Well, ... just no. Shorter string means higher pitch so you loosen it off to compensate.
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Math Problem
- Thread starter Tom Brogan
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Well, ... just no. Shorter string means higher pitch so you loosen it off to compensate.
Okay, let's do the math: Two trains leave Chicago, one traveling at 45 mph.....oh, screw it!
I think you'll see a perceived difference in tension just by changing string brands and/or types. One common example: round core vs hex core.
Riis
Sorry, I worded this wrong. You need a heavier string on the shorter scale to get the lower notes at the same tension. My brain was a bit fried from work yesterday.
For example, my 30" scale needs a .050-.110 set to get similar tension to my 34" scale running .040-.100
That is incorrect. Review the math. The difference in tension in the two scale lengths is approximately 6%. Remember, it is the ratio of 35^2 / 34^2 , not simply 35/34. If any person is unfamiliar with exponential computations, I suggest a refresher course in Algebra I.
The mass per unit length varies with the square of the gauge (diameter) which offsets the square in the length ratios, so 3% is in fact correct.
Have a look at this example on the d'addario site: http://www.daddario.com/DAstringtensionguide.Page?sid=72cd0195-9b3c-45f4-b54c-cde2d844140e
Going from 046 to 052 for the E string (a gauge increase of 13%) increases the tension (and unit weight) by 26%.
No, that is apples and oranges. The OP's question was about scale length. For any given unit weight, the tension of a string for a different scale length varies as I set forth.
The mass varies with the square of the radius, not the diameter, and only if the delta mass per unit stays constant. With different core diameters and hex or round geometry, and different windings diameters or other geometry, like ribbon wrap, there is more or less mass in the finished string for a given overall diameter, so each string must be weighed separately to determine its mass. It cannot be derived or extrapolated. That is a fundamental measurement that must be made directly. Only then can the tension at pitch and scale length for a given string be computed, short of mounting a string on a tension gauge and measuring it directly. Only if the string is solid steel will it vary directly with the square of the radius.
It just so happens by coincidence that the mixing of two variables, scale length and mass of string, did the result end up being a 3% overall difference in the particular example. That is no different than reducing the fraction of 16/64 down to 1/4 by cancelling the 6's. It's not the proper computation.
Now, in addition to taking a refresher course in Algebra I, everyone needs to take a refresher course in Geometry as well.
Let me give a different example. If you order a 10-inch pizza, and it feeds X number of people, assuming constant consumption and constant density of applied toppings, how large of a pizza is needed to feed 2X people? It is not a 20-inch pizza. The computation is to take the square root of 2r^2 times pi, where r is the radius of the pizza, in this case 5 inches. The answer is roughly a 14-inch diameter pizza. Now, to compare it directly with musical instrument strings, the gaps between the windings is analogous to not putting as much toppings on the larger pizza, so you need an even larger pizza to feed 2X people. How much larger? You can only tell by weighing the toppings, which is the same as adjusting the diameter of the core and wrap so there is the proper additional mass to the string because of the gaps in the windings.
This is all basic math and geometry that every person should have learned in high school. I went to high school starting in the fall of 1976 and my diploma is dated 1980. And after all these years, I still remember my basic math, algebra and geometry, even though my college major was liberal arts and my grad school completely unrelated.
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Here's some basic math for you. Going back to Mersenne's law, since mass per unit length equals pi x radius^2 x density, we can write it in terms of the density of the string as:
frequency = sqrt(tension / density) / (diameter x length) / sqrt(pi)
Since we want to keep the tension constant, a decrease of the scale (length) of 3% should be compensated by an increase of the diameter (gauge) of 3%. This is what the OP was asking.
True, density is only approximately constant, but if you look at the Unit Weight column on the D'Addario page, you see that it only varies by a few %. So the correction would be half (because of the sqrt) of a few % of a few %. That's in the 2nd decimal if we're quoting gauge as 45/65/85/105.
That is not the correct function or formula for the OP's question. The difference in the tension is as I described: approximately 6% higher tension for the same string at the 35 inch scale compared to the same string at 34 inch scale, which is solved for the scale length, not the string mass, which is constant because it is the same string referenced, by the formula: 35^2 / 34^2. So knowing that, you find a string that according to the charts has about that much more tension when cross referenced at the 34 inch scale, as defined by the manufacturer. None of what you quoted has anything to do with the OP's question or the computation of the tension, or how to choose which string for the shorter scale bass. Go back to school, and read the OP's question again.
I will go through this one more time: The OP is asking for a string to use on his 34 inch scale bass that has about the same tension as his favorite string on his 35 inch scale bass. But the tension of his favorite string is only published at a 34 inch scale.
If the preferred string of the OP is rated on a 34 inch scale by the manufacturer at, say 40 pounds for the sake of the example, then that same string will have a tension at the same pitch on a 35 inch scale bass of 40 X (35^2 / 34^2), or approximately 42.4 pounds. Then you consult the manufacturer's chart and find out what string at that same pitch has about the same value of higher tension, and mount it on the 34 inch bass. It is that simple: take the rated tension of the favorite string, multiply it by the formula, then chose a string for the 34 inch bass that the manufacturer says has the same higher computed tension on a 34 inch scale.
Because of the different core and wrap diameters for different strings, which means the mass does not vary directly with the diameter, the gauge of the desired string cannot be reliably extrapolated, only estimated. That is the only purpose of my pizza example: "unit weight" must be determined empirically. Therefore, the manufacturer's tension chart must be consulted, because the "unit weight" must be measured empirically, and cannot be derived. My initial example using GHS SS strings was set forth with direct reference to the GHS tension chart, which can be found on line on the GHS web site. The conclusion that it may not matter that much is my own subjective conclusion from experimenting with different gauge strings for almost forty years.
All of the other quoted math, for the OP's question, is not relevant to the OP's question. It is like asking why is the sky blue, and instead of talking about how light refracts through ozone in the upper atmosphere, the person answering the question instead attempts to explain why we see rainbows, with the refraction through rain drops, and if the angle from the sun to the suspended raindrops back to the observer's eye is 42 degrees, the observer sees the rainbow. Equally as interesting, but it does not answer the original question - why is the sky blue.
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HS+CS = T
(Horse Sense + Common Sense= Truth) Put THAT in your gazinta machine and smoke it.
Why don't you answer the OPs question? I gave the answer above based on Mersenne's formula. Rounded to the nearest integer it's:
>>> [int(round(i*1.03)) for i in (45, 65, 85, 105)]
[46, 67, 88, 108]
Very simple, should be accurate enough for all practical purposes and no need to go back to school
Please do go through it one more time and this time actually show us that this is incorrect.I did answer the OP's question. He asked what string on a 34 inch scale bass would have the same or similar tension as his favorite string mounted on his 35 inch scale bass. I showed the way to determine how much tension his favorite string has on his 35 inch scale bass, taking the published 34 inch tension spec. Now the OP can take that computed tension figure and use the manufacturer's chart to cross reference and determine which string has the closest tension, if the manufacturer makes a string that has that tension at pitch.
What is incorrect in that your math assumes string mass varies directly with diameter, which it does not, and assumes the manufacturer makes a string of the derived gauges, which the manufacturer may not, so the manufacturer's tension guide must be consulted.
The error is not in the math, it is in starting with the wrong premise and assumptions and the application of the wrong formula and data.
I started with one piece of empirical data that can be confirmed: the rated tension of the particular string by the manufacturer at a given scale length. Then I applied one straightforward algebraic computation to arrive at an approximate tension figure for reference purposes. I then cross-referenced that computed tension figure to one more piece of empirical data from the manufacturer to find what the manufacturer makes, instead of extrapolating or estimating.
In the real world, this is what we call the difference between practicality and theory. The abstract math may be correct, and many times over the decades, I wondered why the manufacturer would not make a string that conformed to the math. My approach recognizes the practicality of the limits of what is actually commercially available, and helps the OP get to a decision quickly for the next gig. Your approach helps the string designer engineer in developing the next product.
What is incorrect in that your math assumes string mass varies directly with diameter, which it does not, and assumes the manufacturer makes a string of the derived gauges, which the manufacturer may not, so the manufacturer's tension guide must be consulted.
The error is not in the math, it is in starting with the wrong premise and assumptions and the application of the wrong formula and data.
I started with one piece of empirical data that can be confirmed: the rated tension of the particular string by the manufacturer at a given scale length. Then I applied one straightforward algebraic computation to arrive at an approximate tension figure for reference purposes. I then cross-referenced that computed tension figure to one more piece of empirical data from the manufacturer to find what the manufacturer makes, instead of extrapolating or estimating.
In the real world, this is what we call the difference between practicality and theory. The abstract math may be correct, and many times over the decades, I wondered why the manufacturer would not make a string that conformed to the math. My approach recognizes the practicality of the limits of what is actually commercially available, and helps the OP get to a decision quickly for the next gig. Your approach helps the string designer engineer in developing the next product.
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What the formula shows is that you'd be a fool to consult manufacturers' charts trying to find a string that doesn't exist. The difference when going from 35" to 34" is about half a gauge step for the E and A strings and less than that for the D and G strings. The variation in string density is too insignificant to invalidate that conclusion.
You didn't cross reference anything. If you were to do so, you'd surely find out that this is correct.
No, I am not a fool for consulting and cross referencing manufacturers' charts. Unless a person wants to measure the mass and/or tension at scale and pitch directly, that is the only source of reasonably reliable empirical information available. That is the prudent thing to do to see if there is actually a string out there which would meet the OP's question. Where I agree with you, and what I set forth in my initial submission, is that this is at the limit of a discernable difference, and the difference in tension may be so small that it does not matter, or the manufacturer may not make a string that will do what the OP desires, and therefore a compromise on string selection may have to be made: either the same set which will feel slightly softer, or the next available gauge up which will feel slightly stiffer.
Please re-read so you can see where I am actually agreeing with you, and so that you can see exactly where I disagree with you, considering the math and my almost forty years of practical experience of experimenting with strings.
A while ago I developed a string calculator. Hereunder you'll find a link to my DropBox to download it. I could post the formulas I used if you like. It's all rather straight forward. Vibrating length, string density, gauge, tension and desired pitch are the parameters.
https://www.dropbox.com/s/tckc2rjzxs095tx/StringCalculator.exe?dl=0
https://www.dropbox.com/s/tckc2rjzxs095tx/StringCalculator.exe?dl=0
^ By the way, the density used in these calculations is for flatwounds. For most roundwounds 7200 kg/cubical metres will do.
