Changing the value of a pot

[Seven64]

so i searched, but just can't find anything i understand.  i am planning on building the kay tremolo and putting it into an old morley compact wah enclosure that i have laying around.  the morley uses a 50k slider pot along with some extravagant fishing line to slide the pot with the pedal.  i want to convert that 50k into the 1k needed for the speed control.  i understand that it is equal to 50 1k pots in series, but i do not understand how that is relevant.  i tried putting resistors over the 2 points they used for connecting it to the morley circuit, and measured the differences with my voltmeter but don't understand what i found out.  it varies from 44k to 0k normally, with a 100k resistor it goes from 30k to 0k, a 470k resistor and it goes from 40k to 0k, and with a 1k resistor it goes from 1k to 650r.  can someone please teach me something?!

[petemoore]

  Question to explore: How critical is the 1k value of pot to the Kay response? [does the circuit actually care?..looking at the schematic..if there's only 2 wires to the pot...probably Is important that the R range be 1k - 0.0k.
   50k in parallel with 1k is effectively 1k as far as the circuit is concerned, measure the values with DMM for more precise numbers, since the pot is 50x larger than the 1k, the 1k dominates the R value until the pot is turned closer to 1k.
   2 x 1k resistors in parallel = 500ohms [ .5k ].
   A 50k pot does vary between 1k and 0.0k, however this range of it's R sweep will be restricted to a super-small portion at the end of the shaft rotation, the rest of the 'R-curve' will be so close to a flat 1k...all the 'adjustment' will be restricted to say 2% of the pot's shaft travel, so only very 'coarse' tuning would be possible, 98% of the potrange in parallel with the 1k would be ...about 1k.
   Suggestions: look at the circuit schematic or experiment with larger values to replace the 1k.
  If it's a resistive divider [three 'unique' wires connect to the 3 individual pot lugs] the circuit might not care too much about what value pot is doing the adjustable resistive dividing [a pot is essentially 2 resistors that always add to = the pots measured value, as one R on one side of the pot goes up, the R on the other side of the pot goes down by exactly that amont.
 

[petemoore]

  IF it's a a variable reisistor [a pot with only two wires to it] which controls the speed, 1k will be the max or minimum setting, 0.0k would be the opposite [slow or fast], and the taper of the pot will likely be critical to getting a 'smooth adjustment' at the speed knob or treadle control.
  A 50k pot, being 50x the 'optimal value, will be much too large a resistance over most of it's settings..the speed probbly won't change until about 5k [total guess], and that will restrict all of the speed adjustment to a teeny portion of 'twist' at one end of the pots shaft travel...ie super touchy speed control that is say 2% of the shaft travel...closer to a "high/low" speed setter, fine tuned adjustments would be difficult/impossible.
   You want slow to fast settings to be spread evenly across the entire potwafer, so the knob or treadle control has smooth adjustment ability. Ordering the value/taper pot shown on the schematic or BOM one way to do this.
   After some experiments [throw the 50k in the circuit and see how it works] it will become clear whether a smaller value pot is required, whether that translates into a morley treadle-conversion of course depends on whether your pot-pulley [what the control line wraps around to turn the pot] can fit an available smaller value...or whether something can be found which matches the morley 50k's mechanical connections.
  Whether it can be made to do what you want in terms of smooth adjustment when treadle is used depends on...well, it is also ''possible'' to put a 'cam-pulley' on a pot, so that the string pulls make faster motion at a certain point in the travel.
    "Secret life of Pots" @  GEO 'd be a great read.

[Seven64]

ive read that, but you explained it better.

and just sayin its a slider pot, not a rotary lol.

[petemoore]

its a slider pot, not a rotary lol
   ..if you can find a 1k slider that'd be an easy fit, it'd be an easy fit and simple wire-in.
  Otherwise, the parade might go foreward and meet design goals, but without machine shop...for me meant intrusive/risks that involved a few attempts which resulted in very cludgeoned looking, yet functional treadle/conversions [after a lot of mechanical-messing around.
   HAving made 'ernie ball' type volume pedals, the 'spring return' string-drive method for treadle movement = pot rotation was a lot of experimenting...I found the 'right' diameter, flat-sided knob, and wrapped the string around that, tricky, one side to the spring [takes a specific stregth/length spring] the other side of the string to the treadle.

[Seven64]

yea, the problem isn't finding a 1k slider or getting it wired up, its attaching it to the fishing line.  im getting a pic of it so you can see what i am saying.  i cannot slide the spring off the knob of the slider, and have no idea about how i would go about getting it off this one and onto a new one.