The **broken link removed** has an operating voltage range of 28.0V to 53.0V, but I'd still like to regulate it to 48 if I can.
According to the datasheet you linked, it's good for up to 56V.
Is 38V across both secondariness the measured voltage or the voltage rating?
If it's the voltage rating then you're sensible regulating because the voltage will be 10% to 20% higher under light loads.
If it's the measured voltage then a regulator is probably not needed.
If I'm understanding things correctly... would I use R1=240Ω and R2=9KΩ with the ~50V input to get a regulated 48V out?
The input needs to be 3V> than the output to ensure good regulation.
The diodes will drop 1.4V
38√2 = 53.74V - 1.4 = 52.34.
The minimum input voltage for the regulator to regulate properly is:
52.34 - 3 = 49.34V
The maximum allowable ripple on the filter capacitor is:
49.34 - 48 = 1.34V
To calculate the filter capacitor size use the following formula:
[latex]C = \frac{I}{2F \times V_R}[/latex]
Vr is the maximum ripple, I is the current and F is the mains frequency, assuming 50Hz mains frequency:
[latex] C = \frac{I}{2 \times 50 \times 1.34} = 0.0056 = 5600 \mu F[/latex]
I'd recommend using a 6800µF capacitor which needs to be rated to 60V and will be very large.
Also, since the LM317 would be dissipating ≤2W of energy, it should be fine with its built-in heatsink, correct? I may put a small snap-on heatsink on it as well, but just want to make sure I've got things in the right perspective.
No, you need a heatsink.
Look at the datasheet.
https://www.electro-tech-online.com/custompdfs/2009/07/LM117-1.pdf
The maximum temperature rating is 125°C.
The thermal resistance from junction to ambient is 50°C/W, even at 2W the temperature rise would be 100°C which means if the ambient temperature is over 25°C it will overheat.
In this case the power dissipation is:
Voltage across regulator × current through regulator:
52.34 - 48 = 4.34×0.75 = 3.26W
Suppose the maximum ambient is 25°C, the maximum temperature rise is 100°C
The maximum thermal resistance between the IC and ambient is:
100/3.26 = 30.7°C/W
The thermal resistance of the die to the case is 5°C/W (see datasheet).
The maximu thermal resistance of the heat sink plus the thermal resistance of any insulating tab is:
30.7 - 5 = 25.7°C/W.
For example if the insulating tab you want to use has a thermal resistance of 6°C/W then you'll need a heatsink with a thermal resistance of less than 19.7°C/W.
Note that the above calculations assume the regulator is an LM317 in a TO-220 (T package on the datasheet) made by National Semiconductor. Always check the datasheet from the supplier where you've bought your IC from, it might differ slightly.
I assume I should connect the grounds post-rectification (that is, connect on the DC side of things) right?
Yes.