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Please note this is an archived topic, so it is locked and unable to be replied to. You may, however, start a new topic and refer to this topic with a link: http://www.banjohangout.org/archive/370977
fouke - Posted - 12/14/2020: 12:33:48
According to banjo lore, one can both demonstrate the necessity of a compensated bridge and determine its shape by, in the end, (a) measuring or listening to the difference between fretting and “chiming” at the 12th fret. Because the distance of each string between between the nut and the bridge is the same, no matter how the banjo is tuned, I do not understand why this is so. And (b) I do not understand why, apparently, the distance between intervals on a scale is not strictly mathematically proportionate to the distance between between nut, frets, and bridge on a banjo.
Would anyone be kind enough to explain this in an accessible way to me, whose understanding in technical matters is limited?
maneckep - Posted - 12/14/2020: 13:08:37
Not sure I can help with the explanation except to say that the length of the strings between the nut and the bridge are not the same since a normal straight bridge has wider string spacing than the nut. Because of that - the 3rd string is the shortest. 4th and second string are a bit longer, and the first string is the longest. Basically the strings fan-out from the nut to the bridge.
maneckep - Posted - 12/14/2020: 16:30:55
The other thing that makes it all complicated is the only string that is really perpendicular to the frets is the 3rd string - all of the others cross the frets at an angle. So in reality even the distance from the nut to the first fret is different for the strings - although a pretty minute difference.
Kimerer - Posted - 12/14/2020: 17:27:18
When you fret a string, the string stretches, which changes the pitch slightly.
The pitch is changed differently on the different strings because of their differences in thickness and rigidity. The strings go slightly sharp when they stretch. The thicker and stiffer a string is, the more it goes sharp. The third string on a banjo is generally the thickest and stiffest string so it goes sharp more than the first or second string. The 4th string is wound to make it heavier, but the inner core is thinner which makes the 4th string less stiff than the 3rd string, so it actually goes out of tune less than the third string.
To compensate for the strings going sharp when fretted, you can use a compensated bridge or by just angling the bridge out on the bass side. Both of those techniques make some of the strings longer than the others to flatten their pitch slightly. How you do the compensation depends on the height of the action, the characteristics of the strings that you have and some other variables.
This is a very brief summary. The equations are on my web page.
thekimerers.net/brian/music/co...ted.shtml
Richard Hauser - Posted - 12/16/2020: 16:24:13
I have used banjos with a compensated bridge. I now own a banjo with a compensated nut.
I prefer the compensated nut on my Stelling.
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Old Hickory - Posted - 12/23/2020: 10:27:03
quote:
Originally posted by foukeI do not understand why, apparently, the distance between intervals on a scale is not strictly mathematically proportionate to the distance between between nut, frets, and bridge on a banjo.
Would anyone be kind enough to explain this in an accessible way to me, whose understanding in technical matters is limited?
Straight frets, straight bridges and the locations of both are compromises. Our entire system of intonation is a compromise.
But there is in fact a mathematical formula for calculating the location of frets for any desired scale length. It divides the open string length by a mathematical constant, subtracts the result from the open string length, places the fret that distance from the nut, treats the new fret as the nut and repeats the calculation: dividing the new shorter string length by the same constant, subtracting that result from the new length to locate the next fret and so on. I believe the constant is 17.817.
As I said, this is a compromise. The fret locations resulting from this process produce notes consistent with 12-tone equal equal temperment, which I think started becoming the standard in Western music in the 18th and 19th centuries. This system allows instruments with fixed notes to play in multiple keys with a level of accuracy that our ears have learned to accept as in tune and harmonious.
A purer sound - just intonation - puts important intervals at exact ratios. Sounds great. Human voice, fretless stringed instruments and some others can achieve it. But fixed-note instruments would have to be built for each key because over our system of 12-note octaves, the ratios don't work if you start on a different note. Not very practical.
This is the extent of my scratch-the-surface understanding of this. I will not feel slighted at all if someone corrects anything I wrote. My hope was that even this level of understanding would show that intonating a fretted banjo is not a simple matter, no banjo can be perfect, and intonation has been a matter of debate for centuries.
Edited by - Old Hickory on 12/23/2020 10:29:44
Kimerer - Posted - 12/23/2020: 11:11:15
Ken has it pretty much correct. The mathematically correct constant multiplier for each half step in the chromatic scale is the twelfth root of two, approximately 1.05946309436. The pitch of the next semi-tone is the twelfth root of two times the current pitch. Since the string length is inversely proportional to the pitch, the constant becomes a divisor instead of a multiplier.
I have derived all the equations on my fret calculator web page:
thekimerers.net/brian/YAFCalc/YAFCalc.html
My calculator generates the distance from the nut to each fret, not the distance from the last fret to the next fret so the numbers appear a bit different. I always measure directly from the nut to each fret since that minimizes cumulative errors.
The constant 17.817 comes from using it as a divisor and subtracting the division from 1 in order to measure from the nut end of the string. For example, 1/17.817 = 0.0561261716338, which subtracted from 1 is 0.943873828366. The new length of the string is 0.943873828366 times the old length, that is from one fret to the next fret.
We can reach the same approximate multiplier from the twelfth root of two since 1/1.05946309436 = 0.943874312682 which is very close to the other value.
Old Hickory - Posted - 12/23/2020: 12:17:44
quote:
Originally posted by KimererMy calculator generates the distance from the nut to each fret, not the distance from the last fret to the next fret so the numbers appear a bit different. I always measure directly from the nut to each fret since that minimizes cumulative errors.
I've read that as a recommended approach. All the online fret calculators I've seen (Stew-Mac, LMII, others) show both the fret-to-fret increments and the total nut-to-fret distances for each fret.
Old Hickory - Posted - 12/23/2020: 12:35:33
So I think I was correct in there's a mathematical constant that determines fret locations but I incorrectly described what you do with it.
Kimerer - Posted - 12/23/2020: 16:26:02
quote:
Originally posted by Old HickorySo I think I was correct in there's a mathematical constant that determines fret locations but I incorrectly described what you do with it.
You were absolutely correct. I did not mean to imply that you were somehow in error. Different people address the problem in different ways. It's all good.
Peace