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Pitch Interval in Cents

Measures the size of a musical interval in cents.

Formula

c=1200log⁡2 ⁣(f2f1)c = 1200 \log_2\!\left(\frac{f_2}{f_1}\right)

Variables

cInterval (cents)
f_1Lower frequency (Hz)
f_2Higher frequency (Hz)

Example

f_2/f_1 = 1.5 (perfect fifth): c ≈ 702 cents

How it works

This formula converts a frequency ratio into cents, a logarithmic unit that musicians and acousticians use to measure very small or very large pitch differences on a consistent scale. It works because equal musical intervals correspond to equal frequency ratios, not equal frequency differences; using a base-2 logarithm makes an octave, where the frequency doubles, equal to 1200 cents. That means the cents scale matches how pitch is perceived and how musical intervals repeat across octaves.

When to use it

Use this when comparing two pitches, checking tuning accuracy, or analyzing intervals in equal temperament, just intonation, or recorded audio. It is especially useful whenever the frequencies are known and you want the interval size in a standard musical unit.

Worked example, step by step

Suppose a note has frequency f1 = 220 Hz and the higher note has frequency f2 = 330 Hz. Step 1: Form the ratio: f2/f1 = 330/220 = 1.5 Step 2: Take the base-2 logarithm: log2(1.5) ≈ 0.58496 Step 3: Multiply by 1200: c = 1200 × 0.58496 ≈ 701.95 Step 4: Round to the nearest cent: c ≈ 702 cents So the interval between 220 Hz and 330 Hz is about 702 cents.

Common mistakes to avoid

  • Using the frequency difference (f2 - f1) instead of the ratio f2/f1
  • Forgetting that the logarithm is base 2, not base 10 or natural log unless you convert correctly
  • Reversing the frequencies so the ratio becomes less than 1, which gives a negative value instead of the size of an upward interval

Did You Know?

There are 1200 cents in an octave and 100 in a semitone; trained ears can detect differences of about 5 cents.

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Reviewed & maintained by the FormulaWon editorial teamLast reviewed September 2026