String tension

[mandotommy]

What causes varying levels of string tension of different mandolins? I've also noticed this with guitars. Some feel like they have fairly loose strings (at standard pitch of course) and others feel stiff. I have a new Kentucky KM-1000 that feels stiff, with light strings, and my 1982 Stiver A model feels so much easier to play. Intonations are perfect. I'd like the Kentucky to loosen up a bit.

[Dan Voight]

Here's everything I can think of:

(daet=doesnt actually effect tention, in other words, it would create the illusion of more string tention)

Action (daet)

String Gauge

Scale length

Stiffness of the top and how much it gives under the load of the strings. (daet)

Amount of relief in the neck(daet)

Stiffness of the neck(daet)

I'm sure there are more factors.

[mandotommy]

So are you saying only scale length and gauge? Because those are the same on both my mandos and yet the Kentucky is much stiffer. Or are you saying that those other factors only "seem" to effect it? Thanks for taking the time to reply. But I'm still confused.

[Red Henry]

Fine aspects of nut height, bridge height, and fret & fingerboard shape can hugely affect the ease of playing. In my experience, scale-length variation is also significant, although it rarely occurs in Gibson-pattern mandolins (the F-4 scale and F-5 scale are the same). Even string-length (F-4 vs. F-5, for example, where the F-5 strings are longer) does not seem to make much difference.

On the other hand, string gauges will play a very important part in playability, though not always in the way we might think. Too light a set for a particular instrument can make it harder for me to play, because of the lack of response.

[Rick Lindstrom]

I've heard it suggested that "down-bearing" or "bridge break angle" might play into the perceived stiffness of a mando. In other words, given two otherwise identical mandos, the one with a greater neck angle and a commensurately higher bridge (but not a higher action over the fingerboard) will feel stiffer.

I'm not sure I ascribe to this idea, but I've seen it repeated more than once.

Rick

[sunburst]

Yep, only string length (scale), gauge, and pitch change string tension, assuming the same string material. For the wound strings it is the gauge of the core wire that makes the difference, not the string gauge. It is possible for different wound strings to have different core wire diameters for the same string gauge.

The set up of the mandolin is the biggest thing that will affect the perceived stiffness or resistance to playing of the strings, but such things as the response of the instrument, shape of the neck, shape/height of frets will cause a different feel that can feel like a difference in string resistance.

The resistance of a string to being deflected from a straight line, what you do when you fret a string, can be measured fairly easily. As far as I know, nobody has found any difference in resistance to fretting because of greater or lesser break over angle at the nut and/or bridge.

[Rick Lindstrom]

....but such things as the response of the instrument.....will cause a different feel that can feel like a difference in string resistance.
Something tells me that's exactly it- a more responsive instrument will be perceived as looser because you don't have to play it as hard to get the sound out.

Rick

[allenhopkins]

May be a stupid inquiry: wouldn't the overall length of the string make a difference -- the length from where it wraps around the tuning peg, to where it's hooked to the tailpiece?

I was trying to visualize a six-foot-long string, stretching over a mandolin-scale nut-to-bridge interval, but extending a couple feet on either end. Would the tension on this be the same as on a conventional string, hooked a couple inches beyond the nut and the bridge, to achieve the same musical pitch?

I'm probably just revealing my ignorance...

[groveland]

Allen - We have actually had this dispute on older threads. One of us actually ran an experiment that was intended to demonstrate that length of the string behind the bridge and nut don't influence tension. I believe the results were inconclusive, but the consensus was like sunburst indicated above.

I do know that from my 35+ years experience on electric guitar, scales and gauges being equal, the less 'extra' string length you have past the nut and bridge, the easier it is to bend the string up, say, a whole step. My guess is that you don't lose that energy stretching the extra 'inactive' length. Floyd Rose locking systems basically truncate the string length and therefore help out. Steinberger systems have no extra length, so they bend like butter despite long scales. Strats with all that extra headstock can be challenging to bend. Hendrix' headstock was reversed, and a lot of modern electrics reverse the headstock.

But I certainly wouldn't bet the farm on any of this because there are quite a few variables at play here - However, that has been my experience. Maybe I'll email Ned Steinberger. He'll know.

[sunburst]

I was Chris Baird who did the experiment (here's the thread) (http://www.mandolincafe.net/cgi-bin/ikonboard.cgi?act=ST;f=7;t=37199;hl=afterlength), and the results were not what I would call inconclusive, he showed that the length of string beyond the bridge and nut make no difference the pitch of the fretted string, and from that we can conclude that the tension doesn't change with extra string length.
It was, however, only one experiment.

As for electric guitar, well, that's different. When you start doing whole step bends and the like, the extra length does make a difference in the feel for the reasons groveland mentioned, but it makes no difference in the tension of the string or the resistance to normal fretting.

[groveland]

...the results were not what I would call inconclusive, he showed that the length of string beyond the bridge and nut make no difference the pitch of the fretted string, and from that we can conclude that the tension doesn't change with extra string length.
Well, not exactly. He proved that

(Chris) The deflection was the same; .065"

I left the question:

(me) Does the .065" deflection produce the same frequency in both cases?

And this is where it landed:

(Chris) I couldn't measure the increase in pitch due to the deflection because under the circumstances the string wouldn't vibrate due to the weight hanging on it and that it wasn't up against a fret

So it was a good test, but I think that the pitch yielded by that deflection is the final word on the topic. I believe somne of the tension is actually absorbed behind the nut and bridge, so more deflection is required to achieve the same pitch, but that's just a guess. It would be consistent with my experience.

[sunburst]

OK, here's how I read it:

Although the experiment didn't directly measure pitch, it found that the deflection was the same, so that means the string did not pull through the stop at either end. I can assume from that; 1. the string length was the same 2. the string gauge was the same and 3. the string tension was the same.
From there I'm assuming that the pitch was also the same because those things, string length, gauge, and tension determine pitch.

As an aside, (sort of) back in my rock-n-roll days, I knew a guy who got the first Yamaha DX-7 synthesizer I ever saw. For those too young to know, it was one of the first programmable polyphonic keyboard synths, and was the "state of art" for live performance at the time. The owner said that the action of the keyboard felt heavier when he was playing certain sounds than it did for others. There was nothing physically different about the keyboard, but he perceived a difference. I can't help but assume that something similar happens with stringed instruments.

[groveland]

I remember the DX-7. #That was the first synth to have a volume control that went to 11.

http://www.mandolincafe.net/iB_html/non-cgi/emoticons/smile.gif

I'm thinking the tension is a function of string length between two fixed points, that is, the tailpiece and tuning head; and pitch is a function of string length between two fulcrums (if I might call them that), that is, the nut and bridge. The fixed points that affect the tension aren't the same as the fulcrums that affect the pitch, so I'm pretty sure there's some calculation that needs to be applied to the relationship between tension and pitch. It's not straight-line anymore. That's my thoughts on it.

And if my hunch is right, that would affect fretting as well as bending.

[Dave Cohen]

Couple of things:

One is that I hope that Chris isn't taking credit for his "experiment", because Bernoulli took care of that issue, as well as many others, back in the 1720s. The tension is equal in all parts of the string. That is, for a given string, the tension in the afterlength between the bridge and the tailpiece is the same as the tension between the bridge and the nut, which is also the same as the tension in the afterlength between the nut and the tuner post. If anyone remembers the 2001 GAL convention, this very issue came up in a lecture by Bob Benedetto. Bob did a demo experiment in which he concluded erroneously that the tension in the afterlength was different for different afterlengths (He didn't actually measure the tension). Pete Foreman, a retired physicist from Los Alamos, stood up and grilled him on it, adding that his experience as a physicist told him that the tension in the afterlength had to be the same as the tension between the bridge and the nut. The exchange is recorded in one of the back issues of American Lutherie; don't remember which one, but presumably one in 2002 or thereabouts.

The other thing is that you can do the experiment yourself. Use an electronic tuner to get the pitch of a given string (bridge-to-nut), and do it again for the afterlength of the string. Convert the pitches to frequencies, then calculate the tensions for each segment, either from something like Arto's string calculator, or from your own application of Bernoulli's formula. You will see that the tension of a given string is the same as the tension of its' afterlength.

BTW, The Bernoulli formula is

T = 2L(f*2)d

where T is tension, L is the (uncompensated) string length, f is frequency and f*2 is frequency squared, and d is the mass per unit length of the string. For the math-phobic, Arto's calculator makes this easier for you, as it tells you what units to use and actually crunches the numbers for you.

[Frank Ford]

After years of confronting this issue, where one instrument with identical action, scale, fret height, and tuning feels different from another, I finally came to thin that there's a subtle, yet undiscussed difference. That's the neck profile. A different neck profile, no matter how subtle, is likely to cause the left hand to be positioned slightly differently. Even a tiny difference in angle could produce a significant difference in the "mechanical advantage" of the left hand finger joints, and therefore muscle tension needed to get a good grip.


Now, I spend a lot of time talking about action and playability, and I'm constantly surprised at how many players insist on gripping a mandolin neck as though it were a guitar neck. Talk about inefficiency! Better, it should be gripped as a violin neck, with the thumb joint closed and the thumb parallel to the neck. That grip also fits the standard "bluegrass mandolin player stance" with the instrument poked out almost at ninety degrees to the body, and the player facing the audience over the right shoulder. In that position, the back of the instrument is free to vibrate, and the "thumb parallel" position causes no wrist discomfort.

[groveland]

Great stuff, thanks! So we established the tension is the same across the entire string - No dispute there. The experiment you propose would also demonstrate the same regarding tension. But can someone establish that the change in pitch of the shorter length between nut and bridge is equal to the change in tension of that entire string?

If you are expending energy to bend a 25" string, and only 20" of it are creating a pitch, the tension will increase equally over the 25" length (we proved). But will the increase in pitch follow that same curve? Or is there an efficiency gained if, say, the total length is 21" and the oscillating 'pitch' length is 20"? Doesn't the ratio between total length and oscillating length make a huge difference in the amount of pitch change you get for the amount of energy you put into the bend?

If that's already built into the formula, I apologize in advance! http://www.mandolincafe.net/iB_html/non-cgi/emoticons/biggrin.gif

[Ivan Kelsall]

Several weeks ago i raised a similar question re.my 2 Weber Mandolins. My "Fern" is so easy to play,it's a dream when it comes to fretting the strings. My "Beartooth" has exactly the same string height (measured with feeler gauges) at the 1st & 12th frets,but the action 'feels' higher & the strings seem to be under more tension.The scale lengths are the same so this shouldn't be. The ONLY difference between the 2 Mandolins is the frets themselves & the flat fingerboard on the "Beartooth".The Beartooth has narrower frets than the Fern & i'm thinking that it's just the amount of fret 'contact'with my fingers that makes it feel so different. I use J74's on both Mandolins so it's not the strings,
Saska

[mandotommy]

Thanks to everyone that contributed to this discussion!! I'm new to the forum and look forward to discussing other mando topics. I've come to the conclusion that the main thing I'm experiencing with the Kentucky KM-1000 has to do with the responsiveness of the top. The narrower neck might also have something to do with it. I think a tone guard will help and also playing and aging of the instrument. I don't think string length on the non-played sides of the bridge and nut has anything to do with what I'm experiencing. But who knows, I've been wrong before!