Goal = Interior lamp on until start'n car

I see.  So basically when current flows throw the wires leading to the light switch (pins 87 & 30),  the Transistor activates the Relay?

Thanks again for the EE education!

Here is the original Text From Bills website.

The two circuits below illustrate opening a relay contact a short time after the ignition or ligh switch is turned off. The capacitor is charged and the relay is closed when the voltage at the diode anode rises to +12 volts. The circuit on the left is a common collector or emitter follower and has the advantage of one less part since a resistor is not needed in series with the transistor base. However the voltage across the relay coil will be two diode drops less than the supply voltage, or about 11 volts for a 12.5 volt input. The common emitter configuration on the right offers the advantage of the full supply voltage across the load for most of the delay time, which makes the relay pull-in and drop-out voltages less of a concern but requires an extra resistor in series with transistor base. The common emitter (circuit on the right) is the better circuit since the series base resistor can be selected to obtain the desired delay time whereas the capacitor must be selected for the common collector (or an additional resistor used in parallel with the capacitor). The time delay for the common emitter will be approximately 3 time constants or 3*R*C. The capacitor/resistor values can be worked out from the relay coil current and transistor gain. For example a 120 ohm relay coil will draw 100 mA at 12 volts and assumming a transistor gain of 30, the base current will be 100/30 = 3 mA. The voltage across the resistor will be the supply voltage minus two diode drops or 12-1.4 = 10.6. The resistor value will be the voltage/current = 10.6/0.003 = 3533 or about 3.6K. The capacitor value for a 15 second delay will be 15/3R = 1327 uF. We can use a standard 1000 uF capacitor and increase the resistor proportionally to get 15 seconds.