Replace pots with resistor/jumper?

[devnulljp]

Got an odd one -- trying to repair an old Roland keyboard that has a pair of worn out slider pots. Both pots are completely shot and the thing cuts out all the time. Having a real hard time sourcing drop in replacements (contacted Roland and they don't carry the parts any more).
I've bought a few online with the same specs but they all seem to have different pinout configurations, so they're not drop-in replacements.
Thinking about just jumpering them in there, but had a thought. Could I just set them both to 10 (actually one to 0 and one to 10 would work) by jumpering the pins and miss out the whole pot thing altogether? Or throw in a resistor and get the same thing as having the pot in there at a given value?
The things have 3 pins, which I'm assuming are similar to what you'd get in a rotary pot ... they're both 10kΩ, and apparently stereo, whihc probably complicates matters a bit.

They look like this


pins are labelled 1, 2, 3 on both ends but they seem to be a weird layout.

[Dr Tony Balls]

Replacing them with a resistor will work just fine, but you'll have to figure out which two pins to connect and where to connect the resistor, and what the resistor value needs to be. On the pot below when its turned to the clockwise extreme pins 2 and 3 are connected and theres a resistance between 1 and 2. The opposite way and 1 and 2 are connected with the resistance between 2 and 3. Figure out what legs of the sliders need to be bridged, and run the resistor between those and the other leg.

[devnulljp]

Do I need to jumper the other lug or is it just a matter of a resistor between the correct two lugs?
In the above diagram, is 2 ground?
Thanks

[John Lyons]

For a volume type pot pin 1 would be grounded.
The wiper is the active/moving component so typically
it isn't grounded.
A pot in a volume configuration is a voltage divider
so you have a ratio between two resistances. Wiper to ground
and from wiper to signal. When the pot is all the way up it becomes
a single resistance to ground (lug 3 and 2 are shorted). When it's
turned all the way down pins 2 and 1 are shorted which grounds out
the signal which is a single resistance to ground (not between ground
and signal as when it's all the way up).



In the above image the lugs are 1,2, 3 from the top down and
3,2,1 in the schematic top to bottom.