[Healeys] Ballast resistor wiring question

[jim]

 current flowing through the "sawtooth" line (resistance wire) heats up the bi-metallic strip ...which bends closing the contact bypassing the resistance wire ..it cools off as current is no longer flowing through resistance wire ...as the bimetallic strip cools, it bends back opening the ciruit and current flows through the resistance wire heating it up ...etc, etc

on the lucas one, it is really a relay...current flows main armature >-->actuating wire >-->ballast resistor >--> field coil of relay to the bulbs via turn signal switch ...Initially the bulbs are cold with very low resistance, as they heat up the resistance increases
So ..Initial turn-on lots of current through relay field coil, relay operates bulbs heat decreasing current flow . eventually relay released due to field decreasing ..bulbs cool .. current increases, relay operates ..etc, etc
the drawback being if things aren't just right, the relay will act like a vibrator and just "buzz" alternate operate/releaseig\


Jim
 

I can't quite figure out how this type works?It appears to me that there would be constant power to the "L" terminal meaning that the lamps would not flash ... what am I missing?



Interestingly here is the Lucas version.

This type is quite clever for 1953 ... If there isn't sufficient current flow through the coil the coil does not pull the pilot points closed so when 1 bulb fails the pilot lamp does not flash.
M

  
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