when the switch is "off" 2 is connected to 3, and 5 is connected to 6. Therefore because 1 is jumpered to 6, when off the signal just flows from input to output totally bypassing the PCB.
When it is on, the contacts are thrown the other way (think of it as contacts pivoting on the middle row and throwing up or down from there) so that 1 is connected to 2 and 4 is connected to 5. so in that scenario the signal flows input-pcb input-pcb output-output (and obviously between PCB input and PCB output is where the effect circuit is).
Does that help?
So, if I'm understanding this right, 1 and 4 aren't actually connected in any way, they just happen to both serve the same function on the switch. In that case, why in the first picture was the PCB input connected to the same spot as the input and not to a different lug? I guess that is why you can't have true bypass with that switch, correct?
The other question, then, is how the input is connected to ground in the DPDT or 3PDT case, if 3 and 6 and 9 aren't connected in any way. I guess it is connected to the jack, but there was a whole lot of talk about why you would connect it to ground, so I'm assuming that's a different ground we are talking about.
Let me just make sure I understand the 3PDT diagram. In the off state, 2 and 3 are connected, thus the LED is not getting any signal; 5 and 6 are connected, and since 6 is jumpered to 3, the PCB input is grounded (what is the purpose of that? I'm not sure I did that on my pedal); and 8 and 9 are connected, so the jumpered input goes straight to the output. In the on state, 1 and 2 are connected, thus the LED does get signal and is grounded at the switch; 4 and 5 are connected, so the input goes to the PCB input and the jumper goes nowhere, since 9 is not active; and 7 and 8 are connected, so the PCB output goes to the output.