A little math is all it takes to keep your home warm and cozy
R-value describes the thermal resistance of an insulation material.
To calculate R-value, divide the thickness of a material by its thermal conductivity.
Many common insulation materials have already had their R-values calculated.
Whether you’re in the early or late stages of picking out insulation material, odds are you’ve seen the term R-value being used a lot. This number describes how well a type of insulation does or doesn’t help your home retain heat. While most insulation options will tell you up front what you can expect, there are other factors at play that can impact what level of heat resistance you actually get, like the particulars of your local climate. Use this guide to understand how to calculate R-value, so you’re never taken by surprise.
All insulation materials are assigned an R-value, which is a measure of how resistant they are or aren’t to heat. The higher the R-value, the better an insulation type is at both retaining heat when it’s cold out and preventing it from entering your abode when the exterior temps are high. Not only does this ensure that your space is always at an ideal temperature, but it also reduces the need for excessive heating and cooling which can lower your energy bills in turn.
Calculating for R-value requires dividing the thickness of a material by its thermal conductivity. The number you end up with is the sum of the thermal resistance of each layer of insulation, plus surface air film (or a layer of still air that attaches itself to insulation) and any and all bridged layers—or pockets in the walls where heat flows more easily.
Measuring the thickness of a material is relatively simple, but determining its thermal conductivity isn’t so cut and dry. To get a better picture of how a material will perform in the context of your home, you’ll also need to measure the thermal conductivity of steel frames and thermal bridges, as well as surface air film or surface resistance.
First, calculate the thermal resistance of an insulation material by adding insulation thickness to airspace thickness. That could look like this:
4.20 m2 .K/W (insulation) + 0.220 m2 .K/W (airspace)
In which case, R1 (or insulation’s thermal conductivity) would equal 4.420 square meters.
Next, determine the thermal resistance of steel beams and breaks using the following formula.
Rs = a × 1 + Rc1 + Rc2
Here’s a look at what each number and letter represents:
Rs: Steel frame thermal resistance (m².K/W)
A: Flange (or connecting elements) width (m)
l: Total thickness (depth) (m)
d: Web thickness (m)
km: Thermal conductivity of metal (W/m/K)
Rc1: Contact resistance between the steel frame and the thermal break (m².K/W)
Rc2: Contact resistance between the steel frame and any adjacent material below the frame (m².K/W)
When all the math is said and done, that could look like this:
R2 = 0.500 + 0.208 + 0.220 m2 .K/W = 0.928 m2 .K/W
This means that R2 (or the steel beams and breaks thermal conductivity) would be equal to 0.928 square meters.
Next, you’ll need to calculate the area fraction (represented as f), which describes the ratio of cross sections in relation to insulation and steel beams. Solving for that will involve dividing the width of each individual cross-section by the total width of all of them combined. If cross-sections vary in size, you’ll need to calculate them all separately. That could look like this:
The cross-section area of a region divided by the total cross-section area = f1
Here’s how that will break down in practice:
(1200 × 8000) - (79 × 8000) 1200 × 8000 f 1 (region 1) = 0.934 (or f1)
After that, you’re going to divide each region by its respective area fraction and add the two sums. That will look like this:
R1 divided by f1 + R2 divided by f2 or 0.934 divided by 4.420 + 0.066 divided by 0.928 = 0.282
Finally, divide that sum into 1 to get your total R-value for thermal break, insulation, and steel frame. In this case, that will be 3.54 meters squared.
From there, you’ll also need to add in R-values for siding and surface resistance, but, because they’re often small, skipping this part won’t greatly change the number above.
Phew, that was a lot. The good news is that the pros who install insulation will be more than happy to do that math should you prefer not to. So don’t hesitate to call an insulation company near you to spare yourself the headache.
If you’re scratching your head right now, fortunately, many insulation materials have already had their R-values calculated. Here are the R-values for the most common insulation materials.
| Material | R-Value |
|---|---|
| Cellulose | 2.4–2.8 |
| Fiberglass | 3.0–4.0 |
| Mineral wool | 2.5–3.3 |
| Polystyrene | 2.6–3.6 |
The best R-value for your home will depend on your local climate, including how cold, hot, or moist it might be. Areas are broken down into zones, with one being the warmest and eight the coldest. Plus, each different area of a home has a different R-value recommendation. So which R-value do you need? R-38 to 60 is recommended for attics in zone four, while those in zone one must be within the range of R-30 to R-49.
Poor insulation in a home isn’t the only thing that will reduce its R-value. Everything from windy conditions to unexpected leaks from heavy rainfall can damage the system you have in place and make your abode less resistant to heat loss. That’s why it’s so important to evaluate your local climate and select an insulation material that’s up to the task—whether that’s a waterproof option like fiberglass in a notoriously wet place like Seattle or a fire-resistant option like mineral wool in desert areas.
