#1
Hey fellas, it's been a while. Anyways, I had a question that I remember first thinking about back when I frequented this forum, and never really got around to asking:

What's the deal with string tension and normal/reverse headstocks? Longer string = higher tension to achieve the same tone ( basic physics ), therefore a "normal" headstock, as compared to a reverse headstock, with the same scale and same string gauge, will have MORE tension on the high E and LESS tension on the low E.

That's my logic anyways, but I don't have the money/guitars to test this out. My left hand is extremely sensitive due to tendonitis/trigger finger and associated surgical repair, so even a half-inch scale difference ( like PRS 25" vs. Fender 25.5" ) is EXTREMELY apparent to me. Since I don't like high tension in the string I like to bend, that is the top 3, and I want high tension in the lower string for snappy alternate picked metal riffs and whatnot, it seems like the reverse headstock would be a fundamentally better option.

Thoughts? Experiences?



NOTE: For this logic to apply, you have to pretend the locking nuts aren't on the guitars below...this only applies to normal nuts, since a locking one nullifies the tension between it and the winding post, as far as the played portion of the string is concerned.

NORMAL


REVERSE
#2
the tension is the same because the string is ringing from the nut to the bridge.


any length past the nut doesnt affect the pitch
#3
Quote by rickyj
the tension is the same because the string is ringing from the nut to the bridge.


any length past the nut doesnt affect the pitch


/thread
There's no such thing; there never was. Where I am going you cannot follow me now.
#4
Quote by rickyj
the tension is the same because the string is ringing from the nut to the bridge.


any length past the nut doesnt affect the pitch


Can you point me to a source that actually explains the physics of that?? I've been pondering this for a while...according to your perspective, it wouldn't matter if the string was on the tuning post or wrapped around a city block, as long as it meets the nut at the same location; that just seems unlikely to me.
#5
The longer the string, even if it is past the nut, the more tension it'll have, it's not really noticeable to most people, but it will have slightly more tension.

Reverse headstock will have more tension on the low E A and D strings as compared to a normal headstock, because the string is longer past the nut. On normal headstocks, the High E, B and G strings are tight as a mofo, and it's your bass side strings that are loose, which is why guitars like Strats and such need to have larger gauge strings for drop tunings and why bending the high strings is so freaken hard.

I prefer the 3 to 3 style headstocks because everything is balanced and even in tension.

Of course, none of this applies if you have a locking nut, this is only for free gliding strings. Even though the string is only vibrating past the nut, the nut is just like a pivot point, the rest of the string still affects its tension. Your string tension is affected by the length of the string from mounting point to mounting point, in most cases.
#6
Quote by ethan_hanus
The longer the string, even if it is past the nut, the more tension it'll have, it's not really noticeable to most people, but it will have slightly more tension.

Reverse headstock will have more tension on the low E A and D strings as compared to a normal headstock, because the string is longer past the nut. On normal headstocks, the High E, B and G strings are tight as a mofo, and it's your bass side strings that are loose, which is why guitars like Strats and such need to have larger gauge strings for drop tunings and why bending the high strings is so freaken hard.

I prefer the 3 to 3 style headstocks because everything is balanced and even in tension.

Of course, none of this applies if you have a locking nut, this is only for free gliding strings. Even though the string is only vibrating past the nut, the nut is just like a pivot point, the rest of the string still affects its tension. Your string tension is affected by the length of the string from mounting point to mounting point, in most cases.



^ These are my exact thoughts...I'm wondering if the rest of the responders ever had string mechanics/oscillating waves in physics class?
#7
Quote by lumberjack
^ These are my exact thoughts...I'm wondering if the rest of the responders ever had string mechanics/oscillating waves in physics class?



Prolly not, I dropped out physics in the first 2 months

But still, no matter where you block the string, you could block it in 5 different places, it's still gona have the same tension, and it's dependent on it's length. It finds a equilibrium.
#8
Quote by ethan_hanus
Prolly not, I dropped out physics in the first 2 months

But still, no matter where you block the string, you could block it in 5 different places, it's still gona have the same tension, and it's dependent on it's length. It finds a equilibrium.


That's what I'm saying: without a locking nut, the length matters. Lock it = same tension. A pivot point is not the same as a lock/block.
#9
String tension is going to be dependent on scale length and string gauge. The thicker the gauge of wire, the more tension that wire needs to resonate at a desired frequency. You can tune your low E to an A if you crank on the tuner like a mad man. You increase the amount of sting tension over the scale length. That's why a difference in scale length is readily apparent to TS, so long as the string gauge remains constant on each guitar. In other words, to make a sting of W gauge resonate at X frequency, you must apply Y units of force over Z units of length. A locking nut will not change the amount of tension needed to attain your desired resonant frequency, nor will the length of string before the bridge or after the nut.

If you wanted to test this you could construct a simple rig consisting of two bars spaced 15cm apart. Cut a guitar string so that one is 25cm long and the other is 45cm long. attach one end of the sting at the first bar and lay the stings over the second, hanging equal weights off the end of each string. The will both resonate at the same frequency because they have equal tension, even though one is 20cm longer after the posts.
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Last edited by krisschmidt at May 9, 2011,
#10
I don't think a locking nut nullifies the benefits of a reverse headstock. you still have to get up to tune before locking it. so while there will be variance after the fine tuning, you will start with a higher tension, and I believe you will still be higher overall than on a non reversed headstock.
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#11
It has nothing to do with the total length of string, only the portion that is oscillating.

When you pluck the guitar string, the part that is moving is only the part between the nut and the saddle. The part between the nut and tuning peg will only vibrate a very small amount, which is negligible.

The equation for guitar application relates length, mass unit per length, tension and frequency.

With all 3, you can see that mass per unit length will remain constant, the string won't change it's density enough to cause any noticeable change in frequency. If you increase tension, without changing anything in the system, frequency increases.

You just have to get your head around the fact that the string is only vibrating between bridge and nut.

The reason, my guess at least, that they make reverse headstocks is it's a lot easier on your wrist to tune. Instead of wrapping your hand around, you just reach up and tune. Less contortion of your wrist!
#12
i wouldve thought that the tension of the string would be the same whether its reverse or normal simple because its the tension that decides the pitch right? so if you tune a string to an D, its not going to matter how long the string is, because its still tuned to a D. if you change the string gauge, i'd expect it to change.
but yeah i really have no idea. in my mind that works though. if you ever find a website that explains it though id be interested to have a look.
#13
Quote by the_perdestrian
I don't think a locking nut nullifies the benefits of a reverse headstock. you still have to get up to tune before locking it. so while there will be variance after the fine tuning, you will start with a higher tension, and I believe you will still be higher overall than on a non reversed headstock.

I think this explanation actually proves that the length of the string behind the nut makes no difference in tension

say you tune up identical sets of strings on identical guitars (aside from the reverse headstock on one) and get both low strings to E
now if we have locking nuts and tighten them down both strings are still tuned to E (for the sake of argument, let's not worry about fine tuners and assume they both stay at the same pitch, since the variance in tuning is from the locking nut pressing the string out of tune, not from anything relavent to this example)

now the two lengths of identical string are pulled across identical lengths of working space tuned to the identical pitch.

how can they be different tensions?


when you unlock the locking nut, the tension stays the same on both of the strings regardless of the length of string beyond the nut. otherwise when you tune up, lock the nut then unlock it again, you'd have crazy jumps in tuning and tension which just don't make sense
#14
You guys are only thinking about the pitch when it's tuned, not bending. The tension in both is the same when you tune to pitch, because same oscillating length and same string gauge automatically means same tension to tune. It's what happens when you bend that's interesting.

To increase the pitch when you bend, you increase the tension in the string by lengthening it, but it's related to the fractional change in length ie change in length over original length. If the original length is shorter (high e on reverse headstock) then the same distance bend across the fretboard will raise the pitch more, so for say a whole step bend you'd have to displace the string less.
#15
Quote by lozlovesstrats
You guys are only thinking about the pitch when it's tuned, not bending. The tension in both is the same when you tune to pitch, because same oscillating length and same string gauge automatically means same tension to tune. It's what happens when you bend that's interesting.

To increase the pitch when you bend, you increase the tension in the string by lengthening it, but it's related to the fractional change in length ie change in length over original length. If the original length is shorter (high e on reverse headstock) then the same distance bend across the fretboard will raise the pitch more, so for say a whole step bend you'd have to displace the string less.



Ah! This is it.