LED conversion and hazard flasher

[David C]

A voltage drop sufficient to slow your indicators significantly is a little concerning and may point to high resistance in the ignition switch.
I suggest you switch on blower, wipers (wet screen), run engine so fuel pump is working and turn on radio. Then measure voltages of battery and switched ignition feed.

Current should be roughly 10 to 15 Amps.
Measure battery voltage and then voltage on either end of the upper fuse that has white wires one side and green the other. This is fed from the ignition switch.

Then work out power loss in wiring / switch.

If battery voltage is 13.5 and switched supply 12.5V, difference = 1V
For current of 15A, resistance is 1/15 = 0.06 Ohms.
Power is current squared times resistance. 225 x .06 = 13.5 Watts. Roughly the power of a small soldering iron. Although a voltage drop of a Volt doesn't sound much it can have a serious heating effect.

Any significant power loss would suggest you consider fitting a relay or contactor in the switched ignition feed. It can be simply done by removing the wire that feeds the switched fusebox and using that to operate the relay. The main contacts of the relay are then supplied direct rom the battery and fed back to the fuse box. Suitable relay Sterling Power R12120.