I see reference to changing the resistance of the output resistor that is parallel to the headphones. For instance
I didn’t want to reply to this old thread)
A couple of questions:
1. Why doesn’t the headphones resistance serve the function of charging the cap, as it is less resistive than the resistor (usually a 2.49k on a Crack)?
2. Is there a way to calculate optimum resistors in this spot depending on the output capacitor value and headphone impedance? Or can someone explain or link the theory that would be relevant to this question?

Yeah, a lot of the issue was sourcing larger, more expensive caps. Also, the charging time of that capacitor needs to be very short. To achieve this with larger output caps, the 2.49k resistors need to decrease proportionately. If you go to a 300-ish uf cap and roughly a 1k resistor at the output to keep the charging time short and charging voltage low, this resistance isn't such a big deal with low imedance headphones, but suddenly a high impedance set of cans will "appear" to be a lower impedance load, as 600 ohms in parallel with 1k is suddenly a 375 ohm load. With the stock circuit, a 600 ohm set of headphones will appear as a 485 ohm load. Decreasing the load impedance will reduce output levels, so we chose these resistors carefully to provide the best performance with the toughest to drive (normally) headphones.
A couple of questions:
1. Why doesn’t the headphones resistance serve the function of charging the cap, as it is less resistive than the resistor (usually a 2.49k on a Crack)?
2. Is there a way to calculate optimum resistors in this spot depending on the output capacitor value and headphone impedance? Or can someone explain or link the theory that would be relevant to this question?