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Ohms to Box Rise Question
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<blockquote data-quote="hispls" data-source="post: 8832188" data-attributes="member: 614752"><p>Emphasis added.</p><p></p><p>The answer is, because you can also see DCR at other frequencies and because low impedance is extremely torturous on semiconductors. Power is cheap and almost no competition org will give you an advantage using "cheater" amps anymore.</p><p></p><p>Here is a measured graph of the impedance curve of the wall in my Jeep wired to 0.7 ohm. Ratio of Z-max to Z-min to DCR and the amplitude will change but a ported enclosure will always be similar shape with at least two points being possibly extremely close to DCR. If you really care you will need to build the box, install it, then measure impedance.</p><p></p><p>[USER=681010]@some dude[/USER] is right IMO, power is cheap and comes in small packages, there's almost no reason to worry about this sort of thing these days. </p><p></p><p>[ATTACH=full]47790[/ATTACH]</p></blockquote><p></p>
[QUOTE="hispls, post: 8832188, member: 614752"] Emphasis added. The answer is, because you can also see DCR at other frequencies and because low impedance is extremely torturous on semiconductors. Power is cheap and almost no competition org will give you an advantage using "cheater" amps anymore. Here is a measured graph of the impedance curve of the wall in my Jeep wired to 0.7 ohm. Ratio of Z-max to Z-min to DCR and the amplitude will change but a ported enclosure will always be similar shape with at least two points being possibly extremely close to DCR. If you really care you will need to build the box, install it, then measure impedance. [USER=681010]@some dude[/USER] is right IMO, power is cheap and comes in small packages, there's almost no reason to worry about this sort of thing these days. [ATTACH type="full"]47790[/ATTACH] [/QUOTE]
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