# Calculating the Weight of a Coil of Sheet Metal

At work we use generic tables to estimate the weight of coils of sheet metal. The tables are based on the thickness in inches of metal "sidewall" that remains on the coil.

Now upper management says that these estimates are not accurate enough for their new raw materials tracking system, and they want an actual scale weight after each coil use...which is very time consuming.

I need to find the proper formula to more accurately calculate the weight for each coil after each use. The coils are a wide variety of metals, metal thicknesses, coil widths, stamp patterns, surface treatments, core (inside) diameters, sidewall thickness, etc.

For each coil we know the actual original weight of the metal, and we can measure what the original sidewall thickness is and what the core diameter is (the core is what is left when the metal is all used up)

Let's suppose one coil has a core (inside) diameter of 20 inches, a sidewall of 10 inches of metal, and a weight of 10,000 pounds. So we know two data sets for this coil. First, that 10 inches of sidewall metal equals 10,000 lbs of metal and, second, that zero inches of sidewall (when the metal is all used up--the 20 inch core diameter) equals zero pounds of metal.

The generic table we use is a linear extrapolation of weight based on the sidewall thickness. But while the sidewall thickness decreases linearly, the actual weight of metal decreases geometrically as the circumference of the coil decreases. Shouldn't this actually be curvilinear?

10 in. sidewall = 10,000 lbs.

Zero in. sidewall = Zero lbs.

So, given any thickness of sidewall, what would be the way to calculate the theoretical the weight of the remaining metal?

Much more accurate is to use the area of the sidewall times the width of the shet to get the volume of metal. You assume the packing density is constant. In your example the outer radius of the coil is $30$ inches at the start. The area at the start is then $\pi(30^2-20^2)=500\pi$ in$^2$. The weight per sidewall area is then $\frac {10,000}{500\pi}=\frac {20}\pi$ lb/in$^2$. At sidewall thickness $t$ the area is $\pi((t+20)^2-20^2)=\pi(t^2+40t)$ and the weight is $20(t^2+40t)$ lb.