HBP, you are not returning my calls! 
Sorry! Power went out Friday during storms, didn't come back on til Monday (not good when the heat index is over 100). Then lost power again briefly this morning. No internet until this morning.
My next question is: As the grid bias gets more negative, what happens to the grid driving voltage? And what is a knee?
Easy question first. The knee is shown in the attached pic.
Simple answer to "
As the grid bias gets more negative, what happens to the grid driving voltage?" Well, you apply a bigger grid driving voltage.
This really has more to do with "what happens as plate/screen voltage gets bigger?"
If you were designing for a lower power output range from your output tubes, you might use a smaller B+ (like 250v) and select your OT load to provide the maximum output power while running class A. Using the previously linked data sheet's number as an example, 250v plate and screen allows the use of -16v bias, resulting in 60mA of plate current at idle. That's 15w plate dissipation at idle (I tend to believe the 6L6GC data sheet used the same "Typical Condition" numbers in some cases as the original metal 6L6 data sheet). A 5k load yields just under 15w of output power at full tilt. If you look hard at that data sheet example, "Peak AF Grid-to-Grid Voltage" is given as 32v. That's 16v * 2, or enough input to drive each grid momentarily to 0v during that grid's positive peak.
But you want more output power than class A will give. You want to raise plate voltage and plate current to get more power, but if you exceed the class A plate current, you have to dial down the current for part of the cycle to keep from redplating. So you jack the B+ up to 450v plate, increase the bias to reduce idle current, and run class AB. Data sheet says run 450v plate, 400v screen, and 58mA idle plate current (that's 26w, so they're still a hot class AB) by using -37v bias. Note this condition's grid-to-grid driving voltage is given as 70v, just below -37v times 2. Sheet says you'll get 55w output with a 5.6kΩ load.
So in general, bigger outputs require larger bias, but also require more driving signal from the phase inverter to reach full (and now higher) power output.
Getting back to the original question, look again at the other two conditions given for class AB in that data sheet (not the 450v plate condition). Voltages, bias, grid driving voltage are all the same. The difference is one has the load impedance almost halved, and the power drops from 26.5w to 18w. From their numbers, distortion stays the same (THD, so the balance of 2nd and 3rd may have changed somewhat), and the only other significant change is plate current drops slightly while screen current rises slightly.
I think way back I said if you change load impedance (up or down) from the ideal, you'll get either less power, more distortion or both.