OK, I'll wade into this Holy War... (and where's Mark Spaeth when you need him?)
Before discussing the conclusions you reach, I'd like to discuss the electrical logic and calculations used to justify that conclusion--as I feel some aspects are flawed.
The first diagram (
http://bitslicer.tripod.com/images/ar_images/ar_sense_loop1.jpg) is either incorrect or very deceiving. It shows a current path with two different values at different points on the path; which cannot happen (where did the missing current go?) Perhaps you meant to show ~6A going out the +5 line, ~6A returning on the GND line, with ~0A going out the +SENSE line, and ~0A returning on the -SENSE line.
The 2nd diagram, I follow, but I think you meant "I=~6A" to be consistent with the text (which says "...up to 6 Amps for games like..."), and to be consistent with the last diagram which shows the current splitting: 3A going out on each the +5 and the S+ line.
The power calculation, "P = I*V = 6A * 5V = 30 Watts", while mathematically true, does not appear to be electrically correct. In P=IV, the V must be the voltage drop across the component for which the calculation applys. There's no reason to assume that the voltage drop across the sense resistor is 5V (that's instead the desired voltage drop across the _PCB_, so your calculation shows the power consumed by the PCB, not the sense resistor). In fact, following your example that all current must pass through the sense resistor: The voltage drop across a 10ohm resistor with 6A flowing through it must be V = IR = 6A*10ohm = 60V, and then the power would be P=IV=6A*60V=360W. But all of that is pointless as the ARII cannot supply 60V. It would top out well before that. And since the voltage can never reach 60V, the current will never reach 6A and the power dissappated by the sense resistor would never be anywhere near 360W (or even 30W).
My analysis of an ARII with a dirty connector is explained below and in the attached diagrams and spreadsheet.
The supply is assumed to go through two parallel paths: the sense resistor (R29, 10ohm) and the "supply harness." In reality, of course it's more complicated, as there are multiple supply lines, and there is contact resistance in the sense line as well; but we'll assume a simplified case where only one contact resistance is considered (refered to as "supply harness resistance") and everything else is a perfect connection. The attached spreadsheet starts with the supply harness resistance in the first column. A range of assumed harness resistances are shown (one per row) and the remainder of the operating characteristics are calculated for each assumed harness resistance. From there, current passes through the game PCB. We'll assume it acts as just a resistive load. For purposes of this example, it is assumed to draw 6A at nominal 5V operation, and thus has a resistance of 0.833ohm (5V/6A), this is called "PCB load resistance." From there, current again forks, going through parallel paths of the return sense resistor (R30, also 10ohms) and the return harness reistance. While my spreadsheet originally contained columns for these, they are hidden in the screenshot posted here. This is to simplify matters, the return harness resistance is assumed to be 0 so all current passes through the the harness with no loss. So as described before, I'm assuming perfect connections except for the one supply harness resistance. The net resistance for the parallel resistors is calculated, R1*R2/(R1+R2) to get the "supply net resistance." (The same is done for the return, but again that is hidden as it is assumed to be zero). Then the supply net resistance and the PCB resistance are added to get the total circuit resistance.
The next column (I) is the nominal (desired) PCB voltage drop. This is, of course, the desired 5V. Column (J) is the nominal PCB current; i.e. the current needed to cause the nominal PCB voltage drop of 5V. Of course, this is always 6A (since this was the assumption we used to calculate the PCB resistance in the first place). Next, colum K, is the supply voltage required in order to sucessfully generate the desired 5V drop across the PCB. This is calculated by multiplying the nominal current (column J) by the total net resistance (column H), as in V=IR. This shows what the supply voltage would have to be in order to overcome the given harness resistance and provide 5V to the PCB... which is what the ARII tries to do. You can see that some of these supply voltages, as the harness resistance gets higher, are a bit absurd. The ARII board only gets 15V unregulated from the power block, and the LM305 datasheet recommends the output and input be no closer than 3V. So the next column (L) is the "max supply voltage." I've assumed it to be 12V here (15V-3V). Maybe it's higher, maybe it's lower--I don't really know what it'll top out at, but it's likely at least 7V but no more than 15V. Then column M is the "actual supply voltage", which is simply the "supply voltage required" (K) unless that exceeds the "max supply voltage" (L) in which case it is the latter. Column N is the actual supply current, calculated from the actual supply voltage (M) and the total net resistance (H). Note that as the harness resistance increases, the supply voltage increases and the supply current stays the same... until the max supply voltage is reached. Then, the supply voltage stays the same and the supply current drops off. Columns O & P show the voltage drops across each section of the circuit; the supply (both harness and sense resistor), and the PCB. Columns R & S show the currents in each branch of the supply: the sense resistor and the harness. Finally, columns V, W & X show the power consumed in the sense resistor, harness, and PCB.
[EDIT: BAH! The forum is resizing my attached images so they are unreadable... anyone know how to prevent that?]
Temporary links to full-size images:
http://www.finck.net/gfx/diagrams.png
http://www.finck.net/gfx/spreadsheet_calcs.png
