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RE: "If we replaced the wire with foil that had 1/4th the mass and 4X the resistance per inch" ...

Posted by tom_gootee on June 26, 2011 at 14:31:29:
Hi andyr,
I hadn't really gotten around to discussing actual sizes and weights of foil, or the size of the wire that it would be replacing.
The conclusion I reached was only that _IF_ much-lower-mass conductors could be used, then we could still wire-up a driver with a practical impedance AND much less mass (although it would then require a lot more power to achieve the same current through the conductors, which is what determines the force exerted on the conductor by the magnetic field.
The foil that is available could have a thickness of 0.001-inch or 0.0005-inch, which is .0254 mm or .0127 mm thick.
The 31 ga wire has a cross-sectional area of 3.1416 x (.22606/2) x (.22606/2) = .0401 sq mm.
To get the same cross-sectional area using .0254mm-thick foil the width would need to be 1.58 mm (= .0622 inch = about 1/16th inch).
To get the same cross-sectional area using .0127mm-thick foil the width would need to be 3.16 mm (= .124 inch = about 1/8th inch).
So, when replacing 31 AWG wire with foil, it looks like we'd be cutting things pretty close if we tried to get 1/4th the cross-sectional area, because even with the .0005-inch (.0127mm) thickness it would require a width of only 1/32nd inch (.03125 inch), or 0.79375 mm.
Maybe someone can calculate what rating of FUSE that would make, or how hot it might get.
I just measured the wire on the mid/bass panel of my MG-12/QR speakers and it measured as 0.022-inch diameter, which means it is probably 23 AWG with a diameter of .0226 inch, or about 0.57404 mm, with a cross-sectional area of about 0.2588 sq mm. (It would take a foil width of 10.19 mm of .0254mm-thick foil to get the same cross-sectional area.)
So if we used the .0254mm-thick (.001 inch) foil with a width of 2.547 mm (about 0.1 inch) we'd have a cross-sectional area of 0.0647 sq mm. That's 25 PERCENT as much cross-sectional area as the 23 AWG wire. So it would have 4.000 TIMES the resistance of 23 AWG wire, per unit of length.
I need to go back to basics:
Resistance of an aluminum conductor is R = pL/A, where R is in Ohms, p is 2.6 x 10^(-8), i.e. .000000026 Ohm-meters, L is length in meters, and A is cross-sectional area in square meters. From that, the Ohms per meter would be .026 / (area in sq mm).
So one meter of the .0254 mm x 2.547 mm foil would have a resistance of .026/.0647 = .4019 Ohms per meter. So, to get 3.9 Ohms of resistance it would require 9.705 meters of foil, or 31.84 feet.
But since we'd be replacing existing wire, the length would already be decided.
Let's see... Ohms/meter = .026 / (thickness x width). And (Ohms/meter) x length(in meters) = Ohms [so Ohms/meter = Ohms/length(in meters) and length = Ohms / (Ohms/meter)].
So, with aluminum foil length in meters and foil thickness and width both in mm:
(1) width = (.026 x length) / (Ohms x thickness)
and
(2) thickness = (.026 x length) / (Ohms x width)
So, with those equations, we could calculate the needed width or thickness of the foil, if we specified the thickness or width and knew the length and resistance that we were replacing. (Also, remember that the length is in meters but the thickness and width are both in mm.)
Also:
(3) Ohms = (.026 x length) / (thickness x width)
(4) length = Ohms x (thickness x width) / .026
EXAMPLE:
My MG12s have 32 bass/mid wire runs of about 42.5 inches each, which is about 34.544 meters. The driver I measured had a DC resistance of 3.85 Ohms. (Also note that there are only 26 vertical wire positions, but six of those have two wires.)
If I wanted to replace the MG12 bass/mid wires with foil and get 3.9 Ohms total, and wanted to use .001-inch-thick foil (0.0254 mm), then the foil's width would need to be:
width = (.026 x length) / (Ohms x thickness)
= (.026 x 34.54) / (3.85 x 0.0254)
= 9.18 mm
That obviously wouldn't be a practical width but I wanted to see if it came out to about the same cross-sectional area as the original 23 AWG wire. And 9.18mm x .0254mm = .2332 sq mm, which is close to the .2588 sq mm of the original wire. Note that my measurements of the speaker's wire length were not very accurate.
Since length = Ohms x (thickness x width) / .026, the length of the original wire should have been 3.85 x .2588 / .026 = 38.3 meters
----------------
Now armed with some numbers and equations, and GOING BACK to the example in my earlier post, about using 1/4th the cross-sectional area and using two paralleled drivers of half the original panel size and about 8 Ohms each, let's say that the original MG-12 panel has room for 26 vertical runs of about 43 inches each, or 28.4 meters. Half of that would be 14.2 meters.
To get 8 ohms with 14.2 meters of aluminum, the cross-sectional area would have to be .026 x 14.2 / 8 = .04615 sq mm.
The original 23 AWG was .2588 / .04615 = 5.6X larger.
Anyway, to get .04615 sq mm with .0254mm-thick foil the width would have to be 1.83 mm (.072 inch).
THEREFORE, if we replaced the wires of an MG-12/QR's mid/bass panel with .001-inch-thick foil that was 1.83 mm wide, and wired it as two half panels in parallel, we would get a 4 Ohm driver with only 17.8% as much conductor mass as the original wires had.
[EDITED: Had measured the gauge of the MG-12/QR's bass/mid wires incorrectly. Correct diameter is .0226 inch; 23 AWG. Fixed.]
Cheers,
Tom