Insulation Calculator

Written byOleh Soloninupdated 25 September 2026

Area to insulate, m²

Layer thickness, mm

Pack volume, m³

Thermal conductivity λ, W/(m·K)

From the product data sheet: mineral wool and rigid foam differ.

What you are insulating

Heating degree days, °C·day/year

Degree days for your town; 0 skips the comparison. The coefficients come from Russian code SP 50.13330.2024, table 3, for dwellings.

Advanced options

Density, kg/m³

R of the existing structure, m²·°C/W

Thermal resistance of the wall or floor without the insulation, if you are adding to it.

Result

Volume, m³8
Packs27
With a spare28
Mass, kg280
R of the insulation layer, m²·°C/W2.5
How it was calculated
80 m² × 100 mm ÷ 1000 = 8 m³ layer R: 0.1 m ÷ 0.04 W/(m·K) = 2.5 m²·°C/W

Counts the material and its effect

Insulation volume, number of packs, mass, and the thermal resistance the finished layer will deliver.

The volume of insulation is area times thickness: 80 m² with a 100 mm layer is 8 m³, which is 27 packs of 0.3 m³ and 280 kg at a density of 35 kg/m³. That part is pure geometry.

The number that matters, though, is thermal resistance: R is the thickness of the layer divided by the conductivity of the material. A 100 mm layer at λ = 0.04 W/(m·K) gives 2.5 m²·°C/W; the same 100 mm at λ = 0.032 gives 3.13. Enter the heating degree days and the calculator compares the result with the base requirement of Russian code SP 50.13330.2024.

How to use

Enter the area to insulate and the thickness of the layer in millimetres.
Take the conductivity λ and the density from the product data, and the pack volume from the label.
For a comparison with the code, add the heating degree days for your town and what you are insulating; if you are adding to an existing wall, enter its resistance too.

Layer resistance against the requirement

Every row is a separate run of this calculator. The requirement follows the formula in table 3 of SP 50.13330.2024 for dwellings.

Every value was produced by the calculator on this page.
Structure	Thickness	λ	Volume	Packs	Layer R	Degree days	Requirement	Result
Wall	100 mm	0.04	8 m³	27	2.5	—	—	—
Wall	150 mm	0.04	12 m³	40	3.75	—	—	—
Wall	100 mm	0.032	8 m³	27	3.13	—	—	—
Wall	100 mm	0.04	8 m³	27	2.5	6,000	3.5	short by 1
Wall	150 mm	0.04	12 m³	40	3.75	6,000	3.5	above it
Attic floor	200 mm	0.04	16 m³	54	5	6,000	4.6	above it
Wall with R 0.8	100 mm	0.04	8 m³	27	2.5	4,000	2.8	3.3 in total — enough

Where the requirement comes from

From table 3 of SP 50.13330.2024: for dwellings, hotels and hostels the base required thermal resistance is a × degree days + b, where walls take a = 0.00035 and b = 1.4, roofs and floors over drive-throughs a = 0.0005 and b = 2.2, and attic floors and floors over basements a = 0.00045 and b = 1.9.

At 6,000 °C·days that is 3.5 m²·°C/W for a wall, 5.2 for a roof and 4.6 for an attic floor. A 100 mm layer does not get a wall there: the calculator says it is 1 m²·°C/W short and that 140 mm would meet the figure. If you build outside Russia, use your own national requirement — the arithmetic of R does not change.

What the calculator leaves out

The wall itself. Brick, aerated concrete and a floor slab all resist heat too, and if you know their R, put it in the "existing structure" field: 100 mm over a wall with R = 0.8 adds up to 3.3 m²·°C/W and clears the 2.8 required at 4,000 degree days.

Thermal bridges are not in there either — steel brackets, window reveals, corners. The reduced resistance a designer has to verify always sits below the sum of the layers, and by how much is a job for a thermal calculation, not for thickness divided by λ.

What this calculator shows

Volume and pack count, with a spare figure
Mass, which matters for framing and for carrying it upstairs
The layer's R-value, to compare against the requirement where you live

What to keep in mind

R is computed from the conductivity you enter; the default is 0.04 W/(m·K).
Thermal bridges through the frame are not modelled; how much they cost depends on the construction and needs a separate thermal calculation.

FAQs

How much insulation for 80 m²?

A 100 mm layer is 8 m³: 27 packs of 0.3 m³ and 280 kg at 35 kg/m³. With a 5% allowance, 28 packs.

How thick should wall insulation be?

It depends on the climate. Under table 3 of SP 50.13330.2024 the base requirement for a dwelling is 0.00035 × degree days + 1.4: at 6,000 °C·days that is 3.5 m²·°C/W, so a mineral wool layer at λ 0.04 needs about 140 mm before the wall itself is counted.

What are heating degree days?

The difference between the indoor temperature and the mean temperature of the heating season, multiplied by the length of that season. The figure comes from climate data for your town.

How much does conductivity change the answer?

Inversely: 100 mm at λ 0.04 gives R 2.5, at λ 0.032 it gives 3.13. Take λ from the data sheet of the actual product rather than from "mineral wool" in general.

Should I count the existing wall?

Yes, if you are adding to it. Enter its resistance: 100 mm of insulation over a wall with R 0.8 comes to 3.3 m²·°C/W — enough for 4,000 degree days, where the requirement is 2.8.

How much does the insulation weigh?

Volume times density: 8 m³ at 35 kg/m³ is 280 kg. Ventilated facades and roofs take denser boards, and the weight rises in proportion.

Worked example

An 80 m² wall with a 100 mm layer

Input: 80 m², 100 mm, λ 0.04 W/(m·K), 0.3 m³ packs

Output: 8 m³, 27 packs, 280 kg, layer R 2.5 m²·°C/W

Note: At 6,000 degree days the base requirement for a wall is 3.5 m²·°C/W, so the layer is short.

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