Design follows applicable thermal-protection requirements, including Russian SP 50.13330.2024 where relevant. Climate, building use, layers, moisture conditions and thermal bridges matter. A fixed 50 or 100 mm is not a universal solution. СП 50.13330.2024 ↗
Rслоя = δ / λрасч
Rоднородной конструкции = Rsi + Σ(δi / λi) + Rse
- Determine required thermal resistance for the climate and building use.
- Record all layer thicknesses and choose conductivity for the applicable service conditions.
- Assess effective resistance including studs, brackets, lintels and fixings.
- Select thickness, then check moisture accumulation, internal surface temperatures and junctions.
Illustrative arithmetic example
For a hypothetical uniform section requiring R = 3.2, with existing layers and surface resistances providing R = 1.7 m²·K/W, the deficit is 1.5. At a hypothetical design λ = 0.040 W/(m·K), δ = 0.060 m, or 60 mm. This explains the arithmetic only; thermal bridges, moisture and junctions must still be assessed.