I'm developing a software for a stainless steel plant, and I need to calculate the diameter of steel coils so the software assigns a position inside a deposit according to the size of the coil.

I have access to a lot of data from the coil, and tried thinking of a way to calculate using the inner hole diameter and the length and thickness of the steel strip.

In a research I've found this formula:

L = 3,141[(D2/2)² - (D1/2)²]/T


  L = Length of the coil

  D1 = Diameter of the inner hole

  D2 = Diameter of the coil

  T = Thickness of the material

But in the software I have to put in this format:

D2 = (formula)

I'm no good with math, how can I "convert" the formula to suit my needs?

  • 1
    $\begingroup$ $$D2=2\times\sqrt{\frac{2L}{3.141}+\left(\frac{D1}{2}\right)^2}$$ $\endgroup$
    – K. Rmth
    Oct 18, 2015 at 15:39
  • $\begingroup$ I wrote the formula wrong, put 2 instead of T, could you post this comment (corrected with the T) in the answer section so I could accept it? $\endgroup$
    – Bruno Vaz
    Oct 18, 2015 at 16:53

2 Answers 2


What you need to do is called changing the subject of the formula. $$\begin{align} L = \frac{3,141\left[(D2/2)^2 + (D1/2)^2\right]}{T}\\ \\ \hline\\ \text{Multiplying by}\; \frac{T}{3,141}\; \text{on both sides}:\frac{TL}{3,141} &= \left[(D2/2)^2 + (D1/2)^2\right]\\ \text{Adding} \;(D1/2)^2 \;\text{on both sides}:\frac{TL}{3,141}+ (D1/2)^2 &= \left[(D2/2)^2 \right]\\ \text{Taking square roots on both sides}:\sqrt{\frac{TL}{3,141}+ (D1/2)^2 }&= \sqrt{\left[(D2/2)^2 \right]}=D2/2\\ \text{Multiplying both sides by}\; 2:2\times\sqrt{\frac{TL}{3,141}+ (D1/2)^2 }&=D2\\ \end{align}$$


L=([(OD+ID)/2]3.14)[(OD-ID)/2]/T Figure OD being 30 and ID being 20 Figure 24 gauge thick or .024” *note I always Figure the thickness as being slightly thicker Due to wraps not ever being perfectly tight so for .024 I use .026. Example: 30+20=50/2=25 (25 is the average Diameter of coil the multiply by pi) 25x3.14=78.5 then round down to nearest inch Average wrap of coil is 78” long Then 30-20=10/2=5 5” is the thickness of coil 5/ thickness of steel .026 = 192.3 wraps of coil Multiply average wrap length 78” x 192.3 wraps = 14,999.4” of coil / 12 = 1249.9 lineal feet of coil I’ve used this for years in my business and real world application I’m usually within 10 lineal feet. The trick is the rounding down of the average length and the adding about .02” to the thickness of material.

  • $\begingroup$ here is a reference for mathjax that can be used to typeset maths on the site. Also you might like to include some spacings for your answer. $\endgroup$ Sep 1, 2018 at 6:11

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