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Calculate Box Volume Measurements

Discussion in 'Speakers' started by Maximo, Oct 28, 2010.

  1. Maximo

    Maximo Well-Known Member

    So I got a spec from paradigm that a back box for my inceiling speakers should be .42 cu ft or 12 liters. I want to do a back box to keep the insulation out of the speaker and to not let heat escape my house.

    How does one decide the dimensions of the box? the square of .42 is 0.648074069840786 which is 7.77 inches. So would you just make a box that is 7.75" x 7.75" x 12" ? Or a the volume in inches is 725.26 and the cube root is 8.98" so would you build a 9" cube and call it good?

    Or would you use an 8" sonotube cut to 14.45 inches?

    I guess my real answer is, what kind of back box shape is ideal? box, cube or tube? And if tube is the answer can I use a cardboard sonotube?
     
  2. Botch

    Botch I.Y.A.A.Y.A.S! Superstar

    I would design it to fit between the rafters for ease of construction, probably make it a bit long between the rafters so the height is small and you can piles some insulation on top.

    BUT, I'll leave the design of the shape, wrt audio performance, to others on this forum...
     
  3. Flint

    Flint "Do you know who I am?" Superstar

    This is how I do it when designing enclosures as it is the fastest way to get the dimensions you want:

    Take the total volume cubic feet and convert it to cubic inches:

    1 ft^3 = 1728 in^3

    So...

    0.42 ft^3 * 1728 = 725.76 in^3

    This means the internal volume of the enclosure needs to be 725.76 in^3

    Then I add the probable volume the back side of the driver will occupy. You don't provide any data on the driver, not even diameter, so I have no idea what to add to the volume. Just for fun, I'll add 200 in^3 for a total box volume of 925.76 in^3.

    Next, I figure out if there are any limitations on the dimensions I cannot exceed - like perhaps the distance between studs which is usually 16 inches. To allow for a little imperfection in the studs, I'll say the width of the enclosure cannot exceed 15 inches. So, the first dimension is fixes at 15 inches. If you are using 0.75 inch thick MDF or Plywood, then the internal width distance cannot exceed 13.5 inches (1.5 inches used by the two side walls of the box).

    So, you can divide the total internal volume by the first fixed dimension to get the squared distance for the rest of the enclosure dimensions:

    925.76 in^3 / 13.5 in = 68.575 in^2

    If the next fixed dimension is, say 4 inches deep internally (5.5 inches deep externally), then the next equation is:

    68.575 in^2 / 4 in = 17.144 in.

    Thus, the internal dimensions of the enclosure to reach 0.42 ft^3 are:
    13.5" x 4" x 17.15"

    External dimensions (assuming 0.75" thick panels) are:
    15" x 5.5" x 18.65"

    I hope that helps.
     
  4. Maximo

    Maximo Well-Known Member

    it sure does. it tells me that I forgot to factor in the volume of my speaker. I figured the speaker volume would have already been included. BTW they are paradigm AMS-100's which are a coaxial 6.5" speaker. What would be the best way (rule of thumb?) of determining the volume of this speaker?
     
  5. Flint

    Flint "Do you know who I am?" Superstar

  6. yromj

    yromj Well-Known Member

    Just to bust on Flint: :happy-smileygiantred:

    Studs are typically on 16" centers; therefore, the distance between them is actually 14.5" from board face to board face.

    John
     
  7. Flint

    Flint "Do you know who I am?" Superstar

    Good call... the math to get the dimensions is the same.
     
  8. Wardsweb

    Wardsweb Renaissance man

  9. Maximo

    Maximo Well-Known Member

    What about using a thick cardboard sonotube? I like the idea of not having to cut a lot of stuff. I know how to calculate the volume. I am just looking for an oppinion on the quality if the material. I know that some make sub woofers out of these.
     
  10. Wardsweb

    Wardsweb Renaissance man

    Run down to your local home improvement store and get a concrete pillar form.
     
  11. Maximo

    Maximo Well-Known Member

    That's what I intend to do, just as long as someone can support my decision to do so. Are you such a person?
     
  12. Maximo

    Maximo Well-Known Member

    ok, so I am going to put 8" concrete form tubes over the back of the speakers for the back boxes. Any ideas on how to seal the top? Also, I thought about also getting 10" tube and putting them around the 8" and filling the space inbetween the 8" and the 10" with spray foam or sand. Any opinions on this?
     
  13. Flint

    Flint "Do you know who I am?" Superstar

    1) If it were me, I would cut circle disks from MDF to seal the tubes. I would cut two, one which is the same inside diameter as the tube and one which has the same outside diameter, then glue the disks together and slip the tube over the smaller disk so it butts up against the larger disk. Lots of glue at that joint will make it strong and sealed. But, I have a really effective circle cutting guide for my routers. Doing it by hand with a jigsaw is nearly impossible to get good enough, though not impossible.

    2) The dual wall approach filled with sand is the absolute best approach, but it may be overkill for this application. Go for it if you have the time and want to learn from your efforts, but I doubt it will vastly improve the performance for this particular installation.
     
  14. Maximo

    Maximo Well-Known Member

    Thanks flint. I didn't think about it like that. I will definitely do the dual disk thing. As far as the dual wall thing. If I had all the time in the world I would install the speakers and have two pairs hooked to the A/B speaker outs on my receiver and compare them with no back boxes for any discernable differences between the two. Then I would put one set of back boxes on and compare. Then I would put the other set of back boxes on and compare. Then finally I would do the double wall thing on one set and compare.

    I think it would be a very cool deal but may never happen.
     

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