Suppose there is a wall and half of its area has R-value 10, and the other half has R-value 20. Is the total R-value of the wall 15? In other words, can R-values be averaged weighted by area?

Please justify your answer with math. R-value is defined as follows on page 454 of the 2020 Residential Code of NY State.

R-VALUE (THERMAL RESISTANCE). The inverse of the time rate of heat flow through a body from one of its bounding surfaces to the other surface for a unit temperature difference between the two surfaces, under steady state conditions, per unit area (h • ft2 • °F/Btu) [(m2 • K)/W].

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    A) If your AHJ is failing you for some voids in the insulation, he's not going to buy an "average R-value" calculation. B) How can the spray foam not be made the same thickness? Usually, it's sprayed into the wall cavity, allowed to cure, then cut off at the stud level, meaning it's all a stud thickness.
    – FreeMan
    Commented Dec 14, 2023 at 15:47
  • 1
    This isn't something we can answer. In modern wall systems R-value is often averaged across framing, windows, etc., but that generally assumes consistent insulation thickness. Whether your inspector approves of your math is anyone's guess.
    – isherwood
    Commented Dec 14, 2023 at 17:19
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    If the problem is that the insulation company didn't put enough foam into each bay to completely fill them, then they need to come back out and finish the job they started. If they cannot (for whatever reason) apply new foam over old, then they need to scrape out the old and start over. This is what you pay them for. This is all tangential to the question of "can you average R-value" and again, if your AHJ isn't going to let you "average" over a couple of not fully filled areas, he's certainly not going to let you get away with "1/2 at R10 and 1/2 at R20 and call the whole thin g R15"
    – FreeMan
    Commented Dec 14, 2023 at 19:42
  • 1
    You need to know what r value is required by code and have the insulation company provide that. There’s no averaging to be done here: r-13 or more means it. Commented Dec 14, 2023 at 20:23
  • 2
    FYI, foam companies spray to an R-value, not to fill the wall. It's common for foam to not completely take up 5-1/2" because that's overkill in some climates. You should verify what R-value is present. The inspector may have a false expectation that the cavity should be completely full.
    – isherwood
    Commented Dec 14, 2023 at 20:28

2 Answers 2


No, R values can't be averaged to find an assembly's effective R value. R value is a local material property, thickness divided by thermal conductivity, and the physics don't allow for simple averaging.

Given a steady state temperature difference ΔT between an insulation assembly's interior and exterior, the heat flux (W/m²) is ΔT/R. Fixing two assemblies with unique areas and R values, the heat transfer rate for each assembly is A₁·ΔT/R₁ and A₂·ΔT/R₂. To compute an effective R value, I want the R₁₂ that combines with A₁+A₂ to yield the same heat transfer rate as both assemblies combined, i.e.

A₁·ΔT/R₁ + A₂·ΔT/R₂ = (A₁+A₂)·ΔT/R₁₂.

Solving for R₁₂, I get

R₁₂ = (A₁+A₂) / (A₁/R₁ + A₂/R₂).

Note that this is not simply the weighted average of R₁ and R₂.

Your hypothetical wall assembly with half R10 and half R20 has an effective R value of 2/(1/10+1/20) = 13.3. When your building code calls for R20 (or whatever), however, it is understood that thermal bridging by wall studs etc. will reduce the effective R factor below the labeled value on your insulation, and that's okay.

A cleaner alternative to R values:

By introducing U = 1/R, area weighted averaging does work for computing effective U values. Recalling

A₁·ΔT/R₁ + A₂·ΔT/R₂ = (A₁+A₂)·ΔT/R₁₂,

eliminating ΔT, and substituting U values, I get A₁·U₁ + A₂·U₂ = (A₁+A₂)·U₁₂ and therefore

U₁₂ = (A₁·U₁ + A₂·U₂) / (A₁+A₂).

For effective insulation values in the OP's case, then, these U values are much easier to use than R values. U values aren't perfectly superior to R values, however. If, for instance, you were to stack R15 and R19 batts, you can add the R values to determine that the assembly has an R value of 34. For U values you would need a 1/(1/U₁ + 1/U₂) computation.

  • 1
    But I think it does work, averaging that is, if you use U-values instead of R-values.
    – SteveSh
    Commented Dec 16, 2023 at 20:38
  • @SteveSh, yeah, but I think this best answers the OP. I'll add a note to the end, though. The other answer touched on it, but it's worth having tidier notation under this answer.
    – popham
    Commented Dec 16, 2023 at 21:19
  • Great answer. Can you add a small paragraph about why it's not simply the weighted average? Commented Dec 16, 2023 at 21:50
  • But it is the weighted average of the U values, not the weighted average of the R values. This phenomenon happens quite frequently when reciprocals are involved. Here's a typical example. Say you're on a 20 mile trip, and travel the first 10 miles at 30 mph, and the second 10 miles at 10 mph. What's your average speed? No, it's not 20 mph, but rather 15 mph.
    – SteveSh
    Commented Dec 16, 2023 at 22:29
  • @End Anti-Semitic Hate, I guess the "why" is because the steady state heat transfer rate isn't linear with R value, where linearity with U value and area makes the weighted average work for U values. I can't demonstrate it with a small paragraph. Just asserting it makes me feel like I'm "baffling with BS." If you have a small paragraph argument, I would love to hear it.
    – popham
    Commented Dec 16, 2023 at 22:29

Consider a flat panel shape, thickness E, area S. For example, a wall.

For the purposes of this calculation we consider thickness to be small compared to the two other dimensions so edge effects are not taken into account.

DeltaT is the temperature difference on both sides of the wall.

P is the power going through the wall (thermal loss).

The meaning of terms Resistance and Conductance are identical to their electrical equivalent:

Thermal conductance Gth = P/DeltaT in W/°K

Thermal resistance Rth = DeltaT/P in °K/W.

(Conductance is G for electricity and U for thermal stuff, I went with G).

These are both defined for the whole wall. For example a wall with thermal conductance 10W/°K (thermal resistance 0.1°K/W) will lose a power of 10W per every °K of DeltaT. That's for the whole wall, not "per unit area".

In a somewhat misleading misuse of vocabulary, "R-value" is sometimes called "resistance", while it actually means "Thermal resistance of a wall having a surface equal to one unit area".

If our wall has a certain Rvalue, it means each unit area (1m²) has a Rth equal to the Rvalue. Thermal losses of each m² of wall are DeltaT/Rvalue, therefore the whole wall of area S has losses P=S*DeltaT/Rvalue, which means Rth for the whole wall is Rvalue/S. Larger wall = higher losses = lower Rth.

If you want to mix different insulations, it is much easier to use conductance instead.

Conductance is simply the inverse of resistance, so a wall of unit area (1m²) has thermal conductance 1/Rvalue, and a wall of area S has S times more thermal conductance. For the whole wall, Gth=S/Rvalue. Losses are P=DeltaT*Gth.

It's the same thing with electricity: when several conductances are in parallel, the result is the sum of all conductance values.

So it is correct to speak of "conductance per m²" (or "cost per m²") because "per" means the total is divided by the area. "Thermal resistance per m²" doesn't mean anything, because the division is the other way around.

If we have two walls of different area S1,S2 and Rvalue1,Rvalue2 their losses add up. The conductance of each wall is Gth1=S1/Rvalue1 and Gth2=S2/Rvalue2, which gives a total conductance Gth=Gth1+Gth2.

Total losses are P=DeltaT*Gth.

If you use resistances, the formula is the same as parallel resistances in electricity, taking the area into account:

Rth = 1/Gth = 1/( S1/Rvalue1 + S2/Rvalue2 )

"Rvalue for the whole wall" = (S1+S2)/( S1/Rvalue1 + S2/Rvalue2 )

Just like a short circuit in parallel with anything is still a short circuit, the dominant part of this is the area with worst insulation. Basically, if half the wall has R-value 5, and the other half of the wall has R-value 1, the resulting R-value of the whole wall is not the average (ie 3) but rather 1.666...

Note it makes no sense to define R-value for a wall that has different insulation on one half and the other half, because R-value is a property of a homogenous material. A more correct way to say it would be "losses would be the same if the whole wall was insulated with R-value 1.6".

If some readers don't like math, think of cost instead. Thermal loss per m² and conductance are just like cost per m². It is correct to make weighted average (by area) of cost or conductance, because these things add, the total cost is cost per m² multiplied by area. To keep the cost analogy, resistance is like m²/€, you can't add nor average that.

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