Results 1 to 5 of 5

Thread: [RESOLVED] Number Spheres Inside Rectangle

  1. #1

    Thread Starter
    PowerPoster
    Join Date
    Jul 2001
    Location
    Tucson, AZ
    Posts
    2,166

    Resolved [RESOLVED] Number Spheres Inside Rectangle

    I found a live rattlesnake under my solar home yesterday. How he got in??
    I need to crawl under the house, and the only solution we've come up with is to pump concrete (leveling compound) under the house so it fills the voids in between the rock
    and hopefully either drives the snake back through the hole in which he entered or embed him and his friends, if any, in the concrete mix.

    That said, I need to estimate the voided area:

    I have attached an Excel File with a sample of my calculation -- SORRY -- forum would not accept an xls file.

    ======================
    I hope the following is correct, but if someone has the time could they please confirm.

    Volume of rectangle (in inches) = L x W X H
    Volume of sphere (assumed 5 inch diameter) = (4/3) * 3.14159 * (r * r * r)

    I then took the diameter of the Sphere (5 inches) and divided it into the
    Rectangle Length, Width, and Height to determine the number of spheres that would fit in the rectangle volume.

    I then multiplied the number of spheres in each dimension (L x W X H) to derive the total number of spheres.

    I then took the total number of spheres times one spheres volume to get the total volume of all spheres.

    Subracting this total sphere volume from the rectangle volume should equal the voided volume (in cubic inches).

    I then divided this voided volume figure (in inches) by 144 to get the volume in cubic feet.

    I then further divided the cubic feet voided volume figure by 27 to get cubic yards.

    The answer I got appears WRONG by quite an amount.

    Is my calculation correct?



    =======================

    Is there a better approach such as somehow taking a random number of sphere size
    and then doing the calculation and would randomizing it lead to a better estimate?
    Last edited by dw85745; Mar 1st, 2012 at 11:14 PM.

  2. #2
    Hyperactive Member Lenggries's Avatar
    Join Date
    Sep 2009
    Posts
    353

    Re: Number Spheres Inside Rectangle

    Quote Originally Posted by dw85745 View Post
    I then divided this voided volume figure (in inches) by 144 to get the volume in cubic feet.

    ...

    Is my calculation correct?
    Nope: To convert cubic inches to cubic feet you need to divide by 12^3 = 1728.

    Also, I don't quite understand what these spheres are, but I suspect you are miscalculating how many spheres can fit in a rectangular prism... can you give a fuller description of these items for the benefit of those of us who have no handyman DNA whatsoever?

  3. #3

    Thread Starter
    PowerPoster
    Join Date
    Jul 2001
    Location
    Tucson, AZ
    Posts
    2,166

    Re: Number Spheres Inside Rectangle

    Thanks for correction. I knew the figure was to high, just couldn't spot the error.

    don't quite understand what these spheres are
    The house is on a pier and beam foundation, which means the house floor is elevated off the ground.
    Think of the crawl space like a small basement under the house. It is also called a plenum.
    The floor of the crawl space (basement) is washed river rock around 12-18 in thickness (height).
    This acts as a heat and cool sink for heating and cooling the house.
    Hot or cold air is pumped under the house (there is no ductwork) -- and it enters the voids in the river rock,
    which heat and cool the rock.
    Since the air being pumped in underneath must have a place to escape, it enters ductwork in the wall cavities
    of the house that open into the top of the plenum (think of some rectangular holes cut into the top of the cube.)

    The sphere(s) represent the rock that reside in the rectangular cube -- the rock bed area. This was the only way I could think of to calculate a voided space for on object
    (the rock) which would be many different sizes and shapes. I estimated an average 5 inch diameter sphere figure some rock would be smaller, some larger, such that
    this would average out to 5 inches. I also estimated the average height of the rock bed as 15 inches. This way a 5 inch diameter sphere fits into the area without having to deal with partial sphere calculations.
    Last edited by dw85745; Mar 2nd, 2012 at 07:28 AM.

  4. #4
    Hyperactive Member Lenggries's Avatar
    Join Date
    Sep 2009
    Posts
    353

    Re: Number Spheres Inside Rectangle

    I'm guessing that you are significantly overestimating the amount of void between the rocks:

    Assuming an average leads to misleading assumptions: Many smaller rocks, along with debris (and in your case, animal remains) will settle between the cracks, significantly reducing the volume of the void. Furthermore, even larger rocks will tend to settle in the most efficient manner possible, so I think your notion of stacking the spheres is off as well.

    That said, what you have calculated (assuming no other math errors) would constitute a reasonable upper bounds to the problem. Thus, if you purchase that much concrete it should suffice for your needs, but you'll likely have plenty left over.

    From a theoretical perspective, I'm not sure there is any good deterministic method for computing this stuff. I'd go with a Monte Carlo method and then look at the distribution of results. That is one heck of a lot of work, though. Since you already have a decent upper bounds, that should be good enough to act, in my opinion.

    As to the merits of the method... no idea. Couldn't you just smoke the snake out? Also, I can envision a small snake crawling in and then feeding off of small rodents that also crawl in. But as the snake grows, it loses the ability to escape. In that case, is the snake even a threat?

  5. #5

    Thread Starter
    PowerPoster
    Join Date
    Jul 2001
    Location
    Tucson, AZ
    Posts
    2,166

    Re: Number Spheres Inside Rectangle

    Thanks for the other thoughts especially:

    Couldn't you just smoke the snake out?
    Will pass this by the animal expert (reptile). They originally mentioned some type of bomb to kill it, but since this area has the
    air we breath inside the house it was vetoed. Don't know a lot about snakes, just enough to stay clear.
    Have caught a few outside -- rattlesnakes that is -- but don't desire to come eye ball to eye ball with one.
    From what has been explained to me, they have a mind of their own. Not all rattlesnakes rattle before they strike,
    can strike about 2/3 of body length, some are aggressive (will come after you), some not. In my case I could NOT hear it rattling. My biggest concern is even if I catch this one, are their others -- or -- sometime in the future if I need to go under again will I not be so lucky. Maybe misplaced fears, but I've seen the results of someone who got bit on the arm, and their description of all the pain, agony, and $$$$ spent because of it. IMHO hopefully this is money well spent.

    In re:
    From a theoretical perspective, I'm not sure there is any good deterministic method for computing this stuff.
    Sphere idea was best and simplest I could think of. Best guess at best IMHO.
    Last edited by dw85745; Mar 2nd, 2012 at 03:10 PM.

Posting Permissions

  • You may not post new threads
  • You may not post replies
  • You may not post attachments
  • You may not edit your posts
  •  



Click Here to Expand Forum to Full Width