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Yes it is. You might want to look at Nationals datasheet for the LM317 here & typical applications:
https://www.electro-tech-online.com/custompdfs/2011/01/LM117-2.pdf
I wouldn't use a Darlington. To preserve the overcurrent/overtemperature protections built in to the 317, you want it to deliver most of the current it is capable of. For example; if you want 5A, set up the 317T to deliver ~1A, while the bypass PNP transistor delivers ~4A. To do that, select the Emitter to Base resistor so that the drop across it is ~0.65V at the desired 317T current.
The Power calculation in the resistor goes like this: P =E^2/Rb, where E=0.65V
Ok, so the 1 ohm resister both creates a voltage differential between base and emitter to turn on the transistor AND ant the same time limits the current flowing through the LM317?
Yes to the first part; no to the second. Look at the Brown and Red Plots which show the currents through the PNP and the 317, respectively. The value of R5 (and the Hfe of the PNP) is what determines the ratio of Ic(Q2) to I(X). Note the sum of the two currents is the load current.
Cool.
How would I work out how much heat I need to dissipate in the LM317? is it a factor of current flowing through it vs. the voltage differential between in and out?
Yep, now do the calculation as to how many square inches (square cm?) of black aluminum it takes to get rid of 15W.
You can do the regulator dissipation more directly: P = I*E = Iload * (Vin-Vout) = 1 * ( 20-5) = 15W