Dedicated to Really Cheap Audio Enthusiasts
How do you determine the maximum box size for a given sheet of material? First, I constrained the answer to be a box with internal dimensions in the ratio of 0.6 x 1.0 x 1.6
Next, I calculated the total surface area available per half sheet of MDF, or 2304 square inches. I used MS Works Spreadsheet to hasten the iteritive calculations. I kept the cell corresponding to the "1.0" ratio as the input, and used formulae to calculate the other two dimensions, the volume, and the surface area.
I adjusted the one input variable until I had 2047 square inches of surface area. This would not fit on one sheet. I backed down the dimensions such that the volume was 2.73 cubic feet, corresponding to 1826 square inches.
I entered the below drawing, choosing which way the butt joints would lap. I then took its dimensions and created the cut sheet shown below.
I chose the narrowest side to be the baffle, allowing a 10 inch woofer maximum. Actually the box volume may be good for either a 10 or a 12, but the baffle width is only 11.75 inches. I then checked this against a few common low-price woofer. The hands-down winner for low-frequency and low-cost was the Dayton 295-315 10" 70 watt woofer.
This badboy will have a -3dB frequency of 27.9Hz in this enclosure when the box is tuned to 28Hz. The port is 3 inches diameter by 6.84 inches long. The total volume is slightly bigger than what Keele's alignment suggests, but that allows generous room for the internal displacement of the driver, the port and bracing.
