What sets a string's tension
A stretched string's pitch is fixed by three quantities: its length, its mass per unit length, and the tension pulling on it. Mersenne worked this out in the 1630s and Taylor gave it its modern form a century later. Rearranged for tension it says that tension is proportional to the mass per unit length, to the square of the length, and to the square of the frequency.
Both squares matter enormously in practice. Move a string up one semitone and the frequency rises by 5.95%, but the tension rises by 12.25% - because 1.0595 squared is 1.1225. Move it up an octave and the tension quadruples. Fit the same string to a 25.5-inch scale instead of a 24.75-inch one, at the same pitch, and the tension rises by (25.5/24.75)2 = 6.2%. That squared relationship is why a Fender feels tighter than a Gibson with the identical set of strings, and why drop tunings feel so much slacker than they sound.
The mass term is the only one you can choose freely, and it is where the gauge comes in. For a plain steel string, mass per inch is simply the density of the wire times its cross-sectional area, and area goes as the square of the diameter - so going from a .010 to a .011 raises tension by (11/10)2 = 21%. Wound strings break that simple relationship, because a wound string's outer diameter includes a wrap wire with air gaps beneath it and a core that carries almost all the load.
The unit-weight form and why 386.4 appears
String makers publish tension charts in the imperial form T = UW x (2Lf)2 / 386.4, where UW is the string's unit weight in pounds per inch, L is the scale length in inches, f is the frequency in hertz, and T comes out in pounds. The 386.4 is the acceleration due to gravity expressed in inches per second squared - 32.2 ft/s2 times twelve - and it appears purely because unit weight is a weight rather than a mass. Divide weight by g and you get mass; the constant is a unit conversion, not physics.
The SI form has no such constant: T = 4 μ L2 f2, with μ the linear mass density in kg/m, L in metres and T in newtons. The two forms are the same equation.
For a plain steel string you can compute the unit weight yourself, because it is solid wire: UW = ρ x πd2/4, with ρ the density of music wire, about 0.282 lb per cubic inch, or 7.81 g/cm3. That figure reproduces published plain-string tensions to well under half a pound across the common gauges, which is the check worth making before trusting any density value.
Wound strings cannot be handled this way. The outer diameter tells you nothing reliable about the mass, because the wrap wire is helical with gaps beneath it and the core diameter varies between makers even at the same nominal gauge. Treating a wound string as solid steel overstates its unit weight by roughly 20%. This calculator's wound option uses an effective density of 0.235 lb per cubic inch, fitted to published nickel-wound electric sets, and it is a reasonable approximation rather than a specification. When accuracy matters - balancing a set, or specifying for a non-standard tuning - take the unit weight straight from the maker's chart and use the custom option.
The scale length to enter is the nut-to-saddle distance, which is exactly twice the nut-to-twelfth-fret distance; the fret spacing calculator shows why. And to find the frequency of any note at any reference pitch, the note to frequency calculator gives it directly.
Worked example: a .010 high E, and what drop D does to a .046 low E
Start with the high E of a light electric set: a plain .010 tuned to E4 on a 25.5-inch scale, at A4 = 440 Hz.
- Frequency. E4 is four semitones below A4, so f = 440 x 2−5/12 = 329.6276 Hz.
- Cross-sectional area. π/4 x 0.0102 = 0.7853982 x 0.0001 = 7.853982 x 10−5 sq in.
- Unit weight. 0.282 lb/in3 x 7.853982 x 10−5 = 2.21482 x 10−5 lb/in.
- The wave-speed term. 2 x 25.5 x 329.6276 = 16,811.0 in/s. Square it: 2.82610 x 108.
- Divide by g. 2.82610 x 108 / 386.4 = 731,392.
- Multiply by the unit weight. 2.21482 x 10−5 x 731,392 = 16.20 lb. Published charts give 16.2 lb for exactly this string, which is the confirmation that the density figure is right.
- In metric. 16.20 x 0.45359 = 7.35 kgf, or 16.20 x 4.4482 = 72.1 N.
Now take the low E, a wound .046 tuned to E2 at 82.4069 Hz on the same scale, and drop it to D2.
- Standard tension. Published charts give about 17.5 lb for a nickel-wound .046 at E2 on 25.5 inches.
- Apply the detune ratio. Two semitones down multiplies frequency by 2−2/12 = 0.890899, and tension by the square of that: 2−2/6 = 2−1/3 = 0.793701.
- Drop D tension. 17.5 x 0.793701 = 13.89 lb, a reduction of 20.63%.
- Restore the feel with a heavier string. Tension goes as the square of diameter for a solid string, and roughly so for a wound string at fixed construction, so you need the diameter multiplied by 1 / 0.890899 = 1.1225. That takes .046 to .0516, and the nearest common gauge is .052.
Check that last step: a .052 at D2 would carry approximately 17.5 x (52/46)2 x 0.793701 = 17.5 x 1.2779 x 0.793701 = 17.75 lb, essentially back where the .046 sat at E2. That is exactly why drop-tuning sets ship with a fatter bottom string.
What tension figure to aim for
For a six-string electric at standard pitch on a 25.5-inch scale, individual strings in common sets run roughly 15 to 20 lb each, with total set tension around 95 to 110 lb. Acoustic sets run higher - typically 25 to 35 lb per string and 160 to 190 lb total - because a bigger driven top needs more energy. Basses at 34 inches sit around 35 to 45 lb per string. These are the ranges the instruments are designed around, and they are properties of the sets rather than physical limits.
Two figures matter more than the absolute number. The first is balance across the set: whether each string carries a similar tension, so that they respond alike under the fingers. Most factory sets are not balanced - the third string, plain in a light electric set and wound in a heavier one, is usually the outlier. Compute each string in turn and you can pick individual gauges to flatten the profile, which is what balanced-tension sets do.
The second is total load on the instrument, which is the sum across all strings and is what the neck and, on an acoustic, the top and bridge actually feel. Going up one gauge across a six-string set adds roughly 20% to the total, which changes neck relief enough to need a truss-rod adjustment. Going up two gauges on a vintage acoustic designed for light strings is a structural decision, not a tonal one.
Very low tension has its own failure modes. Below roughly 10 lb a plain string buzzes against frets, intonates badly, and loses sustain, because there is not enough restoring force to keep the amplitude in check. Very high tension is limited by the wire itself: music wire has a tensile strength around 300,000 psi, so a .010 plain string with its 7.854 x 10−5 sq in of area breaks somewhere near 300,000 x 7.854 x 10−5 = 23.6 lb - only 45% above the 16.2 lb it normally carries, which is about three semitones of headroom.
Tension of plain steel strings at standard guitar pitches
| Gauge (in) | Note | Frequency (Hz) | 24.75 in scale (lb) | 25.5 in scale (lb) | 27 in scale (lb) |
|---|---|---|---|---|---|
| .009 | E4 | 329.63 | 12.36 | 13.12 | 14.71 |
| .010 | E4 | 329.63 | 15.26 | 16.20 | 18.16 |
| .011 | E4 | 329.63 | 18.46 | 19.60 | 21.97 |
| .012 | E4 | 329.63 | 21.97 | 23.33 | 26.15 |
| .011 | B3 | 246.94 | 10.36 | 11.00 | 12.33 |
| .013 | B3 | 246.94 | 14.47 | 15.36 | 17.22 |
| .016 | B3 | 246.94 | 21.92 | 23.27 | 26.09 |
| .016 | G3 | 196.00 | 13.81 | 14.66 | 16.44 |
| .017 | G3 | 196.00 | 15.59 | 16.55 | 18.56 |
| .018 | G3 | 196.00 | 17.48 | 18.56 | 20.80 |
Only plain steel is listed, because a wound string's tension cannot be derived from its outer diameter. Note the two squares at work: within one pitch, tension scales as the square of the gauge; across the scale-length columns it scales as the square of the length, so 27 in carries (27/25.5)^2 = 12.1% more than 25.5 in.
Pitfalls and assumptions
- Gauge alone does not determine a wound string's tension. Two .046 strings from different makers can differ by several percent in unit weight, and flatwound, half-round and pure-nickel constructions differ more.
- The formula gives static tension, not the load on the neck. The neck sees the vector sum over all strings, and the top of an acoustic also sees a rotational torque from the bridge that the tension figure alone does not describe.
- Scale length means nut to saddle, not nut to bridge pin. Saddle compensation adds a millimetre or two per string, which changes tension by well under a percent and is not worth including.
- A capo shortens the speaking length but does not change the tension in the speaking portion by much - the string is still anchored at the same total tension, which is why the pitch rises rather than the string going slack.
- Stiffness is ignored. The ideal-string law treats the string as perfectly flexible. Real strings are stiff, which raises the partials above exact multiples and matters for intonation, especially on thick plain strings.
- The reference pitch matters. Tuning to A442 instead of A440 raises every tension by (442/440)^2 = 0.91%, which is small but real across a whole set.
- Breaking strength is not part of this calculation. A gauge that gives a comfortable tension at one pitch may be past its tensile limit two semitones higher, and plain strings fail before wound ones.
Balanced sets, extended range and where the law stops
The most useful thing the formula does is let you design a set rather than buy one. Pick a target tension - say 18 lb for an electric - work out the frequency of each open string, and solve for the gauge that hits the target. Because tension scales with the square of the diameter for a solid string, the correction is gentle: a string carrying 15 lb when you want 18 needs its diameter multiplied by sqrt(18/15) = 1.095, so a .011 becomes about a .012. Doing that across a set produces even resistance under the fingers, which many players find changes their touch far more than any change of brand.
Extended-range instruments make the problem acute. A seven-string tuned down to B1 at 61.7 Hz on a 25.5-inch scale would need an enormous gauge to reach a normal tension, which is why multi-scale or fanned-fret designs exist: a longer scale on the bass side lets a sane gauge carry a usable tension, because tension scales with the square of the length. Moving the low B from a 25.5-inch scale to a 27-inch one gains (27/25.5)2 = 12.1% of tension at the same gauge and pitch, which is the difference between floppy and playable. The fret spacing calculator lays out the geometry those designs need.
The law also has limits worth knowing. It assumes a perfectly flexible string, and real wire has bending stiffness that raises the partials above exact integer multiples - the inharmonicity that forces piano tuners to stretch octaves, and that the harmonic series calculator quantifies. Stiffness rises sharply with diameter and falls with length, so the effect is worst on short, thick strings, which is exactly why plain strings above about .024 are avoided and why short-scale basses sound different from long-scale ones.
Finally, remember that changing tuning changes tension, and changing tension changes the instrument's setup. A capo, a drop tuning or a change of reference pitch all shift the load, and after a significant change the neck relief and intonation both need checking. If you are transposing music rather than the instrument, the transposition calculator handles the key arithmetic and leaves the strings alone.