I had a pretty good understanding of everything shift lock when this thread started, but your question made me curious enough to dig in a bit further. I think this is the whole story:


In order to be able to depress the release button on the shifter knob, allowing movement out of the Park position, the following conditions must normally be met:

- battery charged sufficiently
- ignition switch in position II
- brake pedal depressed, and brake switch functional (brake lights coming on verifies this
- shift lock relay functional
- shifter Park position microswitch functional
- shift lock solenoid functional

Note that putting the ignition switch in position II causes 2 things to happen:
- switches through the necessary voltage to drive all the electrical functionality requred above
- releases a mechanical cable running from the ignition switch (4 in diagram) to the shift lock solenoid plunger, that would otherwise prevent any movement of the plunger.



If any of the above electrical components are not functioning correctly (battery, electrical contacts in ignition switch, brake pedal switch, shift lock relay, shifter Park position microswitch, shift lock solenoid, wiring), then the release button on the shifter will still not be able to be depressed with the ignition in position II and the brake pedal depressed.

In this situation, it is still possible to depress the release button on the shifter knob by meeting these conditions:
- ensuring that the ignition switch is in position II to release the otherwise mechanically restricted shift lock solenoid plunger
- pressing on the mechanical shift lock override button (5) to release a locking pin (3) on the shift lock solenoid

This allows the shift lock solenoid plunger to be effectively pushed back by the action of depressing the release button on the shifter knob, allowing movement out of the Park position.
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David Armstrong - '86 240(350k km?), '93 940T(270k km), '89 240(parts source for others) near Toronto