Each 100 mm of mineral wool adds a thermal resistance of about 2.5 m²K/W, enough to take a single leaf of brick from a U-value of about 3.4 to about 0.36.
The resistance of a layer is its thickness divided by its conductivity: 0.1 m ÷ 0.04 W/(m K) = 2.5. Resistances in a row add, and the U-value is one over the total, so adding 2.5 to a wall whose resistance is 0.3 divides its heat loss by about nine. Trapped air does the work: mineral wool is mostly air held still.
It is steady, one-dimensional heat flow. Real walls have studs, mortar joints, lintels and corners that bypass the insulation (thermal bridges), and gaps and compression in the wool cost more. Each extra 100 mm helps less than the one before: from 100 to 200 mm saves only about 0.17 W/(m²K) against 3 for the first.
Estimate with the rule, then check it against the calculator that models it properly.
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