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18 X 18 = 324

324 X 4 = 1296 (4' av depth is probably a little high, but it makes for easier calculations )

1296 X 5.9 = 7646.4 (with a 4' depth Bleachcalc 262 gave me 7633.44 - close enough for me!) (the difference is that Micheal used 5.89 as the multiplier instead of my 5.9 - I'm sure his is more accurate)

I assume that the purpose of wanting to know is for adjusting the chems. If you use 7500 as the gallonage, you should be close enough.

**a note on adding chems**

It is a good idea to add the chems in 2 or 3 doses (letting the system run for a few hours before retesting), as opposed to all at once- the reason being that the dosing tables and gallon estimates can be wrong, it's easier to add a little more of whatever chem later than to have to use another chemical to reduce an overshot mark.

Thanks eveyone!

I don't like this pool calculator as well as the BleachCalc, but this is easy to use if you don't have the bleachcalc program.

dan

Since I'm used to PI*R^2*H *7.48 I was intrigued.

OK: We all know the formula for the area of a circle is PI*R^2 ("Pie are squared").

Volume means multiplying by the height, in feet. This give you the volume in cubic feet...useless.

But there are 7.48 gallons in a cubic foot.

But Dee-squared Times Height time 5.9 seems weird--it's not.

You see, D^2 is ALWAYS 4 times greater than R^2, no matter the value of R. See, D^2 = (R*2) * (R*2) because D=R*2. Simplifying the right side gives: R^2*4 !

So there are now 3 constants: PI--(3.14159.....), 7.48 (gal/cube foot) and 4--the relation between the squares of the diameter and radius.

Multiply PI by 7.48 and divide by 4 to get one constant....

ET VOILA! Approx. 5.9!

So Multiply the square of the diameter, times the height and then by 5.9 and you have a PERFECT calculation of the gallonage of a round pool.

Use this on ovals too...an oval is merely a round that's been cut into 2 halfs with a straight section in the middle..That middle is a rectangle where the width is ALWAYS the same as the diameter. The length of the rectangle is the FULL length of the pool less the diameter. L*W*H*7.48 equals the gallons in the middle. Add that to D^2*H*5.9 and you have your oval figured almost as quickly!

Example: Say you have an 18X33 oval where the water is 4' deep (water, not the walls). The calculation is simple:

1) The two semi circles at each end are really an 18' circle, cut in half:

18*18*4*5.9=7646.4 gallons.

2) the width of the pool is 18', right?

3) the length of the center rectangle is .... 33' - 18', right? That's 15'

4) The rectangle is 18*15*4*7.48 (Gal/cub ft). = 8078.4 gallons.

5) 7646.4 + 8078.4 = 15724.8 gallons

But you can use 16,000 for most figuring. Or if it's lower you may use 15,000 gallons.

Dennis said:

Man, you guys get way to deep for my head. For thirty years now an 18' by 48" pool has held 7500 gallons of water. My guess would fall in the 8200 range.

I have not seen this kind of math since high school.

Later, Dennis

I have not seen this kind of math since high school.

Later, Dennis

I was however going ask Carl that same question. Did you allow for two inches of sand, a six inch cove and the water level a few inches down from the top of the pool. Would that get it closer to 7500 ?

Dennis said:

I was however going ask Carl that same question. Did you allow for two inches of sand, a six inch cove and the water level a few inches down from the top of the pool. Would that get it closer to 7500 ?

Still, if you are within 100 gallons on a 7600 gallon pool, it's not going to screw up you chemical calculations. In fact, it's my contention that if you are off by 5% either way it's TOTALLY irrelevant, and if you are off by 10% either way, it's probably not going to matter much.

Thoughts on calculating the cove (see what you started! :lol: ):

Assume it's 45 degrees, 6" high, and 6" wide.

Let's start by computing a washer shape around the pool that's 6"x6" on the bottom edge, where the cove goes. Yeah, it's got a square profile. I know. Just wait.

Now compute a pool that 18' by 6" and another that's 17' by 6". Using D^2*H*5.9 we get 956gal(18') - 853 gal(17') = 103 gallons for the 6"HxW ring around the pool for the cove. Why 17'? because our washer shape is 6" wide--and it's on both sides of the pool, so you double it to 1'. Thus 17'.

Now, since the cove is a diagonal, it has half of that volume, or 51.5 gallons. I certainly wouldn't worry about that in calculating a 7600 gal pool

Another thing I'd like to point out is that the desired range for any pool chemical parameter is forgiving enough that if you have a good estimate of the gals., you can get away with a little 'slop' in the actual gallonage. (there are no 'magic' or 'perfect' numbers to keep the pH, alk, or any of the other #s at - as long as they 'all play well together' and none are too far out of range, the water will be fine!)