Insulation reference
Thermal conductivity units: Btu-inch versus Btu-foot
Avoid the factor-of-12 error between Btu·in/(h·ft²·°F) and Btu/(h·ft·°F). Compare both conductivity units with SI lambda and R per inch.
Short answer
Btu·in/(h·ft²·°F) and Btu/(h·ft·°F) are different units of conductivity. One unit of the first is about 0.144228 W/(m·K); one unit of the second is about 1.730735 W/(m·K). Copying the same number between them changes the conductivity by a factor of 12.
On this page
Prepare your inputs
- The complete conductivity unit, including the inch or foot convention.
- The Btu definition and stated product/test conditions.
- Thickness in a unit consistent with the selected calculation input.
Read the full unit beside k, not just the symbol
λSI ≈ 0.144227889 × kinch ≈ 1.730734667 × kfoot; kinch = 12 × kfoot
A US data sheet can express conductivity using an inch in the numerator or a foot in the denominator. Cancelling the dimensions does not make the numerical scales identical. Record the Btu basis as well: these examples use the international-table Btu conversion basis used by the linked material-layer calculator.
SI lambda is conductivity in W/(m·K). R per inch is a resistance per thickness, not conductivity. For the same homogeneous material and conditions, R per inch is the reciprocal of k only when k uses Btu·in/(h·ft²·°F).
Example: the same 0.25 means two different conductivities
With 0.25 Btu·in/(h·ft²·°F), lambda is about 0.036057 W/(m·K), and the corresponding US R per inch is 4. With 0.25 Btu/(h·ft·°F), lambda is about 0.432684 W/(m·K). That second conductivity is twelve times larger.
If the first value applies at the stated thickness and conditions, a 4-inch homogeneous layer has US R = 4/0.25 = 16. This illustrates arithmetic, not a promise that any product has constant R per inch across thicknesses, temperatures or densities.
Enter the unit that the existing calculator actually supports
The material-layer calculator accepts SI lambda, US k with an inch in the numerator, or US R per inch. Convert a foot-based k into the inch-based unit by multiplying its number by 12 before choosing the US k input. Check the unit displayed beside the field.
A unit conversion leaves the product, test conditions and declared-versus-design scope unchanged. It does not turn a material value into a complete assembly U-value.
Reference table
Equivalent conductivity numbers on the IT Btu basis; hypothetical values, not product declarations
| k [Btu·in/(h·ft²·°F)] | k [Btu/(h·ft·°F)] | λ [W/(m·K)] | US R per inch |
|---|---|---|---|
| 0.1 | 0.008333 | 0.014423 | 10 |
| 0.2 | 0.016667 | 0.028846 | 5 |
| 0.25 | 0.020833 | 0.036057 | 4 |
| 0.3 | 0.025 | 0.043268 | 3.3333 |
| 0.4 | 0.033333 | 0.057691 | 2.5 |
Common mistakes
- Entering foot-based k into an inch-based field without multiplying by twelve.
- Treating R per inch as if it were a conductivity number.
Check before using the result
- Match the displayed calculator unit to the converted value.
- Record rounding and retain the original document value for traceability.
- Keep conductivity conversion separate from selecting an assembly model.
Method limits
Values are rounded for display. Check the complete unit, Btu definition, product document and conditions before a calculation. No generic lambda or R per inch is assigned to an insulation material here.
Questions and answers
How do I enter foot-based k in the existing US k field?
Multiply the foot-based conductivity number by 12 to obtain Btu·in/(h·ft²·°F), then check that this is the unit displayed beside the input.
Does converting k into R per inch establish the R of a wall?
No. It changes the description of a homogeneous material under the same conditions. A wall calculation still needs its thicknesses, paths and boundary conditions.
Sources and scope
- NIST SP 811, Appendix B.9 — thermal conductivity and insulance units
Unit definitions and conversion factors; no product performance or required insulation thickness.