I was being cute about the speed of light thing.

Similar to the time constant of a capacitor-resistor circuit, t=R*C, the inductance (the coil) has a time Constant. t=L/R, where t is time in seconds, L is the inductance in Henry, R is resistance in Ohms. We don't know what the inductance or time constant is in this direct current circuit. We do know the approximate resistance of the coils. Based on the voltage of 24 volts DC and the typical ITR/IBM time control system, the maximum of 2 amps per circuit is assumed. This includes a master movement wind coil of 43 Ohms (.132 amps), a program magnet coil of 22 Ohms (.545 amps), and sixty-six (66) model type 561-2 secondary coils of 240 Ohms each (* 0.024 amps ea). The lengthy process is to calculate the total resistance of 68 resistances in parallel using the formula of 1/Rtotal = 1/R1+1/R2+1/R3+1/R4+...1/R68. Or we can take a short cut and use Ohms law (R=E/I) that tells us that the total load resistance in this master clock system is the Voltage (24V) divided by the total current, in amps (I=2.0 amps) and is equal 12 Ohms.

**Type Description ___________ 12 Volt 24 volt 12V to 24 V Conversion Resistor**

25 Master's Winding Magnet .273 amps .132 amps 43 ohm 5 watt (actual 3.36 watts)

25 Impulse Accumulator .280 amps .140 amps 43 ohm 5 watt (actual 3.36 watts)

805-2 Program Magnet .545 amps .279 amps 22 ohm 10 watt (actual 6.54 watts)

561 Indicating Clock Movement (20V) *.028 amps. Use R1, 150 ohm 1 watt resistor to

(******20 volt coil measured with 20V applied: I=28 mA) *convert coil from 20 to 24 volts.

561-2 Indicating Clock Movement .050 amps .024 amps 240 ohm 1 watt (actual .6 watts)

563-2 Indicating Clock Movement .333 amps .171 amps 36 ohm 5 watt (actual 3.9 watts)

565-2 Indicating Clock Movement .170 amps .085 amps 68 ohm 2 watt (actual 2.1 watts)

569-2 Tower Clock Movement .050 amps .024 amps 240 ohm 1 watt (actual .6 watts)

When the master relay is opened, the magnetic fields of all the coils suddenly collapses, at the speed of light minus the time constance, in seconds, of L/R. (t=L/C) It has been said the counter electromotive force (CEMF) voltage rise is 10 times the applied voltage (so >240 volts?). But I think this CEFM voltage could be higher because of the fast rise time due to the contacts opening. The diode starts conducting when it's junction reaches 0.7 volts and holds the voltage at 0.7 volts, not >240 volts. This is now a two voltage crop series circuit, a voltage drop of 0.7 volts and a voltage drop of >240 volt drop across the 12 Ohm total load resistance. This looks to me to be better way of protecting the contacts than the suggested threshold voltage of 30 volts from the scheme that mxfrank mentioned. So how is the resistor scheme wired? If the resistor is 10 times the load resistor, it would have to be in series to make a difference. A series resistor would certainly slow the decaying CEMF induced, and probably upset the current to all the coils. If the resistor is in parallel to all the coils, the total resistance would be closer to 12 Ohms. Rt-R1*R2/R1+R2 (or 1/Rt=1/R1+1/R2)

The point is; the contacts are subjected to a lower voltage after the the diode switches on, thous reducing the damage from arcing. The slow decay of the CEMF energy does not matter to the contacts because this happens after the contacts are opened. As long as the time constant is shorter than one second. Why? The two second contact, starting at the 59' 10” pulse from the master clock's movement happens every other second so all this starts over and over for about 21 pulses.