FWIW, I found this wire resistance calculator here:
According to this calculator, for 0.75ft of wire, 16ga is 0.003ohms and 12ga is 0.001ohms. Precision is only 3 decimal points. So, yeah, about 2.5 times the resistance although the total resistance at this length is inconsequential for lower loads.
I think most of the resistance is inside the ammeter rather than the leads, anyway. I measured the resistance of my Extech ammeter with 6” 12 gauge leads at 0.035 ohms. A tailswitch with bypassed spring might have 0.01 ohms, but some switches might have more like 0.03 ohms.
A simple graphical tool I made can predict the resulting current with different ammeter resistances in a direct drive situation. Explained here and here.
For example, a bypassed tailswitch with 0.015ohms with a XPL might result in 6.04A.
graph 4.2-x**(.026.01.015) and (.19**(x/1)+2.74)
With a a 0.06 ohm ammeter instead of the tailcap, you would get about 5.1A.
graph 4.2-x**(.026.01.06) and (.19**(x/1)+2.74)
Using a clamp meter the tail resistance becomes negligible, ~0.001 ohms, and the resulting current would be about 6.44A.
graph 4.2-x**(.026.01.001) and (.19**(x/1)+2.74)
For a triple XPL the difference becomes relatively greater, 9.1A for the 0.06 ohm ammeter compared to 14.5A for the clamp meter.
EasyB, thanks for the detailed information and examples. That’s really insightful.
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