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Myth/science In Lutherie


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Let's all try a little thought experiment. First:

Basic physics problem. You have a wall, a horizontal surface, and a vertical drop. A string is anchored to the wall by means of a tension gauge, which measures tension. The string goes horizontally over the horizontal surface, goes over a pully at the vertical drop, and then goes straight down, with an object that ways X newtons on the end of it. What tension does the spring gauge read?

Answer: For a string, tension is the same on all parts of the string, therefore the gauge should read X newtons.

Now, from here, how can we modify this to be more relevant? BTW, I think the above is essentially Greg's argument...correct me if I'm wrong.

Now, the phenomenon of certain guitars being easier or harder to bend on IS true. Any physical model is either FALSE or INCOMPLETE if it fails to describe what IS KNOWN TO BE TRUE. In other words, no matter how good Greg's argument is, if it doesn't conform to the real world, it's not correct.

So...the question is....what are we missing here?

Consider: what if a string could be approximated by a spring?

IE, the more it is disturbed from its equilibrium length, the more force it will exert.

Now in this analysis there are two forces we are primarily concerned with:

1) The force that we exert on the string (tension, to bring it up to pitch).

2) The force the string exerts in order to get back to its equilibrium position.

My guess is that 2) is what's responsible for the perception of more or less bendability. Think about it, the string wants to be STRAIGHT and at a certain natural equilibrium length. If you put an angle into the headstock, the string is in a non-equilibrium position and will thus exert force (in this case on the nut).

In the case of an angled headstock, fretting or bending will put the string FURTHER from its equilibrium length. And from high school physics, the force the string exerts is directly proportional to the displacement from equilibrium length.

Shoot me down if I missed something.

Cheers

Yike

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Now, the phenomenon of certain guitars being easier or harder to bend on IS true.

I thought that 'phenomenon' had more to do with scale length? --That is, a shorter scale length is easier to bend, because presumably less tension is required to bring the string to pitch.

But you'd think there'd be other variables involved -- personally, I'm more comfortable playing a thin, narrow neck -- bend definitely feel easier on my Melody Maker's pencil neck. Maybe someone else needs a fat neck for bending.

And then there's radius...so a lot of that 'phenomenon' is subjective, and has to do with the player's hand and preferences.

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Now, the phenomenon of certain guitars being easier or harder to bend on IS true.

I thought that 'phenomenon' had more to do with scale length? --That is, a shorter scale length is easier to bend, because presumably less tension is required to bring the string to pitch.

But you'd think there'd be other variables involved -- personally, I'm more comfortable playing a thin, narrow neck -- bend definitely feel easier on my Melody Maker's pencil neck. Maybe someone else needs a fat neck for bending.

And then there's radius...so a lot of that 'phenomenon' is subjective, and has to do with the player's hand and preferences.

True, it is an incredibly complex issue. First let me address that what WE have been concerned with in this debate thus far is whether the string angle has any effect or not on the bendability. I've tried to isolate this issue, and of course there have been some efforts on the experimental end to verify once and for all what is or is not the case.

As far as other variables, I'll mention a few things:

Scale length: Erik mentioned somewhere that he felt for example that his Strat was easier to bend on than his Les Paul. He attributes this to the angles.

Scientifically you are right about scales, but perhaps the angle difference is more significant than the difference in scale? I may do the calculations/further analysis next weekend.

As far as fat vs thin, this is of course because of the difference in each player's hands...what the optimal size so that you have the most leverage advantage.

I personally do not have enough experience with enough guitars to say anything specific...but I know enough about mechanics to offer explanations.

Indeed, my high E string is a 12 gauge! My peculiar preferences diverge pretty far from the norm.

Anyways, the main point is that good experimental evidence is always the FINAL word in science. No matter how beautiful a theory is or however advanced/simple the mathematics are, if it doesn't match up to the real world, it's worthless. Of course, in our case one needs to pay special care in designing the experiment to isolate the variables in question.

While science is concerned with finding out first principles, engineering (in our case lutherie) is concerned with using those principles to accomplish a specific task. If we don't know the first principles, we're essentially wandering blind. And from the very name of this thread "myth/science in lutherie", we always are to some extent. Of course by nature, the making of music/musical instruments IS subjective, so you can't determine from first principles what sounds good or not.

Well actually we could theoretically, but we haven't discovered the first principles of what sounds good (ie the subjectiveness of good vs bad tone)! And that's probably a good thing to some extent.

Edited by boomerlu
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Ok Dokey,

Well first round-

First of all. I have no down force to speak of. This would NOT be playable. I also felt shakey about my tests because the strings felt like they were floating.

Gauge on both strings- .042

Scale length 25"

Distance behind bridge approx 3.5" (slight variance because of intonation)

longer string nut to saddle-25.05", saddle to tuner- 34"

shorter strng nut to saddle -25.11", saddle to tuner-28.0625"

I also compaired distance required to bring the string up a full step at the 12th. fret.

longer string-.41"

shorter string-.325"

I tried to apply 500 grams of pull at the 12th.fret.

My findings shorter string moved slightly less distance. However my method was less than reliable. So I will rig it up in a better way and let you know what I find.

So thats what I got. Thoughts?

Perry is that sounding flawed to you? I am hoping I can count on you as a guide in case I am getting odd results (I would hate to have flawed info passed on to the board).

Peace,Rich

you're getting there.

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Greg, I agree with most everything in your last post. I think no one would disagree with your definition of string tension, I certainly don't, it is exactly what I said in my first post. But perhaps some of us are talking about different things and (thus) talking past each other.

Let me try to clarify. I'm trying to explain what I asserted in my very first post (post #2) on page 1 of this thread.

Hypothesis: For a non-zero break angle over the nut, string tension between nut and tuning post is higher than string tension between nut and saddle, and the string tension between nut and tuning post increases with increasing break angle (keeping string tension between nut and saddle constant).

Why is this important? I think it provides one explanation for why it is easier to bend on some guitars than on others....because when you bend, you're not only stretching the string between the nut and saddle, you're also stretching the string between the nut and tuning post.

String tension force balance across the nut: See post #27 on page 2 of this thread, especially the table of numbers at the end of Break Angles P. 2.

Supporting observations:

(1) Godin's 11° headstock guitar is easier to bend than his 18° headstock guitar (same strings and scale length).

(2) My Strat (3-4° break angle) is easier to bend than my Les Paul (11° break angle) even though the Strat's scale length and (thus) nut-to-saddle string tension is higher than the LP's.

My force balance calculations are rigorous, but all math aside, there is a simple logical explanation that leads to my conclusion. If there is anyone who does not agree on any of these points, speak up!

A For a given scale length, string gauge (and string material) and pitch (in Hz), the string tension between nut and saddle is uniquely defined. Example: If you know the scale length is 25" and the string gauge is 0.040", there is only one string tension that will bring the string to A (110 Hz).

B When you tune the guitar, the tuning post provides the force that pulls the string and provides string tension.

C When the headstock break angle is non-zero (like LP 11°), the string exerts a downward force on the nut.

D When you tune the guitar, BOTH the nut-to-saddle string force (string tension) AND the downward force on the nut must be provided by the tuning post. (where else could it possibly come from?)

E Because the tuning post provides both forces (string tension + nut pressure), the lbs of tension on the short length of string between nut and tuning post is higher than the string tension between nut and saddle.

Simplest possible experimental test of this hypothesis:

Build a tiny 1-string guitar with a 4" scale length, a nut, a 30° headstock tilt, and 4" between nut and tuning post. The exact scale length is not important, the most important thing is that the nut-to-saddle distance be EXACTLY the same as the nut-to-tuning post distance (center of tuning post, actually). Now tune the guitar up to any pitch that provides a decent string tension.

Now pluck the string on either side of the nut (a buzzer-style piezo under the nut would be ideal here). If the string tension on either side of the nut is the same, you will get identical pitches. I predict you will get a sharper (higher Hz) note between the nut and tuning post due to higher string tension.

I suspect Rich and Perry are looking at something else.

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Mick- Yes, These are intonated and tuned to E (440).

Ok, I decided to take the string angle at the bridge down to 13 deg., and I took the headstock break angle to 13 deg..

distance to pull 1 step.

Longer string-.321" (slighter angle= .41"-diff. .080")

shorter string-.306"( slighter angle=.325"-diff. .019")

I have not set up the 500gram pull yet(still need a part).

Side note: This is only my observation (perception). I could definately notice a change in how the strings felt with the angles added. I could strum the string with more force and not have "rattle". Generally the string feel was "stiffer". Godin; I see what you are talking about :D:D .

Peace,Rich

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Side note: This is only my observation (perception). I could definately notice a change in how the strings felt with the angles added. I could strum the string with more force and not have "rattle".

Peace,Rich

There is a very good reason for that. Take a close look, study the nut and bridge area, report back with your findings, then i'll give you more things to try out that relate to what you found :D

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Greg, I agree with most everything in your last post. I think no one would disagree with your definition of string tension, I certainly don't, it is exactly what I said in my first post. But perhaps some of us are talking about different things and (thus) talking past each other.

Let me try to clarify. I'm trying to explain what I asserted in my very first post (post #2) on page 1 of this thread.

Hypothesis: For a non-zero break angle over the nut, string tension between nut and tuning post is higher than string tension between nut and saddle, and the string tension between nut and tuning post increases with increasing break angle (keeping string tension between nut and saddle constant).

Why is this important? I think it provides one explanation for why it is easier to bend on some guitars than on others....because when you bend, you're not only stretching the string between the nut and saddle, you're also stretching the string between the nut and tuning post.

String tension force balance across the nut: See post #27 on page 2 of this thread, especially the table of numbers at the end of Break Angles P. 2.

Supporting observations:

(1) Godin's 11° headstock guitar is easier to bend than his 18° headstock guitar (same strings and scale length).

(2) My Strat (3-4° break angle) is easier to bend than my Les Paul (11° break angle) even though the Strat's scale length and (thus) nut-to-saddle string tension is higher than the LP's.

My force balance calculations are rigorous, but all math aside, there is a simple logical explanation that leads to my conclusion. If there is anyone who does not agree on any of these points, speak up!

A For a given scale length, string gauge (and string material) and pitch (in Hz), the string tension between nut and saddle is uniquely defined. Example: If you know the scale length is 25" and the string gauge is 0.040", there is only one string tension that will bring the string to A (110 Hz).

B When you tune the guitar, the tuning post provides the force that pulls the string and provides string tension.

C When the headstock break angle is non-zero (like LP 11°), the string exerts a downward force on the nut.

D When you tune the guitar, BOTH the nut-to-saddle string force (string tension) AND the downward force on the nut must be provided by the tuning post. (where else could it possibly come from?)

E Because the tuning post provides both forces (string tension + nut pressure), the lbs of tension on the short length of string between nut and tuning post is higher than the string tension between nut and saddle.

Simplest possible experimental test of this hypothesis:

Build a tiny 1-string guitar with a 4" scale length, a nut, a 30° headstock tilt, and 4" between nut and tuning post. The exact scale length is not important, the most important thing is that the nut-to-saddle distance be EXACTLY the same as the nut-to-tuning post distance (center of tuning post, actually). Now tune the guitar up to any pitch that provides a decent string tension.

Now pluck the string on either side of the nut (a buzzer-style piezo under the nut would be ideal here). If the string tension on either side of the nut is the same, you will get identical pitches. I predict you will get a sharper (higher Hz) note between the nut and tuning post due to higher string tension.

I suspect Rich and Perry are looking at something else.

Erik,

I am not really focussed on the point you guys are debating, but your experiment sound like something you could put together in short order. Why not give it a try? Honestly, my belief is that the pitch will prove to be the same given the lengths are in fact identical (the tuner post and string wrap may make that tricky to ensure). After it is at pitch and the tuner is pinning one side and bridge the other. I see no difference in the two sides of the nut. I do see how the downward force on the nut will effect the strings on both sides of the nut. I have seen this effect in my tests. Which brings me to a question for you. Can you help me figure out (on paper). Why I am seeing the decrease in distance I have to pull a string to raise it 1 full step(raising the pitch = increase in actual string tension ) when the only change is an increase in break angles? I believe I am seeing the same work being done either way (low angle=more distance/ less pull, and higher angle= less distance/ pull harder--both will achive same work raising tension enough for 1 step). I am not as bright as most of you guys with the Math/Physics. So I would really appreciate the help.

Peace,Rich

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Side note: This is only my observation (perception). I could definately notice a change in how the strings felt with the angles added. I could strum the string with more force and not have "rattle".

Peace,Rich

There is a very good reason for that. Take a close look, study the nut and bridge area, report back with your findings, then i'll give you more things to try out that relate to what you found :D

Perry,

Got ya. The strings are arching as they pass over the break points. The string height has increased. I did not modify bridge height. So that explains the "rattle".

P.S. Thanks for the help it is much appreciated, and very cool of you. :D

Rich

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Tirapop, you want to make the hypotenuse of an obtuse (scalene) triangle be the horizontal reference, with the other two sides being the "player's" string (between saddle and nut) and the "dead" string (between nut and tuner). Fine, you can do that, and still use trig to calculate the forces parallel to the two lengths of string (although the force will not bisect the break angle....). If you can bring the equations (more difficult than using a right triangle), I am certain you will come to the same conclusion that I did.

Erik,

I'm going to have to play my engineer card... for the past 20 years, I've been an aerospace structural analyst.

This shows how all the trig works for the nut/saddle force to bisect the string force vectors. I apologize for the small size. I don't have an image hosting account. You'll probably have to print it out to read it.

As I said, you're missing the horizontal load component at the nut. It isn't just vertical. You can demonstrate this to yourself with some dental floss and a beer coaster (as I just did).

Tip the coaster on end, on your table, then use the floss to balance it. Coaster is vertical and floss is horizontal.

Give the floss some break angle on one side of the coaster. You can keep the coaster vertical so long as you don't apply tension on the floss that exceeds the friction capability between the floss and the coaster. If you put any tension above that insignicant amount, the coaster collapses. Why does it collapse? The horizontal load component exerted by the floss as it breaks over the coaster.

If you look at archtop guitars with floating (unglued) bridges you'll see that the break angles on either side of the bridge are nearly symmetric. The wide base of the bridge makes it less tippy than the coaster and allows some tolerance.

It's helpful to imagine pulleys as others have suggested. Again, try to apply your method to a 90° string break angle. What answer do you get for string tension? In the solution I gave, where the string tension is equal on both sides of the nut/saddle, for a 90° string break angle, the nut/saddle reaction bisects it at 45°. The nut applies a horizontal reaction equal to the string tension and a vertical load equal to the string tension. It all balances.

I haven't been following the string bending compliance part of the discussion, so I don't know what's being argued. Remember the string is acting like a spring. It follows Hooke's Law: F = kX. "F" is force, "X" is the amount it's stretched, and "k" is the spring rate. "k" is a function of the total length of the string, from the tuner to the ball end (locking nuts and bridge doodads notwithstanding). I think k = AE/L; "A" cross sectional area, "E" is elastic modulus, and "L" is length.

For a given scale, tuned to a given pitch, a longer string will take less force to move a set distance. Compared to a shorter string, the same amount of displacement requires a smaller percentage of stretch for the total length of string. The longer string will also have to be displaced a greater distance to reach the same pitch as a shorter string.

Edited by tirapop
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Tirapop,

Would I be correct to assume that. Length and tension being equal (as I am trying with my mock up), and then increasing the break angles I am actually increasing stretching of the string over these points (explaining the differences I am seeing). I guess that makes sense given the arching of the string Perry pointed out.

Peace,Rich

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NOTE: I've tried finding the "effed up" quoting, but it seems to be fine. Dunno what the problem is-- I'll fix it up later but I'm about to head out the door!

Greg, I agree with most everything in your last post. I think no one would disagree with your definition of string tension, I certainly don't, it is exactly what I said in my first post. But perhaps some of us are talking about different things and (thus) talking past each other.

It's possible. Before I go into the details, it should be clarified that I do think that some guitars are easier to bend than others. I MERELY propose that it's not "string tension" that is the cause. Rather, it is either: 1-friction over the nut; or more importantly 2- the length of string (not the angle) between the nut and tuners and/or bridge and "tailpiece" (or string-through or whatever).

Anybody who's arguing that "real world observations rule" is correct; however, you glanced past the fact that I'm talking with Erik primarily, who proposed a "zero-friction" (ie. NOT real world) math, and I'm discussing things in terms of the nut not having friction. Of course I agree that real world practice is ultimately more important than mere calculations and theory, but that's never been my point.

On to the points at hand:

Hypothesis: For a non-zero break angle over the nut, string tension between nut and tuning post is higher than string tension between nut and saddle, and the string tension between nut and tuning post increases with increasing break angle (keeping string tension between nut and saddle constant).

This is the entire crux of my disagreement. The string tension between nut and tuning post does NOT increase. String tension is distributed evenly across the entire string. Even the force generated by your angle calculation has to be distributed across BOTH sides of the bisector, not just one. The nut isn't a magical area beyond which the rest of the string is ignored. For all the string cares, the "fretboard" side is exactly the same as the "tuner" side. All that "physics" knows is that you have a string, and there's a vertex. That's it. The tension is distributed evenly on both sides. Even with friction as a factor, one can assume that the nut could create an imbalance of tension on EITHER side. The reason that the headstock side FEELS like there's more string tension is because the length (again, my argument is to do with length, not angle or tension) is different. But the ACTUAL tension is distributed across the whole string.

I'm frankly astonished that anyone could think that the tension just kind of 'hangs out' at one side. Whenever you create a vertex in a line (ie., the nut in this case), BOTH sides are treated equally.

Let's look at the angle of the string (as opposed to the headstock) for a moment. The string is a line when there's no nut, but then it's a 168-degree obtuse angle (from the "under" side) when going over a nut at 12 degrees. If you bisect it, the resulting angles are simply an 84-degree angle and another 84-degree angle. This just repeats what I've already said, but in different words: the string doesn't know from tuner or fretboard, it just knows that it's bent at a 168-degree angle.

The tension is still distributed evenly across the entire length.

Why is this important? I think it provides one explanation for why it is easier to bend on some guitars than on others....because when you bend, you're not only stretching the string between the nut and saddle, you're also stretching the string between the nut and tuning post.

Yes, but the incorrect one. Even worthwhile hypotheses can be wrong. It doesn't take any sort of test to know beyond a shadow of a doubt that tension is distributed across the entire string.

String tension force balance across the nut: See post #27 on page 2 of this thread, especially the table of numbers at the end of (pic).

Even this force is still just exerted upon a vertex. It doesn't change the overall tension of the string or divide the string into 2 sections with different tensions. Your calculation that you made MUST apply to BOTH sides of the vertex! You're selectively only applying it to one, but that doesn't make it so! You operate under the assumption that one side (the tuner side) is at a 13-degree angle, which is first of all the incorrect angle (headstock angle isn't actually the angle created!) but then also assumes that the fretboard is "zero degrees" instead of simply the other half of the bissected vertex. It's a false assumption that SEEMS logical because our common sense and background (we always just think of headstock angle as relative to the fretboard) but which isn't actually sound. With that premise nixed, the rest of the maths are meaningless.

Supporting observations:

(1) Godin's 11° headstock guitar is easier to bend than his 18° headstock guitar (same strings and scale length).

OK, but now we're back in the real world. I'm going to assume that the incredible friction generated by an 18-degree angle is a contributing factor. Next, I have to know for sure that the string length is absolutely identical. I'll give a few mm of wiggle room, but even a half centimetre difference in length is by percentage a SIGNIFICANT amount of "material" across which to distribute tension. Furthermore, "easier to bend" can be a subjective observation and it might be psychosematic. Without mechanical scientific tests, there's no way of knowing that it doesn't just seem easier, even if the distance from strings to tuners are identical.

In other words, this anecdotal supporting evidence isn't truly "evidence" at all, unfortunately.

(2) My Strat (3-4° break angle) is easier to bend than my Les Paul (11° break angle) even though the Strat's scale length and (thus) nut-to-saddle string tension is higher than the LP's.

Again, less friction across the nut. Also, MORE string between its "anchor" on the bridge across which to distribute tension, vs. the TOM+tailpiece of the LP. higher strings on a Strat (the ones we usually bend) also have more length of string between the nut and tuning machines than on an LP. Again, it's this LENGTH that's important, not the angle.

My force balance calculations are rigorous, but all math aside, there is a simple logical explanation that leads to my conclusion. If there is anyone who does not agree on any of these points, speak up!

Consider it done! As I said already, the most rigorous of maths can be flawed if they're based on an incorrect premise; in this case, your automatic assumption that on the fretboard side there is "zero" angle and on the tuner side there is a 13-degree one.

A For a given scale length, string gauge (and string material) and pitch (in Hz), the string tension between nut and saddle is uniquely defined. Example: If you know the scale length is 25" and the string gauge is 0.040", there is only one string tension that will bring the string to A (110 Hz).

Agree.

B- When you tune the guitar, the tuning post provides the force that pulls the string and provides string tension.

Well... it's an "anchor". You could do the same thing with a good grip, lots of muscle, and twisting and tying the string around the nail. :D The tuner itself does not exert the force. Your fingers exert the force when you're tuning, and the tuners just keep the potential energy "in place". The tuner itself does not exert any force at all. The resulting energy/force is stored throughout the entire length of the string, and doesn't just end at the nut. This is why we're able to do "behind the nut bends." By adding further tension by pressing down with our fingers, the string on the fretboard side has more tension added to it and the pitch goes up.

C- When the headstock break angle is non-zero (like LP 11°), the string exerts a downward force on the nut.

Yes. On both sides of the nut. Also, the 11-degrees is the "negative" angle and not the resulting angle of the string itself, which is actually 169 degrees.

D- When you tune the guitar, BOTH the nut-to-saddle string force (string tension) AND the downward force on the nut must be provided by the tuning post. (where else could it possibly come from?)

Your fingers. Because the tension/force is distributed evenly across the string, you will find that the tension doesn't actually come "from" the tuner. I don't know where you could be getting that from, but it doesn't "follow" logically. Let's pretend you had a TOM-style bridge, and somehow made your guitar so that the string length from the bridge to the tailpiece is the same as from your high E to the tuning machine. You can and WILL observe that the pitch is the same when you "play" either length of string (from nut to tuner or from bridge to tailpiece). This is because the tension is distributed evenly across the string, and since you've made those 2 LENGTHS (again, nothing to do with angle) the same, they'll play at the same pitch. Heck, since we're on it, even if the angles of each separate length are different, the pitch will be the same, thereby showing that the tension is evenly distributed.

E- Because the tuning post provides both forces (string tension + nut pressure), the lbs of tension on the short length of string between nut and tuning post is higher than the string tension between nut and saddle.

No. The premise is flawed. The tuning post is just an "anchor", and not the source of the force. And even if for a moment I drop that line of argument, and say, "fine... the tuner is providing the force..." it's then providing it to the length from tuner to nut, but also from nut to bridge, and then from bridge to tailpiece.

Simplest possible experimental test of this hypothesis:

Build a tiny 1-string guitar with a 4" scale length, a nut, a 30° headstock tilt, and 4" between nut and tuning post. The exact scale length is not important, the most important thing is that the nut-to-saddle distance be EXACTLY the same as the nut-to-tuning post distance (center of tuning post, actually). Now tune the guitar up to any pitch that provides a decent string tension.

Now pluck the string on either side of the nut (a buzzer-style piezo under the nut would be ideal here). If the string tension on either side of the nut is the same, you will get identical pitches. I predict you will get a sharper (higher Hz) note between the nut and tuning post due to higher string tension.

No, it all depends on the LENGTH, not the tension. :D That's what frets are all about! Pitch changes because we stop the string at different lengths, not because we change the tension. Therefore, I suggest you follow your own model and then discover that the pitch will be the SAME if you have 2 "lengths" of 4 inches each, under the same pressure. B) Your own suggested experiment is one I would suggest to YOU in order to discover the truth of it.

Let me take it a step further:

Put tuning machines on either side of a nut, both at the same length and angle away from the nut. Which side will be higher in pitch? Which one is the surrogate "headstock" and which is the surrogate bridge? Because your premise seems to believe that it's the tuner that's creating the force, and only the angle from tuner to nut matters. If they're both tuners, then are they both therefore equal? OK, if that's your premise, what do you suppose would happen if you had a strong enough stapler to just "staple" those lengths of string in place and remove the tuners altogether? Assuming they didn't slip, wouldn't you expect them to maintain the same pitches? What if you only "stapled" one side? Does that mean that only the tuner side is exerting a force simply because it "looks" like a tuner?

--

I hate to say it, Erik, but your entire premise is flawed.

Sorry for the long post. It was fun, but I don't know who's going to be able to wade through it.

Greg

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Well, before this gets too far off into the realms of the hypothetical, it'd be nice to remember the real world and the point of trying to examine all this.

The hope being that there will be information generated here that's going to help up build our guitars --build 'better' guitars, or more accurately, choose the elements that will work best for what we're trying to achieve.

So choices like scale length, headstock angle, etc. become still more conscious decisions.

As an example, I've already stated that one of the things I'm looking for is the information that's going to help me use the type of bridge I want (non-compensated wraparound) without compromising too much on intonation. So the experiments I'll be running, eventually when I have bit more time, will be oriented toward that goal.

Maybe I'll end up figuring out whatever bridge placement, tuner distance, string angle, even string type and string guage that will help me get that B string in tune. :D

Not that I don't enjoy reading the discussions on the physics of these things --but it's nice to bring it back to the real world.

So, Greg, given that the tension is equal on either side of the nut --what's the use of a headstock angle (other than holding the string in the nut slot) and more interestingly, is there a means of 'dosing' the angle to achieve a desired result?

Another thing, speaking of headstock angles --as erik kind of suggests, it's more accurate to speak of string angle, isn't it? Fenders don't have a headstock angle--but the strings are still angled behind the nut because of the string tree (and not just a little bit either --the angle between the nut and the string tree is pretty extreme, actually).

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[edited to add: I must've missed Tirapop's post at the top of this page, because he explains it much better than I did!]

Just a quick amendment-- using 13-degrees instead of 167 IS the accurate way of looking at it. :D Which doesn't change that either side is the same. And incidentally, I agree that "string angle" is what's really being talked about.

Mick, everything in my post (except for the non-friction part) CAN be used to increase knowledge of building and understanding of what's involved. It's not merely hypothetical, and is in fact simply a counter-argument... a REALLY long-winded way of saying that angle does NOT increase string tension. Another quick observation before moving on to answer your more practical question:

Let's use an example of "no angle" = 100 lbs of tension (or whatever), = the note middle "A" [not the real math... just an easy-to-follow example]. Erik's absolutely right that the force generated by an angle will then increase tension. Say a 13-degree angle generates this force he's mentioning, and I totally agree that the tension IS changed. Let's say you now have 130 lbs of tension. But you no longer have the note middle "A". You have to then tune accordingly so that you're back to 100 lbs of tension = middle "A". The increased tension = higher pitch. Erik unfortunately misses that step and has the increased tension at the same pitch, as far as I can tell.

To the more practical questions:

Headstock angle is still important for at least one major reason:

What people call "positive contact" with the nut. The very reason Fender uses string trees is because very shallow angles allow the string to be too "loose" in their slots. You need angle (and in turn, friction) to create a positive contact and therefore have good tone and non-wobbly strings at the nut.

You could also make a case for aesthetics...if for no other reason than it "looks" good to be able to go without string trees, it's worth having an angled headstock.

As for the non-compensated wraparound, the issue there has little to do with headstock angle. You will minimize the lack of compensation by having LONGER distances from nut to tuners, that much is certain. A really compact headstock won't give enough extra "material" (ie. string length) across which to distribute tension-- particularly since it's a wraparound bridge and therefore has no extra "material" from bridge to tailpiece. I think that with normal guages of string, a non-compensated bridge is a disaster waiting to happen, unless you like the funky vibe of intonation discrepancies; however, the position of the tuners will certainly have an impact.

I might as well say it before someone else says it, but I hope everyone remembers that despite my high post count, I'm a rank amateur. That doesn't diminish my ability to reason things out, and it doesn't take a professional to know that lack of intonation at the bridge WILL cause intonation problems, no matter how you "minimize" them. Anyway, just worth restating that I'm not a very experienced guitar-builder.

Greg

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Tirapop,

Would I be correct to assume that. Length and tension being equal (as I am trying with my mock up), and then increasing the break angles I am actually increasing stretching of the string over these points (explaining the differences I am seeing). I guess that makes sense given the arching of the string Perry pointed out.

Peace,Rich

Hey Rich, cool experimental setup you have going. So, at one end you have a roller bridge and the other end there's a zero fret?

When I was talking about string length and compliance, I should have said that I wasn't considering friction. That extra length of string past the nut and saddle only has an effect if there isn't so much friction that you effectively have a locked nut and saddles.

When you increase the break angle, there's a larger component of load on the string pressing down on the nut/zero-fret/saddle. Friction force is proportional to this "normal" (perpendicular) force. The roller bridge should be less affected by friction, but, it still is. Friction is more a factor on the zero fret.

If you had a roller bridge and roller nut, that used yet to be invented micro frictionless bearings, I would expect that break angle wouldn't have any affect on how far you had to deflect the string to raise the pitch a whole step. Break angle would still have an effect on how rattle-y the strings are when you're strumming. The normal force keeps the strings in their grooves.

Mick,

This is why intonation is adjustable. The correct intonation is going to be a function of scale length, string gage (maybe even brand), action height, string length north of the nut/zero-fret, friction at the nut/zero-fret, and head angle. In the old days, people just accepted imperfection... 3 barrel saddles on Teles, a straight (but slanted) bridge on Gibsons.

If you want your intonation on your homemade wraparound bridge to be perfect, you should probably build your neck, mockup your body, buy an intonator from Stewmac, and determine the actual "correct" intonation. With all the variables in your body and bridge build, it probably won't be perfect when you put it all together, but, close enough.

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If you want your intonation on your homemade wraparound bridge to be perfect, you should probably build your neck, mockup your body, buy an intonator from Stewmac, and determine the actual "correct" intonation. With all the variables in your body and bridge build, it probably won't be perfect when you put it all together, but, close enough.

Well, this isn't the homemade version I'd like to make --that can wait for another day. This one is one of those 'featherweight' aluminum bridges --looks pretty much exactly like the original LP Jr bridges. Since I'm building an Junior type guitar, it just made sense to try this out...

The real reason though is that I have this idea in my head about how a bridge like this might sound --it's made of aluminum, for one thing. And it's a solid piece of metal, not moving parts. So I'm kind of building this guitar around that bridge, just to see.

But not to worry, I'm building two guitars --but the second one has an ultra-adjustable Schaller 3D-6 bridge...so for those times when accurate intonation is necessary, I'll have that...(I also have one of those adjustable Wilkinson wraparounds, if this non-compensated wraparound really doesn't work out).

My interest for right now is to work at compensating this non-compensated wraparound --which is why this discussion might help me. Get things as close as possible. I don't know if it is possible, but it's interesting to me to try.

For one thing, I'm using a zero fret --I believe that will allow some wiggle room (for example, I could possibly bypass the zero fret for the B string and use the nut instead, or just insert an extra shelf).

For another, I'm using a six-in-line headstock. I chose this based on the saddle arrangements on my Strat --which is very nicely intonated. There's a better 'flow' in the line of the saddles that should work with the other bridge.

And I'm also interested in the influence of string guage on this --maybe I can get some tweaking room by mixing and matching guages?

I'm thinking that the more I understand about the different variables involved, the more control I can achieve over this issue. Not just for this guitar, but for the guitars I'll be building down the road. The goal is to experiment a bit, try things out, have fun in the process.

But I'm not looking for 'perfect' intonation-- I don't believe it's all that necessary. I'm not talking about being out of tune, there's a big difference between that and not being perfectly intonated. Depends on the context I suppose --who I'm playing with, what instruments they're using. It's pretty rare that I play up the neck anyway, especially these days.

Could be another point of discussion, I suppose --the facts versus myths of intonation. Too much great music has been made on non-intonated instruments, after all.

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Greg: In response to your LONG post...

Yes, that's basically what I was saying with my little thought experiment. I agree wholeheartedly, you just put it into more detail than I could. The tension is always distributed evenly. The post does NOT exert force, the string is what exerts force on the nut.

So you have to search for another explanation. Of course, my "real world" rant part was simply because it seemed like you were implying that the break angle has no effect at all. You proved me wrong, no harm no foul.

Again, I would wager the break angle effect is because of the string acting like a spring, which is what tirapop proposed. He quoted Hooke's law, just as I was doing.

Re: Friction. Consider a few things: the contact between the string and the nut is extremely small. Also, nuts are designed to minimize friction (think roller nuts, graphite, smooth bone etc). Also, in a well made nut, the string is not supposed to move horizontally at all, because essentially the nut would "kink" the string more or less perfectly. Horizontal movement would be necessary for friction to matter.

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Nah, friction is still taken into account for the sliding motion of the string as it moves along the nut slot during a bend, as well as friction against the "sides" of the grooves during a bend. This is true of all strings, but obviously more true with wound strings than plain ones, where the "drag" will be minimal. I wasn't really thinking of the horizontal factor (though I mentioned it in conjunction with positive contact), but I also didn't really emphasize that I acknowledge the friction on a plain string to be relatively minimal even in the real world.

My main argument, despite long-windedness, comes down primarily to 2 concise premises, which I believe to be true and which can be proven in both theory and practice to be true:

1 - string tension is distributed along the entire length of the string (a locking nut effectively makes the nut the end of the string rather than the tuners, but in a "standard" guitar this includes the length of string from nut to tuner and bridge to tailpiece)

2 - It is the usable length of the entire string rather than break angle that makes a string easier to bend or not. Particularly true in a theoretical "frictionless" environment, but largely practically true as well since a plain string won't really get "caught" in a nut that often, for example.

Given those 2 premises, I believe that the distance from nut to tuners (ie. string length) is far more important than break angle, since the added tension of string-bending will be distributed along this segment as well. Angle becomes important only insofar as creating the aforementioned "positive contact".

Greg

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Given those 2 premises, I believe that the distance from nut to tuners (ie. string length) is far more important than break angle, since the added tension of string-bending will be distributed along this segment as well. Angle becomes important only insofar as creating the aforementioned "positive contact".

Now, there's a great starting place!

And here's another:

A little bit of both actually. The way I'm seeing it the majority of the string volume rings at the base of the guitar where the bridge is, not at the headstock where the tuning pegs are.

The strings going in at a straight angle at the base of the bridge in my theory should resonate better than strings that are bent at a u angle This is just my 2 cents

This comes from the recent BC Rich stopbar thread.

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OK...sound the trumpets, bang the gong, release the doves. It is official...I am a complete and utter IDIOT!

:^B

My math is fine....for ONE side of the nut only (duh), it says nothing about the other side of the nut. Greg, I wish you had just come out and said this from the beginning! :D If I balance it out, then I get tirapop's equation (thanks tp).

I've spent much of the past year doing 2D numerical modelling of particles in convecting fluid, where the velocities are described exactly as I had written them earlier. So, by MY calculations, the nut should be flying off the guitar! :D

Of course, it doesn't.

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Not an idiot at all. Rather, one of the smartest members of the forum who just happened to have overlooked something.

The length of time and amount of words it took me to get to the point shows how much I had to work it out in my OWN head while trying to engage the topic.

Greg

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Tirapop, Greg, Erik,

I was wondering if one of you fellas could post put together a post with a quick summery of what you have agreed upon. It would be cool to "bring the topic up to date". I have been trying to follow your debate, but I don't want to put words in your mouths. I think it is going to be important that we complete the thoughts as we arrive at conclusions in a way that will not require 3-4 pages worth of reading. Most topics like this wind up getting lost in the debate. I want to produce some possitve info from this one that can be re-compiled and used as a possible Tutorial.

I am ready for your next task Perry. Point me in the direction you think will give me the next piece of the puzzle.(assuming I didn't miss your point on the bridge/nut). :D

Peace,Rich

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