I haven't been able to get this info for some reason.
If I used 845s (or equiv) in P/P with 300B (or equiv as drivers), what
kind
of wattage out can I expect

About 110W using full feeding voltage (1250V). Class A will be about
40-45W. A higher anode load will give less power but more class A.
and how much of that will be class A?

Also besides just giving me a figure, can you explain how you go about
to
attain it?
Draw the load line on the anode curves chart. Locate the idle point and
the
points corresponding to the maximum voltage swing. Get the relevant
values
of I and V. Now You have all we need to calculate the amplitudes of two
sine waves, I and V. To get the power, we need to multiply the RMS value
of
voltage times the RMS value of current.
RMS V at full power is: (V at Vg=0) - (V at idle) * 2 ('cos the other half
wave is provided by the other tube) * / (2*sqrt 2)
RMS I at full power is: (I at Vg=0) - (I at idle) * 2 ('cos the other half
wave is provided by the other tube) * / (2*sqrt 2)
To get the Class A power, we need to consider a smaller voltage/current
swing, ie, the one limited by the cut-off point (the one where the
loadline
crosses the zero current axis) and its symmetrical one with reference to
the
idle point.
RMS V class A power is: (V @ "symmetric point") - (V at idle) * 2 (2
tubes)
* / (2*sqrt 2)
RMS I class A power is: (I @ "symmetric point") - (zero) * 2 (2 tubes) * /
(2*sqrt 2)
If You print the anode curves and play a bit with pencil & ruler, it
becomes
apparent.
Ciao
Fabio

Thanks Fabio, that helps me. You are a good friend and I wish you luck in
your project. It sounds ambitious.
west