One other observation : I'm using a V-OFFSET of 0.225 volts. That means that rather than supplying the VID voltage of ~1.145 volts for my particular chip, the motherboard is supply 0.225 more.
While I understand what it is that you are aiming to communicate here, just be aware that the term "V-offset" has a specific meaning in regards to a motherboard unintentionally over-volting or under-volting a cpu.
There are two things that keep your cpu from "getting" the voltage you specify in the BIOS - the first is Voffset and the second is Vdroop.
Voffset would be something akin to voltage regulator error. You specify in the BIOS that you want the CPU to get 1.5V but the actual output by the VRMs is 1.4V or 1.6V (a 0.1V offset in the specified voltage).
What you are referring to as a "V-offset" is not really given a specific term, we just call it the over-voltage. You are over-volting your chip by 0.225V, presumably by specifying a voltage of 1.37V in the bios?
Or do you really mean to say your motherboard actually has a 0.225V Voffset? (that is insanely huge for a Voffset, RMA your board territory)
On your chart, it takes more like 0.45 volts of offset to go from 2.8 ghz to 4.0 ghz. That means that this same graph for my CPU has a shallower slope. (is the rate of rise for a complex polynomial graph called slope? don't remember) Why is that?
Mathematically speaking it is called slope. You are thinking of "linear slope" for which the answer would be no - we don't usually speak of linear slope with functions that have a non-zero second derivative (i.e. it curves, there is a slope to the slope
The graph is CPU specific, my chip will have a different shmoo from yours and from anyone else. Like snowflakes, they will be similar and always have the same general form of increasing from left to right.
