"I take the position that the outside of the string (away from the changer finger)is asked to stretch more than the inside of the string (against the changer finger), thus adding "Shear" forces to the "Tensile" forces. These only happen at the bends. The less the bend, and bending, the less the breakage."

Quote from Hans:

"[snip] My point is that each mm of string that is affected by the bending takes the same bending angle, regardless of how many mm's are bent. Now, if a string had a certain probability of breaking on the next bend, then the probability of string breaking would be proportional to the total length of string being bent, and David would indeed be correct. However, a string subjected to multiple bendings has a lifespan within certain boundaries, and every mm of bend will hold up that long, not affecting each other. Hence, the lifespan of the string will not change. At least, that's my hypothesis, and it may be wrong.
Hans"

Hi Hans,
I agree that each string has a lifespan. And your assumption that the segments of the string are independent is a good place to start since it simplifies analysis. Still, with the longer TSL, from a purely probabistic viewpoint, I believe that you will, on average, experience string breakage a little sooner as compared to the shorter TSL system.

Suppose you know that after 1000 stretches that the probability of a string breaking in the next 100 stretches is 50%. Also lets say that for the short TSL axe 1cm of string bends around the changer and for the long TSL axe 2cm.

For the short TSL axe the probability would simple be .5. But for the long TSL axe the probability of the string breaking is the probability that both 1cm segments break, or that either segment breaks. So, the probability of breakage is the sum of:

(.5x.5) + .5x(1-.5) + (1-.5)x.5 = .75

Joe

Sorry Joe; I wrote this before I saw your post, but hope that it is also a reply to yours.

Hans

[This message was edited by Hans Holzherr on 05 July 2006 at 02:42 PM.]

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Thats why those guys say they cant build a 25 scale key guitar. Only a keyless type.

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'How do I break thee, let me count the ways'…anyone recognize the line?

Agree or disagree: Strings (particularly the approx’ 0.011 G#) tend to break most at the changer finger, second most at the nut, third most at the tuner contact point, and lastly anywhere else along the string. We will ignore the ball and wrap area for now.

What are the forces involved?
1. Tensile forces. These are the same at every “linear” portion of the string, and a tad (very technical term!) where the string bends.
2. Shear forces. These are found where the string bends are at an angle to, and are in addition to the Tensile forces. These Shear forces are aggravated by changing the Bend, as in activating a change, by making the bend tighter/smaller ( as in making the radius smaller), or by wrapping the string tighter (further) around the radius.
3. Even before we get the string, it has been subjected to manufacturing “forces” that affect thin strings differently from thick strings. Manufacturing forces such as heat, and any quenching applied, and so on will affect the surface more than the center of the string. The thinner the string, the greater the surface to volume ratio.
4. The string wire has parameters like Tensile limits, Shear limits, Tortional (twist) limits, Elastic limits, Fatigue limits, and some others…surprising that the strings work at all.
5. The string wire has “Hardness” values. There are “Scratch Hardness” , “Dent Hardness”, etc. These are surface issues mostly. They come into play where the string wire meets the finger/nut radii, and similar.
6. Enough for the forces list for now.

In a tight wrap around a radius there will be more downward forces than in a less tight wrap. Shallow angles would seem to be less stressful to the string wire than steep angles or full 90 degree wraps.

Sharp V grooves for the string wire to be forced against would seem to be more stressful than not so sharp surfaces…think roller nuts, and changer fingers.

Softer materials, like aluminum 6061, would tend to deform before the harder string material. This would cause the aluminum to be dented before the string material. I am sure that you have seen this on some changer fingers. If these materials were harder, and became dinged up etc., they would in turn dent the string wire, encouraging shear failure.

Fatigue refers to the degradation of metals as the result of repeated stress applications, so it sort of sums up all the other stresses. It usually refers to a change in the crystallography of the material under abuse, and at the location of the abuse.

Re the string…it is under Tensile stress all over, and the other stresses at the changer finger, the nut, and the tuner, usually in that order.

The greater the Total String Length to Scale length ratio, the greater the variation in the other stresses at the points of concern. The worst offender is the changer finger re stretch induced motion…the tuner end of the string does not move, the string over the nut may move a small amount, but the string end over the changer finger moves the same amount as the total stretch. = the most localized change in stresses.

What sayest thou?

But supposing that there is a string breakage differential between short overhang and long overhang. There is an idea out there that the long overhang has better sustain and better tone. Likewise, the bigger the radius of the changer top, the less the breakage, but there is an idea I have heard that smaller radius changers sound better. There is probably some corollary to this breakage and tone tradeoff for the shallow bend at the changer of Ed's Beast and Anapegs. There is also a breakage versus sustain and tone tradeoff regarding scale length (independent of overhang differences). So for all these factors there is alleged to be a tradeoff between breakage and tone. At this point, regarding all this, making a pedal steel guitar requires of bunch of choices based on preferences and tradeoffs - so it is more an art than an exact science.

Now my only regret is that we have all this information and thinking and myth busting in a thread without a descriptive title. No one will ever be able to find it again.

[This message was edited by David Doggett on 05 July 2006 at 05:43 PM.]

On string breakage - I can't see how it would affect breaks below the nut. But strings do break at the nut - roller or not - and also at the tuner. Of course, there is no doubt a small probability that a bad piece of string or kink will break above the nut - eliminating seems to reduce that break probability. On the other hand, the clamp holding the string down may damage the string - that seems, to me, to be a possible extra source of breakage for the keyless design. I would never try to put probabilities on such things without running a lot of experiments. IMO, modeling such a stochastic process is way beyond the scope of armchair analysis. Still, if the roller nut is functioning properly and strings are wound on the keys properly, it seems to me that most breaks occur at the changer. I'll have to say that I have yet to break a string on my recently-made Zum keyed universal in 6 months - I play a lot and change strings only when they start to sound bad. On the other hand, my older guitars with worn changer fingers and old roller nuts break 3rds and 5ths periodically. One needs to be careful not to compare apples with oranges on the subject of string breakage.

On the sound aspect - a locking nut has a decoupling effect between the scale and overhang sides of the nut. When the string is free to move laterally, I believe more energy tends to transfer to the overhang side and reflect back. Of course, when there is no overhang at all, like on a keyless guitar, there is no possibility of secondary vibration on the other side of the string. I have certainly noticed significant tonal differences on these 3 different styles of 6-string guitar. Floyd-equipped guitars sound, to my ears, different with locking nut tightened or removed. I have owned about every type of 6-string imaginable, at one time or other.

I also agree with some earlier posters that the keyless design may have a slightly "tighter" feel, since the string can't move laterally at the nut - the guitar may react differently when bending or pulling strings. That is certainly true of locking-nut 6-string guitars - there is a bit more precise feel to them. Many of the virtuoso classical-metal guitar players insist on that design - their emphasis is on precision and speed. Although I like keyless and locking-nut 6-string guitars for some things (like hard rock/metal), for other styles, I still generally prefer traditional keyed guitars. I intend to get a keyless pedal steel some time - I'm not against the concept at all. I sure like the idea of a shorter, lighter, more portable guitar.

Jim, if your talking about strings hanging up on the roller nut. You called it hysterisis? My BMI lot of times after a knee lever pull would not be in tune. Had to bump my knee lever and pedals a couple times. I took it off and cleaned it well, but it still would hang on certain pulls and releas's. Maybe those guitar builders could make a 22 1/2 inch scale length. That way, time you figure in your overhang string length, it would not break those 3rd's as much.

But on guitars, note that locking systems are much less popular than they were during their heyday in the 80s - people found that by using a well-made and lubricated nut and sometimes locking tuners, this problem is much less serious on a keyed guitar. But metalists still use locking systems or keyless guitars a lot. Radical whammy bar tricksters especially - those that tax the nut and tuners - tend to like them. In my experience, at least.

[edited to say I was addressing Jim Peter's post.]

[This message was edited by Dave Mudgett on 05 July 2006 at 09:02 PM.]

1) The segments interact with one another for various mechanical reasons that Ed and others have mentioned, and 2) Even in the simple example I gave, one outcome was that both segments fail together, a highly unlikely event. In fact, the probability of such an event is vanishingly small. Or, in other words, when one segment breaks, that string be bust! Game over.

To illustrate the problem this causes, using my example and Hans's approach one calculates the probability of the string surviving as (.5)^2 = .25. Then the probability of it failing is one minus this or, 1 - .25 = .75. That means that the sum of failure mode probabilities of (not a)(not b) + (a)(not b) + (not a)(b) = 0.75. But the probability of both segments failing is close to zero. The failure mode probabilities are then calculated as 0 + .5x.5 + .5x.5 = .5 which does not agree with the .75 obtained above. The problem is that we are not dealing with independent random variables and are not free to multiply probabilities.

Now, where were we?

Joe

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Bob, I have always been told that the reason builders won't build a 25 inch scale keyhead guitar is because you cant keep a G# on them. That was in some articale with Buddy emmons talking about having to go down on the scale to 24 1/2 because that 25 Sho-bud would not keep a 11 gage G#.

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[This message was edited by Bobby Lee on 06 July 2006 at 07:46 AM.]

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You agree, through your use of this service, that you will not post any material which is knowingly false, defamatory, libelous, inaccurate, abusive, vulgar, hateful, harassing, obscene, threatening, invasive of a person's privacy, racist or illegal.

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"they are multiplied."
Not always.

Adding probabilities only produces "mathematical nonsense".

If you roll a die, the probability of getting a 1 is 1/6. The probability of getting a 1 OR a 2 is 1/6 + 1/6 = 1/3.

If you have a set of mutually exclusive events, a, b, c, etc., the probability of
p(a OR b) = p(a) + p(b).

Joe

Now I get your addition term. But am I interpreting it correctly, if I say that it shows that the probability of string breaking is independent of length undergoing the bend (remaining at .5 in your example)?

Hans

[This message was edited by Hans Holzherr on 06 July 2006 at 01:10 AM.]

Here it is:

1.The Sho Bud has a 24" scale length with a 6.250" overhang (beyond the nut)for string #5.

2. If you pick the 24" scale length you get a pitch that equals the frequency for the calculation using the original equation (that some say is for scale length, and some say is for Total String Length.

3. Now run the calculation again but for the 6.25" overhang (which all agree has the same tension as the 24" section).

4. The pitch value that you will get is what you will hear if you pick the 6.25" overhang!

We have two "scale lengths", 24", & 6.25"; and one Total String Length, no matter which way you want to use them. In one case the 6.25" is the overhang, in the other the 24" is the overhang.

The equation predicts both pitches based upon their scale length, NOT the total string length.

I would have seen the two scale lengths and one total string length thing sooner if I had had the Sho Bud sitting in front of me...all the instruments in front of me are keyless. What is the probability of that?

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did the keyless 25 scale have less tension pounds on it than the longer string Sho-bud? Seems like it would have less tension, since it is a longer total string.

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You agree, through your use of this service, that you will not post any material which is knowingly false, defamatory, libelous, inaccurate, abusive, vulgar, hateful, harassing, obscene, threatening, invasive of a person's privacy, racist or illegal.

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The measurement values are just a tad higher, but in the smae order. The reason for the difference is that I did not correct for the string mass value in the equation. FYI, the 30" scale instrument has a 31.5" total string length.

Earnest....Tensile Force/Cross Sectional Area:

I'm not sure that we have proven anything mathematically. We can discuss this further off line if you like before b0b slaps a lock on this thread!

In summary, I do agree with you that string breakage is pretty much idependent of length, but not entirely.

Suppose we have an anchor chain made out of links. Neglecting the weight of the links themselves, whether the chain is 50 feet long or 100 feet long, the tension on each link equals the weight of the anchor.

But chain links being manufactured are not all identical. Some are a bit weaker than others. The probability of failure at a certain tension is determined by testing sample links.

Suppose that the chain company manufactured 1000 links, and in the lot their strength varies over a range with one being the weakest.

Our chain with the anchor on it will break at the weakest link. If the chain is as short as can be, i.e., one link long, then the chance of that link being the weakest one manufactured is 1/1000. And if the chain were 1000 links long, using up the entire manufactured lot, the chance of getting that weak link is a certainty of probability 1. So, the longer the chain the more the chance of encountering a weak link.

Joe

I'm not trying to be mean, I'm just amused about all the coincidences, and the way you think identically to Curt. And most amazing is how you just happened to have owned the exact guitar Curt claimed would be a string popper, so you can conveniently testify that it was in fact a string popper. But here's a quote from Jim Palenscar, who actually had experience with that guitar:

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Actually, if i'm not mistaken, that is a BMI (also known of as an Erickson in the Great Lakes region) and I have a 10 and a 12 string like that at the shop and neither of them have shown the tendency to break strings any more than any other brand.

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Back in 2000 both Curt Langston and a Chris Lang gave a location in their profile of Alma, AR. Then a few years later, at about the same time, both Curt and Chris switched to an Oklahoma location. Are you saying that was your son who lived in Alma at the same time as Curt? And now Curt lives in Oklahoma near you? Does your son have the same name as you? I'm getting more and more confused by all these coincidences.

Now siccin' b0b on me, that's mean.

[This message was edited by David Doggett on 06 July 2006 at 11:35 AM.]

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Charlie said, The 25" scale would have greater tension along its scale length but less total tension than the 30-1/4" string (24" scale). thats what I think too. It seems that the longer that the total lengh of that g string is the more tension pounds it has and will break faster.

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My son Michael does not have a name on here. he always used my name. Is that ok with you?

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Charlie is right because if you take the roller nut part out of a 24 scale keyed guitar and raise the keys up with washers and tune it to 440, it will have more tension pounds on the whole string, than the keyless 25 scaled guitar, because it is 28 or 29 inchs to the end. People say that the tension pounds are equal on the overhang and scale. Well i say take the nut part out and tune them both to 440 and see which guitar has the most tension punds on the 3rd G# string. it will be the keyed 24 scale guitar. Why do you think keyed type guitars pop the 3rds faster. not rocket sciance here.

[deleted]
And why the red mad face? what are you mad about? i am the one getting harrassed on this forum.

[I took care of the harassment. get over it. I'm mad that you let someone else use your username and password. -b0b- ]

[This message was edited by b0b on 06 July 2006 at 04:17 PM.]