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String length question
http://www-.luthiersforum.com/forum/viewtopic.php?f=10102&t=4757		 Page 1 of 1
Author:  Dave White [ Fri Jan 27, 2006 12:01 am ]
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I'm making a cittern (scale length 605mm, 23.8") which has 5 pairs of unison strings. I was originally going to use a tuning GCGCD, which is like a guitar in CGCGCD tuning with the bottom string omitted. However, as most of my playing is going to be in keys of D, G etc, rather than capo at the second fret all the time, I decided to use ADADE tuning.

Now had I stuck to my original tuning, then the bottom 3 courses would have been wound strings and the top 2 courses plain strings. This, as in a guitar, would require the 2nd pair of strings to have a greater string length (saddle compensation) than the 1st and 3rd pairs around them - usually about 2mm more in length. BUT, moving to ADADE, I have decided to have the top 3 pairs of strings unwound - as this is like the A and D in DADGAD but with an extra E on top. It's more like a 6 string guitar with the 6th and 3rd strings removed and an extra 1st string on the top.

My proposed string gauges then will be 0.036w, 0.027w, 0.016 plain, 0.012 plain and 0.010 plain (in pairs of course). In this configuration the 3rd string is more like the 2nd on a normal guitar. So my question is - is ith the 3rd string that requires the similar extra compensation that would normally apply to the 2nd on a normal guitar.

I am thinking that I will need a saddle angle that can accomodate shaping for string lengths (saddle to break point) of 607mm, 607mm, 609mm, 609mm and 611mm for the 1 - 5 pairs of strings respectively.

Any input, comments, suggestions gratefully received - that's always assuming you can understand the question Dave White38744.3363541667
Author:  tippie53 [ Fri Jan 27, 2006 12:41 am ]
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   Can you do a split saddle? Compenation angles can be trickie. Depending on how you do the numbers I thin a 3 degree will work but the heavier gage will need a touch more.
    Splitting the saddle may help
john hall
Author:  Colin S [ Fri Jan 27, 2006 5:08 am ]
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Dave I'd string it up before cutting for the saddle and use false saddles (like the stew mac intonator) and find the individual saddle positions empirically for each string. With the short scale you may not find it too much of a problem. I have done this using a bent piece of wire as a false saddle.

Colin
Author:  Dave White [ Fri Jan 27, 2006 6:38 am ]
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John,

Thanks for this. On my guitars I normal use a split saddle but don't find I need it on the bouzouki type instruments.

Colin,

I'll do this. I think the saddle angle and width will be such that I can find the "right" profile this way which ever way the compensation turns out.

I thought that on shorter scale guitars you needed a little more compensation - I thought because the string tension is usually a little less. I could be very wrong here though.
Author:  Alan Carruth [ Fri Jan 27, 2006 7:37 am ]
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The compensation length seems to vary depending on how cloes the string is to it's breaking point, sometimes called the '%T'. All else, such as action height, relief, and so on, equal, the lower the %T the more compensation will be needed. Loose strings can stretch a lot. Tight strings don't stretch: they break!

It's possible to calculate the %T of the strings, although with wound strings you need to know how they're constructed: a .036 with a .020 core and .008 wrap is a lot different from a .036 with a .018 core and .009 wrap.

Steel used in strings is usually taken to have a breaking stress of about 325,000 psi. Thin strings will be _slightly_ higher, and there are alloys available now that go up to 500,000 psi, but that 325k figure works pretty well. Steel weighs about .047#/cubic inch.

To find the %T you:
1) divide the density by 386.4, which is the accelleration of gravity in in/sec/sec.,
2) multiply 2*frequency* length of the string, in inches, and square that,
3) take that (2*f*l)^2 and multiply that by the number you got in step one,
4) divide _that_ number by the tensile strength of the string material. That's the %T.

You will note right away that there's no place in the formula for the diameter of the string. That's because the strength of the string varies as the cross sectional area, and so does the mass. A .018" diameter string is four times as heavy as a .009" and also four times as strong. It takes four times as much tension to get the heavier string to a given pitch, and they will both break at the same pitch, pretty much: as I say, really small strings are stronger.

Anyway, you can use the same formula for wound strings, except you have to calculate an equivalent density, and remember that all of the stress is on the core. FYI, the number I've got for the density of bronze is .32, and the breaking stress, if you want to use bronze strings, is 125kpsi.

for those of you who want to fool with other strings, nylon breaks at 44,600 psi, and has a density of .0383. Nylon floss, with about the same density, breaks at 120,000 psi: not only is it a lot smaller than regular nylon, but it's usually a different type too.

It turns out that _lots_ of things in real strings depend on the %T, so it's a nice thing to know.
Author:  tippie53 [ Sat Jan 28, 2006 12:21 am ]
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You never stop amazing us with the techical answers. Thanks agian Alan
   
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