Hello Nathan,
The short answer is - just plug them in and listen!
I'll go for the longer answer though
An operating point for a tube is a set of parameters that completely define the operation of the tube. For a dual triode like the E80CC, it is sufficient to have:
bias (cathode voltage)
current (cathode and plate current are close enough to be identical)
plate voltage
loading resistance
heater voltage (we usually ignore this and assume it's in-spec)
For the 12AU7 in the stock circuit, we have:
bias - 1.57V (LED voltage, the grid is grounded)
current - 3.2mA (will vary a little with the 22K plate loads, or be very well regulated with the speedball)
plate voltage - 75-90V (will vary a little with how potent the 12AU7 is)
loading resistance - 22K stock, 15,000,000 Ohms with the Speedball
Next, we can pop open the data sheet of the E80CC and look at the grid voltage curves on the graph of plate current/plate voltage. In the Crack, the LED fixes the bias at 1.57V, so we know we will be on an imaginary line half way between the 1V and 2V line. With the Speedball, we can then place a dot on the graph at 3.2mA and from that dot determine the plate voltage (looks like ~90V, a little high, but not bad). To bump the voltage down, the current could be reduced by increasing the R1 on the C4S board to something closer to 375 Ohms to slide down that imaginary 1.5V line. If you want to "draw the load line", the 15,000,000 Ohms of the C4S is approximately a horizontal line that starts at the 0V curve and ends just before the available B+ voltage (call it 165V). I have drawn this in red on the picture.
If you have the 22K plate load resistors, the process is a little different. On the graph of curves, start by putting a dot at your raw B+ and 0mA (so on the X-axis). This dot represents the point at which the triode has swung so far that it isn't drawing any current. On the other end, the triode would be pulling all the current it can, drawing the plate voltage to roughly 0V through the 22K resistor. Starting with 165V and knowing that this has to be done through a 22K resistor, we can use V=I*R to determine that the max current is 7.5mA. Since drawing 7.5mA through the tube will give you 0 plate volts, you can put another dot on your graph (this time on the Y-axis). When you connect these dots, you get your load line (though it doesn't go past the 0V curve). Where this load line intersects the 1.5V curve is where your plate voltage will settle. I drew this load line in blue. As you can see, the plate voltage with the 22K resistors might be a hair higher than with the stock Speedball.
If you increase the stock plate load resistors, the line will stay tethered to the point at 0mA/165V, but the max current will decrease (as will your plate voltage). I fiddled around a bit on the drawing and just arbitrarily tossed the green line on there, as it looks to hit the imaginary 1.5V line at around 80V. It draws a maximum of 5mA from 165V, and using V=I*R gives us a resistance of 33K.
Just a word of caution - these are hand drawn lines on a very old drawing that was also hand drawn. If you pop 33K resistors into your crack and get 86V instead of 80V, it wouldn't be that surprising.
