How do I calculate the power that this resister will need to handle? Do I just figure the full rail voltage across the voltage divider formed by this resistor and the load?

Such that:

rails are 20 volts

load is 8 ohms

emitter resistor is .33ohms

8 / (8 + 8.33) = .96

.96 * 20 = 19.2v across load

leaving .8v across the emitter resistor

(.8 * .8) / .33 = ~ 2 watts

since this is peak watts, I imagine this is a pretty conservative value in practice, right? Of course 8 ohms is a light load compared to what may actually be encountered too.

What is the correct way to compute this, and what is a safe margin to use?

Yes that's pretty conservative as it assumes continuous full power, which of course is not realistic.

I would do it like this, assume continuous power of say 5W (which is also already conservative), and from that calculate the load current assuming lowest load impedance like 4 ohms.

This current also goes through the Re's so that gives you the dissipation.

Since there are two Re's each takes half of the calculated power.

Then add a safety factor of your liking.

But aside from the power spec there is also the max current spec and that's really the one you need to select those resistors. Even with low continuous power, they should be able to withstand at least a current of Vsupply/Vload, plus a safety margin.

So it will be the current requirement that is more critical than the power dissipation requirement.

jan