Understanding "grid drive" to drive, 2, 4, 6 power tubes

[Helmholtz]

There is no load required to be specified to produce that line.

You absolutely convinced me, as already confirmed. The formula used shows that it is all in the plate curve + operating point. The max. power loadline is an output not an input.

Thanks for pointing this out! Always glad to learn something new. I am enjoying your methods.

[Helmholtz]

I was wondering why the traditional methods of longest useable loadline and shooting through the middle of the knee seem to completely fail with Nick's example, while usually working satisfactorily. So I drew some more loadlines using different charts and calculated respective power.
I came to the conclusion that it must a geometrical thing. The traditional methods work best when the plate curve looks as square as possible. This apparent geometrical squareness depends on voltage and current scales chosen.
For instance, the knee appears more pronounced when a 100mA step on the current axis is about the same length as a 100V step on the voltage axis.
So I think the graphical methods depend on suitable axes' scaling.

[bea]

This is a lot to learn and think about so I am still checking it out. All the answers are definitely helping me to understand though so thank you

In the meantime I will say that I am not using 6L6. I like to use 6CA7 power tube.

What about sound city 120 or hiwatt DR201?

My DR201 clone - 4xKT88, i.e. even heavier load than 6 EL34 - can be driven from a simple LTPI with ECC83. Although it sounds better with the ECC81.

Hiwatt's 400W output stage envolves DC coupled cathode followers as drivers (which, btw, i do not succeed to simulate in LTSPICE), and i assume that might be a candidate for reals AB2

... based on simple math: 4 KT88 make more than 200W. How to obtain 400W from just 6 KT88 at the same plate voltage?