VOLUMES.HTM --- Part of Manual for Driver Parameter Calculator --- by Claus Futtrup.
Created 2. April 1997, last revised 23. August 2003. Ported to XHTML 1.0 on 2. October 2004. Last modified 25. October 2004.

How to calculate the netto volume for boxes of various shapes

For the purpose of determining the volume of your test box as precise as possible, I have chosen to supply the formulas for various shapes. Especially the rare shapes, like pyramid-sections is usually approximated too inaccurate by speaker builders. Also do not forget to subtract or add volumes from the driver(s) and braces. See later for instructions on how to subtract the volume of the driver(s).

Table of Contents:

  1. Cubic box
  2. Parallelepiped and rectangular box
  3. Cylindrical box
  4. Conical box
  5. Cone-section box
  6. Pyramid box
  7. Pyramid-section box
  8. Spherical box
  9. Subtraction of driver volume
  10. Subtraction of braces

Cubic box

 V = h^3

 where, h = height (any of the three dimensions)

Parallelepiped and rectangular box

 V = h * w * d

 where, h = height

        w = width

        d = depth

Cylindrical box


 V = h * --- * d^2


 where, h = height

        d = diameter (of circular plane)

Conical box

      1     pi

 V = --- * --- * d^2 * h

      3     4

 where, h = height

        d = diameter (of circular plane)

Cone-section box

      1     pi        /                   \

 V = --- * --- * h * (  D^2 + d^2 + D * d  )

      3     4         \                   /

 where, h = height

        D = diameter of the larger circular plane

        d = diameter of the smaller circular plane

Pyramid box


 V = --- * h * a


 where, h = height

        a = area of the (bottom) rectangular plane

Pyramid-section box

      1         /                       \

 V = --- * h * (  A + a + sqrt( A + a )  )

      3         \                       /

 where, h = height

        A = area of the larger (bottom) rectangular plane

        a = area of the smaller (top) rectangular plane

Spherical box

      4     pi

 V = --- * --- * d^3

      3     8

 where, d = diameter of sphere

Subtraction of driver volume

When a woofer driver is mounted appropriately (with magnet pointing into the box and the cone out into the room), the driver will displace a certain amount of air, which must be subtracted from the volume of the box.

Assuming you've cut a flat circular hole for the driver (a cylindrical hole), you must add that volume to the internal boxvolume. Alternatively if you've chosen a conical hole (more likely --- a conesection), you must add that volume. See formulas above.

This is the simplest approach to give you some idea on the volume displaced by the driver. Beranek suggests (L. L. Beranek, Acoustics) the following equation:

        Dvol = 0.4 * d^4, where d is the nominal size of the driver (DPC

        uses Outer as a measuring for this size).

Here follows a more advanced approach. Neglecting the following parts of the driver:

the volume of the driver, pointing into the box, can be calculated as a combination of:

This is going into detail, and frankly I do not think you can get much closer, even if you started adding more details, like the frame, the shape of the diaphragm, a vented pole-piece, air cavities between voice coil and suspension etc.

The above mentioned two main contributors gives you the following equation:

               pi * h                            pi

        Dvol = ------ (  D^2 + d^2 + D * d  ) + ---- * t * dm^2

                 12                               4

As you can see this is somewhat more complicated than Beraneks approximation, but it has the advantage that magnet system and cone size are treated separately.

Driver Parameter Calculator will assist you in calculating Dvol, the volume of the part of the driver occupying your cabinet. Driver Parameter Calculator supports both methods described above.

Subtraction of braces

Bracing can be designed in countless ways, and there is no way I am going to show you how to do, but from the above example, with a combined volume you should be able do subtract volumes in your box occupied by braces.

Most braces are probably just a simple (cylindrical or rectangular) bar, stuck between two sides of the box. Easy job.

More complicated braces, like a plate with a hole that is neither rectangular nor circular can give you some headacke before you end up with your final geometry, simply by adding and/or subtracting volumes from the standard geometries mentioned above.