Seeing some evidence of heat damage here too.
View attachment 1715690524
I would put a wanted add for a good clean '63 connector and ammeter. The ammeter may be the same as other years and models.
The connector may is a little different than '65 but may be the same as other '63 models.
While your fixing the system, I would measure for resistance in the key switch, as well as voltage regulator to ground, the alternator to ground, and the ground connection engine to the battery negative cable.
Any measurable resistance in the ammeter is not good.
If you must reuse the damaged one, it could be checked on the bench by putting a current through.
To do this you'll need a 12 to 14 V power supply (a car battery will work) and something that will draw 5 or 10 amps.
It could be a headlight, a heater blower, or a carbon pile resistor. I dont know what you have available.
Put a voltmeter across the two ammeter studs.
When the circuit is connected, current will flow through. The ammeter should show the amount of current, and if there is resistance in the ammeter it will show as a voltage difference between the studs.
Setup like this
View attachment 1715690532
The complete the circuit if the ammeter has no resistance to the flow (ideal).
View attachment 1715690535
I doubt it will be zero. But it should be pretty close to zero with only 4 to 5 amps through it.
Lets say there is a voltage across the ammeter.
View attachment 1715690538
Having a 0.6 Volts loss with only 5 amps of current is significant.
Besides the heat, it also means that 1.2 Volts would be lost when 8 to 10 amps are flowing through. That's not acceptable.
Remember earlier I wrote that if it shows any resistance on a typical multimeter, its too much?
This is example is just on the edge of the amount of resistance that a multi-meter may be able to reveal.
The digital meter I use generally shows .1 Ohms when its probes are pressed together.
Lets compare that to the resistance in our example.
We'll estimate the current based either on the ammeter reading or the high beam of our lamp.
If the ammeter is showing about correctly and knowing the battery is fully charged (12.5 volts) we can be pretty sure current is around 4 to 5 amps.
V = I x R
0.6 Volts = 4 amps x Resistance
0.6 Volts/4amps = Resistance
0.15 Ohms resistance in the ammeter.
So with my multimeter, it
might just show .2 ohms with the probes clamped to the ammeter's studs.
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