Bare Knuckle Pickups Forum
Forum Ringside => Guitars, Amps and Effects => Topic started by: JDC on October 08, 2010, 05:52:03 PM
-
Had a random thought, I remember a while ago that someone wrote something on how extra string behind the bridge on an archtop gives the strings less tension (if my memory is working right.)
Now I know a longer scale neck helps note definition in a lower tuning, which got me thinking, does inverting a headstock have any effect on tone of lower strings or not if it affects the tension?
Or am I talking utter rubbish?
Cheers
-
I thought the theory was the more length behind the nut (or bridge, presumably), the more tension - I guess because there's a greater length of string overall to be stretched.
Hence the Hendrix-style upside-down Strat is supposed to have tighter bass strings and bendier treble strings, which sounds quite a good idea to me.
I've also heard it said that the 4+2 headstock on Ernie Ball/Music Man guitars makes the top two strings easier to bend, although I have to say I didn't notice a great deal of difference when I owned one.
-
the tension between nut and bridge is exactly the same - otherwise the tuning would be different. thats the first thing to remember. String tensions does not change unless you change pitch, mass of string or scale length
i think the feel is different though.
think about bending a string, when you bend on a guitar with locking bridge (assuming non trem to remove the bridge moving as you bend) and nut you are only affecting the string between bridge and nut and it can feel tight. its the elasticity of the string that is affected - not the tension
the angle over the nut/bridge also affects how much of this you will notice - but some prominent figures are saying its a non-issue as the difference is not something we can perceive :? still not sure about this
http://www.liutaiomottola.com/myth/perception.htm
longer scales is a different issue as it does actually change tension between nut and bridge.
Having said all that i still would not make a 5 string 34" sclae bass without through body stringing as i findthe low B too floppy if its top mounted. I also find a reverse headstock on anything less than 25.5" scale really horrible to play. cant explain either scientifically - but after playing stuff done this way i know what i prefer... so thats my advice, you need to play similar things to see if it will work for you
-
to contradict myself, firebirds seem to work well with a reverse headtsock and shorter scale - i suspect thats the headstock angle at play
-
Yeah, good points about the angle of the headstock and of the strings behind the bridge, all these things have an effect.
And some guitars feel stiffer/looser than others for no apparent reason. I'm sure a full scientific analysis could reveal why (if anyone could be bothered to do such a thing), but I prefer to think sh!t just happens.
-
You dont want to have long lengths of string behind the nut
They end up becoming a nuisance as playing notes will excite resonance in them and this can be heard as unwanted notes and harmonics
The same rules of physics apply to these bits of string that apply to the string that between the nut and bridge and long lengths end up being tuned - almost like a harp and these notes suddenly ringing out are most unwelcome
-
I notice clear differences. My Steinbergers are a piece of piss to bend strings on - headless design, so only a few mm behind the 'nut'. My WezV 23 fret strat has no string trees, which may increase the length behind the nut (?) and its tough to bend strings on - its the only guitar I run as low as 10s on. 25.5" scale alround
-
The total length of the string is the length that determines the tension the string has to be brought to to attain a given pitch.
The bit behind the nut matters, as does the bit below the bridge if relevent on the guitar in quesion.
f=(1/2L).sqrt(T/d)
where f is frequency, L is the length of the string, T is its tension and d is the mass per unit length of the string, f is therefore inversely proportional to L (but we all knew that already, right?) and L has to be the total length. That means that if you increase the length behind the nut for the same pitch, tension as a funtion of length here is
T = 4df^2L^2
But we're keeping f constant d is also constant, and so was '4', last I checked, so may as well write
T~ L^2
= increase length, increase tension needed to attain the same pitch, so philly was right.
Sorry for the maths, but its a physics question, it gets a real physics answer :p.
-
i know you work at windscale mdv, but gives us a break, don't forget the angle of the dangle is proportional to
the heat of the meat!
-
Guitars with string-thru-body seem to have higher tension often, so there must be something true about it, but like Gordij said, the angle comes in there as well.
Since you can always use lighter or heavier strings it doesn't matter a lot though.
-
I'm not about to try and divine the physics behind it. All I know is that my Hendrix strat has seemingly tighter low strings and looser trebles than a regulation one. How much this difference exists only in my head, though, I couldn't begin to tell.
-
Cheers guys, I shouldn't read science when half asleep, so if I had 2 guitars of same spec, same size string, but reverse headstock, do you think the tone on the low E would change, or just a change in tension?
-
The tension will be higher for the same guage and tuning. The tone will be accordingly different, yes.
-
Sorry Mark, but it really is 'gauge' :) Call it something of a specialist spelling correction, though :)
-
Roo, if you really want a job as my online spellchecker, youre going to have your work cut out for you for very little reward or recognition ;)
-
The total length of the string is the length that determines the tension the string has to be brought to to attain a given pitch.
The bit behind the nut matters, as does the bit below the bridge if relevent on the guitar in quesion.
f=(1/2L).sqrt(T/d)
where f is frequency, L is the length of the string, T is its tension and d is the mass per unit length of the string, f is therefore inversely proportional to L (but we all knew that already, right?) and L has to be the total length. That means that if you increase the length behind the nut for the same pitch, tension as a funtion of length here is
T = 4df^2L^2
But we're keeping f constant d is also constant, and so was '4', last I checked, so may as well write
T~ L^2
= increase length, increase tension needed to attain the same pitch, so philly was right.
Sorry for the maths, but its a physics question, it gets a real physics answer :p.
I'm afraid that I can't sign up to this one mate.
What that formula doesn't take into account is the two MASSIVE dampers in the system (being the bridge and the nut). As the length between these two points remains the same, then I would say that for any gauge string, the tension between these two points remains the same (and therefore the tension of the whole string).
I would model the string behind the nut and bridge as a further damper to the system which has the affect Feline talks about.
I'm not a physicist, but that's my engineering opinion :lol:
-
The total length of the string is the length that determines the tension the string has to be brought to to attain a given pitch.
The bit behind the nut matters, as does the bit below the bridge if relevent on the guitar in quesion.
f=(1/2L).sqrt(T/d)
where f is frequency, L is the length of the string, T is its tension and d is the mass per unit length of the string, f is therefore inversely proportional to L (but we all knew that already, right?) and L has to be the total length. That means that if you increase the length behind the nut for the same pitch, tension as a funtion of length here is
T = 4df^2L^2
But we're keeping f constant d is also constant, and so was '4', last I checked, so may as well write
T~ L^2
= increase length, increase tension needed to attain the same pitch, so philly was right.
Sorry for the maths, but its a physics question, it gets a real physics answer :p.
I'm afraid that I can't sign up to this one mate.
What that formula doesn't take into account is the two MASSIVE dampers in the system (being the bridge and the nut). As the length between these two points remains the same, then I would say that for any gauge string, the tension between these two points remains the same (and therefore the tension of the whole string).
I would model the string behind the nut and bridge as a further damper to the system which has the affect Feline talks about.
I'm not a physicist, but that's my engineering opinion :lol:
There are myriad dampers in the system, but the string length in question is definitely the full length, not the scale length.
Anyone that has a floyd rose equipped guitar should know this all too well, and if you've got one you can prove the above with it.
Engineers and physicists both appreciate experiments :)
Tune up the guitar, lock down the nut clamps.
Take the string with the longest total travel and use the fine tuners to change the tuning by a tone/as much as you can
Unlock the clamp for that string.
If you tuned it up, the pitch will drop when the clamp is released because your increase in tension between the nut and bridge is now distributed over the whole string, so the tension per unit length is lower
If you tuned down the pitch will raise when the clamp is released, because your reduction in tension per unit length from the nut and bridge is now over the whole string, and averaged with the higher tension per unit length that the section above the nut was held at by the clamp.
The total load on the string doesnt change when you release the clamp, the distribution of load per unit length above and below the nut is is simply equalised. Ergo, the length of string in question as to what tension one must attain for a given pitch and gauge is the entire length of string, not just the length between the nut and bridge
-
Roo, if you really want a job as my online spellchecker, youre going to have your work cut out for you for very little reward or recognition ;)
Nope, absolutely no problem with this one, go ahead and publish!
-
Roo, if you really want a job as my online spellchecker, youre going to have your work cut out for you for very little reward or recognition ;)
Nope, absolutely no problem with this one, go ahead and publish!
If it were me, I'd put an apostrophe into the "youre". Can I check the checking for little reward or recognition?
In all seriousness though is another consideration not the behaviour of the string as it comes out of the nut over the fretboard? If you bend the string you're changing the angle on the string as it comes out of the nut on the fingerboard side which I'm guessing would cause some sort of friction and change in tension, no? Whether it's a significant amount I don't know but if the string break angle over the nut and onto the headstock is a consideration then a string bend would also be a factor too or am I getting this hopelessly wrong?
-
The total load on the string doesnt change when you release the clamp
You sure?
I'll have a think about this when I haven't just put out a cigarette of the Bob Marley variety PDT_009
We really should have a beer at some point MDV :D
-
The total load on the string doesnt change when you release the clamp
You sure?
I'll have a think about this when I haven't just put out a cigarette of the Bob Marley variety PDT_009
We really should have a beer at some point MDV :D
Reasonably. I've thought about it quite a bit, tested it on 2 FR gutiars with blocked trems, and still sort of expect someone to go 'errr, mark, no, because _insert obvious thing that I should have thought of here_' :lol:
But yeah, by my best estimation, the locking nut gives us a scale length Vs full length variable controlled direct comparison that you can make the testable predictions that I did about when you exclude or include the section of string above the nut where the load, most importanly doesnt change, its just the proportional division of the load above and below the nut
And, yes, we should!
And a week after quitting sellafield, no more random drugs tests, and a bob marley cigarette was one of the first things I wa going to indulge in and I cant find any :(
-
Roo, if you really want a job as my online spellchecker, youre going to have your work cut out for you for very little reward or recognition ;)
Nope, absolutely no problem with this one, go ahead and publish!
If it were me, I'd put an apostrophe into the "youre". Can I check the checking for little reward or recognition?
My sarcasm didn't come over very well there, I take it!
-
Roo, if you really want a job as my online spellchecker, youre going to have your work cut out for you for very little reward or recognition ;)
Nope, absolutely no problem with this one, go ahead and publish!
If it were me, I'd put an apostrophe into the "youre". Can I check the checking for little reward or recognition?
My sarcasm didn't come over very well there, I take it!
No I getcha but I couldn't resist a little recursion. It's a weakness. :)
-
I think the bending of the string isn't affected by the change in tension some are talking about. I think it has more to do with the total amount of free string. Whenever you bend a string, you change the total length of the string. When you bend a string with a theoretically perfect locking nut the total length of the string is from nut to bridge. When you bend a note with a non-locking nut the total length of string is from tuning peg to bridge. So for a set amount of bend (a half-step for example), the force it takes to complete the bend decreases as the length of the string increases. This length is different than the length that is resonating the pitch.
-
If there's a lot of string either above the nut or below the bridge you're going to have to bend the string a lot further to raise the pitch by a tone (for instance) because it takes a great deflection to tension a longer overall string length.
If there's any friction in the system then, obviously, any extension in the string length will cause tuning stability issues simply because more string has to pass over the nut or bridge either when tuning or string bending.
Extra string at either end of the scale length will, as Jonathan said, resonate and add extaneous noise. Depending on the guitar, this will either compliment or detract from the overall sound. Some guitars with extended lengths of string and particularly unique tones have benefited from the additional "overtones" (some jazz archtops, fender jazzmaster/jaguar) but I doubt that there is a hard and fast rule that stipulates how this actually occurs. My feeling is that when you actually look at all the variables in play here it's a case of if it works, great and if it doesn't; well tough, try something else.
My own observations / experience would suggest that the smallest amount of string either side of the nut or bridge is the preferable route. certainly a small stiff headstock will help with sustain and attack and that's why I designed my MG range with only 30mm between tuners and fairly straight string pull.
Inverted headstocks on six-in-a-row layouts probably help to damp extraneous noise in the form of unwanted harmonics and other derivatives of the base frequency from the lighter strings which will be more prone to give off unwanted noise than the bass strings would be.
-
I notice clear differences. My Steinbergers are a piece of piss to bend strings on - headless design, so only a few mm behind the 'nut'. My WezV 23 fret strat has no string trees, which may increase the length behind the nut (?) and its tough to bend strings on - its the only guitar I run as low as 10s on. 25.5" scale alround
was dicking around with my fenders this morning, and i've figured it out..!
nothing todo with the string trees at all i think - seems like my strats with trems need more effort to bend the strings than my hardtails