As I've preached before . . .

[ndutton]

Following the addition of a second electric coolant pump I remeasured the alternator output and coil input voltages (ahead of the ballast resistor) to recalculate voltage drop. With the added load I've crept up to 5% so some engine harness work is added to the project list.

Besides staying on top of the ignition wiring as loads are added, I was directed to make the check by the EWDS. Ignition on before pushing the start button I was getting a coil indicator light, no light when running though.

My current harness is over 15 feet long, hot (red) wire to the panel is 12 gauge, starter solenoid wire (white) back to the engine is 12 gauge and the ignition wire (purple) is 12 gauge. Looks like I'll be replacing those three with 10 gauge.

[ndutton]

Engine harness rework complete and tested

With the larger 10 gauge wire for hot (red) and ignition (purple), measured voltages are now:
Alternator output → 14.0V
Coil input ahead of resistor → 13.6V

VDrop% = [(Valt - Vcoil)÷ Valt] x 100
VDrop% = [(14.0V - 13.6V) ÷ 14.0V] x 100
VDrop% = [0.4V ÷ 14.0V] x 100
VDrop% = 0.0286 x 100
VDrop% = 2.9% (Target drop is 3% or less)

Total ignition system current:
Ignition --------------------------3.24A
Facet fuel pump ----------------1.6A
Two coolant pumps @ 2.2A --- 4.4A
9.24A total, necessary to know so the ignition circuit fuse is sized properly

edit:
I should have mentioned it in the first post, original measurements were:
Alternator output → 14.0V
Coil input ahead of resistor → 13.2V
5.7% Voltage drop

The purpose of this thread is to present a real world example of voltage drop and how easily it is [properly] resolved.

edit 2:
I re-ran the EI calculation with the improved voltage too, 3.24A (3.3Ω coil + 0.885Ω resistor)
Target amperage is less than 4 amps, I arbitrarily prefer 15% less for a target of 3.4 amps max.