Is it really that wide? The MG-3.6/R ribbons are only 5/32 inch wide, 0.00015 inch thick, and 55 inches in length.

Anyway, it depends on the cross-sectional area, the length, and the conductivity of the material.

Resistance of an aluminum conductor is

R = pL/A,

where R is in Ohms, the conductivity of aluminum, p, is 2.6 x 10^(-8) Ohm-meters (where 2.6 x 10^(-8) = 0.000000026), 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, for example, one meter of .0254 mm thick x 2.547 mm wide 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.

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, we have:**

**(1) width (mm) = (.026 x length_m) / (Ohms x thickness_mm) **

and

**(2) thickness (mm) = (.026 x length_m) / (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_m) / (thickness_mm x width_mm) **

**(4) length (m) = Ohms x (thickness_mm x width_mm) / .026**

Handy conversion equations:

mm = inches x 25.4

inches = mm / 25.4

m = inches / 39.37

inches = m x 39.37

**Example: ** MG-3.6/R ribbon is 55 inches long, 5/32 inch wide, and 0.00015 inch thick. What is its resistance?

L = 55 / 39.37 = 1.397 m

W = 5/32 x 25.4 = 3.969 mm

T = 0.00015 x 25.4 = 0.00381 mm

(3) Ohms = (.026 x length_m) / (thickness_mm x width_mm), from above.

Ohms = (0.026 x 1.397) / (0.00381 x 3.969) = 2.402 Ohms