How much room do we need then? There is quite a bit of leeway here for a sealed box (smaller).
![Dimensions : [Click to Enlarge]](wp_350z_audio_woofers/graphics/dimensions.gif)
The recommended sealed enclosure size is .25ft³ but the range in ft³ is from .15 to .35 - according to Rockford Fosgate - (Fig 1)
The box volume will directly impact the performance of the speaker. Larger enclosures will provide flatter response and deeper
bass where smaller boxes will provide a bump in the response curve and generally higher output for greater SPL.
I'm happy to go with the recommended enclosure size and will try for something close to that.
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Let's get to some math. Fig 1 shows the recommended enclosure has dimensions of 9.5"x14.5"x6.25". Multiplying those together we come up with
861 in³. Divide that by 1728 and you get .49 ft³. I was puzzled for a bit until I realized they were measuring to the outside of the box. Earlier
they suggested using ¾" MDF so next I took 1½" off each dimension. That gives 8"x13"x4¾" which then gives 494 in³ or .29 ft³.
Finally, you have to subtract the speaker displacement. The P28S8 has a
displacement of 0.024 ft³ we end up with 0.26 ft³ - close enough by since we had a range of 0.13-0.35 ~
What's my volume then? Take another look at Fig 1. Notice that in the second line where it states "Recommended Sealed .25ft³ (7.08L). There we have
the cubic dimension also in Liters, 7.08L to be exact. Note that this is the recommended enclousure volume after subtracting the speaker displacement.
Therefore, to find the volume I need before subtracting the speaker displacement, we get that speaker displacement in liters which is 0.610L (Fig 3).
Now, add that back to the 7.08L and we get 7.69L ~
By the way, going back to our range for box volume of 0.15-0.35 .. adding back that 0.024 speaker displacement, the range is 0.174-0.374 ~
OK. What is my REAL volume then? All those numbers we came up above are fine. They represent the volume needed inside the box. Now, we need to increase that
number by the thickness of the box we are building. Earlier I said I was going with ½" fiberglass. Therefore, I am going to take a box that is
1ft x 1ft and however high it takes to contain my volume. Since the volume is a constant (for this math), then the surface area is also a constant. Since my
volume is .29 ft³ .. that means a box 1.0'x1.0'x0.29' (about 3½" high).
Now, let's add ½" to all the sides. Our box is now 1.08'x1.08'x0.37'. Multiply that out and you get a volume of .43 ft³ - 12.18L ~ and THAT, is
the final volume that we have to make room for. Oh yeah. That range? The 0.15-0.35? Doing the same thing with those numbers, the range comes out to 0.27-0.5 ft³ ~
or 7.6-14.10L ~
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Edit: This is long after the pods were constructed. The box never did get to the ½" wall thickness. I ended up with ¼" .. and this threw a small wrench into the
mechanism. Remember my REAL volume was caculated by adding ½" to each side of my virtual 1.0'x1.0'x0.29' box .. and the volume caculation proceeded from there. Oops. The
wall thicknes of ¼" changes the outcome. Adding that ¼" to each side gives us a NEW virtual box measuring 1.04'x1.04'x0.33'. Multiply that out and you get a volume
of .37 ft³ - 10.48L ~ and THAT, is the final volume that we should have made room for (assuming that we were shooting for that ¼" wall thickness).
Subtract that original figure of 12.18L with the 10.48L and we find that I made room for an extra 1.7L. I just measured the volume of one of the pods and it came to 12.3L - or
.43 ft³. Subtract the speaker displace of 0.610L and we end up with a volume of 11.69L. This comes out to .41 ft³.
Ok. This means that because the box ended up ¼" less in wall thickness, there was an error of .43 ft³-.37 ft³ or .06 ft³ (the 1.7L) between what was caculated and what
actually was created. This figure can be used in the next step in creating the volume.
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![Dimensions : [Click to Enlarge]](wp_350z_audio_woofers/graphics/dimensions.jpg)
