Simple math question...

Discussion in 'All Things Boats & Boating' started by bntii, Oct 16, 2012.

  1. bntii
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    bntii Senior Member

    I have to do a concrete pour for a shop space and need to calculate the area confined in the this wire frame:

    simplemath.jpg

    All vertical legs are of different height.
    How the heck do I get the volume here?
    Any calculus I knew has receded into the misty past...

    Thanks
     
  2. WestVanHan
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    WestVanHan Not a Senior Member

    Seeing as everything is differenty,just assume it's a box,figure it's volume and divide that by half or so.

    Make extra,and have some step paver molds made up for any excess.
     
  3. bntii
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    bntii Senior Member

    Agree but the variation on this site from halving rectangle might be as much as 3 yards of concrete.. Not a huge amount true but too much to be over on this job.
     
  4. WestVanHan
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    WestVanHan Not a Senior Member

    Formula on that is pretty much impossible.

    If possible,drive in stakes,tack on boards to level,line with poly,and measure how much water to fill.
     
  5. gonzo
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    gonzo Senior Member

    Multiply the area of the floor by the average height of the legs.
     
  6. TeddyDiver
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    TeddyDiver Gollywobbler

    Are you pouring the floor to level? If so you just measure the avarage depth. Highest point having the minimum thicknes, lets say 2". Laser level helps a lot..
    (h1+h2+h3+...hn)/n*A=volume

    BR Teddy
     
  7. WestVanHan
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    WestVanHan Not a Senior Member

    Pretty much what I said,if you look the sides they cancel each other out in differences,so measure the box and x.5 to be close and maybe .4 to be closer,but he wants more accuracy.
     
  8. Tackwise
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    Tackwise Member

    Assuming all sides are flat, the legs are vertical, and several crucial corners (the rectangular corners of B*L and the legs) are 90 degrees.
    (This assumption is based on the thought that you need to create a flat floor or something similar)

    The solution I believe is:
    (H being the height of the legs)
    H1 + H2 = H3

    You could therefore calculate the volume with two triangles:
    1/2*H1*L*B + 1/2*H2*B*L

    It becomes slightly more complicated if the lengths and breadth are not equal, however that is also solvable.
     

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    Last edited: Oct 17, 2012
  9. latestarter
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    latestarter Senior Member

    Building on Teddy's suggestion you need to measure depths at regular spacings.
    How close you make the spacings depends on how uneven the existing surface is. The more uneven, the closer the spacings need to be.
    As an example of a floor 16' by 20' and you chose 4' spacing, you do not start measuring at the edges, but half the spacing in at each end, that is 2' in this case.
    This gives you 4 columns by 5 rows of points to measure. Add them all up and divide by 20.
     

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  10. Tackwise
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    Tackwise Member

    Tsss, I should not make it more complicated than it is....

    Gonzo's solutions is correct, however don't forget you have four legs with one leg height = 0
     
  11. latestarter
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    latestarter Senior Member

    Could you explain where you get H1, H2 and H3 from?

    In the first formula did you mean H1 + H2 + H3?

    How do H1, H2 represent the average depth of concrete of each triangle?

    Is the existing ground a plane?

    Apologies to Tackwise for the above

    My complicated method is based on the real world where it is rare that anything is a nice geometrical shape.
     
    Last edited: Oct 17, 2012
  12. daiquiri
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    daiquiri Engineering and Design

    The above answers are all mathematically correct. :) If the grey surface is a flat plane, then the solid enclosed by the black lines has the same volume as half of the box LxBxH3.

    So the answer is Volume = 0.5 * L * B * H3

    Now that you've got the volume, follow Teddydivers' good advice for the practical realization.

    Cheers
     
  13. bntii
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    bntii Senior Member

    Thanks all.
    I guess West had it right out of the gate.

    I had gotten it into my head that the shallow second leg on high side would diminish volume a great deal given that the deep side is not uniformly deep by half.
    Sort of volume being half of box and reducing this by third or so..

    This is to be a -wee little water closet sized tiny- weld fabrication shop at the yard and as these things go I get volunteered to help due to a long stint in construction in my misspent youth.
    I am always try to get things sort of right on the spec/design side in a vain attempt to hold onto what little credibility I have..

    "The only man who never makes a mistake is the man who never does anything."
    Theodore Roosevelt
     
  14. philSweet
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    philSweet Senior Member

    You are building a shop, not the Hoover dam. Pour footers, check where the steel storage racks and any machine tools like presses are going to be situated and provide for those. Block up (or pour) the walls and backfill. Then pour a suitable slab. If it's a prefab steel building, the supplier should have some info for you. Some dimensions would help.
     

  15. bntii
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    bntii Senior Member

    Oh sure- easy for you to say as history chases me down the road mocking my efforts...

    [​IMG]

    Just for grins:

    The slab is a over-pour on a existing slab poured on grade.

    My foot print is 21'6" x 21'6"

    And the leg heights are clockwise:

    0"-5"-11.5"-7"

    I am getting close to 9 yards which still seems like too much when I am standing in the form....
     
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