Thermal insulation
R-value calculator: thickness and conductivity
Calculate insulation R-value from thickness and lambda, US k-value or R per inch. Enter mm, metres, inches or feet and compare RSI with US R.
How to use this calculator
Method and boundary
Choose thickness in m, mm, in or ft and material data: appropriate λ in W/(m·K), k in Btu·in/(h·ft²·°F) or US R per inch. Do not enter a batt's total R as R per inch. Changing entry type converts a positive value; unit conversion does not correct declared conditions to design conditions.
Layer R = thickness [m] / design λ [W/(m·K)]
Example
For 0.1 m and hypothetical λ = 0.040 W/(m·K), layer R is 2.5 m²·K/W.
The result describes one homogeneous layer, excluding surfaces, bridges and other materials. The tool does not convert declared values to design values.
Calculation privacy
Recent stores calculator ID, route ID, locale, time and unit system only.
How to calculate material R-value
For one homogeneous layer, divide thickness by thermal conductivity. In SI, thickness must be in metres: R = d/λ. With thickness in inches and k in Btu·in/(h·ft²·°F), US R = d/k gives h·ft²·°F/Btu.
Worked example: 100 mm and λ 0.040
100 mm = 0.1 m. Therefore R = 0.1 / 0.040 = 2.5 m²·K/W. In US units this is approximately R-14.196. Entering 100 as metres would overstate the result by a factor of 1,000, so check the unit next to the field.
| Thickness · mm | R SI · m²·K/W | R US · h·ft²·°F/Btu |
|---|---|---|
| 50 | 1.25 | 7.098 |
| 100 | 2.5 | 14.196 |
| 150 | 3.75 | 21.293 |
| 200 | 5 | 28.391 |
How do I use R per inch from a US label?
Select “US R per inch” in the form. Hypothetical R-4 per inch at 5.5 in gives US R = 4 × 5.5 = 22, or approximately 3.874 m²·K/W. The same material data can be written as k = 0.25 Btu·in/(h·ft²·°F). R-22 for a complete batt is not R per inch: do not multiply its total R by thickness again.
Scaling assumes a homogeneous material with unchanged properties. Product R can depend on variant, thickness and conditions; use its declared R in product records rather than your own extrapolation.
Does lower lambda mean better insulation?
At equal thickness and comparable conditions, lower λ gives higher R. For example, 100 mm at λ 0.032 gives R = 3.125, while λ 0.040 gives R = 2.5. R alone does not determine moisture resistance, load capacity or suitability for an application.
What value should I enter for EPS, XPS or mineral wool?
Do not use one conductivity value for an entire material family. Check the exact variant, thickness, product document and declaration basis. For design calculations, establish the applicable conditions and corrections; this tool does not automatically turn declared lambda into design lambda.
Is this the U-value of my wall?
Layer R includes only the material entered. Whole-assembly U requires the remaining layers, surface resistances and applicable corrections, including non-uniformity and thermal bridges. The reciprocal of insulation R alone is not the whole-wall result.
How to read thermal values from product documents · Convert R-value to U-value
NIST SP 1038, §5.3.5–5.3.6 — source for k and U unit conversions (IT Btu).