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Post by dannyhill on Feb 16, 2013 16:06:48 GMT -5
Hi guys,
Will a single volume guitar sound different if in the middle (parallel pup) position I add a bass cut cap before the volume instead of the same cap before each individual pickup? Cheers,
Daniel
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Post by JohnH on Feb 16, 2013 23:22:46 GMT -5
I'd think it would work ok, but it may sound different to having seperate ones for each pickup. Not many guitars have bass cut, so there is not so much experience to draw on
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Post by dannyhill on Feb 17, 2013 4:07:38 GMT -5
Hi John,
Thanks for that. Single volume pot guitars have them usually just before the volume pot and two volumes (and 3?) would have them after the switch, with the more volume drop.
I was trying to get my head around the operator on the waveform, before and each and then combine vs combine and then operator on resulting waveform. Cheers,
Danny
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Post by dannyhill on Mar 3, 2013 16:31:37 GMT -5
I mean how do the two signals combine in parallel? Low end removed? So with a high pass filter even more removed? If we filter BEFORE they combine, less low end to be removed on combining? This is something not clear to me....
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Post by JohnH on Mar 3, 2013 17:32:32 GMT -5
The best way to answer these interesting questions is to run a model.
With the right program, this is easy and quick and free. Personally I use 5spice, and I would have to run that in order to respond to questions such as gretsch wiring and bass cut interactions.
So, instead of me doing that, how about setting that up for yourself? We can work out a generic guitar model that suits your pickups, you can see the response, then a few changes to test the ideas and explore questions.
John
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Post by ashcatlt on Mar 3, 2013 20:59:23 GMT -5
Let's say we have two identical signals. We assign a value to the energy in each at a given (low) frequency. Doesn't matter the actual V level, they're both the same, and this will act as our reference, so we'll call it 0db.
If we use an active mixing scheme to combine them without insertion loss, the output at that given (low) frequency will be +6db = double the V of either input. If we then apply a high pass filter (active again, because insertion losses will complicate the calculations) which is down -3db at that given frequency, the final output will be +3db at that frequency because +6 + (-3) = +3.
OTOH - If we apply the identical filters to each of the signals before the mixer stage, each signal will be at -3db. Mix them together and we get +3db again because -3db x 2 = -3db + 6db = +3db.
This implies that there will be no difference.
But it can't possibly be that simple. The interactions inside a passive guitar are pretty complex. Modeling (as JohnH suggested) will give more reliable answers.
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Post by sumgai on Mar 4, 2013 3:58:03 GMT -5
ash,
I was gonna pick apart your treatise, it needs such, but in point of fact, you were merely illustrating an example. So I'll settle for just pointing out that dB is defined as a ratio between two levels, be they voltage or whatever. But without an initial level in some kind of measurable units, there's nothing to compare. Hence, it wasn't kosher to start your illustration with "Doesn't matter the actual V level.... this will act as our reference, so we'll call it 0db".
Also, after illustrating your calculations with a mixer, you did surmise toward the end that doing those calcs inside of a guitar (no buffering, no signal isolation) is a complex calculation. I beg to differ - it really is a straightforward calculation, and using dB helps, though not necessary.
HTH
sumgai
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Post by dannyhill on Mar 4, 2013 4:14:22 GMT -5
Hi there, JohnH - You're quite right. If I get a few hours to myself I might have a go at it. I just thought anecdotal information might be out here on the ether, so would be a five minute job. ashcatlt - Interesting, I don't understand how you go from 0dB for each pup to 6dB for combined.....is there something subtle I missed here, or is there a numerical oversight? Cheers, D
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Post by ashcatlt on Mar 4, 2013 5:01:27 GMT -5
There's ratios all over the place! The ratio of one input to the other is 1:1 = 0db. I can call anything I want my 0db reference. It could be the voltage drop required to produce 1W through a 600Ω load. It could be 1V or 9V or 16V. But it's not. It's the Voltage of these identical inputs at this given frequency. As long as our circuits are linear the absolutes don't matter. Call it dbAsh.
Mixing the two inputs produces double the output. A 2:1 amplitude ratio is very close to 6db. It takes a small leap of faith to accept that -3 + (-3) = +3 unless you realize that you will be doubling whatever voltage -3db represents, so that the output has a 2:1 ratio to either of the inputs.
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Post by dannyhill on Mar 4, 2013 5:08:22 GMT -5
Hi ashcatlt,
Thanks for pitching in! What is not clear to me, is why voltages in parallel sum in your calcs. Surely, as the pups are in parallel they combination should follow a 1/V relationship?
D
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Post by sumgai on Mar 4, 2013 19:50:04 GMT -5
Danny, ash is correct in all of his math, in so far as active circuitry goes. What I was typing (very) late at night was directed towards getting ash to reconsider his answer. He did get to the part where active and passive are not calculated in the same manner, but then he left out the stuff you wanted/needed to see. I'll not go into any discusison vis-a-vis dB just now, that'd be a good way to derail this topic. If you're curious and want to know more, let me suggest that you trot off to Google-land. If that doesn't work for you, then start a new thread and we'll see what we can do for ya. As it is, you're correct in assuming that 1v from each pickup doesn't amount to 2v at the output jack. But the reason ash said "it's complex" is that we have to consider how that 1v was generated by a passive component... in this case, the pickup. Suffice it to say for now that when we combine pickups, we are actually mating two coils (inductors) (in parallel, in your case) and the interaction between the two coils will effectively reduce the apparent output of each pup. Certainly not down by half or whatever, but down some appreciable value nonetheless. So how come's that? We're talking about 'apparent' values here. That means that more than just we can measure on a meter, or see on a 'scope, but as usual, there's more to the story. In our case, that of a guitarist, we're already familiar with most of the concepts. Consider: When we look at the output of a Neck pup versus the output of a Bridge pup, we see different harmonic factors. Remember, all of the harmonic content as well as the fundamental go towards our reading of "1v" or whatever we got at the outset. So now let's look at the combined output... and we see that some of the harmonics have been reinforced, and some of them have been depressed. By way of example, if the second harmonic is half of the level of the Fundamental on each pup, then no, the two values won't add, due to inductive losses, but they will remain close to the initial level. Whereas, if one of pups has a much reduced second harmonic, then the two values combined will be reduced overall, for a total output level (of the second harmonic) at a very noticibly reduced level. Now, go back and do that "exercise" for say, up to the 20th harmonic. When we see the "plot" or "graph" of the combined output, we notice that total harmonic content is reduced overall. This explains why, when we measure the voltage output at the jack, we see only about the same level as with one pickup. It might vary by 10% or so (up or down), but overall, it didn't change so much. Now, let me address that concept of "inductive losses". Or rather, I'd rather not. This isn't necessarily complicated, but then again, I tend to spout off way too much on deeply technical topics like this. Check me out with Google here: As the field strength builds up in one coil, due to string vibration, the field strength of the other coil is not in step - it may be decreasing, increasing, or staying right with the first coil at any given moment in time. Remember, time is the inverse of frequency, and that's why I stressed that part about 'any given instant'. If we could take a series of snapshots, we'd see that the field strength of each coil is affecting the overal inductance of the total circuit. Knowing that impedance is dependent on inductance (and frequency), and knowing that impedance determines the overall output voltage at the jack, we can put that series of snapshots together and finally see, in almost real time, just why the output voltage did not increase when we switched to two pickups together (be they in series or parallel). And for those who want to say "but in series it is louder". No, I say again, that's a result of apparent voltage. The harmonic content that's been reinforced is, overall, lower in the total spectrum. (IOW, the 2nd through the 8th or 9th harmonics are stronger than when in parallel, and the rest of them are reduced even lower than when in parallel.) This explains why we think the signal is stronger, though with less treble. It's really not noticebly stronger, according to the meter, but we perceive it as such, due to the way our ears are attuned to the frequency spectrum. I leave as an exercise for the reader to work out how inductane is affected when coils are mated in parallel, and again in series. And so on and so forth, blah, blah, woof, woof, yadda, yadda, yadda. Had enough? sumgai
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Post by ashcatlt on Mar 4, 2013 22:16:27 GMT -5
I thought I was pretty careful to specify that I was trying to (over?) simplify the question by specifying that I was talking about active mixing and active filters to avoid insertion losses. Maybe I went a bit too far there?
In order to understand why two pickups mixed passively don't add up to double the voltage of either one, I think it's pretty easy to just look at it like a resistive voltage divider.
Consider the case where we have two identical pickups connected in parallel. We're not adding the low-cut filter yet, just flipping a Tele switch to the middle position.
Analyze the contribution of the neck pickup to the overall output. It generates some voltage Vn. This can be thought of as going into a voltage divider with the impedance of the neck pickup as the "top" resistor, and that of the bridge parallel to any controls and the amp/pedal input as the "bottom". Assuming that the controls and amp impedance are chosen to be acceptably high, they will only reduce the impedance of that "bottom" part by a very small amount, so that our voltage divider is essential made from two equal impedances. This will divide Vn down to half.
Do the same for the bridge pickup and you get the same answer. It's contribution is one half of Vb.
If we assume that for some given frequency Vn = Vb, then Vout= 1/2 Vn + 1/2 Vb = 1/2 Vn + 1/2 Vn = 1/2 (2Vn) = Vn.
Yes, there is an inductive (and also a capacitve) component to each coil's impedance which makes the Z change for different frequencies, but you'll recall that we specified that these two coils were identical, so that the impedance at any given frequency will still be equal, and we still have a 1:2 voltage divider, so inductance and capacitance doesn't really impact this answer in this one specific (and not particularly realistic) instance.
The phase difference between harmonics caused by different positions along the string does impact our perception of how the output changes, but it doesn't actually change the action of the filter, and it changes depending on where on the string you're fretting, and is difficult to predict.
TBH, I think that the actual low-cut action will be approximately the same either way.
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Post by sumgai on Mar 5, 2013 0:03:36 GMT -5
ash, Well, we both complicated it more than necessary.
Danny, Here's the Real Meal Deal. In hooking up a bass-cut control, it goes between the pup "hot" wire and the output (selector switch, volume control, whatever). Now, if you have a cap in series with each pickup, and combine the two pup in parallel, then it stands to reason that both bass-cut capacitors are also in parallel. Caps in parallel add in value (and in series, they divide in value). Hence, a 5nF cap across a Master bass cut will not sound the same as a pair of 5nF caps, each across their respective pickups. The total value of that pair will be 10nF, which as luck would have it, will allow a lower frequency to pass through to the output, when both controls are turned to 'maximum bass cut'. Obviously I've kept the same values for that example. Equally obviously, you can tailor the cap(s) in either application to get the results you want. But consider - if you opt for individual controls per pickup, then you actually have two caps in parallel, either one of which will act upon the entire signal. When rotating the pots, you get various values, and they will affect the tonality, but that's the exact same thing as if you had only the one Master Bass-cut cat with a higher value .... only with an extra knob to twiddle. Without a drawing, I hope I was clear. If not, then I'll whip something up, but you have to ask first. HTH sumgai
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Post by dannyhill on Mar 5, 2013 4:45:34 GMT -5
Hi guys,
Right, just picking bits of my head up after it exploded. So this is obviously non-trivial.
What I didn't realise was that when two pups are in parallel (e.g. middle position of a LP) if only one has a cap between its hot end and its volume pot then the cap affects both it and the other pup. Or did you mean just for single or master volume (Fender) wirings?
So if I put one on each pup, make them switchable I can have: Individual - 0.0033 or off? Parallel - 0.0033 or 0.0017 or off? series - 0.0033 or 0.0066 or off?
Then if I use a triple throw with off/0.01/0.0033 .....oooh!
BTW I didn't realise until last night that in the usual Gibson dependent modern wiring, in the middle position, tone controls are independent once they are both rolled off from 10 a little. Cheers,
Danny
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Post by JohnH on Mar 5, 2013 5:15:28 GMT -5
Might I suggest that what this thread needs is some form of diagram, because Im not convinced that everyone is talking about the same thing!
A gentle suggestion only
John
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Post by dannyhill on Mar 5, 2013 6:22:12 GMT -5
Hi again, I had in mind something like this (excuse the bad drawing), now edited TWICE: The caps labelled 1 and 2 would be switchable using one of these which someone previously drew here from another thread: The "From Vol" would be replaced with "From hot pup end" and the three ends would be tied to another wire, 0.01 cap and 0.0033 caps and the loose ends tied together at 'in' side of volume pot. This would be done for each pup. Does this make sense? Danny
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Post by ashcatlt on Mar 5, 2013 7:10:08 GMT -5
Well there's a number of issues there. The T controls are wired all crazy and will work as bass cut and probably volume to some extent, but not typical hi-cut tone control. For very similar reasons the V controls will not behave in a satisfactory manner.
I could have sworn that we were wondering whether there would be a difference between a cap per pickup and having one master bass cut. A couple of different perspectives have been given. I am prepared to be proven completely wrong, but I definitely think that one of us needs to Spice this up and post the results.
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Post by dannyhill on Mar 5, 2013 7:26:21 GMT -5
Fiddle sticks. I've gone back and updated it now.
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Post by sumgai on Mar 5, 2013 13:52:41 GMT -5
Danny, Fiddlesticks? FIDDLESTICKS? Smoly Hokes, I haven't heard (or read) that one in a coon's age! Are you perhaps showing your age here? I'm looking at your (presumably) edited diagram, so I can't see what bothered ash. (But it must've been something, to confuse him enough to make him swear!) Your latest scheme will work, but I have a tip, and a question. In reverse order - why are you now considering a switch for monkeying with the bass-cut cap? Did you suddenly become afraid of potentiometers? And two: I'd put the "standard treble cut" tone control on the pickup side of the bass-cut cap. The way you have it now, the bass-cut affects the treble cut (they're in series), and that could end up with a weird "cross control" effect, where each of them affects the other. Better would be in parallel, as I suggest, which will tend to isolate the two different tone circuits, and hopefully reduce or eliminate any cross-controlling. Otherwise, your drawing is fine, both electrically and visually. HTH sumgai
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Post by dannyhill on Mar 5, 2013 16:02:19 GMT -5
Hi sumgai,
Something my gran used to say when she was frustrated by something. That and bother. F-iddlesticks and B-other being substitutes for other words, perhaps expletives? :-)
"I'm looking at your (presumably) edited diagram, so I can't see what bothered ash. (But it must've been something, to confuse him enough to makehim swear!)!"
I had the strangle cap on the wrong wires.
"Your latest scheme will work, but I have a tip, and a question. In reverse order - why are you now considering a switch for monkeying with the bass-cut cap? Did you suddenly become afraid of potentiometers?"
Well, these can: Act as bass cuts i.e. strangle caps in a single pup selection or parallel pup selection Give half out of phase if I use a 0.01uF, or does this only work when two pups go to the same volume pot? Give series taming.
I use switches instead of potentiometers for less load and also to enable to jump to preset capacitances. Or did I not understand your question?
"And two: I'd put the "standard treble cut" tone control on the pickup side of the bass-cut cap. The way you have it now, the bass-cut affects the treble cut (they're in series), and that could end up with a weird "cross control" effect, where each of them affects the other. Better would be in parallel, as I suggest, which will tend to isolate the two different tone circuits, and hopefully reduce or eliminate any cross-controlling."
But this is how it is in a Jaguar and a vintage Rick. Bass cuts always come off of the switch or before entering the switch off of a pup. I always understood that a bass cut before a volume pot cuts the volume less then coming after the 3 way gibby switch. ? Cheers,
D
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Post by ashcatlt on Mar 5, 2013 19:47:25 GMT -5
This is the reason we try not to edit things in place, especially not after they've been commented upon. Ruins the continuity of the thread and leaves people wondering what the heck I was talking about.
Are the pots marked T meant as typical high-cut tone controls?
The problem I see is that you've got the T pot, its cap, the presumed bass-cut cap (caps 1 and 2) and the pickup all in series with the V pot parallel to the pickup and its wiper going to the switch. The pickup only finds "ground" through all that crap. With V all the way up and T all the way down you have both caps in series acting as bass cut. With V all the way up and T all the way up you've got the two caps plus the very large resistance of the T pot between the pickup and ground. With V all the way down and T all the way up you've got the pickup hanging "upside down" from hot. With both pots all the way down you'll have silence in those high frequencies passed by both caps, but low frequency noise (like our old friend Buzz) going to the output.
That is all kinds of yuck!
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Post by sumgai on Mar 5, 2013 21:26:17 GMT -5
Danny, When you called it a 'strangle cap', then your wiring scheme made sense. As you note, this is the way it's usually done by other, bigger, companies. But I don't recall that you actually used that term before, nor did I see where you switched horses in the middle of the stream. Perhaps I missed it, and if so, then I apologize. Switches are fine, and if you're willing to experiment a bit with cap values, then you can indeed find two or three useful combinations that are easy to select (that business about using a 6-position rotary on stage scares me no end!). Just be sure to leave a "non-strangle" position, as there will be times when you'll want to go "all out". HTH sumgai
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Post by dannyhill on Mar 6, 2013 5:33:35 GMT -5
sumgai, Bass cut cap=strangle cap=high pass filter cap All the same thing. Yes the idea for each switch is to have no cap (wire)/0.01/0.0033 or 0.0047 (depending on the guitar). ashcatlt, I don't follow you. I have wired it as Fender do for their Jags. Cheers, Danny
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Post by newey on Mar 6, 2013 6:12:27 GMT -5
Your diagram is different than Jag wiring in several respects. In a Jag, the "strangle" cap only operates on the bridge pup. It is wired, in series, to the tone control first, and then to the Vol. The tone pot cap in the Jag is in parallel with the strangle cap, not in series with it. Also note the use of a 56K resistor from the CW lug of the tone pot to the wiper. The negative lead from the pups are grounded, as is the Vol pot CCW lug. I think Ash's problem with your diagram (or part of the problem, anyway) is that you are showing the pickups wired to the CCW side of the Vol pots, without showing those points to be grounded. Did you perhaps "stylize" the diagram and omit showing some of the grounds?
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Post by ashcatlt on Mar 6, 2013 7:10:25 GMT -5
Yeah. I still get confused as to which lug is 10 and which is 0. I guess I had it backwards for the V control? That's probably partly because your original diagram had the strangle caps on the other wires...but I guess only two of us know that...
Frankly, though, that just makes it worse. At least one lug of each V pot and one wire from each pickup needs to reach the jack sleeve via straight wire!
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Post by dannyhill on Mar 6, 2013 7:23:33 GMT -5
I've updated the diagram again. All -ve leads are now grounded. In a jag the strangle acts on one or the other or both pups. To my uneducated eye both the Jag and my setup have the two caps in parallel with each other. But what do I know? EDITED: Good question about the 56kOhm resistor! I looked into this and apparently its there to keep the peak response at 300Hz, similar to a Jazzy with the larger 0.03uF but w/out any mid-range hump.
What is it with the pickup going to the tone pot first and the wiper output going to the in of the volume pot as opposed to the more typical other way around like on a Jazzy? Is this because of the strangle cap? If I wanted to wire a strangle cap into LPs, teles, strats etc should I wire them this way too? How does that affect vol/tone control interaction on a jag? Seems all the Jags are wired that way. ?
Danny
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Post by ashcatlt on Mar 6, 2013 9:11:13 GMT -5
Yep, looks like the Jag strangle cap strangles everything if it strangles anything. The old Rickenbackers supposedly had a cap permanently in series with the bridge pickup.
That potentiometer style of T pot wiring is pretty strange. Don't really see that about anyhere. I'd think that 56K resistor helps keep there from being upwards of 1M resistance between the pickups and the jack!
Neither the Rickenbackers nor the G&L with bass cuts do this, so I don't think it has anything to do with that...
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Post by dannyhill on Mar 6, 2013 10:22:27 GMT -5
"That potentiometer style of T pot wiring is pretty strange. Don't really see that about anyhere. I'd think that 56K resistor helps keep there from being upwards of 1M resistance between the pickups and the jack!"
Can you explain how the 56K stops a resistance from going above 1M? Is that because its in parallel with whatever load there is from the in side and the wiper which always makes it overall less than 1MEG? But then there is 1MEG to go in series afterwards at the volume pot? Or am I still a way from getting my guitar electronics diploma?
"Neither the Rickenbackers nor the G&L with bass cuts do this, so I don't think it has anything to do with that..." So, no problem in putting my strangle caps where I wanted them then?
D
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Post by newey on Mar 6, 2013 16:12:06 GMT -5
OK, we've now had the second diagram change "on the fly". As pointed out above, please stop doing that! This thread is fast losing what little semblance of continuity it had . . .
You are both right, I misspoke on the strangle cap, it can affect both pups.
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Post by sumgai on Mar 6, 2013 21:41:31 GMT -5
Danny, If you edit a mistake such as spelling, or an omitted word that changes the intent of a sentence, that's one thing. But if you edit a diagram, which is the lifeblood of The NutzHouse, then you really need to allow for continuity. While it's hard to make ash look like a dunce (hasn't happened yet, except for that one time when he wore some wet socks on a beer-soaked concrete floor....), the fact is that as soon as we go down the page a few more postings, the diagram in question is 'lost' to view. That means lots of scrolling, back and forth, to try and follow the conversation. It gets really bad when the thread runs to a second page (or longer) then it's flip back and forth. FWIW, I can't recall anyone offhand that has stood up and said "Hey, I like jerking my browser window around all over the place!". IOW, please, don't do that no more, mmmmkay? Just post freshly-minted diagrams as they occur, and that'll be the end of any grousing here in The NutzHouse, I assure you. Now, as to that 56K resistor.... The intent here is to change the taper of the pot, nothing else. When the pot is at max (least amount of treble being cut), the resistor is out of the signal path - the pot is shorting it. But almost as soon as the pot is rotated downward from '10', then it comes into play, remaining in parallel with whatever resistance is found between the two terminals. That combined resistance will then be in series with the remaining portion of the pot's resistance element, and of course, also in series with the cap. Just for drill, compare this setup with the Tone pot found on the Rhythm section of the guitar - what do you notice first? And why didn't Leo just duplicate that Rhythm wiring over into the Lead section, instead of what we see, the 1MA pot with the extra resistor? Finally, about that switch that will allegedly 'multi-task' between bass-cut and HooP.... I want to see a full diagram of everything as you will assemble it, not just bits and pieces. I have to admit that I have my doubts at this point in the game, but I want to make sure that I fully understand your intent before I either pass benediction, or cut you off at the knees. HTH sumgai
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