Is this the correct size box dimensions?
I've got an Alpine Type R 10" sub, it's the SWR-1021D model. The recommended box size for a simple sealed enclosure is .5 to .8 cubic feet. My math isn't the greatest, but I come up with a 10.5" x 12.5" x 12.5" box. Does that sound correct? Just sounds tiny to me
<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by JoshuaVTEC »</TD></TR><TR><TD CLASS="quote">I've got an Alpine Type R 10" sub, it's the SWR-1021D model. The recommended box size for a simple sealed enclosure is .5 to .8 cubic feet. My math isn't the greatest, but I come up with a 10.5" x 12.5" x 12.5" box. Does that sound correct? Just sounds tiny to me
</TD></TR></TABLE> 10.5x12.5x12.5 = 1640.625 cu.in. divide by 1728, [cu.in. in a cu.ft.] = .949 cu.ft. this is an outside volume, if using 3/4" MDF it would be 9x11x11 = 1089 divided by 1728 = .630 cu.ft. internal volume, also these would be gross volumes, you need to add the .092 cu.ft. of speaker displacement, so .592 to .892 cu.ft. gross internal volume, so if they are the outside dimensions you are using and the material is 3/4" MDF the .630 cu.ft. falls in to that .592 and .892 cu.ft. recommendation, however if the dimensions used are inside dimensions then the gross volume of .949 cu.ft. would be a little too big.
</TD></TR></TABLE> 10.5x12.5x12.5 = 1640.625 cu.in. divide by 1728, [cu.in. in a cu.ft.] = .949 cu.ft. this is an outside volume, if using 3/4" MDF it would be 9x11x11 = 1089 divided by 1728 = .630 cu.ft. internal volume, also these would be gross volumes, you need to add the .092 cu.ft. of speaker displacement, so .592 to .892 cu.ft. gross internal volume, so if they are the outside dimensions you are using and the material is 3/4" MDF the .630 cu.ft. falls in to that .592 and .892 cu.ft. recommendation, however if the dimensions used are inside dimensions then the gross volume of .949 cu.ft. would be a little too big.
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