> Let the tree have a diameter 0.5m  and cavity height be 1.5m 

Tree volume = 0.29 cubic meters

> ...Cavity volume 0.04 cubic meters

Assuming a cylindrical cavity, if the height of the cavity is 1.5 m and the cavity volume 0.04 cubic meters, then the cavity diameter would be 0.184 meters.  That leaves ( 0.316 / 2 = 0.158) a 0.158 meter wide thick wall of wood surrounding the hive, a very generous 6 inches all the way 'round.  Such thick-walled trees around rotted cavities would be rare, rarer still in an age when even the (USA) national forests have contractors come through and "harvest" trees of a certain size and condition.    I don't buy that 6 inch thick wall as being "typical" at all, and this is adding both excess mass and excess thermal resistance to the mix.

> the height of the thermal envelope= cavity height  + 0.5m up and down =2.5m 

That's a lot of wood above and below - its not clear to me what part all that extra wood plays in being part of a "thermal envelope", nor is it clear what mechanisms are at work to heat and cool that mass of wood that would overcome ambient forces.  Again, this seems excess mass and excess thermal resistance, and this is not thermal mass that is easy to "harness" and "utilize".

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