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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 (A) on July 2, 2011 at 13:51:32

In Reply to: RE: "If we replaced the wire with foil that had 1/4th the mass and 4X the resistance per inch" ... posted by josh358 on July 1, 2011 at 09:49:12:

Since the total available foil length should be known (based on the panel where the wires are being replaced), I derived an equation that can be used to find, from a desired total length of foil, the cross-sectional area needed to get two identical parallel foil runs that result in a 4 Ohm load:

Area = thickness x width = length x .001625

where length is in meters, Area is in square millimeters, and thickness and width are in millimeters.

If the desired width or thickness were known, then that could be rearranged to one of the following:

thickness = length x .001625 / width

or

width = length x .001625 / thickness

Or, if you had foil with a known thickness and width, you could calculate what total length would need to be divided in half to get 4 Ohms when the two halves were run in parallel, with:

length = thickness x width / .001625

The following conversion formulas can be used:
mm / 25.4 = inches
inches x 25.4 = mm
meters x 39.37 = inches
inches / 39.37 = meters

Example:

If the optimal width that Wendell mentioned turns out to be 0.1 inch, i.e. 2.54 mm, we can calculate what total lengths would give 4 Ohms when divided into two parallel runs, for each of several different available thicknesses:

.0005" (.0127 mm) x 0.1" foil: .0127x2.54/.001625 = 19.85 m (65.13 ft)

.001" (.0254 mm) x 0.1" foil: .0254 x 2.54 / .001625 = 39.7 m (130.25 ft)

I'll stop there, since .0015-inch by 0.1-inch foil would need 59.55 meters split into two parallel runs, to get 4 Ohms.

-------------------------

With only a tiny bit more work, I was able to derive the equations for n aluminum foil runs in parallel:

(1) Rn = .026 x Ltot / nnTW

where Rn is the resistance in Ohms of n parallel foil runs, each with length Ltot/n (in meters), and Ltot is the total foil length in meters, T is the foil thickness in millimeters, and W is the foil width in millimeters. Note that nn is n x n which is "n squared".

The derivation for n parallel foil runs was easy, if we can assume that we want the same current through all runs, and therefore we have to assume that their resistances are equal. For equal resistances, the parallel resistances equation [Rpar = 1 / (1/R1 + 1/R2 + 1/R3 + ... 1/Rn)] boils down to Rpar = R/n.

(We could also use shorter lengths of thicker foil, for example, to get the same current, which would give the same force. But then we'd be accelerating a different mass, so the acceleration would be different, even though the current was the same. We could calculate the difference in mass needed for a different current, to get the same acceleration from a different length, but then other aspects of the behavior of the membrane would probably still be changed. Maybe someone should figure that out, though, because it would open up an infinite number of options, if it could be done that way while still retaining satisfactory characteristics.)

If we assume that we want to end up with 4 Ohms, that gives us the following different rearranged versions of the same equation:

(2) Area = T x W = .026 L / 4nn

(3) T = .026 L / 4nnW

(4) W = .026 L / 4nnT

(5) L = 4nnTW / .026

Examples:

1) What total lengths of .0005 x 0.1 inch foil would give 4 Ohms when divided into different numbers of parallel runs?

TW = .0127 mm x 2.54 mm = .032258 sq mm

n = 2: L = 4 x 2 x 2 x .032258 / .026 = 19.85 m (whew, same as before)
n = 3: L = 4 x 3 x 3 x .032258 / .026 = 44.665 m

2) What thicknesses would work to get 4 Ohms from n parallel runs of 0.1-inch-wide foil that totaled 28.4 meters in length?

T = .026 L / 4nnW

n = 2: T = (.026 x 28.4) / (4 x 2 x 2 x 2.54) = .01817 mm (.0007153 inch)
n = 3: T = (.026 x 28.4) / (4 x 3 x 3 x 2.54) = .008075 mm (.000318 inch)

It doesn't look too good for the 28.4 meters needed for the MG12s, SO far.

3) What widths of foil would work to get 4 Ohms from n parallel runs totaling 28.4 meters, for a couple of different thicknesses?

W = .026 L / 4nnT

T = .001 inch (.0254 mm):

n = 2: W = .026 x 28.4 / 4 x 2 x 2 x .0254 = 1.817 mm (.0715 inch)
n = 3: W = .026 x 28.4 / 4 x 3 x 3 x .0254 = .8075 mm (.0318 inch)

T = .0005 inch (.0127 mm):

n = 2: W = .026 x 28.4 / 4 x 2 x 2 x .0127 = 3.634 mm (.143 inch)
n = 3: W = .026 x 28.4 / 4 x 3 x 3 x .0127 = 1.615 mm (.0636 inch)

---------------------------

It's not looking overly-convenient for the MG12s, yet. But I guess it depends on what optimal foil width would be.

We want the current through all runs to be equal, I assume. So if we can't get what we need with n identical parallel runs of foil, we're out of luck? It looks that way, to me, at the moment. The next-best thing would probably be to change the desired resistance to something slightly different than 4 Ohms. And I guess another option would be to sacrifice some power and add a resistance in series or parallel with the whole shebang.

Cheers,

Tom

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