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RE: here is a Wendell Diller follow-up on hot-rodding Ferdinand Posche's "VW", REALLY

The resistance per inch changes as you say. But by using smaller driver sections in parallel, the resistance seen by the amp can be lowered.

Say a full panel with original wires is 4 Ohms. If we replaced the wire with foil that had 1/4th the mass and 4X the resistance per inch, then the full panel would be 16 Ohms.

We "could" do it that way and if we had an amp that could drive 16 Ohms with 4X the original power it would give us the same current in each part of the conductor as we had with the original wires at 1X power, except now the foil only has 1/4th as much mass. (It would require relatively high voltage at the amp output.)

BUT, if we divide the foil into two parts, and do the left half of the panel as if it's a separate driver, and do the right half the same way, then each one of those is only 8 Ohms, because each has only half the foil of the 16-Ohm full panel. (We could also run them next to each other, interleaving them in alternating columns, instead of as left and right halves on the membrane. But it doesn't seem like it would make any difference in the sound.)

Then we just connect those two 8-Ohm foil sections in parallel and the amp will see 4 Ohms, just like the original wire.

We'd still need 4X the power, I think, to get the same current as when we used 1X power with the original wire. But the foil's mass would only be 1/4th of the original wire's mass.

So we'd have the same current as before but with only 1/4th as much conductor mass, with the same impedance seen by the amp as before, but the amp would need to provide four times as much power.

The same current as before would give the same force on the membrane as before. But the _acceleration_ of the membrane, due to the same force (F = ma so a = F/m), would probably be slightly less than 4X the original acceleration, depending on how much of the mass is due to the mylar, in each case.

Zoom Zoom!

Cheers,

Tom



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