Octave reduction

Reduction is the process of replacing an interval by the unique equivalent interval situated between the unison and the equave. In practice, this is done by adding or subtracting equaves from the starting interval as necessary.

Octave reduction is the application of this process in an octave-equivalent tuning (eg. 12edo), where the equave is the octave. Therefore, an octave-reduced interval is always obtained through transposition by octaves, and the reduced interval lies between the unison (1/1) and the octave (2/1).

Tritave reduction is the application of this process in a tritave-equivalent tuning (eg. Bohlen-Pierce), where the equave is the tritave.

Practical methods

An easy way to find a reduced interval is to use a specialized calculator (see External links). This is especially useful when working with very complex ratios.

There are also simple algorithms one can follow to reduce an interval. The choice of the appropriate algorithm depends on the interval size measure being used: linear measures (e.g. frequency ratios), or logarithmic measures (e.g. scale steps or cents).

Linear measures

1. Find the linear measure of the equave; e.g. the octave is 2/1 (or 2), the tritave is 3/1 (or 3), the just perfect fifth is 3/2 (or 1.5), etc.
2. If the starting interval is less than the unison, 1/1 (or 1), multiply it by the equave. Repeat until the resulting interval is greater than the unison.
3. If the starting interval is greater than the equave, divide it by the equave. Repeat until the resulting interval is less than the equave.

Examples (octave-reduction)

• 3/4 is less than 1, so multiply by 2 to get 3/2.
• 7/2 is greater than 2, so divide by 2 to get 7/4.
• 4/1 is greater than 2, so divide by 2 to get 2/1, which is equal to 2, so divide by 2 to get 1/1.
• Adding 4 just perfect fifths (3/2 corresponds to (3/2)⁴, thus 81/16 (or 5.0625), which is greater than 2 octaves (2² = 4), but less than 3 octaves (2³ = 8), so divide by 2 twice to get 81/64.
• Subtracting a just perfect fourth (4/3) from a classic minor third 6/5 corresponds to 6/5 divided by 4/3, thus 9/10 (or 0.9). This interval is less than a unison (2⁰ = 1) but greater than one octave down (2^-1 = 1/2), so multiply by 2 once to get 9/5.

Examples (other equaves)

• Consider a tritave-equivalent tuning; 7/9 is less than 3, so multiply by 3 to get 7/3.
• Consider a just perfect fifth-equivalent tuning; 7/4 is greater than 3/2, so divide by 3/2

Logarithmic measures

1. Find the logarithmic measure of the equave in the same unit as the one used for your starting interval; e.g. an octave in 19edo can be expressed as 19 edosteps, 1200 ¢, 1900 r¢, etc.
2. If the interval is less than the unison (0), add the equave. Repeat until the result is greater than the unison (0).
3. If the interval is greater than the equave, subtract the equave. Repeat until the result is less than the equave.

Examples (octave-reduction)

• 1442¢ is greater than 1200 ¢, so subtract 1200 ¢ to get 242 ¢.
• In 12edo, the octave is 12 steps and the patent val of the fifth harmonic is 28 (steps). This interval is greater than the octave, so subtract 12 to get 16, so subtract 12 again to get 4.

Examples (other equaves)

• In the equal-tempered Bohlen-Pierce tuning, the tritave can be expressed as 1300 hekts and a BP fifth down as -500 hekts. This interval is less than the unison, so add 1300 hekts to get 800 hekts.

See also

• Octave complement

External links

• https://www.yacavone.net/xen-calc/ (web calculator with reduction functions)