formula help

Randy you were not far off. You had volume so simply divide by the average depth and you have average area.

(maybe) - I like the calc and planimeter myself.

Martin Martin H. Eastburn @ home at Lions' Lair with our computer lionslair at consolidated dot net NRA LOH & Endowment Member NRA Second Amendment Task Force Charter Founder IHMSA and NRA Metallic Silhouette maker & member

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Randy Replogle wrote:

Reply to
Martin H. Eastburn
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I was actually thinking volume at the time. I'm no math whiz but after reading some other responses...what if one poured a cubic foot of oil into/onto the pool then measured the thickness of it floating on the surface? I'm not sure where to go from there. Randy

Reply to
Randy Replogle

Cut a bubble-pack, pool cover to fit inside the edge. Cut 1 sq-ft of the same material. Weigh them both. Divide the big one by the little one.

This sounds like an "Angels on a Pin" question.

Reply to
Gary A. Gorgen

Get yourself a math book and look up Simpson's #1 Rule or Simpson's #2 Rule.

Reply to
Robert Swinney

You can't do it mathematically from just the perimeter. The formula for area used by pool companies is to add the width of the narrow and wide ends together. Multiply by the overall length and then by .45.

For example Narrow end = 12' wide, Wide end = 18' total = 30' Length =40' ((12+18)* 40)/.45 = 540 sq.ft (approximately)

Reply to
Glenn Ashmore

See if you can spot your pool on a Google satellite map, clip the pool image, enlarge it, and print it.

Then cut out the pool shape, weigh it on a lab balance and calculate the area by comparing the weight of the "pool shape" to the weight of a known rectangle of the same paper you printed the image on. (Ignore the addrd weight of the ink.)

Jeff

Reply to
Jeff Wisnia

I"d stay away from the oil myself - once a pool owner -

And remember that won't do it - since likely you have a deep end and a shallow end. So it isn't so easy even that way.

Martin

Martin H. Eastburn @ home at Lions' Lair with our computer lionslair at consolidated dot net NRA LOH & Endowment Member NRA Second Amendment Task Force Charter Founder IHMSA and NRA Metallic Silhouette maker & member

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Randy Replogle wrote:

Reply to
Martin H. Eastburn

Yep - you integrate 1/2(x*dy-y*dx) along the perimeter. A special case of the "generalized" fundamental theorem of calculus.

Reply to
nick

The best suggestion has been to divide the pool into 5' or 2' etc segments and measure the width across the end of each segment (Roy I think). Each segment is a rectangle x' wide and poolwidth' long. If you need more accuracy, either use short segments or triangles at the ends on the curvy bits.

Yep. And if the kids with buckets lower the pool by exactly 1/2" you can find that area, because you know the depth of the sample.

Reply to
OldNick

This question brings up the image oaf a hatchet planimeter, which gets my vote as the coolest, and most elegantly simple device I've ever seen to do a complex job.

It is bent piece of wire, that is traced over the outline, and yields the area.

Take a look at this link for more information:

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Pete

Reply to
3t3d

Wow. Never heard of that. Very interesting and non-intuitive (at least to me). Took me a while to understand what they were doing. Looks like an interesting project for a metal worker.

Building and setting one up to measure a swimming pool might be a little impractical, though.

Reply to
xray

Yeah, that is ONE Cool invention! Not practical to use on a 1:1 layout of a pool, but for a smaller picture, it will work OK. And, the longer the arm, the more accurate it is according to the website. My vote for the most complex task performed by the simplest machine possible. Pete

Reply to
3t3d

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