Alrighty YOU engineering types... I heard that heat-loss can be generalized by multiplying a constant we'll call K by the square of the temperature differential of the current inside/outside temps. This being the case, the function representing this generalization for my home (heat loss figured at 50K at zero degrees F outside) is f(x) = (50000/70^2)(70-x)^2. If this is accurate, one could use this to determine daily coal needs depending on the temps outside (not a windy day of course) to see if one is burning (ballpark figure) what they should be or maybe more than they should be.

Any takers on this thread?

I determined (roughly) that I'll need 6600 lbs of coal for the year including about 2000 lbs for DHW.