Not much to add to what has already been stated here. Here's how I break this stuff down in my own way:
The Voice pot is a variable resistor, i.e. it changes resistance as it is turned. Since lugs 1 and 2 are connected, this means that when it is fully counter-clockwise it produces the full resistance of the pot, 10k. When fully clockwise it is 0k. Since it is linear, then you know the "in-between" points more or less fall in even steps. IOW, turning the pot 10 degrees in the bottom part of the sweep produces the same rate of change in resistance as it does in the top part of the sweep, or any other are for that matter.
There are two "states" to consider with the variable pot:
One state is the amount of gain produced by the negative feedback loop which is all the stuff between pins 1 and 2 of IC1. It's helpful here if you just forget the amount of feedback there is variable by the Drive pot and consider it fixed. So, think of Drive as just a 500k resistor for the moment and ignore R9.
When the Voice pot is all the way up, it produces 0 resistance and therefore we can consider it removed from the circuit. Using the formula for calculating gain in a non-inverted feedback loop, we see
Gain = 1+ (Drive/R3) = 1+(500/1) = 501
When the Voice pot is fully counter-clockwise, we get the full resistance of 10k, so the formula for gain becomes
Gain = 1+ (Drive/(R3+Voice)) = 1+(500/11) = 46
So, just by turning the Voice pot we've varied the gain output by a factor of about 11. IOW, our gain goes DOWN when the Voice pot is all the way down and vice-verse and it will be quite noticeable since it the difference between the two is an order of magnitude.
Now consider the second "state" which is the filter created by connecting R3, the Voice pot and C3 to Vb. Since Vb is virtual ground, just consider it as a simple low pass filter. The cut-off point of this filter depends on two values: the total resistance and total capacitance connected to ground. You should also consider impedance, but since C3 is fixed I don't think you have to worry about it (maybe someone knows better). Anyway, we already know that Voice is a variable resistor, so that means this filter is also variable.
When the Voice pot is up, we have a 1k resistor and 100n cap. Using the formula for single pole HP/LP type filters we have
f(cutoff) = 1/(2*pi*R*C) = 1/(6.28*.001*.1) = 1592 Hz
When the Voice pot is down, the formula becomes
f(cutoff) = 1/(6.28*.011*.1) = 145 Hz
Keep in mind that the "cutoff" here is not a sharp knee. Rather, it is -6dB per octave. So, the further away from the cutoff the more drastically the frequency response is reduced. This means when our cutoff is 1592 Hz, we have more mid-frequency content with the high frequencies being steadily reduced by the filter. When the cutoff is at 145Hz, we reduce both mid and high frequency content.
The two states here are what make the Voice pot what it is. It simultaneously changes the mid-frequency content and overall gain. Remember this example kept the Drive pot as a fixed resistor. When you remove this restriction, you can see now that the Drive pot also influences the total amount of gain produced it also works with C2 to produce another variable filter but that is for high frequency content only) . IOW, the two controls are interactive, as mentioned in earlier posts.
Here's a much better (and probably more correct) technical explanation of how this works: http://www.geofex.com/Article_Folders/TStech/tsxtech.htm
Look at the "clipping" section.
The Voice pot is a variable resistor, i.e. it changes resistance as it is turned. Since lugs 1 and 2 are connected, this means that when it is fully counter-clockwise it produces the full resistance of the pot, 10k. When fully clockwise it is 0k. Since it is linear, then you know the "in-between" points more or less fall in even steps. IOW, turning the pot 10 degrees in the bottom part of the sweep produces the same rate of change in resistance as it does in the top part of the sweep, or any other are for that matter.
There are two "states" to consider with the variable pot:
One state is the amount of gain produced by the negative feedback loop which is all the stuff between pins 1 and 2 of IC1. It's helpful here if you just forget the amount of feedback there is variable by the Drive pot and consider it fixed. So, think of Drive as just a 500k resistor for the moment and ignore R9.
When the Voice pot is all the way up, it produces 0 resistance and therefore we can consider it removed from the circuit. Using the formula for calculating gain in a non-inverted feedback loop, we see
Gain = 1+ (Drive/R3) = 1+(500/1) = 501
When the Voice pot is fully counter-clockwise, we get the full resistance of 10k, so the formula for gain becomes
Gain = 1+ (Drive/(R3+Voice)) = 1+(500/11) = 46
So, just by turning the Voice pot we've varied the gain output by a factor of about 11. IOW, our gain goes DOWN when the Voice pot is all the way down and vice-verse and it will be quite noticeable since it the difference between the two is an order of magnitude.
Now consider the second "state" which is the filter created by connecting R3, the Voice pot and C3 to Vb. Since Vb is virtual ground, just consider it as a simple low pass filter. The cut-off point of this filter depends on two values: the total resistance and total capacitance connected to ground. You should also consider impedance, but since C3 is fixed I don't think you have to worry about it (maybe someone knows better). Anyway, we already know that Voice is a variable resistor, so that means this filter is also variable.
When the Voice pot is up, we have a 1k resistor and 100n cap. Using the formula for single pole HP/LP type filters we have
f(cutoff) = 1/(2*pi*R*C) = 1/(6.28*.001*.1) = 1592 Hz
When the Voice pot is down, the formula becomes
f(cutoff) = 1/(6.28*.011*.1) = 145 Hz
Keep in mind that the "cutoff" here is not a sharp knee. Rather, it is -6dB per octave. So, the further away from the cutoff the more drastically the frequency response is reduced. This means when our cutoff is 1592 Hz, we have more mid-frequency content with the high frequencies being steadily reduced by the filter. When the cutoff is at 145Hz, we reduce both mid and high frequency content.
The two states here are what make the Voice pot what it is. It simultaneously changes the mid-frequency content and overall gain. Remember this example kept the Drive pot as a fixed resistor. When you remove this restriction, you can see now that the Drive pot also influences the total amount of gain produced it also works with C2 to produce another variable filter but that is for high frequency content only) . IOW, the two controls are interactive, as mentioned in earlier posts.
Here's a much better (and probably more correct) technical explanation of how this works: http://www.geofex.com/Article_Folders/TStech/tsxtech.htm
Look at the "clipping" section.