For short welds, 4AWG will be fine.  However, if you want, you can use aluminium (or if you can get it, copper) bar stock to make your secondary windings (make them in pieces that bolt together).  You only need flexibility for the connection between one side of the secondary winding and the movable welder jaw.  Design for the highest current that you plan to use, but keep in mind what the current will be on the primary side so that you don't go blowing breakers every time you weld (secondary current / turns ratio = primary current).  Current limiting is going to be done on your primary side by limiting the input voltage (a rheostat will do the job nicely).  Likewise, measure your output current on the PRIMARY side, and simply multiply it by the turns ratio.  That way, the actual current that you will be measuring will be a safer level.

-Chris

I see where you are coming from and your train of thought, but the more you reduce the volts the more resistance works against you. Your example of 600A at 1V is impossible to achieve, that would require you to have 1/600 of an ohm in the low voltage circuit. Consider my example with the extension cored where I got 1/20 of an ohm, a far cry from 1/600 of an ohm. Your circuit will have thicker copper wires and possibly a shorter distance, so 1/50 of an ohm is a fair estimate. with the resistance constant we can calculate the amperage with ohms law then the watts.

 5V:   5V / 1/50Ω = 250A  5V x 250A = 1250W

 4V:   4V / 1/50Ω = 200A  4V x 200A = 800W

 3V:   3V / 1/50Ω = 150A  3V x 150A = 450W

 2V:   2V / 1/50Ω = 100A  2V x 100A = 200W 

 1V:   1V / 1/50Ω = 50A    1V x 50A   = 50W

.5V:   .5V/ 1/50Ω = 25A   .5V x 25A   = 12.5W

If there was no resistance then your reasoning is absolutely correct, but its resistance that eats away all your amps and the only way over come this is with increased volts. As said before that 1/50 of an ohm is just an estimate, so you will need to make it and then experiment to find the resistance. To settle this once and for all make the machine with one volt, see how well it makes the spot weld and use the extension cord amp meter to measure the load, then rebuild it with three volts and test again. The guidelines that make a good spot welder are low resistance in the clamp arms, and make it with as high a voltage as possible without exceeding your ingoing Watt limit. Say you were to make a spot welder with 120 volts with the arms straight out of the wall.

120V:   120V / 1/50Ω = 6,000A  120V x 6,000A = 720,000W

In this case you will certainly blow a fuse, but if you had access to that much power then you could weld cars together.           

The poor contact would most likely be the culprit, perhaps it would be best to machine two solid block of copper that would be crimped around the rods. Also could take photos of the complete assembly so that we can get some ideas for troubleshooting?     

Tyler, when I was talking about the turns ratio in your transformer, I didn't mean for you to actually count primary windings, I'm far too lazy to do that! ;-) .  Rather, connect the primary to 120V and measure the open circuit voltage at the secondary.  The ratio of these two voltages IS your turns ratio.  Since you know the secondary, you can calculate (estimate) the actual number of turns in the primary.  You     don't need to be exact with the primary, since it has a lot of turns (plus or minus 5-10 turns is fine).  By the same token, since the number of windings in your secondary is small, adding or subtracting a winding or two, even though it's just one winding, is a HUGE change in the spot welder tuning.  At any rate, once you have an idea of the turns ratio of your transformer, put an ammeter in series with the primary (dear god, DON'T put an ammeter on the secondary!!!).  Measure the current during your test welds and use your measured turns ratio to compute the weld current.  Now, you can start to fine tune your machine by measuring weld current as a function of clamping pressure, welded material thickness, how clean the weld is, etc.

PS, putting transformers in series will not help you here.  At the end of the day, you are producing an AC voltage between a pair of electrodes.  The current that will flow is determined by Ohm's law, just like it is for all basic electricity.  All the transformer is doing is increasing the potential current that is available for the weld.  Just because you hook a 1:100 transformer up to a 15A circuit doesn't mean that you'll get 1500A in your weld.  Don't work forward from the wall plug, work backwards from the metal to be welded.  That circuit has a resistance, and the weld will require a given current to make a successful weld.  The product of the resistance and the minimum necessary current is the minimum voltage that you must put across the weld.  Ignoring losses in the electrodes, this is the same as necessary secondary voltage from transformer.  Since your primary will be energized at 120V, you know what turns ratio you require.  Final step, verify that the current available at the secondary will not blow the breaker on the primary side.  e.g., if you find that you need 20V across your metal to get sufficient current , then your turns ratio will be 20/120 = 1/6.  Assuming that you have a 15A circuit, you would have only 15*6 = 90A available for the weld before you blew a fuse.  If this is too low, then you have two (only one is simple) options.  Option 1: get a more powerful primary power source.  This could be the 240V circuit in your house, or a 20A 120V source.  Option 2: build a more sophisticated circuit that will store energy in a capacitor bank and use this to kickstart your welds (once the weld heats up, it's resistance increases, so your welder will draw the most current at the very beginning of the weld).