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Everything posted by djjdnap
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http://www.youtube.com/watch?v=rA7ICvs9VxQ haven't tried it though
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wondering if sealing all of those holes would make it sound better
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Probably was talking about this thread and i probably do, but those alts for those older cars are so cheap brand new there'd be no point lol.. Then again there only like 60-90 Amps i believe
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Question about power, and impedance.
djjdnap replied to ford302redneck's topic in Advanced Discussion
That sounds weird man like m5 said, ur only increasing wattage, so yea the subs are going to move more, however if the sub is moving more it should technically be louder.... It would be the same as switching out a 1500w amp for a 3000w amp and saying that it only made ur subs move more, but not change the volume.. however they do say u need to double volume or cone area in order hear an audible difference. So maybe it is louder on the mic just not to ur ear. -
Question about power, and impedance.
djjdnap replied to ford302redneck's topic in Advanced Discussion
that is weird imo... when i switch to a lower impedance without doing anything else u can hear an audible difference regardless of freq. because your getting more watts. -
Question about power, and impedance.
djjdnap replied to ford302redneck's topic in Advanced Discussion
i say its wired wrong.... but if u put the dmm on it who am i to say. set ur system to a good volume, wire it to 2 ohms... leave the system the same and wire it to 1 ohm... if it gets quieter u sir have something wired wrong. easiest way to check imo -
Higher voltage/lower impedance?
djjdnap replied to Wood.'s topic in Amplifiers / Head Units / Processors / Electrical
in a basic world ohms law says e/r=i so 12v/100ohms = .12amps 14v/100ohms = .14amps and so on so at higher voltage you can produce more amps at the same impedance. so then you say ok what about power or watts p=1^2 *r or e*i .12a^2 *100=1.44watts .14a^2 *100=1.68watts or .12 * 12= 1.44watts .14*12= 1.68watts so we also see that if we reduce the impedance it gives us a higher amperage and when you go back in to use it in the power formula it works out accordingly.