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Post by fobits on May 16, 2006 14:31:22 GMT -5
Oh-oh. Two Gods are going to fight. Didn't the Ancient Greeks have a saying about that? Something like "Ruuuuun for the hills, folks!" I'll keep my mouth shut except to repeat, for convenience, the link that ashcatlt gave. The graph is about half-way down the page. www.geofex.com/Article_Folders/potsecrets/potscret.htm
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Post by UnklMickey on May 16, 2006 14:44:28 GMT -5
Oh-oh. Two Gods are going to fight. Didn't the Ancient Greeks have a saying about that? Something like "Ruuuuun for the hills, folks!" ... calm thyself, my son.
no fight need ensue here.
although aspersions have been cast, i will gladly light the way.
i have no desire to smack-down another deity.
i will simply guide him gently to the truth.
have patience with Sumgai, he's new to this strata. Sumgai, the ball is in your court, dude.
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Post by sumgai on May 16, 2006 16:20:07 GMT -5
EditEdited to change terminology. At all times hereafter, the word 'input' means the wiper of the pot, and the word 'output' means the non-grounded end of the resistance material. I don't think I need to define the remaining end of the resistance material, do I? /editunk and Frank, Sorry for being tardy, couldn't be helped. Lessee, where'd I leave off (have to go round up data from earlier posts)...... OK, the start of the problem is where Frank read a pot, and found: My question was to the effect that I wanted to know which two terminals Frank used. I stated that he was measuring a reverse-audio taper, if he was measuring between the input and what we'd normally refer to as the ground end of the pot. BTW, I don't see an exact answer to that yet, Frank. Not to worry though, unk has taken up your cause. And that's where I stepped in, asking unk to re-think his proposition. unk, I'm sorry to be a pita here, especially in public, but the idea of resistance is to inhibit the flow of electrons, isn't it? So, by your lights, if there is only 12% of the resistance between ground and the input (at midpoint), then there has to be 88% of the total resistance between that same point and the output, right? When I was in school, they took pains to teach me that electricity will flow along the path of least resistance, and that particular little gem of knowledge has yet to fail me in all my years. So if I see eight times the resistance in one side of the pot as I see in the other side, where should I think the signal is going to flow more easily? How could it not be towards the ground terminal? No, my fine God, the resistance of an audio taper pot tends to increase more rapidly between ground and input than does a linear pot, when viewed from the perspective of the ground terminal. Conversely, it tends to decrease in like manner between the input and output terminals when viewed from the perspective of the output terminal. And that bit about 12% at the midpoint is a bit off. My experience has been more along the lines of about quadruple or quintuple that percentage (again, in re the ground-to-input perspective). By way of example, I find my Strat's 'A' taper volume pot (marked as 250K) to be: Starting at full-stop CCW, and measuring from ground terminal to input terminal: 25% rotation - 94K, about 41% of the total resistance 50% rotation - 187K, about 70% of the total resistance 75% rotation - 229K, about 86% of the total resistance 100% (full-stop CW) - 266K, more than 99% of the total resisance (267K ohms) As you can see, much more than half the total resistance between the input and ground terminals was passed by the time I got to 50% of the rotation. It shouldn't take a rocket scientist to see that almost twice as much signal is heading from the input towards the output as is going towards ground, at that point. All of which points up that to one's ears, a large change in volume level takes place very early on in the rotation of the pot, and not much at all at the upper end. As I said before, this is the effect that makes one device appear to be 'more powerful' than one that's using a linear pot. But we guitar players nearly always work from the upper end of the range and progress downards, so much of this discussion is moot. Turning down from the least amount of resistance between the input and output, a standard logarithmic taper pot will tend to not act as quickly to cut off signal as would a linear pot. See my reasoning above, and interpolate as needed. I believe I've put paid to this account. Make mine Mortons, please. ;D sumgai
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Post by sumgai on May 16, 2006 16:31:04 GMT -5
Frank,
One more thing. I have to say, in all time I've spent looking at R.G. Keen's site, I've never before noticed any mistakes. On the page you linked, I finally found one, sad but true. The graphs showing the tapers are inverted. I don't mean the axis are swapped, I mean that the labels are backwards - the so-called reverse taper should be identified as the normal audio taper, etc.
And no, I'm not gonna write to the guy and straighten him out. If it hasn't been corrected by now, then either no one is paying attention, or he doesn't care to respond to readers who do point out the obvious error.
I haven't taken the time to check the rest of his page. Perhaps, having now correctly learned the most basic functions of pot arithmetic, you could do so? It might be nice to have that data in hand when we get around to setting up a FAQ with off-site links.
Thanks.
sumgai
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Post by fobits on May 16, 2006 16:44:56 GMT -5
Ahh, all right, everybody calm down, folks.
Just a simple misunderstanding about the two ways to wire pots.
That depends on where you connect the wires, right? I was thinking (and I'm sure Unk was thinking) of 12% between ground and output, and the other 88% between the wiper and input.
Fight's over folks, nothing to see here ;D
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Post by UnklMickey on May 16, 2006 16:57:29 GMT -5
And no, I'm not gonna write to the guy and straighten him out. If it hasn't been corrected by now, then either no one is paying attention, or he doesn't care to respond to readers who do point out the obvious error. Sumgai, if you ever do plan to write, you could get a bit busy. it seems that "error" has proliferated: sound.westhost.com/pots.htmit has even spread internationally: swingguitars.com/column_pot.htmlare you still sure it's an error? BTW: if those numbers are correct, you don't have an audio taper pot. (or the commercial two-slope approximation of one) maybe a reverse semi-log pot. unk
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Post by sumgai on May 16, 2006 17:13:33 GMT -5
Oh, man, Frank, you were doing so beautifully. Tsk, tsk, tsk. 12% between ground and output If that's the case, then where is the total pot resistance being measured? By definition, the total resistance is measured between the two non-wiper terminals, yes? Then we have to have labels for them that are easily understood. Right and left don't work, they could be pointing either up, down or sideways. From that, it develops that calling one 'ground' is easy - it is always connected to ground, no matter how you orient the pot to look at it. All that remains is to call the other terminal 'hot', the opposite of ground. But that can cause some viewers to think in terms of 'input' instead of the pot's normal usage, which is, of course, the output. You refer to 'the two ways to wire pots'. I had hoped to avoid that by consistently using 'output' to mean the non-grounded end of the resistance material. It appears that I should have called the wiper the input in order to be consistent, or else to use the word 'hot' to denote the non-ground end of the resistance material. Looks like I better go edit my post, doesn't it. This will cause your message immediately above to go strange, so modify it accordingly, eh? ;D sumgai
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Post by UnklMickey on May 16, 2006 17:20:04 GMT -5
Sumgai,
i think you and i are debating within the same terminology.
CCW (volume=0) terminal as the reference point.
resistance measured between the CCW terminal and the wiper.
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Post by sumgai on May 16, 2006 17:37:21 GMT -5
unk, Why, yes, we are on the same page, terminology-wise. So why aren't we coming up with the same percentage figures? Why do I differ from Frank's results, and not coincidentally, R. G. Keen's charts?
That factor is what's confusing to newbies, and why I'm being so bleeping picky about it. We already get enough repeat questions around here that re-awaken fresh debates, so I'm hoping to head off yet more of them (discussions, not the newbie questions) by being able to refer to a FAQ that is commonly accepted and agreed upon by all of us. Fair enough? ;D
sumgai
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Post by fobits on May 16, 2006 19:53:22 GMT -5
Things get confusing around here, when people are talking at the same time.
I wrote a reply arguing about "input", "output", and Fender vs Gibson methods of wiring volume pots. While I was writing it, Sumgai added an edit to an earlier post defining his terms. So I deleted that one and will try again.
All right, so
CCW == ground wiper == input CW == output
To make a quibble, I wouldn't vote for this terminology, which can cause great confusion. The word "output" has another definition, the signal that goes down the cable and whichever terminal it comes from. The same is true of "input". These terms lead to explanations like "The output signal is taken from the input terminal of the potentiometer." Say again?
Here I'll use CCW, CW and wiper, viewed from the front.
I assume you mean the the resistance curve rises most steeply, between CCW and wiper, near the beginning of the arc, when the volume is low.
The reason for a log pot, as I understand it, is that human perception of volume is a more-or-less exponential function. In order for a sound to be perceived as twice as loud, the voltage has to be roughly squared. So instead of the output rising like 1-2-3-4 as the pot is turned up, it's more like 21-22-23-24. Not exactly, let's not argue about details, but roughly.
That's a curve which has a small slope in the beginning and becomes steeper near the end, as the graph shows.
Whether that is the most desirable in this particular application is another question altogether.
That's a good point. A log pot, as defined above, may not be the best for this application. Players probably spend 98% of the time in the upper half of the arc, and it might be better to have the flatter part of the curve there, for more precise control.
Still, I stick with the contention that a logarithmic curve is a fancy name for an exponential curve, and it goes up most steeply at higher values.
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Post by sumgai on May 16, 2006 21:47:35 GMT -5
Frank, You're correct in pointing out that 'input' and 'output' are still confusing in this context, even when explained up front. I won't assign blame for this one, I'm just as much at fault, if for no other reason than I don't like to use the word 'hot' to describe the non-grounded end. And most definitely yes, as I've applied the terms 'input' and 'output', this is more to the way of wiring used by Gibson, which is in the minority of cases for guitars. I really should have thought that one through before editing my post, but now that you've made your points, I don't wanna start any "Editing Wars". For this reason, I am going to 'drop' this for a day or two, and let things simmer down. (Not that we're hot or anything. ) Then I'm gonna come back in on a new thread, and use diagrams as much as needed to make the lesson complete. I will also 'revert' to using the normal way of wiring a pot, what we've all been calling "forward". This should help to get past any misinterpretations. I hope. Ciao for now. sumgai
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Post by UnklMickey on May 23, 2006 9:39:14 GMT -5
... So why aren't we coming up with the same percentage figures? Why do I differ from Frank's results, and not coincidentally, R. G. Keen's charts?...sumgai in fact, Franks results are well supported by R.G. Keen's charts. yours are not. your results seem consistant with a reverse semi-log taper. unk
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Post by sumgai on May 23, 2006 15:56:36 GMT -5
unk, Ah, ah, ah! I said Keen was off base, and you didn't question me as to how that might be true. The same way that Frank is out of order, above. (I've written and re-written this 4 times now, so I'm getting a little tired. If I seem to blow it, forgive me, and make a note of it. I'll then explain myself more thoroughly, or else try to dig myself out of the hole. There seems to be no simple quicky answer here, so bear with me, please.) Put as simply as possible, Frank, you were speaking of voltage levels when you gave the increasing squares formula - you wanted to illustrate the resulting volume levels when a log pot wiper was rotated. But we're dealing with a resistance problem here, which turns your formula on its ear. In fact, we want to deal with the square-root of the resistance at any given point along the track. After all, you did measure the resistance of a Stew-Mac pot in your tests, you didn't measure any voltages at all, right? Let's play around with a 500K linear taper pot. When you rotate the wiper halfway, it's sitting at 250K ohms, right? This is important: there are 250K ohms between each end terminal and the wiper, or to be more emphatic, there are 250K between the "hot' and the wiper. Keep that in mind..... Now, let's say that you wish to produce a pot that makes your amplifier sound like it's more powerful - what do you do? You move that 250K point to the position where 3 on the knob will encounter that value, don't you? You just made the pot so that it will decrease it's wiper-to-hot value more quickly, because so much more of the total resistance was scrunched into the first 3 numbers on the dial - the remaining 250K between "hot" and the wiper are spread across 3 to 10 (they are no longer confined to between 5 to 10, as in the linear pot). This is the definition of an audio taper, and it has nothing to do with logarithmic functions, but it turns out that in formulating just where to position the half-way resistance point on the wiper's track, the resultant curve was very close to logarithmic. (BTW, Keen and Westone show this as the sharp-cornered curve labeled the 'straight-line audio taper'.) With a log or audio taper pot, turning the knob from about 3 or so, up towards 10, the amp's volume will not increase more than several percentage points - the greatest part of the total resistance has already passed under the wiper! To recap, the wiper on a volume pot sees near-infinite resistance to signal flow at 1 on the knob. As the wiper is turned up towards the max level (10 on the knob), the resistance it sees is reduced. The amount of change in resistance (for a log taper) is such that the resistance must drop to the square-root of the previously measured value, at each point along the track. This won't happen if the taper is linear, and it sure won't happen if the taper is such that the wiper encounters most of the overall resistance between the mid-point and the "hot" end of the track. I ask you, how could a wiper that sees much more than 50% of the total resistance between midpoint and the 'hot' end work to give you a louder amp at a lower knob setting? sumgai p.s. For the record, the Westone page on pots uses the same graph as Keen, but he labels it correctly - the first few degrees of wiper rotation yields a much greater amount of change in resistance than movement over the remaining track area. Keen involves himself with a ratio, and that's not conclusive of anything. The Westone chart more or less duplicates my results, and refutes Frank's measurements, check it out at Frank's measurements. (The swingguitars site is amost incomprehnsible to me, anyone else get gibberish on that page?) p.p.s. In going over previous posts here, I expressed such things as "viewed from the ground terminal". I see now that this may have caused some confusion to you viewers, for which I must apologize.
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Post by UnklMickey on May 23, 2006 17:44:25 GMT -5
....(I've written and re-written this 4 times now, so I'm getting a little tired....
....Let's play around with a 500K linear taper pot. When you rotate the wiper halfway, it's sitting at 250K ohms, right? This is important: there are 250K ohms between each end terminal and the wiper, or to be more emphatic, there are 250K between the "hot' and the wiper. Keep that in mind.....
...With a log or audio taper pot, turning the knob from about 3 or so, up towards 10, the amp's volume will not increase more than several percentage points - the greatest part of the total resistance has already passed under the wiper!...
......The Westone chart more or less duplicates my results, and refutes Frank's measurements, check it out at Frank's measurements.... 4 times and you still just don't seem to get it right. both charts support Franks results.what they refute is YOUR results. on Rod Elliots chart we will be using either the yellow or the red line NOT the purple one. on R.G. Keen's we will be using either the line marked 2 or 3. NOT the one marked 4. let's take YOUR EXAMPLE of a linear pot set a 5. equal resistance on both sides of the wiper. that give a division ratio of 50% on audio pot the 50% division ratio occurs at 80% ~ 90% of the rotation. that would be about 8 1/ 2 on a 0 to 10 knob. Franks results show 43% of the resistance at 7 1/ 2. so by the time the rotation is continued to 8 1/ 2 the division ratio should be considerably more than 50%. but the bottom line here is: the 50% point on an audio taper pot occurs at 8 ~9
not at 3 as you incorrectly suggestyou incorrectly infer that an audio taper pot "makes your amplifier sound like it's more powerful" on the contrary, it is designed to give you better control. the point where a linear pot is set at 5, won't be reached on an audio pot until it is set at 8 ~ 9. as i previously stated your results look like they come from a reve rse semi-log pot. (it is not listed on these charts, but the curve would be between those for the linear and the REVERSE or ANTILOG curves. so again i suggest you try an AUDIO TAPER pot. if you ever do try one, you might find that you like it. unk
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Post by JohnH on May 23, 2006 18:06:19 GMT -5
Interesting discussion. I must say that every pot that I have in my spares pot is either linear, or (if called 'log' or 'volume') follows a curve somewhat like number 3 on Keens chart. Typically based on my measurements, if wired in the usual way as a volume control, at mid position (5 on the dial) there is about 10% of the total resistance from the wiper to the grounded lug, and 90% from wiper to the other outer lug.
John
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Post by sumgai on May 24, 2006 16:24:14 GMT -5
unk,
Well, that would mean that someone has gone and changed the definition of an audio taper since the days it was invented (back in the 1920's, I believe).
But all things considered, that's entirely possible..... after all, nobody died so that I could be put in charge, which means that no one ever asks me if it's OK to make wholesale changes like this. Too bad, I'd elect to maintain continuity between the generations of electronics techs and engineers, instead of turning the industry on it's ear every so often, just to see everyone scurry around like rats on a sinking ship.
More precise control, indeed. unk, in the 1920's, they didn't give a d@mn about control, they felt blessed-by-the-Lord lucky to get a radio to even work at all - precision control was the least of their worries! And Fletcher and Munson weren't even teenagers yet, so no one had even an inkling of a clue about how the ear responds to volume level changes..... Bell's work was still in the throws of peer review, and it was by no means being univerally accepted as fact.
But, since neither your no I were there, I don't expect anyone to believe me on any of this. Either.
Maybe someone should go fetch back that pail of dirt and dust they brushed off of this old fossil, and just bury me like they found me. You gotta admit, it would make life a lot easier for you youngsters.
sumgai
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Post by UnklMickey on May 24, 2006 16:35:22 GMT -5
...Maybe someone should go fetch back that pail of dirt and dust they brushed off of this old fossil, and just bury me like they found me. You gotta admit, it would make life a lot easier for you youngsters.
sumgai no need to "throw the baby out with the bathwater." unk
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Post by ashcatlt on May 25, 2006 12:18:59 GMT -5
wow, this sure has gone way out there, hasn't it. It's interesting reading, but I think I've gotten the answers that I was looking for.
But now I've got a new question. I went and did some testing on this guitar. Extended the wires from the humbucker out the jack plate so I could try it out in series, parallel, and single-coil mode without having to take the thing apart each time. I recorded three very similar guitar tracks side-by-side so I could do A-B-C comparisons and see what I liked best.
What I found in this test is that the parallel and single-coil positions sound very similar, with the parallel being a bit hotter and just a touch darker, which is to be expected. The series setup was a little hotter than the parallel but quite noticeably darker. So I decided that I would probably wire this guitar to have the HB positions on the switches be in parallel. This way it doesn't actually change the tone of the guitar, but gives me a bit more crunch. And if I want that series HB sound I can always just grab my LP.
Now, the question comes in when I put the thing back together. (My budget doesn't allow me to purchase the parts I need to actually wire this thing yet, and I want to be able to play it in the meantime) I wired the HB in parallel with itself and reconected it to the switch, screwed the whole thing back down and started playing. Now I'm finding that the HB is noticeably quieter than the two singles which came standard in the guitar. When it was in series it was quite a bit louder. Does this make sense?
I didn't actually solder the thing back together, just twisted the wires and taped them up, so it is possible that one of them came undone while stuffing it back into the cavity and I'm only getting one coil out of the thing.
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Post by UnklMickey on May 25, 2006 13:11:48 GMT -5
Ashcatlt,
who ARE you?
and what do these questions have to do with audio taper pots?
oh ... oh yeah ... 3 Humbuckers, No Pots ... yeah, i remember now!
if you lost connection to one of the coils of your HB it will sound even quieter than in parallel.
if you made an error and had the 2 coils OoP, that would make it REAL quiet, and extremely thin. but that would have been so obvious, you would have known it right from the start.
you can do the "screwdriver pull-off test to make sure both coils of the HB are contributing to the signal.
if you connect a D.C. voltmeter to your cable, you can even verify that the phase is the same on both of them.
i would expect a paralleled HB to be about the same volume as a SC, all other things being equal.
but all other things are rarely equal.
some SCs blow away a vintage wound HB, and some HBs make a vintage wound SC seem like a whisper.
also, is it possible any of your pickup heights have changed?
that can a big difference in the relative volumes.
unk
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Post by ashcatlt on May 25, 2006 13:28:51 GMT -5
I don't think the heights have changed, I never took any of them out. This pickup is supposed to be over wound and extra hot (if you believe the hype) and - as I stated - was very much louder than the original SC's when I had it wired in series with itself. I don't think it's out of phase. I guess what I really need to do is just take the thing back apart and verify that everything is wired correctly and making good connections.
Thanks.
I'll let y'all get back to your argument now...
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Post by spitfire23bc on May 25, 2006 13:51:13 GMT -5
Out of interest, what sort of thing are the anti-log pots used for?
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Post by UnklMickey on May 25, 2006 14:28:30 GMT -5
...I'll let y'all get back to your argument now... let's not, and say we did.
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Post by UnklMickey on May 25, 2006 14:30:12 GMT -5
Out of interest, what sort of thing are the anti-log pots used for? good question. i don't know. unk
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Post by sumgai on May 28, 2006 14:56:06 GMT -5
Well, It took awhile, but I've mended well enough to return to what became a sore spot for awhile. I'd be fooling only myself if I tried to say that I wasn't affected by all this negativity, and the rather lop-sided "discussion" we've been having. But I'm back, and I'm gonna give it another shot. The reason? spitfire's question about what an anti-log pot is user for. Thanks, spit, I owe you one, you made me wake up the $#%^ up, and get off my keister. So here's the real-meal-deal, as can be verified only by asking a very, very old former radio technician, possbily an old Electrical Engineer, or best of all options, an older HAM operator. And since I know that at least one of you is going to ask, yes, at the tender age of 13, I was W7OLN. I am now KA7ZPS, inactive. What I've been talking about was a volume control as described for that purpose in the teens and 20's of the 20th century. When radios were first brought out, there was mighty little danger of receiving a signal so strong that one needed to reduce the volume. As tubes were introduced, radios became much more powerful, and they went from barely audible to blaring at painful levels (or nearly so). Enter the need for a volume control. In those days of early tube radio sets, the volume control was a natural adjunct to the On/Off switch, as that switch was already a rotary knob anyway - toggles as we know them having yet to be invented, and exposed knife switches being dangerous in the average home. This was done just to make sure that the next time the radio was turned on, it wouldn't be blasting out at full volume. Now, for purposes of general discussion, this is a "consumer oriented" volume control. It's purpose was to give the listener an enjoyable experience at a comfortable level above the threshold of audibility - not to reduce the volume from ear-splitting levels. That's important - it was the mindset of both listeners and radio designers of that day and age, and it still holds true right up to this very moment. Flash-forward to today. We now have a plethora of devices that need to be controlled, but they can be broken down into two main groups: those that need to be turned up to be usable, and those that need to be turned down for the same reason. A very brief glimpse at your own equipment will give you an example of each - the guitar needs to be up in order to be useful, and the amplifier needs to be down, if we want to keep our hearing intact for a few more years. So, to cap it off, there are two needs for attenuation, regardless of how we, or I, label those needs. One requires that we think in terms of reducing the current volume level (the guitar), and some folks think that a log taper (or an audio taper) will work best here. The other job requires that we increase a volume level from inaudibility (the amp), and that requires what is being called by everyone here an anti-log (or reverse audio) taper. It so happens, as you might have guessed, that I come from the old school, where the consumer version of things was the way we all looked at the attenuation scene. In that reality, the audio taper pot was used to make a radio appear to be more powerful, because it got louder much more quickly than a radio using a linear pot. (The same volume level was reached at a lower number on the knob.) It doesn't take a rocket scientist to see that louder equals more volume, which equals less resistance between the wiper and the hot terminal. To seal that particular deal, why else do you think the engineers of the early 20th century would have called it an 'audio' taper?! Sadly, in today's terms, we have forgotten this little factoid. In fact, I am forced to admit that no one has posted any history like this on the web at all, so it would appear to the casual researcher that I have been blathering myself blue in the face because I like to hear my jaw muscles squeak. Which is why I said at the outset, a field trip to the old-folks home in your area should set you straight. I now suspect that all this "new-think" in pot tapers changed about the same time as the labeling did, some 20 or 30 years ago. I suspect that the manufacturers were getting tired of making and stocking two different things, when from their standpoint, they could do the same job with just one part. Of course, I don't know that, I'm just speculating here, but it does make sense, in a perverse kind of way. Now, what you've all been waiting for.....FWIW, I have adjusted my ways of thinking to the modern reality. It was just another wake-up call, one that I didn't particularly like (yet another familiar old friend, gone and soon to be forgotten), but that's just the way of it. I hereby recant all previous references to linear versus logarithmic potentiometers, and everything I have ever said about the topic is hereby declared null and void. That should put paid to the topic. sumgai
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Post by spitfire23bc on May 28, 2006 17:12:51 GMT -5
A nice new LP should do the trick....! ;D
No worries; interesting answer! Cheers.
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Post by sumgai on May 29, 2006 6:00:02 GMT -5
spit, OK, what artist do you want on that 33 1/3 rpm LP recording?
;D
sumgai
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Post by spitfire23bc on May 29, 2006 9:38:27 GMT -5
I did set myself up for that one particularly well... PS: Dire Straits
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Post by sumgai on May 29, 2006 14:17:28 GMT -5
spit,
The Post Office won't insure mailing a vinyl record anymore! So, howza about I send you a nice 'Round Tuit' as a consolation prize, will that help?
;D
sumgai
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Post by spitfire23bc on May 30, 2006 13:12:02 GMT -5
Ack - I've already got one
What's postage like on one of those...?
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Post by ashcatlt on Oct 30, 2006 1:22:28 GMT -5
I finally did it! Thanks to everybody for all your help. It's not really anything like what I originally posted above.
I abandoned the idea of the stepped attenuator because I thought it was a silly reason to spend $20 on a switch.
I used 2p5t rotaries for each pickkup so I have 1-off 2-series 3-north 4-parallel 5-south for each HB.
The middle pickup has the opposite coils in 3 and 5 so that position 3 on middle and neck or bridge give humbucking. Position 5 is for combining neck+bridge. I know I really only needed that on one pickup, but I decided it was easier to buy 6 of the same switch (got a couple other projects coming up) than several different.
The original strat 5-way is nothing but a kill switch. Position 1-2 on, 3-5 off.I've changed this so that it includes a "tone kill" cap on one end, and kills the sound by shorting, so it's now 1 on, 2-4 off, 5 one w/tone on "0"
I originally had the shields from each of the pickups connected to the string ground, the body of the strat switch, and the shield of the cable running to the jack. They met signal ground at the jack. I lifted all the shields when I originally found that none of my kill positions were working (they weren't killing, I could still hear the signal). Do you think I should reconnect them? Will the scheme I originally had for this cause any problems?
The L'il Killers are pretty awesome, btw. Sound great in all the switch positions. Of course, this guitar has no pots, so part of the wide open "alive" sound I'm getting might have something to do with that.
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