# Help with this calculation...

Discussion in 'General Math' started by justbob, Jun 4, 2006.

1. ### justbobGuest

Hello,

I'm wondering if someone can help me calculate the amount of dirt in cubic
yards that was removed for a pool I recently had installed. Should be
simple, but since one end was dug deeper than the other, the math escapes me

A 21' circle was dug and since our yard from front to back has steep grade
at the highest portion of the grade, 14" of dirt was removed on the opposite
end (lowest point) only 6" of dirt was removed.

Bob thanks YOU!

justbob, Jun 4, 2006

2. ### [Mr.] Lynn KurtzGuest

When you say a 21' circle was dug, you need to be more specific.
It matters whether that 21' is the diameter, circumference, or radius
of the circle.

Also, how was the circle drawn? Was it drawn on the slanted lawn so it
looked like a circle looking straight at it or if you looked at it
from straight above it. One of those views would be elliptical and it
matters which one.

--Lynn

[Mr.] Lynn Kurtz, Jun 4, 2006

3. ### [Mr.] Lynn KurtzGuest

If we assume that the yard was a flat slanted surface and we assume
that the finished pool is circular i.e., the the "cool-deck" is flat,
horizontal, and circular with radius "a" measured in feet, I get the
volume removed is:

V(a) = ( 7 pi a^2) / 8 cubic feet.

So depending on whether the 21' represents diameter, radius, or
circumference:

21' diameter: V(21/2) = 303

21' circumference: V(21/(2pi)) = 30.7

--Lynn

[Mr.] Lynn Kurtz, Jun 4, 2006
4. ### justbobGuest

Oh so sorry, that is correct - it was 21' in diameter. Thanks for your
help!

Bob

justbob, Jun 5, 2006
5. ### MikeGuest

Then the pool has a surface area of pi*21^2/4 = 346.36 square feet.

Assuming the slope of the lawn is constant, the average depth of the pool is just equal to the average of the depths at
the highest and lowest points ao average depth is (14+6)/2 = 10.

Now assuming you meant 14' and 6' rather than 14'' and 6'' (in which case it would have been a paddling pool rather
than a swimming pool) then the volume of the pool is depth x area so vol is ~346*10 = 3464 cubic feet.

Converting to cubic yards we get 3464/27 = 128 cubic yards. If the depths were actually in inches then you need to
divide this final figure by 12 - I will leave that to you.

Mike

Mike, Jun 6, 2006
6. ### justbobGuest

Thank you Lynn and Mike for your help with this!

Regards,
Bob

justbob, Jun 6, 2006