It all derives from the 3/2 power law for tubes - the current goes as the voltage to the 3/2 power. It's a theoretical rule, based on good physics but assuming a lot of real-world effects are not important. From that, if you double the voltage then the current goes up by 2.828 times, hence 2mA becomes 4.25mA. Since resistance is voltage divided by current (Ohm's Law), the 8K load becomes 5.6K.
I post my rule for SE outputs every so often in various forums - the formula is:
RL = Rb - 2.38*rp
RL is load impedance, Rb is beam resistance (plate to cathode voltage divided by plate current), rp is plate resistance. The triode-mode 3S4 at 22v/2mA has a plate resistance of about 4K ohms.
This rule looks odd, but the physics behind it is fairly simple. I assume you have a graph of the plate curves at hand, and you have a decent idea of what a load line looks like. Draw a load line, and note the current at zero grid bias. Then find the place on the load line where the current is half that - that is the quiesvcent operating point. At the low-current end of the load line, you can find the point where the grid bias is twice that of the quiescent point; it will be a bit more than zero current - this keeps operation out of the region near zero current where the curves are all scrunched up together. That's it; I just did the algebra using the 3/2 power law.
