Apple Logic Pro 9 - Manual (Page 1227)

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.
Notes Frequency (Hz) Note
x 1.5 divided by 2. 100 C
Divide by 2 to stay in octave. 106.7871 C#
Divide by 2 to stay in octave. 112.5 D
Divide by 2 to stay in octave. 120.1355 D#
Divide by 2 to stay in octave. 126.5625 E
135.1524 F (E#)
Divide by 2 to stay in octave. 142.3828 F#
(x 1.5) divided by 2. 150 G
160.1807 G#
168.75 A
180.2032 A#
189.8438 B
202.7287 C
As you can see from the table above, there’s 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.
1227 Chapter 43 Project Settings in Logic Pro