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Need help designing a round enclosure
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<blockquote data-quote="Subless" data-source="post: 8495620" data-attributes="member: 669685"><p>Now getting to more finalized volume calculations. I'll go ahead and post this stuff</p><p></p><p>in case someone who's done a sonotube style sub sees anything funky about my calculations.</p><p></p><p>Here are the specs for what's required for this particular sub woofer for a sealed enclosure.</p><p></p><p>Net volume required after the volume of everything else in the box including the driver</p><p></p><p>are taken into consideration is .35 cu feet. JL recommended for my Jeep to bump that up by 10%</p><p></p><p>if possible. So, I will shoot for .385 cu feet or 665 cubic inches of free air space after it's all said and</p><p></p><p>done but realizing if I get into some trouble with space, I could go as low as .35 cu feet which is 605 cu inches</p><p></p><p>of free airspace, so we'll say between 605 to optimal 665 cubic inches.</p><p></p><p>TARGET FREE SPACE = 605-665 cu in</p><p></p><p>This drum shell I'm looking at is 13 7/8" in diameter.</p><p></p><p>Wall thickness for an 8-ply shell is .21" or about 7/16" for both walls</p><p></p><p>So the inner diameter will be 13 7/8" - 7/16" = 13 7/16" diameter</p><p></p><p>The radius of that is approx 6.7"</p><p></p><p>To get our cu in multiplier</p><p></p><p>pi x r x r or</p><p></p><p>3.14 x 6.7 x 6.7 = abt 142 cu inches per each inch length of tube.</p><p></p><p>So, starting with a tube length of 7", I get</p><p></p><p>7 x 142 = 987 cubic inches of air space before adding on the ends</p><p></p><p>or inserting the driver.</p><p></p><p>Subtract the displacement of the driver which is .047 cu feet or 81 cubic inches</p><p></p><p>987 - 81 = 905 cu inches of free space remaining</p><p></p><p>If the two ends are 3/4" deep, then the amount of cubic inches they will take up</p><p></p><p>will be 3/4+3/4=1.5" x our multiplier above which is 141 cu inches per each inch of tube length</p><p></p><p>1.5 x 141 = 212 cu inches will be used up by the bottom solid end and the top (sort of).</p><p></p><p>The top is a combination of about 2 5/8" of 3/4" thick wood and the top part of the driver</p><p></p><p>whose space is already subtracted. But I'll just consider the top to be a solid piece of wood</p><p></p><p>keeping in mind I have some fluff I can use if necessary.</p><p></p><p>So, the total remaining free space is:</p><p></p><p>905 - 212 = 693 cu inches of free air space. The minimum required space is 605 ci to optimum which is 665 ci</p><p></p><p>So I have between 88 ci and 28 ci remaining for extra support, wiring connectors, bolds for mounting, dynamat, etc.</p><p></p><p>Seems a 7" deep drum shell should suffice which is really good news considering my worse case scenario</p><p></p><p>photos above showing a 9" deep model which takes up more space than I care for. 2" less will be fantastic.</p><p></p><p>Any flaws in my logic ?</p></blockquote><p></p>
[QUOTE="Subless, post: 8495620, member: 669685"] Now getting to more finalized volume calculations. I'll go ahead and post this stuff in case someone who's done a sonotube style sub sees anything funky about my calculations. Here are the specs for what's required for this particular sub woofer for a sealed enclosure. Net volume required after the volume of everything else in the box including the driver are taken into consideration is .35 cu feet. JL recommended for my Jeep to bump that up by 10% if possible. So, I will shoot for .385 cu feet or 665 cubic inches of free air space after it's all said and done but realizing if I get into some trouble with space, I could go as low as .35 cu feet which is 605 cu inches of free airspace, so we'll say between 605 to optimal 665 cubic inches. TARGET FREE SPACE = 605-665 cu in This drum shell I'm looking at is 13 7/8" in diameter. Wall thickness for an 8-ply shell is .21" or about 7/16" for both walls So the inner diameter will be 13 7/8" - 7/16" = 13 7/16" diameter The radius of that is approx 6.7" To get our cu in multiplier pi x r x r or 3.14 x 6.7 x 6.7 = abt 142 cu inches per each inch length of tube. So, starting with a tube length of 7", I get 7 x 142 = 987 cubic inches of air space before adding on the ends or inserting the driver. Subtract the displacement of the driver which is .047 cu feet or 81 cubic inches 987 - 81 = 905 cu inches of free space remaining If the two ends are 3/4" deep, then the amount of cubic inches they will take up will be 3/4+3/4=1.5" x our multiplier above which is 141 cu inches per each inch of tube length 1.5 x 141 = 212 cu inches will be used up by the bottom solid end and the top (sort of). The top is a combination of about 2 5/8" of 3/4" thick wood and the top part of the driver whose space is already subtracted. But I'll just consider the top to be a solid piece of wood keeping in mind I have some fluff I can use if necessary. So, the total remaining free space is: 905 - 212 = 693 cu inches of free air space. The minimum required space is 605 ci to optimum which is 665 ci So I have between 88 ci and 28 ci remaining for extra support, wiring connectors, bolds for mounting, dynamat, etc. Seems a 7" deep drum shell should suffice which is really good news considering my worse case scenario photos above showing a 9" deep model which takes up more space than I care for. 2" less will be fantastic. Any flaws in my logic ? [/QUOTE]
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