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In this circuit and explanation, it all makes sense except I'm having a hard time picturing Q2 not turning on until C1 has discharged first.
When C1 is discharging through Q1 why doesn't Q2 turn on as it's discharging since it's base is higher than 0.6v through R2?
Also is C1 discharging only by the + plate having a path to ground rather than both + and - plates of C1 being connected together?
Suppose C1 is charged with 4 Volts with the positive voltage at side marked +.
Now suppose that Q1 switches on pulling the collector of Q1 to 0 V (ground). Since C1 is still charged with 4 V, the base of Q2 will be at - 4 V. So Q2 is forced to be off as Vbe < 0.6 V. It is actually negative !
The base of Q2 needs to be above 0.6 V before it can conduct so C1 has to discharge to at least -0.6 V for that to happen (still assuming Q1 to be on so C1's + plate is 0 V).
So in reality C1 first discharges completely (to 0 Volt) and then charges with opposite polarity to at least 0.6 V.
For discharging a capacitor it is not really needed to short the capacitor plates. As long as the voltage on both plates is the same then the capacitor is discharged. It does not matter how this is achieved (making the voltages the same).
