HOW MANY GALLONS?

divnkd101

LifeTime Supporter
#3
OK. If I have done my calcs correct, I come up with roughly 8247 gallons. I am sure there are other math wizards on the forum as well if I am not correct. Good Luck :D

Volume of a cylinder = pi * radius2 * height
where pi (Ï€)=3.1415926535 ...
 

waste

TFP Expert
LifeTime Supporter
Mar 29, 2007
4,160
Coastalish 'down easter'
#4
Using the formula for a round pool as diameter X diameter X av depth X 5.9

18 X 18 = 324
324 X 4 = 1296 (4' av depth is probably a little high, but it makes for easier calculations :p )
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.
 

waterbug

LifeTime Supporter
Mar 28, 2007
182
Richmond, Va
#6
Okay thanks, this sounds about right to me according to what the pool installer told us. I asked a chemical guy on Saturday how much of .....we would need - how many gallons was our pool etc. and he told me 15, 000 and that just didn't sound right to me.........I am beginning to understand "Pool stored" very well.

Thanks eveyone!
 

CarlD

Well-known member
Apr 8, 2007
104
#10
I was fascinated by this formula for capacity of round pools: D^2*H*5.9

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

In The Industry
Apr 15, 2007
42
Missouri
#11
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
 

CarlD

Well-known member
Apr 8, 2007
104
#13
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
Nope, 7646 by the D-Squared-H-5.9 formula, 7614 using Pi. 7500 gallons is MORE than good enough... (unless you have to carry those extra 114-146 gallons! that's 975 - 1250 lbs of water... :cry: )
 

Dennis

In The Industry
Apr 15, 2007
42
Missouri
#16
Hi waste, I am not doing any math here so making a mistake would be a little difficult. :) My fiqure came from an installation manual I read over twenty years ago.

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 ? :)
 

CarlD

Well-known member
Apr 8, 2007
104
#17
Dennis said:
Hi waste, I am not doing any math here so making a mistake would be a little difficult. :) My fiqure came from an installation manual I read over twenty years ago.

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, I ALWAYS consider the WATER level, not the wall height. I never thought about the cove (the angle would be fun to figure in--must be a twist on the volume of a cone--but I'm WAGging it that it would mean no more than 1%, if even that.). By using the water level, you don't need to figure the 2" of sand--it's already there. Normally, if the WALL is 4', I assume a water level of 3.5'. But I also assume if someone tells me the water depth is 4', it's 4'. Just like I assume if they tell me the diameter is 18', it's 18', not 17' 6".

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
 

waste

TFP Expert
LifeTime Supporter
Mar 29, 2007
4,160
Coastalish 'down easter'
#18
Carl, thanks for the extra math ! :D I've just always used the diameter ^2 * av depth * 5.9 because it happens to work for estimating gallonage, when dosing a 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!)