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Tuning box freq.
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<blockquote data-quote="watts" data-source="post: 1722377" data-attributes="member: 567836"><p>Off the top of my head. To calculate a complex box volume you usualy have a expensive audio measurement system like LEAP or CLIO. You can measure the impedance function of the enclosure with the ports sealed of and then measure it again with the ports open then calculate the box compliance and use the volume to air compliance of sub and calculate the box volume. This is the most acurate way to do it. If you realy want to get technical about it and shell out some bucks get you a bag of packing penuts and a half a cubic foot box and measure out how many penuts fit into you enclosure.</p><p></p><p>You need to add to the box volume to expand for the space ocupied by the drivers magnet and cone structure and you do the opposite if you invert the driver (account for cone and dust cap).</p><p></p><p>Be careful about using the term driver displacement. Typical driver displacement (represented by Vd) is the effective cross sectional area (Sd) of the cone multiplied by the maximum displacement of the woofer (Xmax).</p><p></p><p>What I am referring to is if you submerge the driver in water to the top of the rim magnet down and measure the volume of water displaced.</p><p></p><p>etc.</p><p></p><p>Vd=Sd*Xmax</p></blockquote><p></p>
[QUOTE="watts, post: 1722377, member: 567836"] Off the top of my head. To calculate a complex box volume you usualy have a expensive audio measurement system like LEAP or CLIO. You can measure the impedance function of the enclosure with the ports sealed of and then measure it again with the ports open then calculate the box compliance and use the volume to air compliance of sub and calculate the box volume. This is the most acurate way to do it. If you realy want to get technical about it and shell out some bucks get you a bag of packing penuts and a half a cubic foot box and measure out how many penuts fit into you enclosure. You need to add to the box volume to expand for the space ocupied by the drivers magnet and cone structure and you do the opposite if you invert the driver (account for cone and dust cap). Be careful about using the term driver displacement. Typical driver displacement (represented by Vd) is the effective cross sectional area (Sd) of the cone multiplied by the maximum displacement of the woofer (Xmax). What I am referring to is if you submerge the driver in water to the top of the rim magnet down and measure the volume of water displaced. etc. Vd=Sd*Xmax [/QUOTE]
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