The TSL myth is in contradiction to physics books, which, based on rigorous reproducible experimentation, say that it is solely the scale length that determines the tension required to achieve a given pitch with a given gauge string. The length of "overhang" behind the nut or on the other side of the bridge is not even in the equation at all, because it has no effect on the tension required, and it has the same tension (if nut friction is negligible) as the string within the scale length.

Ed Packer has done experiments with actual pedal steel guitars that confirm what the physics books say.

Curt has done no experiments, and refuses to do the simple experiments suggested. The one worthwhile thing Curt has done is to ferret out every statement of the myth he can find. This may be the single greatest service this thread has done - to gather all these erroneous assertions into one place where they can all be dismissed at once.

All of Curt's "proofs" are either circular, just plain wrong, or anecdotal statements backed up by no acceptable evidence. Where is the evidence that guitars with longer overhang and the same scale length have more string breakage, or that guitars with shorter overhang and longer scales have less breakage? These baseless assestions are part of the myth, not evidence for it. Ed's keyless guitar with short overhang can't tune an 0.011 to A without excess breakage because the scale is 30", almost 5" longer than most pedal steels. Ed has said himself that, if he puts a capo or nut at 24 1/4", he can easily lower the tension and tune to A. Notice that he has to detune or lower the tension to get down to A with the shorter scale. This obervation alone completely disproves the TSL myth. The distance from the new short-scale nut to the end of the string has increased, yet the tension has decreased. This is the exact opposite of what the TSL myth predicts. Yet this is exactly what the standard physics equation predicts. That is the point of good science, it allows predictions that prove true.

I don't know why there are no 25" scale keyed guitars. I'm guessing like so many other things in pedalsteel manufacture it is simple conformation to a tradition established half a century ago when strings weren't as durable. This conformity is certainly no proof of anything. Keyless designs are mostly newer inventions. If they use longer necks, they are simply taking advantage of the better strings we have today. Are all keyless scales longer and all keyed scales shorter? Do the keyless guitars have less string breakage? We don't know if any of this is true, but Curt cites them as proofs for the TSL myth. Even if they are true, that could be for other reasons.

So, Curt, I'm not mad, but the others and myself are getting very frustrated. No matter how many different ways we explain what the physics books say, you don't understand the explanations. No matter how many thought experiments we suggest, you can't think them through to their logical conclusions. No matter how much actual experimental evidence we present, you deny it all, and present no evidence of your own. You're like those toys with the round bottom that no matter how many times you knock them over, they keep springing back upright with the same goofy grin.

Anyone who has taken the time to understand the terminology and think through the physics, or has done any of the suggested experiments for themself, has been convinced by now that Ed and the others are correct, and Curt and the myth are wrong. So I don't know what value continuing this longer will have. Unfortunately, we can't all assemble in front of a lab bench and perform the definitive experiment for all to see. But Curt has become like a hangnail, like that one unbroken window pane in the abandonned house. We can't walk away without trying one more time to introduce an argument that Curt can understand.

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So, G# TSL Show Bud = 28.5”,Scale length = 24.0”
..G# TSL Bill’s Sierra = 27.0”, Scale length = 25.5”

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Simple. The Sho-Bud is under more tension at 24 inch scale, because of more TSL. It is a full 1.5 inches longer. Per your charts above,(or someone's) one inch is roughly a two pound increase in tension.(remember TSL must be measured, since the keyhead portion is under the same tension)

Thats why it will break more strings.

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Why does the shorter TSL have the higher tension?

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It doesn't

Bill is referring to "attack" or "feel" when he says "tighter". (you called it street terms or something)

When the keyless is tethered at the nut, there is no way to draw upon(stretch) the extra length of string in the keyhead.

NOW, where is my 25 inch scale keyed guitar?

[This message was edited by Curt Langston on 03 July 2006 at 02:32 PM.]

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I don't know why there are no 25" scale keyed guitars. I'm guessing like so many other things in pedalsteel manufacture it is simple conformation to a tradition established half a century ago when strings weren't as durable.

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I'm guessing

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Well, I will agree on that.

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I don't know why there are no 25" scale keyed guitars.

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I do. I have explained it to you, and others. TSL will not allow it. (unless you are into changing strings)

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Ed's keyless guitar with short overhang can't tune an 0.011 to A without excess breakage because the scale is 30", almost 5" longer than most pedal steels.

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Yes, but only about 1.5 inch longer,(at tether point or keypeg) than the Sho-Bud Ed used in comparison.

I guess we can agree to disagree!

[This message was edited by Curt Langston on 03 July 2006 at 03:01 PM.]

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Curt...Sorry, I thought I made it clear that I measured the Tension and it was greater on The Long Scale, short TSL Sierra, as well as the other measurements that were put in table form.

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HMMMMM interesting.......BUT!

I don't mean to imply the you are stretching the truth. Don't get me wrong, thats not what I am saying. Thats a little hard to swallow.
Nevertheless, back to the heart:

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As far as scales lengths go,most keyed guitars max out at 24.5" before you run into excessive string breakage because the section of string under tension(changer to tuning post)is as much as 27" on the middle strings of a keyed guitar.Conversely on a keyless design,say on a 25" scale,the section of string under tension is only 25" in total. So there is actually more tension on most of the strings on a keyed 24.5" guitar than all the strings of a keyless 25" guitar. Less tension means less string breakage and a shorter pedal pull/string stretch to achieve a given pitch change - so strings last longer and don't break nearly as much.

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Ed. I want to know where my 25 inch scale keyed guitar is?

Can you tell me why they are not building those?

Come on dude!

LOL

Still, all in good fun.

Thanks for ALL of the inputs. Whether agreed upon or not!

I don't know what the import of not finding a 25" with key head is so I have ignored the subject, however, since it seems to mean something to you:

1. Not enough folk asked for one.
2. It is not traditional so the market is not worth the effort for a manufacturer.
3. Folk that believe what DD calls the TSL myth would mean mouth it.
4. Instruments are getting shorter for those that will accept the keyless...25" plus standard key head for a 10, 12, 14 string unit goes in the other direction...heavier too for a given design.
5. Same kind of business reasons that Carter is not making keyless.

The 25" Scale key head could have a G# TSL of 25+0.750+3.250" = 29.0".

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Curt...Yeah...I only thing that I stretched is the string!

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And Ed, my email is the same as shown in my profile: curtlangston@hotmail.com

Try it again, as I want to see the docs.

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1. Not enough folk asked for one.
2. It is not traditional so the market is not worth the effort for a manufacturer.
3. Folk that believe what DD calls the TSL myth would mean mouth it.
4. Instruments are getting shorter for those that will accept the keyless...25" plus standard key head for a 10, 12, 14 string unit goes in the other direction...heavier too for a given design.
5. Same kind of business reasons that Carter is not making keyless.

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(1) Maybe it is because folks realize that it is asking too much of the strings.

(2)Why is it not traditional? I'm talking a KEYED guitar here, not KEYLESS. (can't get more traditional than that)

(3) Well, thats just being silly, and a way to drift off topic

(4) We're talking about adding 3/4 inch to the guitar! Seriously. And the weight? What, about a pound more?

(5) What does a keyed 25 inch scale have to do with a keyless? If the old Stringmaster had such a good sound, then why don't pedal steel builders make a 25 inch scale keyed?
Business reasons?........I think not.

Here is the answer:

It does not matter whether it be keyed or keyless. What matters is the distance from changer to tuning anchor(keypeg or keyless)

Yes, once again. The longer that distance is, the more tension the strings will be under, to be tuned to E9th. The more tension strings are under, the closer they are to their respective breaking points.

That is why there will be NO 25 inch scale keyed guitars. BE already tried it. He tried "experiments", and dropped the idea, when he saw that the 0.011 G# will not hold up, when stretched out over the 25 inch scale, then 3 1/2 more inches to the tuning peg. For a total of close to 29 inches.

Your Beast will not do it either.

I'm sorry you do not see this.

Thats ok though. I still consider you to be a valuable source of info. Even when we disagree.

I can't explain it any simpler.

[This message was edited by Curt Langston on 03 July 2006 at 05:50 PM.]

PLEASE send me the data!

My email is working fine!

Many thanks friend.

CML

[This message was edited by Curt Langston on 03 July 2006 at 03:55 PM.]

As far as I can tell Ralph Mooney was the first one to use a high G# pulled to an A on pedal steel. The definitive experiment you cite as proof that this can not be done with a 25" scale keyed pedal steel was done by BE and Shot Jackson half a century ago with strings that were not of the same strength and consistency as the modern strings used on modern 25" scale keyless pedal steels. Until somebody does the same experiment again with modern strings on a 25" scale keyed pedal steel, you have no proof it wont work just as well as it does on the keyless instruments.

You have resorted to questioning whether Ed read his measuring instrument correctly. Other than the fact that you don't like his results, there is no reason to question his readings, because they agree perfectly with the expectations of the known laws of physics.

Here are Ed's facts on the relation of scale length to tension, given certain specified gauges and a constant pitch:

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We will use three string gauges = 0.0090”, 0.0110”, and 0.0115”.
We will use three scale (nut to bridge) lengths = 24.0”, 25.0”, and 30.0”.

The tensions required to get A with a 0.0090” string are:
24.0” scale = 20.7 pounds pull.
25.0” scale = 22.5 pounds pull.
30.0” scale = 32.5 pounds pull.

The tensions required to get A with a 0.0110” string are:
24.0” scale = 31.0 pounds pull.
25.0” scale = 33.5 pounds pull.
30.0” scale = 48.5 pounds pull.

The tensions required to get A with a 0.0115” string are:
24.0” scale = 34.0 pounds pull.
25.0” scale = 37.0 pounds pull.
30.0” scale = 53.0 pounds pull.

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The "overhang" on either side outside of the scale is not given, nor is it even stated that it is held constant. That is because it is not considered relevant. It is known to be a variable that does not affect the scale tension, and therefore, does not affect the results.

In other words, Curt and the others he quotes may be right about the increased string breakage with longer overhang, but their explanation of this supposed phenomenon relating to string tension is wrong. Of course, if this supposed differential string breakage is true, it is completely independent of whether the tuner is keyed or keyless, and only relates to the length of overhang. However, it is of course possible to have less overhang with keyless tuners.

So with some evidence on string breakage, we may all end up agreeing that shorter overhang leads to less string breakage, all else being equal. But this means that the only proof acceptable for the "longer overhang causes greater string tension" hypothesis would be readings from a force gauge attached to one end of the string. Evidence of differential string breakage can not prove the tension hypothesis, because there is at least one other explanation that logically fits known laws of physics.

[This message was edited by David Doggett on 04 July 2006 at 01:57 PM.]

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In other words, Curt and the others he quotes may be right about the increased string breakage with longer overhang, but their explanation of this supposed phenomenon relating to string tension is wrong. Of course, if this supposed differential string breakage is true, it is completely independent of whether the tuner is keyed or keyless, and only relates to the length of overhang. However, it is of course possible to have less overhang with keyless tuners.

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but their explanation of this supposed phenomenon relating to string tension is wrong.

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David, tension is the greatest factor in string breakage..........

I looked at Ed's calculations and charts. They did not compare keyed to keyless. All of the string tension charts do not mention keyed or keyless. They do not mention overhang. They are referring to scale tension. Not Total String Length, and the added tension of the overhang. That is why it is hard for some to grasp my point. But there ARE some that do.

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it is completely independent of whether the tuner is keyed or keyless, and only relates to the length of overhang.

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BINGO. I agree!

Here is my WHOLE point. One last time. (I promise)

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It does not matter whether it be keyed or keyless. What matters is the distance from changer to tuning anchor(keypeg or keyless)

Yes, once again. The longer that distance is, the more tension the strings will be under, to be tuned to E9th. The more tension strings are under, the closer they are to their respective breaking points.

That is why there will be NO 25 inch scale keyed guitars. BE already tried it. He tried "experiments", and dropped the idea, when he saw that the 0.011 G# will not hold up, when stretched out over the 25 inch scale, then 3 1/2 more inches to the tuning peg. For a total of close to 29 inches.

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Thats it. Thats all of it. That is what I know to be true.

Say what you will...........

Thank you David and Ed. And all the rest that put up with my hard headedness. If I caused any tension,(pun intended) I did not mean to. I look forward to further discussions on different topics with you guys.

b0b...From my standpoint, the tension on a string is as per the Equation given far above. This equation has the parameters of Scale Length = L, Tension = T, Mass = m, and Pitch = Hz....that is four.

The string mass is in weight per unit of volume...this varies with the precise type of processed material that the string is made of.

The amount of elongation (stretch) for a given Applied Tension, String Diameter, and Total String Length = TSL, and will be the same amount of stretch in each inch of string, just as the tension is the same at all points in the TSL...friction effects being ignored.

The tensile limit of the string wire (the breaking point) is a function of the precise string material used, and the Cross Sectional Area of the string (ignoring bends and wraps), and is given in terms of thousand pounds per square inch = KPSI per square inch of string (US type units) = the area of a circle having the diameter of the string.

Before the Tensile limit is reached, there is a "plastisity area" in which the string will not return to its oringinal length when the tension is reduced.

There is also a failure mode from Shear Strength (sort of like tearing a cloth). This failure mode can be initiated by burrs and nicks that the tensioned string is pulled against, such as the top of a finger, or nut.

Th string material also has Hardness as a parameter, as do the fingers and nuts/rollers.

Or an Ex-wife that I have a good lawyer to protect myself and my assets against..

Intelligence has finite limits.

We've all just peered into the infinite abyss..

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I hope you are not saying the the overhang portion, (keyhead portion of string) does not move. Because if you are then, you too are wrong.

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Curt, that is not what I am saying. I assume that you are not familiar with differential equations, which are the basis for solving the motion of vibrating strings. The boundary condition for this vibrating string problem is that at the nut and the bridge/changer, the string does not move. It is true to a high degree of accuracy - string movement is highly inhibited at these two points. If there is string overhang outside the scale length, then that part of the string vibrates at the frequency given by the equation Ed gave, based on the tension, effective string length, and mass per unit length of string along that piece of overhang.

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Of course not Dave. You have already tuned it to pitch, and the keyhead portion was under the same tension as the scale, until you loosened or cut it. The locking nut holds the tension on the scale, that was pulled up, which included the keyhead portion
You are proving my point here....The longer the Total String Length, the greater the tension.

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I believe you missed my point. Before locking down the nut, the tension on either side of the nut is equal. Visualize the point of contact at the nut. The tension (force) on the scale-side of the nut is equal and exactly opposite to the tension on the overhang side - otherwise, the string would move laterally according to Newton's law F = (d/dt)(mv), where F is the difference in tension, m is the string mass, v is the resulting velocity of said motion, and (d/dt) means time rate of change. Now lock the nut down, and look for any change in tension on the scale side of the nut - there isn't any. Until you slack the traditional tuner, the tensions are still equal. Now slack off the regular tuner and cut off the string. Again, the tension in the scale portion of the string doesn't change. Conclusion: The string overhang doesn't impact the tension required to sound a note in the scale length part of the string. You can go around and around all you want, but this is the only reasonable conclusion.

Well conceived and well said.

This subject has been discussed at length on just about every 6-string board I've been on. It's been demonstrated over and over by engineers and guitar manufacturers...and string makers...that total string length is completely irrelevant.

"Regular" guitarists just consider this an engineering fact. It's also an easy thing to "feel" when you bend strings (and fret them) *by hand*, not with pedals, bars and levers.

Curt, call Fender Gibson, PRS, Martin, Santa Cruz, D'Addario, GHS, Ernie Ball - you'll get the same answer from every darned one, and most of them will likely have the data on hand that they could email or fax to you. I remember when someone brought up the idea that Steinberger headless guitars would have lower string tension than "regular" guitars - that argument lasted mere minutes. Even Ned Steinberger agrees, and he's the engineer who designed and patented the headless 6-string.

Here's the simplest experiment I can think of, and it doesn't require a force meter. Take a guitar with no fixed nut, but either a movable nut, or no nut but some type of capo that doesn't clamp the string to the neck. This could be like one of those floating Dobro capos that clamps the strings but doesn't touch the neck. Or it could be a simple bar that you can move under the string, but that allows the string to move freely over the top of the bar, so that string tension remains the same on each side. We will call this a movable nut. Say the bridge to tuner length is 27". Put the movable nut at 24" (3" overhang behind the nut), and tune the string to a given pitch, say an 0.011 string tuned to A. Now, without touching the tuner to change the tension, move the nut to 25" (2" overhang behind the nut). The pitch will drop, because the scale length from bridge to nut is longer. The only way to get the pitch back up to A is to crank the tuner to increase the tension. Thus, we have a longer scale with shorter overhang requires more tension to tune to the same pitch. This is exactly opposite to what Curt claims. Now reverse the experiment. Move the nut back to 24" (3" overhang). The pitch rises, because the scale length has shortened. The only way to get the pitch back down to A is to slack the tuner and lower the tension. Thus, a shorter scale with longer overhang requires less tension to get the same pitch. Again, exactly opposite of what Curt claims.

But the truly definitive test requires a force meter to measure tension at the end of the string. Take the guitar with a movable nut. Put on two strings of the same gauge, say 0.011. Attach on string to a near tuning post, and the other to a far tuning post, so that the total string lengths are different. Now remove the nut, so that the scale length becomes the same as total string length. Use the force meter to adjust to the same tension on both strings. They will play different notes, because the scale lengths (now the total string lengths) are different. Now put the nut back under both strings at the proper place for the nut. The strings will both play the same note, the tensions will both be the same, but the overhang lengths behind the nut as well as total string lengths will be different. Or do the reverse experiment. With the bridge in place and the two strings attached to posts at different lengths behind the nut, tune both strings to A. Then look at the force meter measuring tension. The tension will be the same on both strings, even though they have different total string lengths, and different lengths of overhang behind the nut.

[This message was edited by David Doggett on 04 July 2006 at 10:44 PM.]

Aarrrghhh...

EJL

PS. I used to marvel at the folks that feel that they can and have to minorize their major chords to "get the beats out"...

[This message was edited by Eric West on 05 July 2006 at 05:47 AM.]

Thus the TSL (total string length) may vary for a given pitch, but the tension per inch is fixed, all other factors (mass, or guage) being equal. It seems to rectify the equation.

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This means the changer finger must roll the string further around its circumference. This requires bending over a greater string length at the changer. This will produce more heat and may fatigue the string to the breaking point at the changer sooner.

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But let's give Curt the concession that - theoretically - string breaking does indeed increase with overhang, because the more overhang, the greater the risk that the string will be hit by a meteorite...

"Ed, I'm wondering if your research can be applied to the following problem.
At one time I was able to duplicate the major chord, constructed by using the A pedal and fourth string raise lever, by using the C pedal and fourth string lower. This no longer works properly on my guitar. I assume that this is because I have changed strings. I don't remember what string configuration I had when the changes worked properly.
Does your research give a guide to string gauges which will enable this substitution to function accurately?"

This is probably not a string gauge problem. It takes a greater amount of string stretch to raise a string a halftone from pitch, than to lower the string a halftone from pitch. Your 4 string raise takes E to F = one halftone raise. Your E to F# via the C pedal, combined with the E to Eb change activated to get back to F can be made to work if you tune the E to Eb change to get the F note with it and the C pedal activated, but then the E to Eb change by itself will not give an Eb.

If we can agree that most G# breakage happens at the top of the changer finger, then what causes that as you see it?

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.

You don’t have to look at keyed and keyless and long overhang versus short overhang to find guitars that pop more strings. Many people have had a keyed guitar that popped a lot of 3rd strings, and they switched to another make of keyed guitar with the same scale length and same overhang, and the new guitar didn’t pop strings as often. So the mere fact that one guitar pops strings more often than another is not necessarily proof that any one particular aspect of the design is the cause. There are a number of possibilities, some known (such as changer radius) and some unknown.

Nevertheless, if your BMI did pop strings more often, one possible explanation that results from the longer overhang is the one I gave above and that Hans quoted. Because of the extra stretch in the longer overhang, the string has to be pulled around further on the changer to reach the new pitch. It would not surprise me if that caused more string breakage at the changer. But notice in the quote that I said it “may” cause more breakage. As Hans points out, there are possibly mediating factors. Although the string is bending over more of its length and creating more heat, that heat is dissipated over a longer portion of the string. Maybe it’s a wash. The amount of bending and heat per inch of string is the same, but it extends over a longer portion of the string. But even if that is technically a wash, there is another practical matter that could lead to more string breakage. Suppose there are imperfections in the string. If you apply the same amount of bending and heat to longer portions of string, there is an increased likelihood that you will encounter an imperfect spot in the string that will break sooner (this is a variation of Hans' meteorite theory).

So I’m thinking where there is smoke there is fire. IF (notice the big if) some people’s impression that longer overhang behind the nut leads to more string breakage is true, then there must be some other reason besides the bogus idea of increased tension. I have suggested what seems an obvious culprit to investigate: a longer throw and wrap-around at the changer (which is where the string usually breaks). There may be other explanations.

I think the “total string length causes more tension” myth came about because everyone knows that extending the scale length requires more tension to reach the given pitch, and they erroneously extrapolated that to the idea that extending the total string length by adding overhang also required more tension to reach the given pitch. But the tension is the same on both sides of the nut. And if the scale length stays the same, the tension over the scale length cannot change if the pitch remains the same. And since the tension is the same on both sides of the nut, that means the tension over the whole string cannot change if the pitch is the same. If there is negligible friction in the roller nut, you cannot change the tension anywhere in the string without changing the tension over the scale and changing the pitch. Therefore, if the pitch does not change (we are always tuning the open sting to G# and the raise to A), the tension cannot change over the scale, behind the nut, or over the total string length. Adding more length behind the nut and increasing the total string length does not change that. You cannot introduce more tension without changing the pitch. And if the pitch doesn’t change, then the tension has not changed.

Now that the “longer total string length means more tension” myth has been laid to rest (after being beaten do death well past any life, except for Curt, who apparently will be buried alive screaming), we are free to move on to the question of whether longer overhang and the consequent longer throw on raises really do cause more string breakage, and if so, why?

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Student of the Steel: Zum uni, Fender tube amps, squareneck and roundneck resos, tenor sax, keyboards

My C69 tuning is essentially the same as the 10 string E9, except for string 9 and the extra strings. The G# to A change is there, except in C it is E to F. No breakage problems so far (about a year and a half on the same strings).

If you wish to see the instrument (the BEAST), you can see it on the SIERRA website, or do a Forum search for BEAST under Per Berner's posts.

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Hans...welcome to the wars!
If we can agree that most G# breakage happens at the top of the changer finger, then what causes that as you see it?

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.

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Hans

[This message was edited by Hans Holzherr on 05 July 2006 at 11:06 AM.]