changing the OT primary impedance would alter the idle current of a cathode-biased tube
He did indeed, here's where I made the left turn;
looking at Tubewell's loadlines, Instead of looking right at the intersection of the 3 examples, I "followed" each example to the left side where each line "points" to a different plate current. ...
Got it! Now this, we can work with.
As I'm using the term, "operating point" is the same for all 3 loadlines: 340v plate, 250v screen (these curves are plate currents obtained when the screen is at 250v; high/lower screen volts results in higher/lower plate current), and -15.5v bias (however we got there).
Unless the 6V6 is driven by a transformer or a
power amp able to drive current to the grid, the 0v grid curve is our boundary for max clean output power.
Now look along the light green loadline:
It runs into the 0v grid curve somewhere between 25 & 50 volts, maybe 38v. So the 6V6 can only pull its plate as-low as 38v, or 340v - 38v = 302 volts peak. Plate current rises from idle at 35mA to ~95mA, or 60mA peak. CHECK: 302v / 0.06A = 5033Ω, so pretty close.
From here we can estimate the power output, 302v peak x 0.06A peak / 2 = ~9 watts. EXCEPT! This is Class A, so the current-change for higher plate current should be equal to the current-change for lower plate current. But 35mA-to-90mA is much more than 35mA-to-0mA, so we know the output tube will distort well before reaching 9w.
We can take a shortcut to power output by calculating the entire peak-to-peak plate current swing from 0mA up to 90mA: 90mA * 5kΩ = 450v peak to peak. So 450v p-p x 0.09A p-p / 8 = ~5 watts and much more reasonable.
We would pull back in our memory-banks to know that a waveform squashed on one side has even harmonic distortion, and estimate this 5kΩ load has a lot of even harmonic.
Odd harmonic distortion is a sine wave squashed equally on both sides, which would happen when our drive signal runs into the 0v grid curve, as well as the 0mA axis. The idle bias is -15.5v so the 0v grid curve is when the grid input signal is +15.5v peak. The 0mA axis is not labeled, but is surely -40v or a bit more: 40-15.5v = -24.5v peak. So distortion starts with a grid input of 0.7071 x 15.5v = ~11v RMS, and odd harmonic gets significant beyond 0.7071 x 24.5v = 17.3v RMS
You can also take a shortcut to the realization above by seeing the peak positive plate current happens when the grid receives a +15.5v peak signal (to reach the 0v grid curve). So look to see what plate current does with a -15.5v peak, or the grid at -31v: the green load line is down to 4 or 5mA at -31v on the grid, but is not cut off. This fact plus the dissimilar current-change indicates distortion, and a "non-optimum load."
Again, our lines & calculations are somewhat useless, because the actual amp has a screen voltage other than 250v. We'd have to make our own custom plate curves using the existing graph and estimating plate current for the different grid voltages at a different, higher screen voltage. RCA published an Application Notes on how to do that (back in the 30's I think).