9

Table Of Contents
To simplify this example: Imagine that you are tuning an instrument, beginning with a
note called C at a frequency of 100 Hz. (A real C would be closer to 130 Hz.) The first fifth
would be tuned by adjusting the pitch until a completely clear tone was produced, with
no beats. (Beats are cyclic modulations in the tone.) This would result in a G at exactly
150 Hz, and is derived from the following calculation:
The fundamental (100 Hz) x 3 (= 300 Hz for the second harmonic).
Divided by 2 (to drop it back into the same octave as your starting pitch).
This frequency relationship is often expressed as a ratio of 3:2.
For the rest of the scale: Tune the next fifth up: 150 x 3 = 450. Divide this by 2 to get 225
(which is more than an octave above the starting pitch, so you need to drop it another
octave to 112.5).
The following table provides a summary of the various calculations.
NotesFrequency (Hz)Note
x 1.5 divided by 2.100C
Divide by 2 to stay in octave.106.7871C#
Divide by 2 to stay in octave.112.5D
Divide by 2 to stay in octave.120.1355D#
Divide by 2 to stay in octave.126.5625E
135.1524F (E#)
Divide by 2 to stay in octave.142.3828F#
(x 1.5) divided by 2.150G
160.1807G#
168.75A
180.2032A#
189.8438B
202.7287C
As you can see from the table above, theres a problem.
Although the laws of physics dictate that the octave above C (100 Hz) is C (at 200 Hz),
the practical exercise of a (C to C) circle of perfectly tuned fifths results in a C at
202.7287 Hz. This is not a mathematical error. If this were a real instrument, the results
would be clear.
To work around the problem, you need to choose between the following options:
Each fifth is perfectly tuned, with octaves out of tune.
Each octave is perfectly tuned, with the final fifth (F to C) out of tune.
1227Chapter 43 Project Settings in Logic Pro