Re rad output a 30 deg rad does not emit 30/50 or 60% of a 50 deg rad, it emits (30/50)^1.3 or "only" 51.5% of that of a 50 deg rad.
Ah, right. So 30/50 or 60% but then hat 1.3. What does hat 1.3 mean? You're losing me I've already stated that Stelrad uses 52% (rounded from a correction factor of 0.515), but I don't understand how they get that figure. Not saying they are wrong, but what I'm saying is I fail to understand how they calculate that figure as it conflicts with the logic of heat loss calculations used in U values. Perhaps U values are only a guideline are are known to be imprecise?
To be pedantic, W or Wh/h (flow of energy) is power, whereas Wh (quantity) is energy and I think this might be the source of the confusion. So obviously it will take more than twice the energy to heat a room/house by 10 degrees that it would take to heat it by 5 as it's losing heat the whole time. Obviously, if you could stick the house in a Thermos flask and heat it, 10° would only take a little over double as there would be almost no heat lost. But the energy required to maintain the higher temperature (Wh/h) is only double.
So, yes, I agree with you, basically.
Obviously we also overlook the fact that the heating system will struggle as the delta T of emitters will fall as the rooms get warmer and that since even an unheated house will be a few degrees above ambient due to solar gain and incidental gains from the body heat of the occupants and any electrical appliances in use, the first few degrees of warming are essentially already compensated for heat loss.