Basically, yes. But to be somewhat pedantic about it, the breaker gets sized based on the LESSER of either the wire AWG or the circuit's intended draw. In turn, the wire AWG should be chosen based on the load
AND the run length..
Here's some pix
The white cable is marked 8/2 with ground.
I don't see info on the black cable
Unless I'm misunderstanding your application, the single-conductor black cable is unnecessary and irrelevant.
Per this handy-dandy chart Google kindly dug up for me:
http://lugsdirect.com/WireCurrentAmpacitiesNEC-Table-301-16.htm
the AWG 8/2 cable you already have
MIGHT be just barely adequate, depending on two other factors: Namely, the aforementioned run length, AND the insulation rating of that particular cable (which should also be marked on the jacket -- if it isn't, you have to assume the worst). As you will note from the chart, that temperature rating needs to be 75-deg. C. or better for an AWG 8 cable to be acceptable for a 50A load. The run-length issue is a little fuzzier; while there are probably tables and such for this sort of thing SOMEWHERE, the simple rule of thumb is that you don't want to allow more than 3% voltage drop over the length of the run at full load (and less is always better). The standard formula to calculate this is:
CM = K x I x L / E
where:
CM = Circular Mil area of Conductors
K = 10.75 (a constant representing the resistance of copper)
I = Current in Amperes
L = Length in Feet
E = Voltage drop at load (in Volts)
The cross-sectional area of all standard AWG sizes can be found at
http://en.wikipedia.org/wiki/AWG#Tables_of_AWG_wire_sizes. Looking there, we find that AWG 8 has a cross-sectional area of 16,500 CM.
Re-arranging the formula slightly (since we're solving for Length in this case and have a known AWG), we get:
CM * E / K / I = L
So, for AWG 8 with a 50 Amp load, we get:
16,500 * 7.2 / 10.75 / 50 = L
118,800 / 10.75 / 50 = L
11,051.2 / 50 = L
221.0 = L
VERY IMPORTANT NOTE: This calculated maximum length is for the entire ROUND TRIP that the electrons must take from the distribution panel, to the load, and back again. So for anything resembling "normal" wiring, you need to divide that by two to get the maximum one-way run length. In your case, that's 110 feet. Arbitrarily allowing about ten feet for the welder's own power cord, call it 100 feet (unless you're hard-wiring it, of course).
30 amp all you need for a tomb welder
Too broad a generalization; and in this case, dead wrong. The model cited by the OP requires at least a 50A circuit.
I guess it's not romex then
It could be.
As already noted, "Romex" is a brand name which often gets (mis-)used as a generic term for all similar products, like the aforementined "Kleenex" and "Xerox" (also "Band-Aid", "AstroTurf,", "Baggies", "Dumpster", "Kitty Litter", and many more). The correct generic term is "NM-x", where the "NM" stands for "Non-Metallic" (in reference to the outer jacket), and the "-x" is a letter code indicating the temperature rating of the insulation.
Further, and as also noted by others, while the smaller gauges of NM-x use solid conductors, the larger ones DO use stranded wire, at least for the main current-carrying conductors. "Norcal" said that the switchover point is at AWG 8, and I have no reason to disbelieve him.