Anyone ever reamed out a 6mm aluminium knob for a 6.35mm shaft? Good idea or not??

xefned

xefned

Well-known member

I wanna order these groovy ass knobs for my next envelope filter, but I can't find a 6mm shaft pot in reverse log 100k (not even from Tayda.)

groovy.jpg

I have a harbor freight drill press and below-average skills with it. Can I ream these out? Is it worth trying? Or am I gonna end up with a case of regret and an emergency room bill?
 
Just a heads up— I have a set of those knobs, and they’re incredibly tiny. I’ll snap a pic in a bit, but I was expecting them to be larger (when I bought them no dimensions were listed. Idk if that’s still the case).
They’re definitely cool, but yeah.


Anyway… if you want a reverse log 100k pot but can’t find it, worry not— if you can combine a 1M linear pot with a 110k resistor to get a 100Kc pot. As long as you can find 1M linear pots with a 6mm shaft, you’ll be golden!
 
6.35 mm is 1/4", so a 1/4" bit will get you there. I use an oversize reamer when I make knobs but a straight up 1/4" drill will almost always drill a hole slightly larger so you should be fine (the knob hole needs to be slightly larger than 6.35 to actually fit on the pot), just make sure to have a small file to de-burr.

I think the hardest part will be finding a secure way to reliably hold it.
 
Bricksnbeatles said:
Anyway… if you want a reverse log 100k pot but can’t find it, worry not— if you can combine a 1M linear pot with a 110k resistor to get a 100Kc pot. As long as you can find 1M linear pots with a 6mm shaft, you’ll be golden!
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Super helpful dude. TYSM!
I was always told, “you can only change the value for a linear pot” to another (smaller) value. I had no idea you could do log or anti-log. I've clearly been misinformed! Thanks. I need to research this and update my knowledge for this century. Jeez. What a breakthrough.

Thanks for the picture. Wow. They might be actually too small for a 125b. Maybe better on a tiny 1590a build. How do you like them? Are they still usable? (If you've tried them on a pedal.)

Thanks duder.
 
swelchy said:
I've drilled aluminum knobs out to fit 6.35 in a pinch and it worked fine
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I'll try it. Thank you!

temol said:
A-2577 and A-1952
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I'd originally written those 2 off for 1 being not-PCB mount (important) and the other for having a knurled shaft. But I think I can make that work.

I've used the knurled shaft pots before with set-screw knobs and they usually work. The only time it was really bad was when the set screw was perfectly parallel to the split middle seam. I tightened it too much and crushed the two halves together. But if I'm careful, I can avoid that.

Thanks for making me give those a 2nd look.
 
xefned said:
Super helpful dude. TYSM!
I was always told, “you can only change the value for a linear pot” to another (smaller) value. I had no idea you could do log or anti-log. I've clearly been misinformed! Thanks. I need to research this and update my knowledge for this century. Jeez. What a breakthrough.
Click to expand...
I’m trying to create a resource for displaying the effect of tapering resistors on pots in the same style as the Duncan Tone Stack Calculator. Adding a resistor in parallel with lugs 1 and 3 of a pot will always bump the middle of the range, so a log pot will become more linear and a linear pot will become more anti-log. The basic formula I use for linear to anti log (90% resistance @ 50% rotation, since that’s what a C-taper pot is) is 10XΩ Lin pot and (11/10)XΩ resistor. It has the benefit of being a true anti-log curve too, instead of the dual-slope curve of most log and anti-log pots.

xefned said:
Thanks for the picture. Wow. They might be actually too small for a 125b. Maybe better on a tiny 1590a build. How do you like them? Are they still usable? (If you've tried them on a pedal.)

Thanks duder.
Click to expand...
Yeah, they’re teeny tiny! I dig them. Haven’t used them on a pedal at all (since no indicator dots), but they’re pretty looking knobs and nice for drone synths. I actually planned on using them on a guitar, but they were way too tiny for that 😂
 
Done it a few times on drill press...Hardest part for me is finding a way to securely hold the knob AND keep it at a perfect 90 deg... Have more than my fair share with a slightly tilted hole so that it spins a bit crooked. :P
 
How about wrapping the knob with masking tape and putting it in the drill press? (Assuming you have a big enough chuck.)

First, you’d take the 1/4 inch drill, and drill a hole into a thick (1 inch plus) small scrap of wood. Then, stick the shaft of the bit into the wood. Basically, it would be as if you were doing this on a vertical lathe—spin the part, into a stationary bit. The bit will self center. You’re opening up the diameter 0.35 mm, which is under 1/64th inch, so it’s not by much.
 
Bricksnbeatles said:
I’m trying to create a resource for displaying the effect of tapering resistors on pots in the same style as the Duncan Tone Stack Calculator.
Click to expand...
You might find this worth checking out:

It doesn’t have the amount of detail you were talking about nor a visual representation of the sweep, but it’s still handy if you approach it with the understanding that there will be idiosyncrasies involved with altering a pot value with parallel resistors.
 
I've also done this before (with a handheld drill) and ended up with slightly crooked knobs because despite my best efforts it was not drilled at a perfect 90 degree angle. In this case, it's hardly noticeable unless you're looking for it but I'd still recommend using a drill press or some kind of jig that would allow you to drill it at a perfect 90.
 
fydo said:
…despite my best efforts…
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Mine are guaranteed to be crooked given my experience level but, that's how you learn. 🙃

benny_profane said:
It doesn’t have the amount of detail you were talking about nor a visual representation of the sweep, but it’s still handy if you approach it with the understanding that there will be idiosyncrasies involved with altering a pot value with parallel resistors.
Click to expand...

inverse-log.png

I'm guessing we can use this graph along with this calculator to do a 2-resistor solution.
So in my case, a 110kΩ resistor + a 10kΩ resistor if I'm understanding right. (I'll experiment.)
 
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xefned said:
Mine are guaranteed to be crooked given my experience level but, that's how you learn. 🙃



View attachment 32945

I'm guessing we can use this graph along with this calculator to do a 2-resistor solution.
So in my case, a 110Ω resistor + a 10Ω resistor if I'm understanding right. (I'll experiment.)
Click to expand...
If you want to end up at 100K you'll need to start with something like a 1 Meg linear pot and a 110K resistor.

I tend to just throw the whole mess in LTspice, although doing accurate pot log taper sims is sometimes not its strongest point.
 
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