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Specifically, I have a free form pool. I want to know the correct gallons and not continue to incorrectly estimate.

2006-12-31 05:47:26 · 7 answers · asked by Tim 1 in Science & Mathematics Mathematics

7 answers

A polygon is a plane figure. Therefore, it doesn't make sense when you say the "volume of a polygon".

To estimate the volume of a pool, you can estimate its average surface area and its average depth. So the volume = the average surface area x the average depth.

2006-12-31 05:55:52 · answer #1 · answered by sahsjing 7 · 0 0

First, a polygon is a flat figure - without any volume
but if your swimming pool bottom or top is in the shape of a polygon and you want to know the volume then find the area of the top or bottom polygon presuming that they are equal in area,
this can be done by dividing the polygon into several areas and then calculating all of the different areas and adding them together. Then multiply this area by the depth of the pool to obtain the volume of water. This would be in cubic feet and then, if you wish, you can covert it to gallons
A gallon of water = 277cu inches and a cubic foot = 1728 cu in
Therefore divide 1728 by 277 to get the number of gallons in a cubic foot and then multiply by the number of cubic foot in your swimming pool to get the number of gallons required. However if you are using USA measurement then there are only 231 cu in in a gallon. so calculate accordingly

2007-01-01 19:54:15 · answer #2 · answered by David C 2 · 0 0

2solutions:

Practical:

Use a CAD soft (like AutoCAD), vectorise (if you don't know what that is, it means just drawing basic shapes, lines&arcs over a scaned image) a scaned image of the transversal section of the pool (for better precision use simple , not command); form a polyline with the resulting lines; select the polyline, then look in the properties list and it will display the area of the polyline.
Now, if the depth of the pool is a constant, just multiply the area with the deepth.
Otherwise (this is the tricky part) you have to calculate the deepth in each specific point of the polilyne, from a certain known depth and the depth angle; after that, you have to calculate the volume yourself.

Theoretical: same as last time, find specific points of the curve of the perimeter of the pool, find simple equations for each part (arc: y=ax^2+bx+c;line: y=ax+b), and integrate on the perimeter the obtained functions. this way you'll get the area. for volume follow previous example

2006-12-31 14:20:26 · answer #3 · answered by Zeus 1 · 1 0

If it's free-form, then you won't be able to use a formula for the volume.

You can find the surface area by putting in a known amount of water and (carefully and accurately) measuring the change in depth. Dividing A by B will give you the surface area.

Even then you'll have problems calculating the volume if the depth isn't constant.

2006-12-31 13:57:10 · answer #4 · answered by Tim P. 5 · 0 0

A free form pool is not a polygon. You need to estimate area on top with shapes that we are familiar with like circles or rectangles and multiply these areas by the varying depths of the pool.

2006-12-31 13:50:46 · answer #5 · answered by Professor Maddie 4 · 1 0

Really, out of experience, the only way to ACCURATELY measure the total volume is to put a water meter on a hose and fill it up. BUT, you can estimate by dividing the pool into rectangular and triangular prisms and adding their volumes, assuming the pool is a single depth. If you have different depths, you can estimate by stairstepping small prisms. make marks on the sides of the pool somehow to mark where each part is.

2006-12-31 14:16:04 · answer #6 · answered by Rockstar 6 · 0 0

assuming it has a uniform (constant) depth, then have lines from one point to every other point. this will divide it into many triangles. then find the area of each of these triangles using the formula (1/2)*base*height, and then add each of these areas together to get the total area. to find the volume, multiply this sum by the depth

2006-12-31 13:51:49 · answer #7 · answered by Anonymous · 0 0

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