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<blockquote data-quote="samson" data-source="post: 2960924" data-attributes="member: 571279">&quot;Efficiency = ( B^2 * L^2 ) / ( R * Sd^2 * Mms^2 )B = magnetic field strengthL = length of wireR = resistanceSd = surface areaMms = massSo for your 8 ohm voice coil, using the same wire as a 4 ohm voice coil, you would need twice the L or length to get an 8 ohm impedance. That makes sense doesn&#039;t it? A longer wire will have more resistance. Now, looking at the formula above, doubling L actually causes your efficiency to rise, even though the impedance also rises. So in this very oversimplified example, raising the impedance actually causes efficiency to go up and lowering the impedance actually causes a loss of efficiency.&quot;Not sure but I think he&#039;s overlooking mass.&nbsp; As a denominator, it would at least partially offset the his length of wire.&nbsp; Seems like more wire in the voice coil = more mass you have to move.</blockquote>

[QUOTE="samson, post: 2960924, member: 571279"] "Efficiency = ( B^2 * L^2 ) / ( R * Sd^2 * Mms^2 ) B = magnetic field strength L = length of wire R = resistance Sd = surface area Mms = mass So for your 8 ohm voice coil, using the same wire as a 4 ohm voice coil, you would need twice the L or length to get an 8 ohm impedance. That makes sense doesn't it? A longer wire will have more resistance. Now, looking at the formula above, doubling L actually causes your efficiency to rise, even though the impedance also rises. So in this very oversimplified example, raising the impedance actually causes efficiency to go up and lowering the impedance actually causes a loss of efficiency." Not sure but I think he's overlooking mass. As a denominator, it would at least partially offset the his length of wire. Seems like more wire in the voice coil = more mass you have to move. [/QUOTE]

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