How do I figure the area of a pool from the perimeter? It is a kidney shaped (exaggerated) pool.

Steve

Alone, you don't.

Example of magnitude difference depending on shape, multiplier is pi for a circle, 4 for a square of the same "radius" so square would bound 4/pi

--

MAXIMUM not minimum is bounded by a circle

NO. Circle has area: pi * r^2 and perimeter 2

And a square with equivalent perimeter has area: pi^2 * r^2/4 So square is pi/4 as large - or about 22% SMALLER than a circle with equivalent perimeter

blueman wrote:
*...> NO.*

Brain fart... :(

circle has minimum perimeter for given area, and i turned it around w/o thinking....

--

Brain fart... :(

circle has minimum perimeter for given area, and i turned it around w/o thinking....

--

No problem - we all suffer from them - and the older we get, the more frequent they become, just like real gas ;)

The old joke. Country yokel's son goes off to college.

Back at xmas vacation. "Well, son, what'd they teach you at college?"

"Pi r^2"

"Sure am wasting my money sending you off -- everyone with any sense knows that pie are square!"

Has anyone NOT heard that one?

David

^^^^^^ round

Only I could screw that one up! :-)

David

On Nov 1, 9:48 pm, snipped-for-privacy@panix.com (David Combs) wrote:

*m> wrote:*

You__ _almost_ __got it. Pi R Square...No pi are not square, cake are
square, pi are round.

Harry K

You

Harry K

David Combs wrote:

Lim time!

Said a rather dense yokel named Pete, "Mathematics has fair got me beat. I thought a square root, Is some sort of fruit, And Pi is a nice thing to eat."

Jeff

Lim time!

Said a rather dense yokel named Pete, "Mathematics has fair got me beat. I thought a square root, Is some sort of fruit, And Pi is a nice thing to eat."

Jeff

--

Jeffry Wisnia

(W1BSV + Brass Rat \'57 EE)

Jeffry Wisnia

(W1BSV + Brass Rat \'57 EE)

Click to see the full signature.

On Mon, 2 Nov 2009 05:48:15 +0000 (UTC), snipped-for-privacy@panix.com (David
Combs) wrote:

Cornbread are square :-)

Cornbread are square :-)

wrote:

0.45 x (A+B) x length x average depth x 7.5 = volume (in gallons) of kidney or irregular-shaped pool http://www.1paramount.com/poolcare/formulas.php Google is your friend

0.45 x (A+B) x length x average depth x 7.5 = volume (in gallons) of kidney or irregular-shaped pool http://www.1paramount.com/poolcare/formulas.php Google is your friend

The OP asked for AREA not volume in gallons. Also your formula at best is some vague type of approximation since there is no standard kidney-shape and certainly irregular-shaped is even less well-defined. Although since the site doesn't define what A and B are, the formula will by definition be true for some values of A and B ;)

Learned this once (and yes, it has a name that I don't remember):

For symmetrical even if then leaned over -- something like that, sphere, cone, etc:

Volume = area of top + 4 * area of middle (cross section, I guess) + area of bottom, all divided by 6.

David

snipped-for-privacy@panix.com (David Combs) writes:

Your formula is certainly in general FALSE since volume is 3-dimensional and grows to first order as the cube of some notion of "radius" while your formula is just a weighted sum of 3 cross-sections.

Volume is volume. Area is area. Irregular shapes don't in general have nice formulas and require numerical approximation/integration.

Your formula is certainly in general FALSE since volume is 3-dimensional and grows to first order as the cube of some notion of "radius" while your formula is just a weighted sum of 3 cross-sections.

Volume is volume. Area is area. Irregular shapes don't in general have nice formulas and require numerical approximation/integration.

SteveB wrote:

How critical is the measurement? I would sketch out the length and width, cut off triangles for the belly of the kidney and outside the curves....area of the rectangle less the areas (roughly triangular) outside of the curves should give a fairly close measurement.

How critical is the measurement? I would sketch out the length and width, cut off triangles for the belly of the kidney and outside the curves....area of the rectangle less the areas (roughly triangular) outside of the curves should give a fairly close measurement.

This is probably the best simple way if an approximation is OK. You can get as precise as you want by making the sketch more precise and projecting it on a fine grid and counting the "squares" and fractions of "squares" covered by the pool.

writes:

For my use, I took four widths, averaged them, then multiplied by the length.

Close enough.

Steve

For my use, I took four widths, averaged them, then multiplied by the length.

Close enough.

Steve

SteveB wrote:

You can't. That's what Integral Calculus is for.

You can't. That's what Integral Calculus is for.

So what is the formula then, or how would one use integral calculus to derive the area of the pool?

MikeB wrote:

First you write the equation for the curve as a function of x: f(x) equation.

Area = the integral [from 0 to max x] f(x)dx. Turning the crank gives the answer. http://hyperphysics.phy-astr.gsu.edu/Hbase/integ.html#c3

An alternative is the Monte Carlo method.

Surround the curve with a box. Generate random points that will land inside the box. Determine whether each generated point is inside the curve or outside. If 62% of the random points lie within the curve, the area of the curve is 62% of the area of the box. Obviously precision grows as a function of the sheer number of points.

First you write the equation for the curve as a function of x: f(x) equation.

Area = the integral [from 0 to max x] f(x)dx. Turning the crank gives the answer. http://hyperphysics.phy-astr.gsu.edu/Hbase/integ.html#c3

An alternative is the Monte Carlo method.

Surround the curve with a box. Generate random points that will land inside the box. Determine whether each generated point is inside the curve or outside. If 62% of the random points lie within the curve, the area of the curve is 62% of the area of the box. Obviously precision grows as a function of the sheer number of points.

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