formula help

What's the formula for area when you only know the perimeter?
Say a kidney shaped pool is 132' around the edge. What's the area?
Steve
Reply to
Steve B
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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:
Reply to
RoyJ
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 easily-measured 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.
Reply to
Don Foreman
Use the principles of integral calculus.
In simplicity, measure the area of 1-inche 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 1/16-inch 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:
Reply to
hhc314
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'.
Reply to
xray
Could we please see an example of a simple equation describing a kidney shape of given or arbitrary parameters?
Reply to
Don Foreman
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.
Reply to
Glenn Ashmore
By using two formulas, C = #D, and A = #R(squared) , (# is pi in this formula), I come up with 1385.48 feet of area That should put you on track.
Reply to
BULLDOG
Hire some kids with 5 gallon buckets. Have each kid bail out the pool and keep track of how buckets full of water he bailed. Do the math. :) Randy
Reply to
Randy Replogle
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
Reply to
Lloyd E. Sponenburgh
AREA, Randy. We're looking for AREA!
Steve
Reply to
Steve B
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
Reply to
ED
Well heck. I thought I had the answer. :) Randy
Reply to
Randy Replogle
--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..
Reply to
steamer
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.
Reply to
Don Foreman
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 2-D of the pool layout. Then you could do it on graph paper, the easy (approximate) way.
LLoyd
Reply to
Lloyd E. Sponenburgh
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 managably-sized float.
Reply to
Don Foreman
| --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 a-wash 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
Reply to
Voltes34
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 real-world units. A quick google seems to indicate that most kidney-shaped 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
Reply to
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 back-flush 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 set-up. 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
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Reply to
wfhabicher

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