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Old 01-27-2018, 02:03 PM
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simplyconnected simplyconnected is offline
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You guys know how a relay works. When energized, the clapper immediately draws-in magnetically. Ford clocks are identical BUT as soon as the clock 'clapper' draws up the coil contacts break connection. So, here we have the coil wire, fed by a normally closed relay contact.

How long does this take? It's so fast that six volt clocks work equally as well on 12-volt systems because the contacts open even sooner. The windings don't have enough time to burn out because they are only used for a fraction of a second.

The purpose of the 'clapper' (which is really an armature) is to wind the clock spring. Ford attached the electrical contacts to the clock spring. The contacts close when the spring depletes. As noted, the spring takes ~75 seconds to deplete before the next 'bump'.

Ok so, how many milliamps per time unit? This is tough because inrush current is huge but only for about 1/10th of a second every 75" or about 1,150 times per day. This looks like a lot but these intervals are intermittent.

Current drain would be worse if the drain was constant because lead-acid batteries partially recover during 'rest time'. Did you ever crank the starter until the battery was so dead that the starter solenoid would only click, only to try later and find the battery has more cranking power? That is the type of intermittent drain and recovery I'm referring to.

A healthy battery should last a few weeks of idle time before the clock affects cranking power. - Dave
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