Slippery
"My explanation from page 1 until page 5 has remained unchanged, and until the laws of physics change - will remain so."
It certainly isn't the laws of physics at issue, so let's have another look at your first post:
"If you consider two cups of water - 1x 50 degrees celsius, 1x 100 degrees celsius... if you put them over a flame of 200 degrees for exactly one second, the cooler cup of water will absorb more heat (because the temperature split between the two is larger)... The same applies in an engine cylinder."
No. In a cylinder, when you burn a given quantity of fuel 'Q', you get:
ΔU = Q - W
The idealised cycle has a constant volume combustion process and looks like this:
The time may vary by a millisecond or two but ALL the heat of combustion goes into the working fluid because W = 0 during this idealised process [#2 to #3 on the diagram].
Bearing that in mind it is clear that your first post is ill-founded:
"When the ignition occurs, a lower compression ratio engine will have a cooler air/fuel charge in the cylinder - and so it will absorb more energy (which is wasted as exhaust gas heat).
A high compression ratio engine will ignite a hotter air/fuel charge which will absorb less heat. Less energy wasted as heat = more energy transferred to the crank."
To be clear, the energy absorbed during this idealised heat addition process will NOT vary with CR in the way you suggest. No useful work can be done - in theory or in practice - with any of the heat until it HAS been absorbed. The clue is in the phrase 'heat addition process'.
"Of course, it follows then that if you were to have two almost identical piston engines (one low/one high compression) burning exactly the same amount of fuel, the exhaust gases from the higher compression engine would be slightly cooler than the low compression engine."
The exhaust gas temperature would vary with the change in PdV work during the power stroke. Not because "A high compression ratio engine will ignite a hotter air/fuel charge which will absorb less heat".