In the electrical industry, we use a unique unit to describe the area of a conductor – the circular mil (CM). Standardizing on the CM gives us a simple way to describe both the cross-sectional area (CM) and the specific resistance (Ohms per CM/ft) of conductors. Indeed, the CM/ft forms the basis of voltage drop calculations (at least in the United States).

Most electricians are familiar with the standard voltage drop formulas:

Single Phase Voltage Drop:

Three Phase Voltage Drop:

Where:

- V = Voltage Drop
- K = Constant for the specific resistance of the conductor in Ohms per CM/ft at a given temperature (for example 12.9 is the approximate resistance of 1 CM/ft of copper at 75° C).
- I = Current in Amps (intensity).
- L = Length of the conductors in feet from the origin to the point of termination.
- A = Area of the conductor in CM.

The multipliers (2 for single phase and 1.73 for 3-phase) are used because the current must travel out AND back. The full current travels on both conductors in a single-phase system, doubling the length (and resistance) of the run, while the current in a 3-phase system divides among 3 conductors, effectively lowering the resistance of the run to about 1.73 times the single wire resistance.

Back to the Circular Mill. A circular mill is defined by the simple formula:

Where:

- A = Area in CM
- d = Diameter in mils (a mil is 1/1000 of an inch)

A common point of confusion is that the diameter of the circle also defines the length of the sides of the square that would encompass the circle, so it seems that this formula should give us square mills (see illustration below).

Although the formulas for calculating CM and mils^{2} are essentially the same, the results are not equal because the *units* are not equal. Circular mils define the area of a circle that would fit inside of a square with sides equal to its diameter.

To illustrate this, let’s assume that the diameter of the circle in the illustration above is 1 mil. That would make each side of the square 1 mil in length, so the area of the square surrounding the circle would be:

If we want to describe the circle in terms of square mils then the formula would be:

or

If we want to go the other way and describe the square in terms of circular mils then we can apply our newly calculated ratio of 1:0.785 and derive the number of circular mills in our 1 mil square by taking the number of square mils divided by 0.785 giving us:

(and now we also have the multiplier for converting square mils to CM).

So, why would you need to know how to convert between square mils and circular mils? One of the most common reasons is to determine the current capacity of a square or rectangular conductor, such as a buss bar. For example:

Let’s say that we have a buss bar that is 2” wide and ¼” thick. The area in square mills is:

Some of you may be wondering why I used 1000^{2} to calculate square mils. The dimensions of the buss bar needed to be converted to mils. Since there are 1000 mils/inch the conversion would be:

Since it is necessary to multiply each side by 1000, it is simpler to take the inch units times 1000^{2}.

Now to convert to CM we use our new multiplier of 1.273 and get:

The actual conversion formulas to convert between CM and Square Mils are:

Mils^{2} to CM:

CM to Mils^{2}:

However, it is much easier (though technically less accurate) to use conversion multipliers:

CM to Mils^{2}:

Mils^{2} to CM:

Purists may point out that the multipliers are rounded (and therefore approximate) values; however, using the conversion multipliers will suffice for all general wiring applications and are much easier to use.