[Presumably the question was, "How do you figure out the correct operating point for a triode?"]
Perhaps surprisingly, that’s not a really easy question to answer. I’ve been playing with the theory part of the question (with only slight forays into the subjective sonics part) for several years now. I’ll try to get you started though.
First, I think it’s important to include load impedance as well as plate current and plate-to-cathode voltage in what we call “operating point”, at least for circuits where the output signal is a large fraction of the quiescent voltage or current.
I’ll restrict the rest of this to triodes with resistive loads, to keep it simple.
The specs will give a maximum plate voltage, dissipation, and current. You’ll have to stay within these limits or risk a shortened tube life. Operating well below the maximums will increase life somewhat but reduce available power, voltage, or current – a good trade for preamps but not power amps, in general.
The parameter I find most useful is the “operating resistance”, plate voltage divided by current. It must be larger than 1.5 times the plate resistance if the grid is to be kept negative, and in practice a minimum of about 3 times plate resistance is desirable. For tubes that are not very linear, a low value (such as 3) for this ratio will keep the operation in the high current area where distortion is relatively low.
Highly linear tubes are less prone to distortion at high voltage and low current. A higher ratio of operating resistance to plate resistance will increase the available output voltage and power efficiency. The disadvantages are reduced current drive available, and a higher impedance load (more susceptible to capacitive problems such as loss of highs, RF pickup, etc). A very linear tube such as a 300B may be operated at a ratio as high as 10, but 5 to 6 is usually as high as you want to go with most tubes.
Once you’ve chosen your operating voltage and current, you need to find a load impedance. A good approximation is the operating resistance minus 2.4 times the plate resistance. I derived this in a 3-part article in VALVE; I won’t try to cram the math into this post though!
You can look up the bias on the plate curves. There’s a formula but it’s not simple. A rough estimate for the cathode bias resistor is load impedance divided by mu.
[edit: here are links to the three parts of the article paul mentions:
http://www.bottlehead.com/VALVEarchive/1998/va031998.pdf (starts on page 6)
http://www.bottlehead.com/VALVEarchive/1998/va041998.pdf (also starts on page 6)
http://www.bottlehead.com/VALVEarchive/1998/va051998.pdf (starts on page 7)]