There is no generalized fomula for the area based on the perimeriter.
You need some sort of additional information to fully constrain the
answer. ie "oval" and either major or minor dimension, rectangular and
one side, etc etc.
Easy way to approximate it is to measure the width at some reasonable
spacing, say 5'. Your accuracy will improve as the segments get shorter.
And, as the 'delta' goes to zero, your accuracy goes to perfect. Issac
Newton parphased? :)
Get one of the accoustic/laser distance measuring devices. Accurate to a
couple inches at your distances.
Steve B wrote:
You need to define the shape more quantitatively. A "fatter" shape
will have more area for same periphery, the limiting maximum case
being a circle.
If you have or can get a scale drawing of the pool, you might be able
to find a draftsman, surveyor or architecht with a planimeter. That's
a device which can tell the area by tracing the periphery. Planimeters
are purely mechanical devices that were around long before computers.
If you can get or make a CAD drawing (or vectorize an aerial photo),
then about any CAD drafting program will compute the area for you.
If there is a meter on the water supply, you could see how many
gallons it takes to raise the water some easilymeasured amount, like
half an inch maybe. The area in square feet would be 3.208 * gallons
for 1/2" rise. A square having that periphery would take about 3500
gallons; a kidney shape would take somewhat less, probably between
1/3 and 2/3 as much.
Use the principles of integral calculus.
In simplicity, measure the area of 1inche slices across the width of
the pool and total each area over the length of the pool. This will
give you an approximate answer.
Reduce the width of the slices to 1/16inch and you get a very close
answer.
As the width of the slices approach zero and you continue totalling
them, you'll obtain the precise answer. This is a simple mathematical
tool taught to engineering students.
In doing so you will begin to understand the concept of integral
calculus.
If you can write a simple equation describing the contours of the pool
perimeter, it becomes a simple matter to compute the pool's area by
simply manipulating the numbers Same thing with it's volume. Simply
grinding out a few numbers. When in doubt you approximate...which is
usually good enough for 'government work'.
No engineer will tell you this, but problems like this are what
calculus is useful for, and trust me as a math dummy, it isn't exactly
rocket science limited to some engineers. I've known contractors that
do this stuff, without realizing what they are doing, in their head.
Harry C.
Steve B wrote:
As mentioned in other messages, somewhere between 1386.5 sq ft and 0 sq
ft. Probably in the vicinity of 1000.
The max is a circle 42' in diameter. The min is two straight sides 66'
long with nothing between them. Both have perimeters of 132'.
Areas of odd shapes can be calculated using bearings and distances
mathematically by using a process called a "traverse". It is a rather
complex process and requires that curves be broken up into small segments
and if you make a small mistake in either a bearing or distance the shape
will not close up and you are SOL.
You can also draw it out to scale and use a planimeter.
Irregular shapes can be measured by planimeter, or by successive
approximations by "filling" the irregular area with regular polygons. There
isn't a "formula" for an irregular plane shape's area.
LLoyd
When I have the FSA certify different fields they used to
use a arial photo and a grid overlay that is reasonabally accurate,
+ . to a tenth of an acre
The last time it was on a computer said to be nuts on. Used a graphic
program of some sort.
If you had a scale drawing/photo and a overlay grid you would get
very close. How accurate do you have to be?
ED
There is a formula: a line integral. That's how planimeters work.
There are planimeters available on Ebay, and it doesn't look like
it'd be hard to make one. A homemade one wouldn't be calibrated, but
one could sweep a circle or square of known area, then sweep the
unknown area and take the ratio of the readings.
Well... yeah... but how many 'normal' people know even trig, much less all
the trigonometric identities commensurate with understanding The Calculus?
Machinists should, but...
Yep... and if a person had frequent need to do such measurements, it would
be a nice tool to have. It would be much easier to approximate the regular
polygons in such a scale as a pool. Best would be to have an overhead photo
in 2D of the pool layout. Then you could do it on graph paper, the easy
(approximate) way.
LLoyd
Stick a machinist's scale to the swim ladder so part of it is
submerged. Read waterline when water is calm.
Bail out or add 50 gallons. Wait for water to calm, read scale
again.
area = 80.2/depth_change, area in square feet and depth in inches
example: if the depth changes .080" the area is 1003 sq ft.
If you change volume by 100 gallons you'll get twice the depth change
or about 0.160" per thousand square feet, and so on.
I thought about using a dial indicator and a float, but the spring
force of the DI introduces significant error for a managablysized
float.
 Not enough info to give an answer. Now if you could meter flow and
 add an inch of depth to the water you'd have the info you need..

 
 "Steamboat Ed" Haas : Proud to be the
 Hacking the Trailing Edge! : family crackpot!

formatting link
 Decks awash in a sea of words
If the depth of the pool is pretty much constant you can fill the pool with
water and find out how much volume it holds with an accurate water meter.
With that information you can use the formula:
V = Area X Height
then,
Area = Volume / Height
If you have a reasonably accurate scale drawing or aerial photograph
(Google Earth?) of the pool, you could trace it onto graph paper and
count the squares (round squares of less than half down, and more than
half up), then scale by a single measureable dimension to get the
total area in realworld units.
A quick google seems to indicate that most kidneyshaped pools have
about a 2:1 ratio of lengths, so checking a generic pool drawing might
be fairly close. Then you could scale it to other sizes by the square
of the perimeter. Eg. if you have the area of a pool with a perimeter
of 100', and you need to know the area of a 132' perimeter pool it
would have area of: X * (132/100)^2 = 1.74 * X , where X is the area
of the 100' perimeter pool.
Best regards,
Spehro Pefhany
To get a VERY accurate measurement on the change of waterlevel use a
depth micrometer mounted to the pool ladder. (metal working content!
:))
Ideally the measuring rod would have a sharp point, but you can turn
one up from brass and stick it onto the end of the measuring rod.
Securely mount the micrometer (preferrably of stainless steel) onto the
ladder.
On a calm night lower the measuring spindle until the point just
touches the water surface and draws up a meniscus. Do this a couple of
times and determine the average level reading.
Pump out a known volume of water using the backflush line. A couple
of plastic barrels, 55 gal. each would do nicely; but you gotta know
where the calibration mark is on the barrels.
After having pumped out a KNOWN volume (gallons) of water, measure the
drop in water level heigth with the micrometer setup. This will give
you a very accurate measurement in the change of waterlevel.
Using Don's calculation methodology will establish the surface area of
the pool.
You can also add a known volume of water using your watermeter, and
measure the level rise as before. With this method only a small rise
or drop in level is required, but you must know the volume change
accurately.
With accurate volumetric knowledge this method can give excellent
results. I believe the repeatability of the micrometer method depends
primarily on the calmness of the water surface.
I've used this level measurement to calculate volumetric flow changes
in open channels.
Wolfgang
D> >
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