analog alchemy linear taper calculator wrong?

[pappasmurfsharem]

I need two 150K linear pots and have two 200k pots. So I plug the numbers in and it says I need two 300K resistors on Lugs 1-2 and 2-3.

This gives me a read out of 120K. Which makes sense being a parallel resistor.

So what gives?


PS.
Also in actuality my pots measure 188K~ putting 680K on each set of lugs gives me 146K which is close enough for my liking

[slacker]

Like it says on the site, the value fluctuates a bit with pot settings. The value will be 150k with the pot turned to 50% going down to 120k with the pot turned fully one way or the other. Using the tapering resistors you can't have it be the same value at all settings, for whatever reason Joe chose to quote the value with the pot at 50%.

[pappasmurfsharem]

It measured 120 at all settings though...

Well not exactly... rounded of course.

It was 120 between lugs 1&3 at all values. Between 60 and 70 on lugs 1-2 and 2-3, Fully clockwise or counterclockwise. @ 50% Lugs one and two , And two and three Was also 120

[pappasmurfsharem]

It looks like I was a bit quick to type. It does raise a bit at50% although it's not near 150K. However upon looking at the layout I'm using... The two 150k knobs are wired as  variable resistors. With only lugs 1&2 used. The 680K gives me a range of 0 to 147K from lug 1 to the wiper. So I guess that is plenty good.