#1
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how do you find the break tension of a string mathmatically?
Is their a formula? I espacally want to know how to do it with wound strings. I don't want to mess with my instruments.
Last edited by guitar344; 10-22-2014 at 01:35 PM. |
#2
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1 + T(tension) = 1 divided by 2
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#3
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You need to know the yield strength of the string material, usually expressed in megapaschals. Multiply that by the string's cross-sectional area, and that's your breaking tension in theory. For wound strings, ignore the windings and calculate for the material and area of the core only.
A string will most likely break before it reaches this theoretical limit, because it will have a weak spot somewhere along its length, or its strength will be compromised by additional strain where it bends over the nut or saddle.
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Herb Proud owner of only one guitar --- https://soundcloud.com/bucc5207 "Science is the belief in the ignorance of experts." - Richard Feynman, 1966 |
#4
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A long time ago, I went through the math with you.
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#5
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This is what I wrote in an archived post:
You seem to have a fascination of string tension. Why not learn the science behind it? To find the tension of a string in Newtons use: Tension = 4* (frequency)^2 (length)^2 (mass/unit length) frequency = frequency of string in Hz (can be googled) length = length of the string (scale length) in meters (google the conversion to meters) mass/unit length = mass of a string in kg divided by it's length in meters (this can also be googled) To find the pressure exerted on the string use: Pressure = Tension/(3.1415 radius^2) radius = half the string diameter in meters (it would be 0.004 inches for 0.008 gauge string, but convert to meters) Compare this pressure to the ultimate strength of steel which can also be found by a google search. If the pressure you calculate is greater or close to this value the string will most likely break. What is also fun is examining the equation for tension again: Tension = 4* (frequency)^2 (length)^2 (mass/unit length) It's proportional to the frequency squared. That means doubling the frequency without changing any other variable will quadruple the tension. Increasing the frequency by 41% or 1.41 times the original frequency, will double the tension. It's also proportional to the length squared, but the length does not change drastically compared to the frequency. Also note, each half step corresponds to a 2^(1/12) increase in frequency. So from hi e to hi a, which is 5 semitones, the increase is 2^(5/12) = 1.33. So (1.33)^2 = 1.78, and the tension will increase by 78% compared to high e. And then this: I'm bumping this up, because I just realized something when it comes to the stress (what I called "pressure" before, but it should really be termed "stress") on a string. The tension on the string is: Tension = 4*frequency^2*length^2*(mass/unit length) The mass of the string is: mass = density*volume = density*(cross sectional area)*Length, So, the mass unit length is: mass/length = density*(cross sectional area)*Length/Length = density*(cross sectional area) So, the tension is: Tension = 4*frequency^2*length^2* density*(cross sectional area) In other words, if two strings are the same length, tuned to the same frequency, and are made out of the same material, the one with the greatest cross sectional area will have the greater tension. Since the high e and b strings both have the same densities, but different cross sectional areas, lengths, and are tuned to different frequencies, they will have different tensions. However, the stress is: stress = Tension/(cross sectional area) = 4*frequency^2*length^2* density*(cross sectional area)/(cross sectional area) So, stress = 4*frequency^2*length^2* density It does not depend on the string gauge at all... just the density of the string, the frequency, and length of the string. In other words, if the high e and b strings were tuned to the same frequency and were the same length, the tension, would be different because the cross sectional area's are different, but the stress on each would be the same. If they were both the same length, the would both break at the same frequency. Sorry for the math, but I never realized that before.
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Guild CO-2 Guild JF30-12 Guild D55 Goodall Grand Concert Cutaway Walnut/Italian Spruce Santa Cruz Brazilian VJ Taylor 8 String Baritone Blueberry - Grand Concert Magnum Opus J450 Eastman AJ815 Parker PA-24 Babicz Jumbo Identity Walden G730 Silvercreek T170 Charvell 150 SC Takimine G406s |
#6
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I thought you said you were over this stuff ?
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#7
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Wow you are challenging me to think.I want to figure it out mathematically rather than test the tension. This is much better.
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#8
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I don't get the physics. I see the scale length in inches and tension in pounds. In physics classes length is done in centimeters and tension is in newtons.
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#9
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Quote:
Tension to break = 2200MPa x 145psi/MPa x Pi x (.012"/2)^2 = 36 pounds. Exercise for the student: To what pitch is the string tuned when it reaches its yield strength? [Hint: now you need to know the scale length and the density of the wire.]
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Herb Proud owner of only one guitar --- https://soundcloud.com/bucc5207 "Science is the belief in the ignorance of experts." - Richard Feynman, 1966 |
#10
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this one again?
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www.friendincrises.blogspot.com Old age and treachery will outsmart youth and skill every time. - My dad... |
#11
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Does anyone know the diameter core for the wound strings. Anywhere I can find that info. It will help me find the breaking tension.
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#12
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OK. Get the strings you "need" to know about, tie one end to an overhead bar. Add weights incrementally 'til it busts. Q.E.D
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#13
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Quote:
Back to the original question, tho'... How do you hope to use this information? What do you hope to change? Where are your strings breaking? When I see strings break, it is usually at a stress riser, most often at the saddle. This takes the situation of a failure to a different issue - failure occurs at stresses far below the Ultimate Tensile Strength.
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#14
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I'm sure the windings affect the result.
f-d
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#15
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Why don't you find an engineering forum and ask there?
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