On stringed instruments, the scale length needs to be slightly adjusted (at the bridge) for each string so that the first octave happens at the 12th fret/position, the 2nd at the 24th, etc. My physical intuition tells me that larger strings need a slightly longer scale to compensate for their thickness, and this is indeed what I observe when I intonate my electric bass: the bridge saddle positions are monotonously further away from the nut as you go from smallest to largest string.

However, when I look at pictures of bridges on the internet, they often make some wacky pattern with one or two seemingly random strings longer than the others. Even "compensated" fixed saddles for acoustic guitars often have one string in the middle longer than the others. I've also heard electric guitar bridges tend to fall into a "double stairs" pattern.

So, what's going on with all that?

Edit: here are some examples: enter image description here enter image description here enter image description here

  • Good question! +1. On most of my basses, there's a pretty well straight line (diagonally) across, as you say. Approximately doubling the extra length per string, as it gets to the E, or B.Though a couple have a strange spacing, but intonated.Something to do with the % extra mass of each string? My 6 string bass is almost a dead straight diagonal line. Guitars usually have that diagonal line, but the 2nd string is usually not in that line, being longer. Compensated bridges reflect that. Comparative string mass/gauge is usually the reason for the stagger, and maybe 12tet tuning is B string prob?
    – Tim
    Commented May 14, 2021 at 14:03
  • when I look at pictures of bridges on the internet -- Examples? Commented May 14, 2021 at 14:03
  • Interestingly,. if you use skinny top/heavy bottom strings you get almost a straight line. I have this on several guitars strung the same - i.sstatic.net/DK31G.jpg
    – Tetsujin
    Commented May 14, 2021 at 14:30
  • You set the bridge positions to accommodate the combo of string, neck, etc. (maybe even asymmetric frets) you have on your axe. By comparison, bowed instruments never offset the bridge but some people like the tailpiece (where the strings are attached) to be asymmetric to change the down-bridge resonances. // Guess I should add that real-world strings, not to mention the nuts & bridge components, vary from the "ideal" behavior of simple models of harmonic series. Commented May 14, 2021 at 14:53
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    @ToddWilcox fretless bass guitars are usually set up with a bridge similar to what you'd have on fretted ones. Commented May 14, 2021 at 17:32

3 Answers 3


I'm actually planning to make a YouTube video about intonation to delve into this phenomenon.

The short answer is that is depends upon the cross-sectional area of the tension-bearing part of the string and also the formulation of the metal.

First, why do we have to intonate at all? One would think that we could just set the bridge at 2x the distance from the nut as the 12th fret and start the music, right? Well, it turns out that, when you fret the string, you're deflecting it to the fretboard. Imagine that you went between two power line poles and pulled the power line all the way to the ground. You took a wire which was (more or less) straight between two raised points (the nut and the bridge, for example) and forced some point in between to be displaced from that line.

That stretches the string. Now, when you stretch something, the tension in it goes up... but how much? It turns out that is determined by the Young's modulus of the material and the cross-sectional area of it. It shouldn't be difficult to understand why the area matters. Imagine how hard you'd need to pull on a piece of wire to lengthen it by 1%. Now, put two of those wires together and try to lengthen them both by 1%, and you're going to have to pull 2x as hard. 2 wires are no different (in this case) from a single wire of 2x the area. You're trying to pull 2x as many atoms away from their neighbors.

So, if you double the cross-sectional area of the string, you'll double the increase in tension when you displace the string by a certain amount. So, when you fret a string, you didn't just increase its pitch by shortening it; you also increased its pitch a little more by increasing the tension in it (no different than if you had just turned the tuning peg a little).

Now, when we intonate, we lengthen the remaining distance of the string (from that fret to the bridge) to compensate for the increase in tension from deflecting the string. So, the 12th fret is no longer at the mid-point between the nut and bridge, but it's a little closer to the nut so that leaves the 12th-fretted string length a little longer than half, to compensate for the increase in tension. The point is: the larger the tension-bearing cross-sectional area of the string, the more the pitch would increase when we fret it, and the more we're going to have to move the bridge saddle back to compensate.

Your question specifically asked about the stair-step appearance on guitar strings. That's why I was careful to mention tension-bearing cross-section of the string (since that's the part that is providing the tension). On the wound strings, you have a smaller string wound by a long coil of wire. The outer wire is just to provide more mass to get the string to vibrate at a low pitch, and it's not providing any of the tension in the string.

What you'll find, if you pull the winding off of a wound string, is that the wire inside of a light wound string (like the D string on an electric guitar) will be thinner than the heaviest solid string (like the G string). That is why the bridge positions jump forward when you transition from the solid to the wound strings. In fact, if you actually take some calipers or a micrometer to measure the diameter of the strings, you'll discover that the saddle for a wound string will be in almost the exact same position as a solid string of the same diameter as the one that's inside the windings.

Bonus Section : Young's modulus

As I mentioned earlier, different materials increase in tension by a different amount when you stretch them (or, they require a different amount of tension in order to stretch them a certain amount... same phenomenon). Pull on a rubber band with 1 pound of force and it'll stretch a lot. Pull on a piece of steel cable (of the same cross-section) with that same 1 pound of force, and it'll hardly stretch at all. The characteristic which describes this about a material is called "Young's modulus". Put simply, it's the amount of force, per area, that you need to stretch something by a certain fraction. You want to stretch that steel cable by 1%, you'll need a certain amount of force. Need to stretch it 2%, you'll need 2x as much force. Need to stretch a cable with 2x the area, you'll also need 2x as much force. If you make that cable out of something with a Young's modulus that 1.5x higher, you'll need 1.5x as much force, as well.

That last aspect matters to us guitarists. Guitar strings come in a variety of metallurgical formulations: stainless steel, differing amounts of nickel, maybe some bronze, etc. All of this gives a particular brand/model of string a different Young's modulus, which affects the proper position of the bridge saddle just like the cross-section of the string does. This is why I tell people, when teaching them to intonate their guitar: step #1 is deciding what brand/type of strings you plan to use for a long time (not just the gauges, but the actual brand and product name). If you change gauges, you'll need to re-intonate. Even if you get the same gauge strings from a different manufacturer (or a different formulation from the same manufacturer), you might need to re-intonate due to the Young's modulus being different in the different metal. #ErnieBallSuperSlinkysForLife

  • Thanks! Actually, after sleeping on it, I realized that, two other things happen when you fret the string: 1) You lengthen it slightly (which is what causes the stretching, after all), and 2) that also decreases the mass-per-length ("linear density", the physicists call it). The first should lower the pitch (since it's like moving the bridge saddle back), and the second should raise the pitch (since the tension doesn't have as much mass, per section of string, to sling back and forth). I'll have to work out the math before I make the video.
    – Jemenake
    Commented May 15, 2021 at 16:26
  • So, theoretically, the two cancel each other out, roughly! Sounds like a fun project - and there's a new question on at least part of the conundrum...
    – Tim
    Commented May 15, 2021 at 16:31
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    Well, they certainly don't cancel each other out (or we wouldn't have to intonate our guitars at all). Pitch varies inversely with the length, but it varies with the square-root of the tension, so some of those factors are increasing/decreasing faster than others.
    – Jemenake
    Commented May 15, 2021 at 17:40
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    Certainly looks that way. The change in length of a 1m string deflected in the middle by 3mm (and 3mm is pretty high action) looks to be a factor of about 0.00001. I was just curious whether it would be possible to account for all of the factors and accurately predict what the un-intonated error in pitch would be. Indeed, it's probably not possible to measure the effect of density change with a typical tuner, as even a cent of a semitone is going to be a factor of about 0.0006; over an order of magnitude larger than the expected change in freq from the density change.
    – Jemenake
    Commented May 15, 2021 at 23:30
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    This is a very good answer, but it needs a TL;DR at the top that says something to the effect of: "Notice that the unexpected jump in saddle position occurs where the transition between wound and plain strings occurs."
    – Theodore
    Commented Mar 24, 2022 at 19:30

tl;dr: the saddle position depends on the action, the thickness of the string's core and its Young's modulus, because these factors govern how much the string tension goes up in fingered notes.

My physical intuition tells me that larger strings need a slightly longer scale to compensate for their thickness

There's no need to compensate “for thickness” per se. This can be seen from a thought experiment: consider a “thick string” that actually consists of multiple thin strings in parallel (like in mandolin etc.). Clearly, the intonation for each of these strings will behave roughly the same way, i.e. you don't need more compensation just because you have more string-thickness.

What actually needs to be compensated for is two effects:


A real string is not quite the idealised one-dimensional object as which we like to model it in physics classes. Actually a string, even with no tuning-tension at all applied, has a certain stiffness to bending movements. The thicker the string, the more influence this has. One result of that is that the overtones on a guitar aren't exactly integer multiples of the fundamental, like you get from a synthesizer, but they get slightly sharper the higher you go in the overtone series. However, the fact that it's possible to use high flageolett harmonics in bass playing and generally stay in tune shows that this is actually not such a strong contribution.

(Note that harmonics are not influenced by the bridge compensation.) I think stiffness is largely negligible as far as bridge setup is concerned.


The string must have a certain action, i.e. height over the fretboard. That means you need to stretch it some way to finger notes. By the same effect you'd exploit when bending up tones, this stretching causes the pitch to rise, and that really is what the bridge needs to compensate.

How much the pitch goes up depends on two factors:

  • How high the action is. High action means the string length has to increase more to touch the fret. The bass strings are usually set to higher action than the treble ones, so that alone is a reason to have the bass saddles further down than the treble ones. Note that this has nothing to do with string thickness whatsoever.

  • How much the tension responds to stretching. This is governed by the Young's modulus of the string's core (this is why bending is more effective on steel strings than on nylon, which has a much lower Young's modulus), multiplied by its thickness. The windings don't contribute, because they're not under longitudinal tension. Therefore, strings with a thicker core need to be compensated more, but a wound D-string needs to be compensated less than a plain G-string.

  • The 1st factor - if the string thickness was less, then surely the string itself wouldn't need to be higher off the fretboard! So I believe it must be a contributory factor.
    – Tim
    Commented May 14, 2021 at 18:56
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    Nice answer! To the last paragraph I would add that the string tension also matters, i.e. note pitch change due to fretting on a low tension string will be higher than for the same string tuned at higher tension. Perhaps some math formulas would help to quantify those various dependencies, which sometimes go opposite way. Commented May 14, 2021 at 20:46
  • @Tim “surely”? No. The bass strings need to be higher off the fretboard because they have lower tension, and because bass parts need to be played with more power to produce same volume as treble parts. None of this has to do with the thickness of the bass strings. In fact, the strings being so thick is to bring up the tension, so they have less need for high action! Commented May 14, 2021 at 21:23
  • Thanks for the answer! I get what you're saying, and I agree that it isn't directly the thickness that requires scale adjustment. I'm still a bit baffled by some of the actual instances of adjustment necessary. If you look at the 1st pic in my question, all strings are flatwound, and I would guess their stiffness and action vary pretty smoothly, yet there's a huge step between the 3rd and 4th string adjustment.
    – Alexbib
    Commented May 14, 2021 at 21:27
  • Can't say much about the flatwound. IMHO it's pretty difficult to even define proper intonation on flatwounds, because all tones sound just dead... but anyway, my points should still apply: the bass strings have a winding that makes up most of the weight around a thin core, whereas the treble strings have a thin tape winding that adds only a little bit of mass to the core. Commented May 14, 2021 at 22:35

It's because of intonation, and it's not related to string thickness by itself but stiffness. The step you see is between thinnest wound string and thickest plain string. It is there because the core wire of the wound string is thinner and more flexible than the largest plain string.

  • Agree that thickness per se isn't really what matters, but I'm not convinced stiffness (to the vibration modes) is the reason either. If that had such a strong influence, the inharmonicity in the overtones would be unbearable. I think the main factor is actually increase of tension under stretching. Commented May 14, 2021 at 17:37
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    You're technically correct. The bending stiffness affects the frequency of higher harmonics. While we're being technical, I'd like point out that you really mean Hooke's constant. Both it and bending modulus are derived from Young's modulus.
    – ojs
    Commented May 14, 2021 at 19:22
  • I use (like most ) plain 3rds. So this theory doesn't ring true. If it did, the 2nd string wouldn't need to be relatively longer, but it still does.
    – Tim
    Commented May 15, 2021 at 6:28
  • @Tim I have seen bridges set up like yours, but it is uncommon and I've never played one, so I can only speculate. It could be lower action on 3rd string, a stretchier string from a different set, different shaped saddle so that the contact points do form the double stairs pattern when it doesn't look like it, or anything.
    – ojs
    Commented May 15, 2021 at 9:17

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