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 Cristofori meets Newton (Posted on 2006-04-22)
```261.626.523.251.261.626.1108.73.415.305
554.365.369.994.369.994.698.456.261.626
293.665.329.628.391.995.587.330.932.328
739.989.932.328.329.628.329.628.783.991
783.991.391.995.329.628.739.989.587.330
830.609.554.365.311.127.783.991.391.995
261.626.783.991.739.989.261.626.880.000
329.628.311.127.261.626.932.328.698.456
329.628.783.991.293.665.391.995.493.883
415.305.466.164.329.628.523.251.329.628
```

 No Solution Yet Submitted by Highway6 Rating: 4.1429 (7 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
 re(3): Half a Hint? (Full Solution) | Comment 4 of 6 |
(In reply to re(2): Half a Hint? by Dustin)

The numbers are the frequencies of 19 of the notes from C4 (also known as Middle C) to C#6, separated by a period.
Each note (frequency) corresponds to a letter of the alphabet.
Substituting the letter of the alphabet with the frequency value  reveals the first few lyrics of a famous hymn:

Amazing Grace how sweet the sound that saved a wretch like me

(4th Octave)
A = 261.626           B = 277.180  C#          C = 293.665  D
D = 311.127  D#          E = 329.628  E          F = 349.23    F
G = 369.994  F#        H = 391.995  G         I = 415.305  G#
J = 440.000   A          K = 466.164  A#       L = 493.883  B

(5th Octave)
M = 523.251  C            N = 554.365   C#       O = 587.330  D
P = 622.25    D#       Q = 659.26             R = 698.456  F
S = 739.989  F#       T = 783.991           U = 830.609  G#
V = 880.000  A         W = 932.323  A#       X = 987.77    B

(6th Octave)
Y = 1046.5    C         Z = 1108.73  C#

Edited on April 23, 2006, 7:03 am
 Posted by Dej Mar on 2006-04-23 05:25:54

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