Vtyjkyuk

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

Where

 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?

What you need to do is called changing the subject of the formula.
$$beginalign
L = frac3,141left[(D2/2)^2 + (D1/2)^2right]T\
\
hline\
textMultiplying by; fracT3,141; texton both sides:fracTL3,141 &= left[(D2/2)^2 + (D1/2)^2right]\
textAdding ;(D1/2)^2 ;texton both sides:fracTL3,141+ (D1/2)^2 &= left[(D2/2)^2 right]\
textTaking square roots on both sides:sqrtfracTL3,141+ (D1/2)^2 &= sqrtleft[(D2/2)^2 right]=D2/2\
textMultiplying both sides by; 2:2timessqrtfracTL3,141+ (D1/2)^2 &=D2\
endalign$$

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.