Coil / Roll Stock Length
How much strip is left on the coil without unrolling it.
Example
You enter
- Coil outside diameter OD (in) 48
- Core / mandrel diameter ID (in) 16
- Material thickness t (in) 0.024
You get
- Coil length (in) 67021
- Coil length 5585.1 ft
Checked against published worked examples from: Dallan (Andrea Dallan).
Details, formula, and sources
ScopeThe wound material is an annulus, so L = pi (OD^2 - ID^2) / (4 t) for coil outside diameter OD, core diameter ID, and thickness t. A 48 in coil on a 16 in core at 0.024 in (24 ga) holds 5,585 ft; halve the thickness and the same coil OD holds twice the length. Exact for a tight coil with no telescoping; the last wrap and core stub trim it slightly. A layout aid; weigh or measure-off before a critical cut.
Formula or tableL = pi (OD^2 - ID^2) / (4 t); L_ft = L / 12.
Source and editionThe coil / roll stock length annulus identity, first-principles; the coil OD, core ID, and thickness come from the material.
Free accessThe coil-length relation is a first-principles geometric identity (annulus cross-section = length x thickness); the coil OD, core diameter, and material thickness come from the coil.
What governsEstimate. AHJ and licensed professional govern.
API field namesoutside_diameter_in, inside_diameter_in, material_thickness_in, length_in, length_ft
- Annulus identity wound cross-section pi/4 (OD^2 - ID^2) equals unwound length x thicknessgeometry
- Tight coil assumes no air gaps or telescoping between wrapsscope of this tile
- Usable length the last wrap and core stub trim the usable length slightlyshop practice