Ok being 41 now I have had no use for advanced math for a lot of years now and forget most of it. What I need to work out is how many litres are in a small pool the kids have so I can add the chemicals. The water is 60cm deep and the diameter of the round pool is 330cm.

So what is the equation to work this out?

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05:56 PM

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AntiKev Member

Posts: 2333 From: Windsor, Ontario, Canada Registered: May 2004

Never saw it divided by four before, but it still works. I always just used pi*R^2. Once you get that, your answer will be in cubic centimeters. One cubic centimeter equals one mililiter. Then just move your decimal point to the left three spaces for your answer. Should be somewhere around 5131.7916 Litres.

Radius: 330/2 = 165cm R^2 = 165^2 = 27225 cm Multiply times pi (3.141592654...) and get 85,529.86 square cm take that and multiply it by the depth (60 cm) and get 5131791.6 cubic centimeters That also equals 5131791.6 mililiters Divide by 1000 and get 5131.7916 Litres

Got it? Good. There'll be a test later.

------------------ Whade' "The Duck Formerly Known As Wade" Duck '87 GT Auto '88 Ferrario '84 Indy (8/26/06)

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06:14 PM

topher_time Member

Posts: 3231 From: Bailey's Harbor, for now. Registered: Sep 2005

Ok now class we didn't read the entire equation now did we? He was asking for how many litres. So most of you only get partial credit. No soup for you! One year!

------------------ Whade' "The Duck Formerly Known As Wade" Duck '87 GT Auto '88 Ferrario '84 Indy (8/26/06)

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06:22 PM

Mike Marden Member

Posts: 432 From: Fernandina Beach, FL Registered: Aug 2006

AntiKev, I've never seen Pi * Dia ^ 2 / 4 before. Works fine, just my daddy always told me to use the fewest steps to get the right answer. As poorly as I did in high school, the simpler the better worked for me.

AntiKev, I've never seen Pi * Dia ^ 2 / 4 before. Works fine, just my daddy always told me to use the fewest steps to get the right answer. As poorly as I did in high school, the simpler the better worked for me.

A = Pi * Dia ^ 2 / 4 works because = Pi * (2 * Rad) ^ 2 / 4 = Pi * 4 * Rad ^ 2 / 4 = Pi * Rad ^ 2

It's actually the same number of steps if you start with diameter instead of radius You either divide the diameter by 2 to get radius, or you just plug the diameter into the equation and divide the result by 4. If you mean using less complicated (aka harder to remember) formulas, then yeah I agree

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08:48 PM

Dec 19th, 2006

Pyrthian Member

Posts: 29541 From: Detroit, MI Registered: Jul 2002

Originally posted by Mike Marden: AntiKev, I've never seen Pi * Dia ^ 2 / 4 before. Works fine, just my daddy always told me to use the fewest steps to get the right answer. As poorly as I did in high school, the simpler the better worked for me.

He gave me diameter, I gave him the formula with diameter. Quite often in engineering we're given a diameter, and it's just quicker to plug it directly in than to use the extra step and divide by two. It's efficiency not laziness.

He gave me diameter, I gave him the formula with diameter. Quite often in engineering we're given a diameter, and it's just quicker to plug it directly in than to use the extra step and divide by two. It's efficiency not laziness.

Exactly! If the problem is stated in terms of diameter, there is less chance of error if it's solved in terms of diameter. (Note that at least one person above got the wrong answer because he plugged the diameter into the formula that requires a radius.)

R ^ 2 = (D / 2) ^ 2 = (D ^ 2) / 4 ... They're all exactly the same, and equally correct. Use whatever works best, is easiest to understand, and is least prone to error.

quote

Originally posted by Pyrthian:

or, RTFM most pools come with a booklet....with a capacity listing

But what if the "FM" is wrong ... as often happens, especially with products from Asia? If you understand the basic principles, you can always figure it out for yourself.

[This message has been edited by Marvin McInnis (edited 12-19-2006).]

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12:13 PM

84fiero123 Member

Posts: 29950 From: farmington, maine usa Registered: Oct 2004

OK "Math Nerds" - since you clearly have a grasp on the simpler problems... how about this one (this was a question posed on a local radio show a couple of months back)?

Using whatever resources you can (site them if necessary)... what would be the average depth (in feet), if the Volume of Earth's water covered the entire surface of the earth (basically, assume a smooth sphere with the water filled "equally" all around it)?

Now I didn't "win" anything, but they did use my answer as reference for others... I'll share my details after a couple of other people take some educated stabs at it

Do we need to include the water in the atmosphere too or just the surface water?

Good Question!

Doesn't make much difference to me... as long as you can site your source. The source I used had a number of other sources listed... all with approximately the same volume shown.

quote

Originally posted by Pyrthian:

dunno how deep it is right now... i will throw my guess at 7,000 feet

You're actually not too far off!

[This message has been edited by MinnGreenGT (edited 12-19-2006).]

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Dec 20th, 2006

AusFiero Member

Posts: 11500 From: Dapto NSW Australia Registered: Feb 2001