# Just intonation point

(Redirected from JIP)

The just intonation point (JIP) is a special tuning map that maps every monzo in some subgroup to its span in cents (or any other logarithmic unit), relative to the point 1/1 (which maps to 0 cents).

For instance, in 5-limit JI, the JIP is 1200.000 1901.955 2786.314]; if we take the bracket product of this tuning map with any monzo, we get its size in cents. Of course, one can always build the JIP using different units than cents.

For prime limits, the JIP has a particularly simple definition in Tenney-weighted coordinates, where it is always the all-ones vector, 1 1 1 ...].

## Units

It may be helpful to think of the units of each entry of the JIP — as with a normal (temperament) tuning map — as $\mathsf{¢}\small /𝗽$ (read "cents per prime"), $\small \mathsf{oct}/𝗽$ (read "octaves per prime"), or any other logarithmic pitch unit per prime. For more information, see Dave Keenan & Douglas Blumeyer's guide to RTT: units analysis.

## Mathematical definition

The JIP, commonly denoted J, is a point in p-limit tuning space which represents untempered p-limit JI. Specifically, it is equal to log22 log23 log25 … log2p], meaning that each prime q in the p-prime limit is tuned to log2q octaves (which is exactly the just value of the prime q).

The JIP is the target of optimization in optimized tunings including TOP and TE tuning. If m is a monzo, then <J|m> is the untempered JI value of m measured in octaves. In Tenney-weighted coordinates, where m = [m2 m3 m5mp is represented by the ket vector [e2log22 e3log23 e5log25 … eplog2p, then J becomes correspondingly the bra vector 1 1 1 … 1].

As seen in the 5-limit projective tuning space diagram, it is the red hexagram in the center. ET maps which are relatively close to this hexagram, which means low tuning error, such as 53 84 123 …], have integer elements which are in proportions relatively similar to the proportions of the corresponding elements in J = log22 log23 log25 …]1.000 1.585 2.322 …], e.g. $\frac{84}{53} ≈ \frac{1.585}{1.000}$ and $\frac{123}{53} ≈ \frac{2.322}{1.000}$.