Anyone in for some electronic theory?
Assuming the ammeter (actually a tiny voltmeter) needs .5 volts for a full deflection (-.5 for one way, and .5 the other way--remember there are 2 extremes--discharge and charge)
Anyway, about .5 volts for full deflection. This means at the "extreme" condition, you need .5 volts going to the ammeter.
Assuming "full-load" in your car is 60 amps, the shunt resistor needs to have .5 volts across it when 60 amps are flowing thru it to provide the meter in your dash with enough potential.
OHM's LAW E=I*R
.5 volts divided by 60 AMPS = .0083 ohms
The shunt resistor would have to be about .0083 ohms--barely any resistance.
You also have to consider the equation for power for safety on that resistor.
P= I*E 60 amps (max) X .5 volts (max) =
30 WATTS MINIMUM rating on the resistor.
So you'd need a .0083 ohm resistor rated at at least 30 watts.
On the later chevelles the ammeter connected to the battery junction and the horn relay for each wire. The wiring between them served as the "resistor" that had a voltage drop across it to feed the ammeter.
I DON'T know the specifications on you guy's cars. Those #'s are examples--I take no responsibility on the accuracy. Just to give you an idea.