Summary

R-value and U-value are two sides of the same calculation, but they answer different questions. U-value (covered in detail in u value calculator) tells you whether a finished wall, roof, or floor build-up meets a Building Regulations Part L target. R-value tells you how much resistance to heat flow a single material or layer provides — which is the number on the back of a roll of mineral wool, the spec sheet for a PIR board, or the figure you need when working out "how much more insulation do I need to add" or "can I swap this product for that one and get the same performance."

For a tradesperson on site, R-value calculations come up more often than full U-value calculations: topping up loft insulation, deciding whether a thinner PIR board can replace a thicker mineral wool layer in a stud cavity, working out how many layers of a given product are needed to hit a target, or explaining to a customer why "thicker" isn't always "better value" once you compare pound-per-R-value across products.

This article covers the R-value formula, how R-values combine across layers (the part most people get wrong), how to convert between R and U, and three fully worked examples using real UK insulation products and real numbers. For full wall/roof/floor build-ups against Part L targets, see u value calculator; for loft-specific depth tables, see loft insulation depth.

Key Facts

Quick Reference Table — R-Value per 100mm for Common UK Insulation Materials

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Material Typical λ (W/mK) R per 100mm (m²K/W) R per 50mm (m²K/W) Notes
PIR insulation board (Celotex, Kingspan Kooltherm/K-range) 0.022–0.023 4.35–4.55 2.17–2.27 Best R per mm; higher cost per m²
PUR rigid board 0.024 4.17 2.08 Similar performance to PIR, slightly higher λ
Phenolic foam board (Kingspan Kooltherm K5/K8) 0.019–0.021 4.76–5.26 2.38–2.63 Premium performance, premium price
Mineral wool (rock wool, e.g. Rockwool) 0.034–0.044 2.27–2.94 1.14–1.47 Standard loft/cavity/stud fill
Glass wool quilt (standard grade) 0.038–0.044 2.27–2.63 1.14–1.32 Common budget loft roll
Premium glass wool (e.g. Knauf 32/Earthwool) 0.032–0.035 2.86–3.13 1.43–1.56 Better R per mm than standard glass wool
EPS (expanded polystyrene) 0.030–0.038 2.63–3.33 1.32–1.67 Cavity wall bead, floor slab insulation
XPS (extruded polystyrene) 0.033–0.038 2.63–3.03 1.32–1.52 Ground contact / high compressive load applications
Woodfibre board (breathable) 0.038–0.050 2.00–2.63 1.00–1.32 Vapour-open, heritage/breathable builds
Sheep's wool 0.038–0.040 2.50–2.63 1.25–1.32 Natural, moisture-buffering
Blown cellulose 0.038 2.63 1.32 Loft topping, retrofit cavity fill
Blown mineral fibre 0.040 2.50 1.25 Cavity fill retrofit
Unventilated air cavity (25mm+) 0.18 (fixed, not thickness-scaled) Not a material R — a fixed convention value

Detailed Guidance

The core calculation — R = d ÷ λ

Take the thickness in metres and divide by the material's thermal conductivity:

R = d (m) ÷ λ (W/mK)

Example: 120mm PIR board with λ = 0.022 W/mK:

The same calculation for 200mm of standard mineral wool at λ = 0.040 W/mK:

This immediately shows why PIR is specified where depth is limited (loft hatches, floor build-ups over doorways, dormer cheeks) — it delivers a similar R-value in roughly 40% less thickness, at a materially higher cost per m².

How R-values combine — the calculation most people get wrong

R-values in a build-up add in series, layer by layer, in the direction heat is flowing through them:

R_total = R_layer1 + R_layer2 + R_layer3 + ... (+ Rsi + Rse if converting to U-value)

This is different from λ, which does NOT average or combine directly across different materials — you must calculate each layer's individual R first, then add the R-values together. A common site error is trying to average two λ values before calculating R; this gives the wrong answer unless the two layers happen to be the same thickness.

Worked example — topping up existing loft insulation

A loft currently has 100mm of old mineral wool (λ = 0.044 W/mK, a lower-grade older product). The homeowner wants to know how much additional insulation, laid crosswise over the joists, is needed to bring the loft up to the Part L1B renovation target of 0.16 W/m²K (see loft insulation depth for the full target table).

Step 1 — calculate existing R: R_existing = 0.100 ÷ 0.044 = 2.27 m²K/W

Step 2 — calculate the R needed to hit the target U-value: Target U = 0.16 W/m²K → R_total required = 1 ÷ 0.16 = 6.25 m²K/W Subtract surface resistances and the plasterboard ceiling (12.5mm, λ = 0.21, R = 0.06): R_insulation_required = 6.25 − 0.13 (Rsi) − 0.04 (Rse) − 0.06 (plasterboard) = 6.02 m²K/W

Step 3 — subtract the existing insulation's R: R_still_needed = 6.02 − 2.27 = 3.75 m²K/W

Step 4 — convert to a thickness of the new product being used to top up. If topping up with a premium glass wool roll at λ = 0.035 W/mK: d = R × λ = 3.75 × 0.035 = 0.131m = 131mm, round up to the next standard roll thickness — 150mm, which is standard stock for a cross-laid top-up roll.

Result: lay 150mm of premium glass wool crosswise over the existing 100mm of old mineral wool (crosswise both to cover the joists themselves, which are otherwise a thermal bridge, and to avoid compressing the existing layer). This combined build-up comfortably exceeds the 0.16 W/m²K target.

Worked example — matching R-value when substituting a different product

A stud wall cavity is designed for 100mm of standard mineral wool (λ = 0.040 W/mK, R = 2.50 m²K/W), but the merchant is out of stock and the only mineral wool available is a slightly different grade at λ = 0.044 W/mK. The customer doesn't want to lose thermal performance. What thickness of the substitute product is needed to match the original R-value?

Step 1 — target R to match: 2.50 m²K/W (from the original spec)

Step 2 — rearrange R = d ÷ λ to solve for d: d = R × λ = 2.50 × 0.044 = 0.110m (110mm)

Since the stud cavity is 100mm, the substitute product at a slightly worse λ cannot quite match the original R-value within the same 100mm cavity (it would only achieve R = 0.100 ÷ 0.044 = 2.27 m²K/W, roughly 9% less than the original spec). In practice, either accept the small shortfall, use a batt product designed to be compressed slightly into a service void, or switch to a higher-performance product (e.g. a premium glass wool at λ = 0.032, which gives R = 0.100 ÷ 0.032 = 3.13 m²K/W — comfortably exceeding the original spec in the same 100mm).

Worked example — cost-per-R-value comparison for a customer quote

A customer is choosing between two loft insulation options and asks which is "better value." Real indicative UK prices (check current supplier pricing before quoting):

Product Thickness λ (W/mK) R-value Price per m² (indicative) Cost per unit of R
Standard glass wool roll 270mm 0.044 6.14 m²K/W £3.50 £0.57 per m²K/W
Premium glass wool roll 200mm 0.032 6.25 m²K/W £5.80 £0.93 per m²K/W

Both options deliver a near-identical R-value (and therefore near-identical thermal performance and running-cost saving), but the standard product is cheaper per unit of thermal resistance achieved — the premium product's advantage is the reduced thickness (200mm vs 270mm), which matters where loft headroom, joist depth, or hatch clearance is restricted, not where performance-per-pound is the only criterion. Present both figures to the customer rather than defaulting to "premium is better."

When simple R-addition breaks down

The R = d/λ and R_total = ΣR method assumes homogeneous layers with heat flow perpendicular to the layer (the "combined method" in BS EN ISO 6946). It does not directly apply where:

Frequently Asked Questions

What's the difference between R-value and U-value?

R-value (thermal resistance) measures how much a layer or build-up resists heat flow — higher is better, units m²K/W. U-value (thermal transmittance) is the reciprocal of the total R-value including surface resistances — lower is better, units W/m²K. UK Building Regulations Part L set targets in U-value; insulation products are more often compared and specified by R-value because it's the additive, layer-by-layer figure.

If a roll of insulation says "R 2.65" on the packaging, is that the whole story?

It's the R-value of that product at that specific thickness, tested to a declared λ. It does not account for surface resistances, other layers in the build-up, compression once installed, or thermal bridging from structural elements. Use the packaging R-value as one input into the full calculation, not as the final answer for whether a wall or roof complies with Part L.

Can I just add up the R-values on the packaging of two different products to get my total?

Yes — this is exactly how R-values are meant to be combined, provided both products are correctly installed at their stated thickness (no compression) and are genuinely in series in the direction of heat flow. This is the calculation used in the loft top-up worked example above.

Why does PIR cost more than mineral wool if they can achieve the same R-value?

PIR's lower λ means it achieves a given R-value in roughly 40-45% less thickness than standard mineral wool, which matters where space is constrained (flat roof upstands, floor build-ups, loft hatches, dormer cheeks). Per unit of R-value achieved, PIR is usually more expensive than mineral wool — the premium buys thinner construction, not necessarily better ultimate performance per pound spent.

Does R-value account for air leakage (draughts)?

No. R-value and U-value are both steady-state conductive heat loss calculations — they say nothing about air leakage through gaps, poorly sealed loft hatches, or uncontrolled ventilation. A well-insulated but draughty building can still lose significant heat that no R-value calculation captures; airtightness is assessed separately (air permeability testing, Part L requirement for new dwellings).

Regulations & Standards