R-Value Calculator
Professional R-value calculator for building engineers, architects, insulation specialists, and energy assessors. Calculate thermal resistance, insulation thickness, wall buildup R-values, and U-value conversion. Includes insulation material comparison tables, thermal conductivity data, and comprehensive building physics reference for energy-efficient design.
Interactive R-Value Calculator
📏 Enter insulation thickness and thermal conductivity. R = d / λ.
🧱 Build up multiple layers to calculate total R-value and U-value. Add internal and external surface resistances.
📐 Enter target R-value and thermal conductivity to find required insulation thickness.
🔄 Convert between R-value (m²K/W) and U-value (W/m²K). U = 1/R.
📘 The R-Value Formula
The fundamental equation for thermal resistance in building physics is:
Where:
- R = Thermal resistance (m²K/W in metric; ft²·°F·hr/BTU in imperial)
- d = Material thickness (metres; 100 mm = 0.1 m)
- λ = Thermal conductivity (W/(m·K)) — the rate at which heat flows through a material
For multiple layers, R-values simply add together:
Where Rsi and Rse are the internal and external surface resistances (typically 0.13 and 0.04 m²K/W for walls).
🔄 R-Value vs U-Value Explained
The U-value (thermal transmittance) is the mathematical reciprocal of the total R-value:
- R-value = thermal resistance — higher is better (more resistance to heat flow)
- U-value = thermal transmittance — lower is better (less heat passes through)
Example: A wall with Rtotal = 5.0 m²K/W has a U-value of 1/5.0 = 0.20 W/m²K. UK Building Regulations Part L typically requires wall U-values of 0.18–0.30 W/m²K, equivalent to R-values of 3.3–5.6 m²K/W.
Imperial to Metric R-Value Conversion
US R-19 batt insulation ≈ R-3.35 m²K/W metric. UK Passive House walls typically require R-30 to R-50+ imperial (R-5.3 to R-8.8 metric).
🧪 Insulation Materials & Thermal Conductivity
The thermal conductivity (λ-value) determines how effective an insulation material is. Lower λ = better insulation per unit thickness.
Common Insulation Materials Comparison
| Material | λ (W/m·K) | R per 100mm (m²K/W) | R per inch (imperial) | Typical Applications |
|---|---|---|---|---|
| Phenolic Foam | 0.018 – 0.022 | 4.5 – 5.6 | R-6.5 – R-8.0 | High-performance walls, roofs |
| PIR Board | 0.022 – 0.026 | 3.8 – 4.5 | R-5.5 – R-6.5 | Walls, flat roofs, floors |
| Closed-Cell Spray Foam | 0.024 – 0.028 | 3.6 – 4.2 | R-5.0 – R-6.0 | Retrofit, irregular cavities |
| XPS (Extruded Polystyrene) | 0.030 – 0.036 | 2.8 – 3.3 | R-4.0 – R-4.7 | Below-grade, inverted roofs |
| EPS (Expanded Polystyrene) | 0.035 – 0.040 | 2.5 – 2.9 | R-3.6 – R-4.1 | Cavity walls, EIFS |
| Mineral Wool (Rock/Glass) | 0.032 – 0.040 | 2.5 – 3.1 | R-3.6 – R-4.4 | Lofts, cavity walls, fire protection |
| Fiberglass Batt | 0.035 – 0.044 | 2.3 – 2.9 | R-3.2 – R-4.1 | Lofts, timber frame walls |
| Cellulose (blown) | 0.038 – 0.042 | 2.4 – 2.6 | R-3.4 – R-3.8 | Lofts, timber frame retrofit |
| Wood Fibre Board | 0.040 – 0.048 | 2.1 – 2.5 | R-3.0 – R-3.6 | Breathable construction |
| Aerogel Blanket | 0.014 – 0.016 | 6.3 – 7.1 | R-9.0 – R-10.2 | Space-constrained, heritage |
🧱 Wall R-Value Calculations
A typical external wall comprises multiple layers, each contributing to the total thermal resistance:
Example: Typical UK Cavity Wall (Post-2006)
| Layer | Thickness (mm) | λ (W/m·K) | R (m²K/W) |
|---|---|---|---|
| External surface (Rse) | — | — | 0.040 |
| Brick outer leaf | 102 | 0.77 | 0.132 |
| Cavity (PIR insulation) | 90 | 0.022 | 4.091 |
| Block inner leaf | 100 | 0.15 | 0.667 |
| Plaster | 13 | 0.57 | 0.023 |
| Internal surface (Rsi) | — | — | 0.130 |
| Total | — | — | 5.083 |
U-value = 1/5.083 = 0.197 W/m²K — compliant with UK Part L new-build standard (≤0.18–0.26 depending on construction type). Use our Wall Buildup calculator tab above to model your own wall construction.
🏠 Roof & Loft Insulation R-Values
Roof insulation is critical — typically 25–30% of heat loss occurs through the roof in uninsulated buildings.
Recommended Loft Insulation Thickness (UK)
| Material | To achieve U=0.16 | To achieve U=0.13 | To achieve Passive House (U=0.10) |
|---|---|---|---|
| Mineral Wool (λ=0.040) | 250 mm | 310 mm | 400 mm |
| PIR Board (λ=0.022) | 138 mm | 170 mm | 220 mm |
| Cellulose (λ=0.040) | 250 mm | 310 mm | 400 mm |
UK Building Regulations recommend a minimum of 270 mm of mineral wool loft insulation (U-value ≤0.16). Our Required Thickness calculator tab computes the exact thickness needed for any target R-value or U-value.
🌉 Thermal Bridging & Condensation Risk
Thermal bridges are localized areas of reduced thermal resistance where heat flows more easily — typically at junctions between building elements (wall/roof, wall/floor, around windows).
- Repeating thermal bridges: Timber studs in insulated frames, wall ties in cavity walls — accounted for in U-value calculations.
- Non-repeating thermal bridges: Junctions, corners, reveals — accounted for via linear thermal transmittance (ψ-value) in SAP/Passive House calculations.
Thermal bridging can reduce the effective R-value of a wall by 10–50% if not properly addressed through continuous insulation and careful detailing. The internal surface temperature drops at thermal bridges, increasing condensation and mould risk.
🌍 Climate Zones & Insulation Requirements
Insulation requirements vary dramatically by climate:
| Standard / Climate | Wall R-Value (m²K/W) | Roof R-Value (m²K/W) | Equivalent U-Value |
|---|---|---|---|
| UK Part L (new build) | 3.3 – 5.6 | 5.0 – 6.25 | U=0.18–0.30 / U=0.13–0.20 |
| Passive House (cold climate) | 6.0 – 10.0 | 8.0 – 12.0 | U=0.10–0.17 / U=0.08–0.13 |
| US IECC Zone 5 (cold) | R-20 cavity + R-5 ci | R-49 | U≈0.22 / U≈0.12 |
| US IECC Zone 2 (warm) | R-13 | R-38 | U≈0.44 / U≈0.15 |
| EnerPHit (retrofit) | 2.5 – 4.0 | 3.3 – 5.0 | U=0.25–0.40 / U=0.20–0.30 |
📋 Worked Engineering Examples
Example 1: Loft Insulation Upgrade
Scenario: Existing loft has 100 mm mineral wool (λ=0.040). Homeowner wants to achieve U=0.16. How much additional insulation is needed?
- Existing R = 0.1/0.040 = 2.50 m²K/W
- Target R = 1/0.16 = 6.25 m²K/W
- Additional R needed = 6.25 − 2.50 = 3.75 m²K/W
- Additional thickness = 3.75 × 0.040 = 0.150 m = 150 mm additional mineral wool
Example 2: Passive House Wall Design
Scenario: Achieve U=0.15 W/m²K using PIR insulation (λ=0.022) between brick and block leaves. Rsi=0.13, Rse=0.04, brick=0.132, block=0.667, plaster=0.023.
- Target Rtotal = 1/0.15 = 6.667 m²K/W
- R from non-insulation layers = 0.13+0.04+0.132+0.667+0.023 = 0.992 m²K/W
- R needed from insulation = 6.667 − 0.992 = 5.675 m²K/W
- Insulation thickness = 5.675 × 0.022 = 0.125 m = 125 mm PIR