The capacitor was allowed to discharge completely when the transistor was activated with the 3.4 V.
The capacitor is allowed to charge again when the conduction from the 5 V line through the resistor to the ground of the IC is open circuited because the transistor deactivated.
The transistor must deactivate at 0V to have the 5 second cycle. In the case that the transistor deactivated at 1.4 V across the capacitor, the 5 s cycle would be reduced to the time it takes to charge the capacitor from 1.4 V to 3.4 V.
So my question!:
If the transistor was activated at 3.4 Vin (3.4 V across the capacitor) and allowed the capacitor to discharge through it, then the voltage at the capacitor would be decreasing. V in of the transistor would not be 3.4 V anymore..
There is the mystery.
If The transistor activates at 3.4 V, then when does it close as it has current running through it from the capacitor to ground. The voltage across the capacitor would be 1 V at some time 't' , therefore the V in of the transistor would be 1V, which is less than 3.4 V, which is when it activates.
Does it deactivate at 0 V in - ~0 V in AFTER it has activated (opened/activated), and only activate at 3.4 V in?
A second question: If I replaced the cap with a resistor the IC would switch rapidly correct?
The schematic on the top left is the one I redrew on the wide ruled paper.