(While I was writing this, Bob/bsee posted his comments. I completely agree with his thoughts and advice, but I'm going to post my more exhaustive discussion on the peghead style issue anyway, in hopes that we may finally lay this to rest.)
About all I can really address is the speaking length issue (I've only ever had one Alembic, and never played others more than a few minutes at a time).
This is going to be a bit hard to explain, especially after the next glass of wine... so let me get a few things out of the way before I forget.
- Yes, I am absolutely convinced, especially after fooling with a spreadsheet (again) for an hour or so, that the speaking length theory is in fact correct.
You may prefer the look of a reversed peghead; you may even convince yourself that it will somehow improve the tone, and (I'm not joking here) end up getting better sound out of it... or at least believing that you do. But it will not be because of the physics :-)
- Yes, I am aware that some people claim Hendrix played upside down because of the peghead effect on tension (in his case, perhaps to loosen the high E rather than tension the low?). While I believe he was an absolute genius on guitar, I don't think anyone ever accused him (or the people who make such claims) of being a brilliant physicist.
- Assuming you go with a 33 scale, do you intend to use 32 or 34 strings on this? Unless you plan to fool around with custom gauges, this is a fairly important factor. If you put 34's on a 33, they'll be looser than on a 34; 32's will be tighter. I think it is *really* important that you think about this up front. You can have a bass built any length you want, at no extra charge within reason, but getting custom strings is a completely different story.
- I recall a fairly interesting discussion about laminate complexity and combinations a few years back, which you could probably turn up by searching on hippie sandwich. Without any practical experience to back it up, my gut feeling is that by the time you have a mahogany core with coco bolo top, and a nice set of neck laminates (as you propose), I suspect you can pretty much treat the back laminate and accent layers as decoration.
Again, with no real experience to base this on, if you want the most benefit from the front coco bolo, I would be careful not to interpose too many, or thick, accent layers (plus the glue that goes with them). And if you want some combination of different factors, I sort of feel like you'd have a better chance of accomplishing that with different wood for the front and back, rather than sticking laminates between the core and outer laminates.
But I really don't know.
- Though somewhat off-topic, while looking for the fret spacing formula I stumbled onto a site that might be of more general interest:
http://www.bsharp.org/physics/stuff/guitar.html Haven't tried out the applets yet, but the discussion of overtones, and in particular how they are affected by plucking at different positions, is one of the clearest that I've seen so far.
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Now it gets ugly. I will eventually address the capo theory, which I agree has an intuitive appeal, but I believe it is flawed and it will take a while to demonstrate.
For those who believe in the speaking length theory, the magic formula is this:
T = 4 x (FxF) x L x M / 980621
where,
T = tension in kilograms
F = frequency in Hz
L = length in cm
M = mass of length L in grams
(I've offered a reference for this before, perhaps in my first ever post, but I'm not going to waste my limited energy looking it up right now.)
For the moment, let's just assume this applies to the open string frequency F, such that L is the distance between nut and bridge. And to keep things (relatively) short, let's consider a low B string, with a frequency of 30.868 Hz.
We know F and L, but T and M can be a bit of a problem - unless we happen to have a string for which we know one or the other. So I chose to use a set of TI Jazz Flats as an example, specifically the JF345 set, 34 scale with a low B of .136 gauge, which has a specified tension of 15.8 kp (assuming it is used on a 34 scale and tuned to low B).
(By the way, kp seems to mean kilopond, which has no U in it so do not confuse it with pounds. I'm a little unclear on the precise conversion factor, but you can at least get really close by thinking of 1 kp as 1 kilogram as 2.2 lbs of tension. I'm going to do everything below with metric units, just because they're better.)
With a little algebraic rearrangement, we can say that:
M = 980621 * T / (4 * F*F * L),
and substituting the values for the TI low B we get:
M = 980621 * 15.8 / (4 * 30.868*30.868 * 86.36)
M = 47.07 grams, for the speaking length of the string
As a sanity check, I weighed a brand new one, still in the paper envelope and untrimmed, and it was 62 grams. Considering how much I would need to cut off, plus the amount beyond the nut, we're in the right ballpark. As we go on, it becomes useful to know the weight per centimeter, which is 0.545 grams (divide by 86.36 cm = 34).
For those of you following along at home with your own spreadsheets, a good warm up exercise might be to see what happens to the tension, if we use this same string, tuned to the same pitch, on scale lengths of 33, 34, and 35 inches. Using the first formula I mentioned above, to solve for T, I come up with tensions of 14.88, 15.80, and 16.74 kp, respectively. In other words, same string and pitch, is looser on a shorter scale, and tighter on a longer scale. I don't think we have any arguments so far, right?
It's also somewhat interesting to see what happens if for some reason we wanted to keep the tension the same. Rearranging again, and eventually taking the square root to get just F itself,
FxF = (980621 * T) / (4 * L * M)
To keep the tension at a constant 15.8 (nominal for the 34 scale), we end up with a frequency of 31.803 on the 33 scale, or 29.986 on the 35. In other words, same string, same tension, on a shorter scale yields a higher pitch. This is consistent with the results above - to use the string on a shorter scale, we would have to tune it down a little, thereby reducing the tension - same string and pitch on a shorter scale is looser.
One last point before getting into the capo theory... so far, all of this discussion assumes we only care about the speaking length, and we have only been talking about playing the open string. So what happens if we play fretted notes?
I'm going to simply state as fact that within the speaking length, the tension is the same at any point on the string, and further, that simply fretting the string (assuming reasonably low action) will not significantly change the tension or effective length. True, you will ever so slightly stretch the string, and make it a tad bit longer, but we would be out at least a few decimal places in overall effect.
To use the simplest example, if you play a note at the 12th fret, then you reduce the length and mass of the remaining speaking length, each by half. Since tension remains the same, you must therefore increase (FxF) by a factor of four, which is the same as doubling the frequency. One octave doubles the frequency, so it works.
(I sure hope someone takes the time to check me on this stuff, because I'm at a point where I could easily make a typo or accidentally state something backwards...)
Okay... let's play the peghead game.
First, assume the bridge and whatever string length extends beyond it doesn't matter. We have some wonderful bridge design that locks the strings right at the bridge saddles, and can ignore any length or potential stretch beyond that.
Next, we are using exactly the same string we have been discussing so far. However, we now eliminate the nut, and just extend the fretboard up to the tuner (okay, the open string isn't going to sound very good, but just play along).
Here's the thought experiment, and the resulting numbers. Let's pretend we want to build a really long neck, so we can get some extra tension on the strings, and then put a capo on it where the nut would normally be. To emphasize the distance between tuner and nut, let's make it so that if we considered the full length from tuner to bridge, we would be placing the capo (nut) at the 5th fret.
If we build an instrument in which the 5th fret is 33 from the bridge (to go with the proposed design), the formula for fret spacing (look it up) says that we would need the nut to be at 44 (more than an upright!). That's slightly approximated, and the 5th fret would actually be at 32.957, but close enough, eh?
Now somewhere in here is where the confusion arises, I think. We want the note at this capo/nut position to still be a low B. That means we have to tune the open string 5 half steps lower, to an F# at 23.1245 Hz.
Take a frequency of 23.1245, scale length of 44 (convert to cm), grams per cm as specified above, plug them into the formula... and you come up with a tension of 14.85 kp.
That's the tension of this string, if we tune it straight from the tuner to the bridge over a distance of 44, such that we can get a low B at the equivalent of the 5th fret position.
If you scroll up a few pages, you will find that using this same string on a 33 scale instrument results in a tension of 14.88. So we are off by a whopping 0.03 kp, which I am convinced is due to rounding or minor errors in the constants by the time we get out to several decimal places.
So by extending the peghead something on the order of 8 inches or more(!), we end up with an essentially immeasurable difference in string tension. In contrast, just using the same string at pitch on scale lengths of 33, 34, and 35, results in a a full 1.0 kp difference per inch.
Honestly, I thought the capo theory was quite clever, and intuitively convincing - but I don't believe it holds up. The best way I can explain it so far, is that if you want to pretend the nut isn't there, so that you can somehow take advantage of extra peghead length, then you have to tune the string to a lower pitch to play the same notes in the same positions, relative to the bridge.
This is really, totally, separate from the physics of how the string itself delivers a particular pitch. What matters is how the string vibrates, given its mass, tension, and length, and that only applies to the speaking length. For a given set of those values (plus a mysterious constant), you will get a particular pitch.
It seems that people hope they can somehow change the tension, independently of the length and mass of the string, and get better tone out of it. No, this will just give you a different pitch.
You can change the distance beyond the nut as you please, but the bottom line is that you still have to get the tension of the speaking length to be the same. And - somewhat to my surprise, having never gone through the exercise before - if you approach it as if using a capo on a much longer neck, the math happens to work out very nicely.
Again, I could be wrong, and would appreciate it if someone would run through a similar exercise. Until then, you should probably just consider this my opinion.
Regardless... congratulations on your plans for a new instrument. It's a very exciting time, but the best is yet to come.
-Bob
