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RicodJour wrote:

I used to calibrate Simpson 260s. The accuracy left something to be desired, especially on AC. Cheaper meters may have been equally accurate, but the meter movement and mirrored scale made the 260 more precise. Precision was very desirable even if the accuracy didn't match it.

Pushing a ruler against the skimmer throat won't tell me accurately where the water level is, but as long as it gives precise, repeatable readings, it will tell me the change in level. Change can be very useful for calculations.

Don't you ever use stacks of playing cards as shims or gages? That gives you a resolution of about .25mm.

If you're repairing something and need the piece to fit, it may be worthwhile to read a rule to the nearest .5mm. It's roughly 1/32".

I decided decades ago that 16ths and 32nds were anal. They require working with fractions of five denominators. Harder to read the rule, harder to remember, harder to write, harder to add and subtract.

The correct way of doing it would be to draw out the two circles that
make the upper and lower part. Find the area of those two spots, then
subtract out the arcs that are over lapping.

yes you can

using calculus...... simpsons method

I figger it's 0.

0+0+0=0 0x0x0=0 0^0^0=0 (logic) 0v0v0=0 (logic)

Go the the mfr; he'll have all the figures AND usually a method to calculate. Try other mfrs if yours doesn't have the calculators. They do exist.

Still got the paperwork? It's in there too.

Or, lay it out on grid paper and count the grids, remembering to take into account any depth changes. Might need a little algebra for a shallow to deep slope.

HTH,

Twayne`

#### Site Timeline

- posted on October 9, 2009, 8:38 pm

RicodJour wrote:

I used to calibrate Simpson 260s. The accuracy left something to be desired, especially on AC. Cheaper meters may have been equally accurate, but the meter movement and mirrored scale made the 260 more precise. Precision was very desirable even if the accuracy didn't match it.

Pushing a ruler against the skimmer throat won't tell me accurately where the water level is, but as long as it gives precise, repeatable readings, it will tell me the change in level. Change can be very useful for calculations.

Don't you ever use stacks of playing cards as shims or gages? That gives you a resolution of about .25mm.

If you're repairing something and need the piece to fit, it may be worthwhile to read a rule to the nearest .5mm. It's roughly 1/32".

I decided decades ago that 16ths and 32nds were anal. They require working with fractions of five denominators. Harder to read the rule, harder to remember, harder to write, harder to add and subtract.

- posted on October 9, 2009, 9:22 pm

--

Dymphna

Message origin: www.TRAVEL.com

Dymphna

Message origin: www.TRAVEL.com

Click to see the full signature.

- posted on October 10, 2009, 3:35 am

mike wrote:

I checked again to day. The owner had cleaned leaves from the skimmer basket, so there was more rippling from the return jets. It looked like 1mm peak-to-peak in the skimmer throat.

Measuring at the bottom peaks gave repeatable results, but it wouldn't work to measure a change in level at the skimmer with the pump going. Water velocity would affect the reading, and the velocity in the skimmer throat would slow down as the level in the skimmer rose.

It looks as if I couldn't guarantee to calculate the area of the pool within 1% this way. I don't even know how accurate a municipal water meter is. However, it requires only two ruler readings and two meter readings, and the results could be accurate to 1% or so.

Now suppose it's a rectangular pool 5 x 10 meters, and your tape readings are 1% high because of sagging of the tape and rounded edges on the pool. Your calculation will be 2% too high. If in addition the corners of the pool are rounded, the calculation will be more than 2% too high. Even for a simple rectangle, it might be more accurate to calculate area by measuring a change in depth of 100mm or so.

I checked again to day. The owner had cleaned leaves from the skimmer basket, so there was more rippling from the return jets. It looked like 1mm peak-to-peak in the skimmer throat.

Measuring at the bottom peaks gave repeatable results, but it wouldn't work to measure a change in level at the skimmer with the pump going. Water velocity would affect the reading, and the velocity in the skimmer throat would slow down as the level in the skimmer rose.

It looks as if I couldn't guarantee to calculate the area of the pool within 1% this way. I don't even know how accurate a municipal water meter is. However, it requires only two ruler readings and two meter readings, and the results could be accurate to 1% or so.

Now suppose it's a rectangular pool 5 x 10 meters, and your tape readings are 1% high because of sagging of the tape and rounded edges on the pool. Your calculation will be 2% too high. If in addition the corners of the pool are rounded, the calculation will be more than 2% too high. Even for a simple rectangle, it might be more accurate to calculate area by measuring a change in depth of 100mm or so.

- posted on October 9, 2009, 4:46 pm

E Z Peaces wrote:

I don't have a pool so I may be missing what you are saying. I am reading that you are measuring the difference in the pool level when you add water.

I really doubt you can measure to the nearest mm. If you can the possible error is plus/minus 0.5 mm. To measure the height increase to 1% you would have to add 50 mm of water (0.5 mm possible error is 1% of 50 mm). (That excludes temperature effects, evaporation and accuracy of the water measurement.) Converting, 50 mm is about 2 inches. Might be more practical when you are filling the pool.

I don't have a pool so I may be missing what you are saying. I am reading that you are measuring the difference in the pool level when you add water.

I really doubt you can measure to the nearest mm. If you can the possible error is plus/minus 0.5 mm. To measure the height increase to 1% you would have to add 50 mm of water (0.5 mm possible error is 1% of 50 mm). (That excludes temperature effects, evaporation and accuracy of the water measurement.) Converting, 50 mm is about 2 inches. Might be more practical when you are filling the pool.

- posted on October 9, 2009, 7:56 pm

bud-- wrote:

The skimmer throat is 145mm high. If I wait until the level is 25mm, I can easily add 100mm. If my reading is within .5mm, that would be .5%.

The water comes from a reservoir. Tap temperature is about the same as pool temperature. Anyway, things balance out. The expansion of water is about 3*10^-4 per C. If I added 5,000 liters to 95,000 liters that was 10C cooler, the added water would contract 14.25 liters. However, in warming .5C, the 95,000 liters already in the pool would expand 14.25 liters.

My concern about temperature would be not to make measurements while the sun was high. If while I was filling, the sun warmed the pool 1C, that would raise the level about 1/2 mm.

In summer, the neighbors' pool could easily lose 25mm a week. It can be calculated, based on water temperature, dew point, and wind. At that rate, it could lose about 1/2 mm during a 3-hour filling process.

On a given October morning (and probably most mornings of the year), calculations would probably show a negligible gain or loss.

The skimmer throat is 145mm high. If I wait until the level is 25mm, I can easily add 100mm. If my reading is within .5mm, that would be .5%.

The water comes from a reservoir. Tap temperature is about the same as pool temperature. Anyway, things balance out. The expansion of water is about 3*10^-4 per C. If I added 5,000 liters to 95,000 liters that was 10C cooler, the added water would contract 14.25 liters. However, in warming .5C, the 95,000 liters already in the pool would expand 14.25 liters.

My concern about temperature would be not to make measurements while the sun was high. If while I was filling, the sun warmed the pool 1C, that would raise the level about 1/2 mm.

In summer, the neighbors' pool could easily lose 25mm a week. It can be calculated, based on water temperature, dew point, and wind. At that rate, it could lose about 1/2 mm during a 3-hour filling process.

On a given October morning (and probably most mornings of the year), calculations would probably show a negligible gain or loss.

- posted on October 10, 2009, 4:06 pm

bud-- wrote:

Below are the basic calculations to work out your swimming pool volume. These figures are useful when chemicalizing your swimming pool.

If measured in feet: Rectangular swimming pools - Length x Width x Average Depth x 6.25 Pool volume in gallons Circles/Kidney shaped swimming pools - Length x Width x Average Depth x 4.54 = Pool volume in litres For all shapes measured in feet - Gallons x 4.54 = Volume in litres

If measured in Metres: Rectangular swimming pools - Length x Width x Average Depth x 1000 Pool volume in litres Circles/Kidney shaped swimming pools - Length x Width x Average Depth x 1000 x 0.79 = Pool volume in litres For all shapes measured in feet - Gallons x 4.54 = Volume in litres If measured in meters - 1 cubic meter = 1000 litres

Above Ground Swimming Pools Fast Set / Steel Frame

10' x 30" = 1000 gallons 12' x 32" = 1185* / 1520 gallons
15' x 36" = 2240 / *2780 gallons
18' x 42" = 4200 / 6076 gallons

Below are the basic calculations to work out your swimming pool volume. These figures are useful when chemicalizing your swimming pool.

If measured in feet: Rectangular swimming pools - Length x Width x Average Depth x 6.25 Pool volume in gallons Circles/Kidney shaped swimming pools - Length x Width x Average Depth x 4.54 = Pool volume in litres For all shapes measured in feet - Gallons x 4.54 = Volume in litres

If measured in Metres: Rectangular swimming pools - Length x Width x Average Depth x 1000 Pool volume in litres Circles/Kidney shaped swimming pools - Length x Width x Average Depth x 1000 x 0.79 = Pool volume in litres For all shapes measured in feet - Gallons x 4.54 = Volume in litres If measured in meters - 1 cubic meter = 1000 litres

Above Ground Swimming Pools Fast Set / Steel Frame

10' x 30" = 1000 gallons 12' x 32" = 1185

- posted on October 8, 2009, 10:07 pm

wrote:

You can not calculate area from the perimeter of an irregular shape. Draw the shape on graph paper, then add up the squares. Try to guess the percentage included for those squares cut off by the pool boundaries. The smaller the squares, the greater the accuracy.

You can not calculate area from the perimeter of an irregular shape. Draw the shape on graph paper, then add up the squares. Try to guess the percentage included for those squares cut off by the pool boundaries. The smaller the squares, the greater the accuracy.

- posted on October 8, 2009, 11:09 pm

yes you can

using calculus...... simpsons method

- posted on October 8, 2009, 10:26 pm

wrote:

Look at the blueprints. I don't do math :-/

My pool is shaped like a snowman ( oOo ) Go figger.

Look at the blueprints. I don't do math :-/

My pool is shaped like a snowman ( oOo ) Go figger.

- posted on October 9, 2009, 2:24 am

I figger it's 0.

0+0+0=0 0x0x0=0 0^0^0=0 (logic) 0v0v0=0 (logic)

- posted on October 9, 2009, 7:59 pm

Go the the mfr; he'll have all the figures AND usually a method to calculate. Try other mfrs if yours doesn't have the calculators. They do exist.

Still got the paperwork? It's in there too.

Or, lay it out on grid paper and count the grids, remembering to take into account any depth changes. Might need a little algebra for a shallow to deep slope.

HTH,

Twayne`

- posted on October 12, 2009, 5:29 am

SteveB wrote:

Kidney shaped. Do you just***want*** people to pee in the pool? (the power
of suggestion)

Bob

Kidney shaped. Do you just

Bob

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