers in other strings have compained that the MOA system is far too complicated to be practical and, in truth, I think they may have a point. Their complaints have included the following:

Computing IFUs by mouth to base of tail is too difficult.
Medium per gallon is simply an outrageously complex idea.
The MOA doesn't account for fish behavior (Actually, it does, but I haven't gotten to that yet).
The MOA doesn't work well for Dutch aquariums (planted tanks).

If you agree with these points, please let me know. Also, tell me if you are simply not interested in the MOA so that I don't waste my time explaining it to people who don't care.

RE: MOA's: Advanced Mathematics for Aquarists

math-only aquarium — Mon, 17 Dec 2007 09:59:56 GMT

Hmm, that's an interesting idea.

If I understand you correctly, you are asking if the MOA's filtration capacity equations can be used to predict the bioload that an El Natural tank requires with regard to the requirements of the plants? If this is what your asking, the simple answer is no--the MOA is designed with respect to fish and not plants. However, it should be possible to get a reasonable estimate of the minimum IFUs such a setup would require, but I can't garantee appreciable precision in the result. In fact, since the MOA is designed to make predictions on the safe side, the result could be an underestimate by up to 20%. So long as you understand that the results will be an underestimate rather than overestimate, you can use the following process:

El Natural IFU Stocking Potential

1) First, find the turnover of the system and divide by 5. If, for example, your aquarium uses a flow pump rated at 120 gph and the tank is a 30-gallon, then the turnover of the system is 4 (120 / 30 = 4 cycles per hour). Dividing this number by 5 produces 0.8 (4 / 5 = 0.8), which indicates the efficiency of aerobic bacteria in the system.

2) Second, measure the number of pounds of substrate (leaf litter or other soil) and multiply by three (x3) to achieve the IFUs that the aerobic bacteria can handle and multiply by the corresponding temperature factor. Using the 30-gallon again, let us say that it uses 45 lbs. of leaf litter and it has a temperature of 75 degrees Fahrenheit. Thus, the aerobic bacteria can handle up to 130.36 IFUs (45 x 3 x 0.9656 = 130.356 = about 130.36 IFUs).

3) Third, multiply the number IFUs the aerobic bacteria can handle by the efficiency factor. In our example, the factor is 0.8 (found in the first step) and the IFUs that the aerobic bacteria can handle is 130.36. Subsequently, the actual IFUs that the aerobic bacteria can handle is 104.29 IFUs (130.36 x 0.8 = 104.288 = about 104.29).

4) Fourth, count complex plants and height of each one and compute their corresponding IFUs. For our example we will say that we are using 4 complex plants and half of them are 14" and the other two are 16". Using the edited ratios from an earlier reply, the process would be:

First Plants: 2 x 14 x 0.27 = 7.56

Second Plants: 2 x 16 x 0.27 = 8.64

First and Second Plants: 7.56 + 8.64 = 16.2 IFUs for Complex Plants

5) Fifth, count stem plants, height of each stem, and compute appropriate IFUs. If we let our 30-gallon aquarium hold 16 stems each 15" long, then the process would be thus:

16 x 15 x .14 = 33.6 IFUs for Stem Plants

6) Add up the IFUs of aerobic bacteria, complex plants, and stem plants:

104.29 + 16.2 + 33.6 = 154.09 IFUs total = about 154 IFUs

Thus, the El Natural 30-gallon aquarium is going to require at least 154 IFUs to keep the plants alive. This is the equivalent of about 10 fish with an overall length of 2.75 inches or 5 fish with an overall length of 3.5 inches. That is a decent fish population for an aquarium with no independent filter system. Please remember, though, that this is an underestimate, but it will get you in the ballpark.

RE: MOA's: Advanced Mathematics for Aquarists

ILuvMyGoldBarb — Sun, 16 Dec 2007 09:40:57 GMT

The stocking level of an El Natural setup needs to be higher than your standard aquarium since your source of nitrate and CO2 is entirely dependent on the fish. So how is the formula adjusted to take is added need into account?

RE: MOA's: Advanced Mathematics for Aquarists

math-only aquarium — Sun, 16 Dec 2007 08:06:00 GMT

As you may have noticed, planted tanks are not my specialty (obviously) and I am enjoying your responses in that they are forcing me out of my comfort zone--good job! However, I simply do not have enough resources or experience to to be able to answer some of your questions with any real authority in the subject. Thus, when you read my equations regarding plants, you have to consider the source.

That being said, I think you may may have misunderstood the way I was measuring stem plants. In my calculations I was referring to a single stem and since most people do not plant just one stem of such a plant the total IFUs they represent will be proportional to their size and number. Thus, stem plants do actually represent quite a bit of stocking potential. Nonetheless, I will look into it further and edit that reply when I have a more substantial answer. (If you good provide me with some growth stats on complex plants and stem plants, that would be helpful)

One note on the El Natural aquarium: it actually isn't a new concept at all. In fact, experiments were done on the basic concept as early as the 1890's and have continued to the present day. The form you talked about is simply the latest (and, thankfully, most reliable) version of a very old idea. With regard to a stocking formula for the El Natural aquarium, the basic equations already listed should sufice provided you use the proper substitutions. To be honest, though, I'm not really sure by what you mean about a formula for the El Natural aquarium--I thought I already answered that. So please tell me what specifically you're having a problem with that I might give you a better reply.

RE: MOA's: Advanced Mathematics for Aquarists

math-only aquarium — Sat, 15 Dec 2007 10:58:08 GMT

One qualifier on the new equations:

No aquarium, regardless of how large its calculated filtration capacity is, can hold more than about 43 IFUs per gallon without risking that the fish won't have enough dissolved oxygen. To put that in perspective, a ten-gallon aquarium at the max capacity of 43 IFUs / gal. could hold no more than 10 fish that measure 4 inches overall ( (4x0.85)^3 = 39.304; 10 gal x 43 IFUs / gal. = 430 IFUs total; 430 / 39.304 = 10.9404 fish total).

RE: MOA's: Advanced Mathematics for Aquarists

ILuvMyGoldBarb — Sat, 15 Dec 2007 10:51:16 GMT

Any thoughts on how the formula would work in the El Natural tank? One addition to that is that they are normally stocked heavier than your standard planted tank.

One addition you should consider in the Stem Plant section is that very often these plants are much more efficient at using the nutrients in the water. For this reason they are often added to a brand new setup to help alleviate the stress caused by the presence of NH3/NH4. I would think the rate of uptake would play quite heavily into this part of the formula.

RE: MOA's: Advanced Mathematics for Aquarists

math-only aquarium — Sat, 15 Dec 2007 10:05:15 GMT

Wow, you have an exception for every generality!

You're a great at disproving me--I love it!

You make an excellent point regarding planted tanks in that if they are maintained appropriately, they can indeed function independent of additional biological filtration. I, personally, really don't like the idea, but the numbers are certainly there and justified. As such, I must conceede that plants can be relied upon (no kidding, right). However, the numbers I provide will only be averages as I do not have the resources to gather stats on every plant species out there.

Complex Plants

For the purposes of the MOA, a complex plant is one similar to aponogeton, sword plants, or crypts. On average, such a plant requires the equivalent of 0.27 IFUs per inch of height. Thus, if you have a complex plant that measures 16 inches from base to leaf tip, then it will require at least 4.32 IFUs (16" x 0.27 IFUs / in. = 4.32 IFUs).

Stem Plants

Stem plants, according to the MOA, are those similar to hornwort, moneywort, combomba, etc. Since these plants do not have the mass that more complex plants do, they require less than their more complex counterparts. A stem plant requires 0.14 IFUs/gal. on average. Thus, if a stem plant measures 20 inches, it will require about 2.8 IFUs (20 x .14 = 2.8 IFUs).

Thus, the original two equations need to be rewritten:

Equation 1: IFUs total = G x [f(T) x (f(MAT) + bio-substrate IFUs)] + C + R + Plant IFUs

Equation 2: p = a x f(T) x [f(MAT) + bio-substrate IFUs] / n

Great response, let me know if you see anything else. PLEASE

RE: MOA's: Advanced Mathematics for Aquarists

ILuvMyGoldBarb — Fri, 14 Dec 2007 19:13:34 GMT

It seems to me for this formula to be accurate and useful you need to take into consideration the proper maintenance being done. I have had a number of planted tanks and one of them was actually quite heavily stocked. My planted tanks are very consistent in their lighting and CO2 addition. Most planted tank keepers actually run their lights on timers so that the plants receive the needed light even while the owners are away. So in a system such as that, the plants play a major role in filtration and must be added into the calculation. You see, my planted tanks actually require the addition of nitrate and phosphate. Even with a heavily stocked tank the plants were up to the task due to being given the needed CO2 and Light for processing the nutrients.

If you want a really interesting addition to your formula, try to get your hands on a book by Diane Walstad on El Natural Planted Tanks. This is a fairly new concept in keeping planted tanks. It is, as the name suggests, an attempt to recreate a natural environment in the tank. Substrate is a rich soil/leaf mixture, lighting is done almost entirely by the sun and the only "filtration" that is used is that of a powerhead to move the water around and bring it into contact with the nitrifying bacteria. The biofilter is located almost entirely in the substrate, dead leaves are not removed, excess food is used as fertilizer, fish waste is left to break down and provide nutrients for the plants, and water changes are only done every 6 months to a year. Sounds crazy but these systems are incredibly stable.

RE: MOA's: Advanced Mathematics for Aquarists

math-only aquarium — Fri, 14 Dec 2007 11:50:07 GMT

Wow, GREAT questions!

Your question about tanks that house cichlids is excellent in that it is something I had actually neglected completely--hadn't even crossed my mind. This forum is great because of the experiences that it brings together. I really appreciate your feedback in that it helps me get it right.

In answering your question, I talked to some old friends who have more experience with cichlids than I do (I've actually kept most common species but only under commercial conditions, which isn't the same as keeping them well) and they were able to provide me with some reliable (in that about 80% of them agreed) general stocking guidlines for cichlids. I, of course, converted those stocking limits into MOA formating and the result is that your average cichlid has a biomass signature that is about 31% greater than its IFU size. Therefore, when you compute the IFU size of a cichlid, multiply its IFU size by 1.31 to make it reflect what is really going on. Thus, if a particular cichlid measures 4 inches from mouth to the base of its tail, its true bio-footprint would be about 83.84 IFUs (4 x 4 x 4 = 64; 64 x 1.31 = 83.84 IFUs).

Since this basic principle that you described for cichlids applies to all messy eaters and predators, I composed the following table to help you make the necessary adjustments:

General Fish Type	Adjustment Factor
African Cichlids	1.27
South American Cichlids	1.35
Predatory Catfish	1.42
Herbivorous Catfish	1.13
Bichirs	1.17
Spiny Eels	1.09
General Mid-Water Predators	1.21
Knifefish	1.19
Butterfly Fish	1.07

If you dissagree with these numbers please let me know, but if you do please also give me some stats to work with (I need number and size of fish, temperature of tank, filtration stats (everything), and other mitigating factors like plant mass).

Additionally, you mentioned the role of aquatic plants in tank filtration and that is something I have considered. In my experience, while plants make an appreciable impact on filtration capacity, they do not have a consistent impact. That is, if you leave for a few days and leave the tank lights off, then the impact of your plants will be greatly ruduced. Another thing to consider about plants is a well-planted tank almost never coincides with a heavily stocked tank (stocked with fish that is). The added CO2 that the plants require inhibits the stocking potential and plant tanks generally cannot support for than 25 IFUs per gallon. Because of these items, considering plants in your bioload calculations can be a little foolhardy in that you can not rely on them to provide consistent numbers at all times. Thus, it is safer to look at plants as added security and not a consistent dynamic (a major dynamic, but just not consistent).

Despite all that, if you want to give me the stats on your tanks (every stat you can think of, please), I can crunch the numbers and come up with an add-on for the filtration capacity equation that will approximate what is going on. Especially, if you could provide the actual mass of your plants in a given tank and the numbers and sizes of the fish in it, then I can provide a reasonably accurate result.

Lastly, you mentioned a bio-substrate. Technically, I addressed such substrates in the section on residual flow IFUs. I know I only made mention of denitrifying bacteria in that case, but the numbers are almost identical. However, if you want to be more accurate than using pound-to-IFU ratios, then you can use this equation:

IFUs / gal. = .000071 x M of bio-substrate x A of aquarium as a whole x T of aquarium as a whole x f(T)

This equation will tell you specifically how many IFUs your bio-substrate adds to your filtration capacity. (By the way, it isn't a ton because of a lack of forced flow through the medium).

RE: MOA's: Advanced Mathematics for Aquarists

ILuvMyGoldBarb — Thu, 13 Dec 2007 19:01:02 GMT

How does you formula account for high bioload fish like Cichlids. The bioload from most Cichlids is not limited to the waste they produce from what they actually eat, it also includes the fine particles of food that end up in the water column and break down also contributing to the bioload. So the biomass of the fish alone can not account for all bioload in the tank.

The other question also involves Live Plants. Where do they play in the equation with the removal of things like NH3/NH4 and NO3?

Finally, what about special substrates like 100% organic Soil? I use soil in one of my tanks and the soil is already heavily populated with nitrifying bacteria.

RE: MOA's: Advanced Mathematics for Aquarists

math-only aquarium — Thu, 13 Dec 2007 14:38:41 GMT

Greetings!

Remember that the base equation that MOA uses for estimating aerobic bacteria filtration assumes that the aerobic bacteria do not break wastes down any further than nitrate? Well, it is finally time to address that. The MOA system accounts for this by attaching water change percentages and frequencies to aerobic filtration capacities. In other words, the more fish you keep, the more water will need to be changed out for a given span of time.

I know that water changed are a fairly charged topic and so I want lay some ground rules first. When it comes to water changes there are two schools of thought--the "I just want to keep my fish alive" group and the "I want it to be impossible for my fish to die" group. Neither group is ever going to agree with each other. Thus, I am posting the appropriate equations for each group:

"I Just Want to Keep my Fish Alive"

If you fall into this category, the appropriate equation for finding out how much and how often you need to change your aquarium water is

p = (3 x .0005 x MAT x temperature-specific factor)/n

where p is the percentage of water that you exchange per water change and n is the number of times per month that you change the water. As an example of this, let's say that you have a twenty-nine-gallon aquarium with an M of 460 square inches per gallon, an A of 12.83 square inches per gallon, a T of 8 cycles per hour, and a constant temperature of 71 degrees Fahreheit. Additionally, lets suppose that you want to change the water every two weeks. The process for such a circumstance would be as follows:

M = 460;A = 12.83; T = 8;

Temperature Factor = .9079;

n = 30 days in a month /14 days in two weeks = 2.1429

p = (3 x .0005 x 460 x 12.83 x 8 x .9079) / 2.1429

p = 64.2989 / 2.1429

p = 30.0056 = about 30% every two weeks

"I Want it to be Impossible for my Fish to Die"

If you are in this group (which I expect many of you are), then the approriate equation is as follows:

p = (8 x .0005 x MAT x temperature-specific factor) / n

Notice that the only difference between the two is the leading coefficient. The reason why this equation has a factor of 8 while the other had a 3 is that, for a standard condition tank, 80% of the water needs to be removed per month to keep fish happy while 30% is sufficient to keep the fish alive.

Using the previous example, let's make a alteration: from changing the water once every two weeks to once a week and stating that we want the fish to be happy. The process would be as follows:

M = 460;A = 12.83; T = 8;

Temperature Factor = .9079;

n = 30 days per month / 7 days in a week = 4.2857

p = (8 x .0005 x 460 x 12.83 x 8 x .9079) / 4.2857

p = 171.4638 / 4.2857

p = 40.0084 = about 40% every week

Filtration Capacity in the Net

Because the MOA uses a separate equation to deal with water changes, two equations determine the filtration capacity of an aquarium according to the MOA. Also, both of them must be satisfied if you plan on stocking your aquarium to the indicated IFU total:

Equation 1: IFUs total = G x f(MAT) x f(T) + C + R

Equation 2: p = (a x f(MAT) x f(T))/ n

Key:
G = gallon size of aquarium f(MAT) = .0005 x M x A x T
f(T) = temperature-specific factor C = chemical purification IFUs
R = Residual Flow IFUs p = percentage removed
a = applicable constant (3 or 8) n = number of times per month

RE: MOA's: Advanced Mathematics for Aquarists

math-only aquarium — Thu, 13 Dec 2007 12:26:52 GMT

While I'm at It,

I might as well provide you all with a table that lists a fish's overall length and its corresponding average IFU size so that you know what I'm talking about when I relate filtration capacity to biomass:

Overall Length (Inches)	IFU Size if Fish is of Average Proportions	Overall Length (Inches)	IFU Size if Fish is of Average Proportions
1.00	0.61	10.25	661.35
1.25	1.20	10.50	710.93
1.50	2.07	10.75	762.93
1.75	3.29	11.00	817.40
2.00	4.91	11.25	874.41
2.25	7.00	11.50	934.01
2.50	9.60	11.75	996.25
2.75	12.77	12.00	1061.21
3.00	16.58	12.25	1128.92
3.25	21.08	12.50	1199.46
3.50	26.33	12.75	1272.88
3.75	32.39	13.00	1349.23
4.00	39.30	13.25	1428.58
4.25	47.14	13.50	1510.98
4.50	55.96	13.75	1596.49
4.75	65.82	14.00	1685.16
5.00	76.77	14.25	1777.06
5.25	88.87	14.50	1872.24
5.50	102.18	14.75	1970.76
5.75	116.75	15.00	2072.67
6.00	132.65	15.25	2178.04
6.25	149.93	15.50	2286.92
6.50	168.65	15.75	2399.38
6.75	188.87	16.00	2515.46
7.00	210.64	16.25	2635.22
7.25	234.03	16.50	2758.73
7.50	259.08	16.75	2886.03
7.75	285.87	17.00	3017.20
8.00	314.43	17.25	3152.27
8.25	344.84	17.50	3291.33
8.50	377.15	17.75	3434.41
8.75	411.42	18.00	3581.58
9.00	447.70	18.25	3732.89
9.25	486.05	18.50	3888.41
9.50	526.54	18.75	4048.19
9.75	569.21	19.00	4212.28
10.00	614.13	19.25	4380.76

Additionally, here's a table that relates MAT product to IFUs/gal.:

MAT Product	IFUs/Gal.	MAT Product	IFUs/Gal.
2,000	1	41,000	20.5
3,000	1.5	42,000	21
4,000	2	43,000	21.5
5,000	2.5	44,000	22
6,000	3	45,000	22.5
7,000	3.5	46,000	23
8,000	4	47,000	23.5
9,000	4.5	48,000	24
10,000	5	49,000	24.5
11,000	5.5	50,000	25
12,000	6	51,000	25.5
13,000	6.5	52,000	26
14,000	7	53,000	26.5
15,000	7.5	54,000	27
16,000	8	55,000	27.5
17,000	8.5	56,000	28
18,000	9	57,000	28.5
19,000	9.5	58,000	29
20,000	10	59,000	29.5
21,000	10.5	60,000	30
22,000	11	61,000	30.5
23,000	11.5	62,000	31
24,000	12	63,000	31.5
25,000	12.5	64,000	32
26,000	13	65,000	32.5
27,000	13.5	66,000	33
28,000	14	67,000	33.5
29,000	14.5	68,000	34
30,000	15	69,000	34.5
31,000	15.5	70,000	35
32,000	16	71,000	35.5
33,000	16.5	72,000	36
34,000	17	73,000	36.5
35,000	17.5	74,000	37
36,000	18	75,000	37.5
37,000	18.5	76,000	38
38,000	19	77,000	38.5
39,000	19.5	78,000	39
40,000	20	79,000	39.5
		80,000	40

RE: MOA's: Advanced Mathematics for Aquarists

math-only aquarium — Thu, 13 Dec 2007 11:15:08 GMT

Hey Again,

Thinking about the exception that princessotfu (sp?) found made me think about some of the other exceptions regarding M,A, &T. To help you see what I'm talking about, I made a graph that roughly models the data gather in my original testing (temperature, C, R, and other such factors were held constant; tests were done on a fifty-gallon aquarium and two ten-gallon aquariums):

In the above graph, you can see that the MAT product and IFUs/gal. share a linear relationship from 2,000 - 80,000 MAT, but behave asymptotically on either side of those points. As was previously discussed, the graph indicates that an aquarium without a filter (0 MAT product) does have some biological filter potential due to the bacteria that colonize the remainder of the tank. However, the limit of the actions of these non-filter bacteria is about 0.7 IFUs/gal., which is almost ridiculously small. In fact, if you had a ten gallon aquarium with no filter, you could only keep one fish that displaces 7 IFUs, or about a 2.25 inch fish (.7 x 10 gal. = 7 IFUs total; 2.25 x .85 = 1.9125; 1.9125 x 1.9125 x 1.9125 = 6.9953 IFUs = about 7 IFUs).

With regard to the upper limit indicated by the graph, 43 IFUs/gal. is the maximum an aquarium can support because the water can only absorb so much oxygen--regardless of how much you agitate it. Thus, the equation that I gave does not account for MAT products below 2,000 or above 80,000. Realistically, though, these values are pretty rare.

For the MAT equation to be correct, the values of the variable must fall within the following boundaries:

MAT Product	2,000 – 80,000 MAT
M	31 – 1,300 sq. in. / gal.
A	1.3 – 51.3 sq. in. / gal.
T	0.5 – 20 cycles / hr.

Additionally, no aquarium, regardless of how much filtration capacity it has, can hold more than 43 IFUs / gallon and every normal aquarium can hold at least 0.7 IFus / gallon.

Sorry about not posting this earlier. Hopefully it will be helpful. Please tell me if I missed something or if you have a question.