Help with this calculation...

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

  1. justbob

    justbob Guest

    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
    #1
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  2. 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.

    More details please.

    --Lynn
     
    [Mr.] Lynn Kurtz, Jun 4, 2006
    #2
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  3. 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' radius: V(21) = 1212

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

    --Lynn
     
    [Mr.] Lynn Kurtz, Jun 4, 2006
    #3
  4. justbob

    justbob Guest

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

    Bob

     
    justbob, Jun 5, 2006
    #4
  5. justbob

    Mike Guest

    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
    #5
  6. justbob

    justbob Guest

    Thank you Lynn and Mike for your help with this!

    Regards,
    Bob
     
    justbob, Jun 6, 2006
    #6
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