Honors Physics is still learning about resistors in circuits, but now regular Physics -- a chapter or two behind the honors class -- is starting the lesson on electrical current. Besides the mathematical calculation that's a natural part of physics problems, I do help one student in math during tutorial. She is taking AP Calculus BC and learning about arc length. The function whose arc length she's trying to find is a nasty rational function. The formula is the integral of sqrt(1 + (y')^2), and as it turns out, the function y is chosen so that sqrt(1 + (y')^2) works out to be something that's easy to integrate.
OK, let's get back to music. Today's task is to compose music in 14EDL. I'm seeking out music that is easy to play on both Mocha and the guitar.
The 14EDL scale:
Degree Ratio Note
14 1/1 red F#
13 14/13 ocher G
12 7/6 white A
11 14/11 amber B
10 7/5 green C
9 14/9 white D
8 7/4 white E
7 2/1 red F#
For 12EDL, we started with the main triad, which happens to be an A minor triad. The corresponding triad in 14EDL uses Degrees 14, 12, 10. This reduces to 7:6:5 and plays as an F# diminished triad.
But how would we finger F#dim on the guitar? The rule of thumb is that in order to have an open chord, at least two notes of the chord need to be open strings (E, A, D, G, B, E). The F#dim chord contains notes F#, A, C, and only C is open.
Well, the common trick is to add a seventh to this chord. This is called a half-diminished chord -- guitarists are more likely to write this chord as F#m7b5, "F# minor 7th with a flat 5." That this chord would appear makes sense, as the 14EDL scale is similar to the Locrian mode, and the Locrian mode is strongly associated with the m7b5 chord.
OK, so let's try to finger the F#m7b5 chord. The bottom string must play the root note, which is F# at the second fret. The next string can be open A. Then the next three strings are fretted as E, A, C, and then the top string is open E. Notice that this is simply an A minor chord played over an F# bass, and so the chord can also be written Am/F#. It's easiest to use the middle finger on the bottom string and then adjust the other fingers from Am accordingly.
This fingering is still awkward, so we might wonder whether a bar F#m7b5 is available here. The following link shows several chord shapes, but the one that feels the least awkward on my hand is the sixth one, where the bar is on the second fret and the other fingers are C (fifth string), F# (fourth string), and E (second string).
But now we want to play other chords. Starting from an F#m7b5 chord, the easiest chord to move to is Am -- just leave out the bottom string. But I'd like to play a chord that involves the three missing notes -- G, B, and D. (We ignore "ocher" and "amber," even though "ocher G" is more like G# and "amber B" is more like Bb.)
The notes G, B, and D obviously spell out a G major chord -- even though a true 13:11:9 chord sounds more like a minor chord than a major chord. Since we're ignoring the colors, I must play the chord as G major.
Now fortunately, G major is an easy open chord to play. The bad news is that it's not easy for me to move my hand from F#m7b5 to G. It's slightly easier to move from open F#m7b5 to open G if the index finger is not used for open G. Ironically, I find it easier to move from the bar F#m7b5 to open G with the usual fingering,
Don't forget that I'm an amateur musician. Professional guitarists can probably make all of these chord changes without thinking about it. But if I were a professional guitarist, I obviously wouldn't be a math teacher! I must focus on all of these chord changes deliberately to make sure that I compose something that I can actually play in the classroom.
Also, all of these chord changes are relevant to playing harmony (rhythm guitar). It's also possible to play a melody on the guitar (lead guitar). But I'm not talented enough to play lead guitar and sing the melody at the same time. My goal is to have Mocha play a melody while I sing along the lyrics, and then once I've memorized the lyrics, I play rhythm guitar to accompany my voice. (Once again, if I were more talented at guitar, I wouldn't need to teach math!)
At one of the links from yesterday, Seymour Duncan suggests playing the following progression in B Locrian mode: Bm7b5 - G - Am - F. In F# Locrian, this becomes F#m7b5 - D - Em - C. I tried playing this sequence on the guitar, and I think I can master it with some practice.
And so let's program Mocha to compose a random tune using this sequence:
10 CLS
20 N=16
30 FOR A=0 TO 7
40 B=4
50 X=A-INT(A/4)*4
60 IF X=0 THEN D=14
61 IF X=1 THEN D=9
62 IF X=2 THEN D=8
63 IF X=3 THEN D=10
70 PRINT D;
80 L=RND(B)
90 SOUND 261-N*D,4*L
100 IF L>1 THEN FOR I=1 TO L-1:PRINT " ";:NEXT I
110 B=B-L
120 IF B>0 THEN D=15-RND(8):GOTO 70
130 PRINT
140 NEXT A
150 PRINT 14
160 SOUND 261-N*14,16
I admit that Lines 60-63 are clumsy here, but I can't see an easier way to make the degree D have the correct value depending on the measure A. The song repeats F#m7b5 - D - Em - C twice until ending on a whole note F#. Notice that in true 14EDL, the C major chord sounds slightly like Caug, while the Em chord is completely off -- it sounds either like E major or Edim. The D chord becomes D supermajor, so it should still sound like a D major triad.
We can change N, which can range from 1 to 18. As usual, smaller numbers represent the higher pitches (the Yanny range) while large numbers represent the lower pitches (the Laurel range). (That's right -- I just had to make a Yanny/Laurel reference there!)
Just as I did last week with 12EDL, let's try to convert 14EDL to an EDO scale that might be more suitable than 12EDO. Using the program I wrote last week, 14EDL produces the following EDO's:
1, 2, 5, 6, 9, 27, 31, 58, 66, 72, 270, ...
The first nontrivial EDO on this list is 27EDO. As it turns out, 27EDO has some strange properties that make it very different from 12EDO.
For example, we know that C-E is a major third in 12EDO. At 400 cents, this interval is a close approximation of a just 5/4. As it turns out, a major third in 27EDO is also 400 cents -- it's the ninth step of 27EDO, so it's exactly a third of an octave. But the name of the note that's a major third above C isn't E -- instead, this same note is called D#!
The problem is that C-E is a major third only in meantone (that is, a scale in which the syntonic comma 81/80 is tempered out). Now 12EDO is a meantone scale, but not 27EDO. Therefore the major third in 27EDO isn't C-E. In fact, here is the 14EDL scale converted to 27EDO, starting from both C and F#:
The 14EDL scale (in 27EDO):
Degree Ratio Note (C) Note (F#)
14 1/1 C F#
13 14/13 C#v Ab^ (Ab-up)
12 7/6 Eb A
11 14/11 E A#
10 7/5 Gb^ Db
9 14/9 Ab D
8 7/4 Bb E
7 2/1 C F#
The distance from C-D is five steps of 27EDO, while sharps and flats raise notes by four steps. The rule is that C-G (Bb-F, B-F#) must represent a perfect fifth in this EDO. Since perfect fifths are slightly sharp in this EDO (711.1 cents), it makes C-D# into a major third. Notice that the simple looking intervals tend to be septimal (C-Eb a subminor third, F#-E a harmonic seventh, and even C-E representing either 9/7 or 14/11).
Here is some more information on 27EDO, preserved from the Xenharmonic site. (Recall that all Xenharmonic links will be dead after July!)
http://xenharmonic.wikispaces.com/27edo
If octaves are kept pure, 27edo divides the octave in 27 equal parts each exactly 44.444... cents in size. However, 27 is a prime candidate for octave shrinking, and a step size of 44.3 to 44.35 cents would be reasonable. The reason for this is that 27edo tunes the third, fifth and 7/4 sharply.
Assuming however pure octaves, 27 has a fifth sharp by slightly more than nine cents and a 7/4 sharp by slightly less, and the same 400 cent major third as 12edo, sharp 13 2/3 cents. The result is that 6/5, 7/5 and especially 7/6 are all tuned more accurately than this.
Combining ups and downs notation with color notation, qualities can be loosely associated with colors:
| quality | color | monzo format | examples |
|---|---|---|---|
| minor | blue | {a, b, 0, 1} | 7/6, 7/4 |
| " | fourthward white | {a, b}, b < -1 | 32/27, 16/9 |
| upminor | green | {a, b, -1} | 6/5, 9/5 |
| mid | emerald | {a, b, 0, 0, 0, 1} | 13/12, 13/8 |
| " | ochre | {a, b, 0, 0, 0, -1} | 16/13, 24/13 |
| downmajor | yellow | {a, b, 1} | 5/4, 5/3 |
| major | fifthward white | {a, b}, b > 1 | 9/8, 27/16 |
| " | red | {a, b, 0, -1} | 9/7, 12/7 |
| color of the 3rd | JI chord | notes as edosteps | notes of C chord | written name | spoken name |
|---|---|---|---|---|---|
| blue | 6:7:9 | 0-6-16 | C Eb G | Cm | C minor |
| green | 10:12:15 | 0-7-16 | C Eb^ G | C.^m | C upminor |
| jade | 18:22:27 | 0-8-16 | C Evv G | C~ | C mid |
| yellow | 4:5:6 | 0-9-16 | C Ev G | C.v | C downmajor or C dot down |
| red | 14:18:27 | 0-10-16 | C E G | C | C major or C |
Question 7 of the SBAC Practice Exam is on inequalities:
Which inequality represents all possible solutions of -6n < -12?
A) n < 72
B) n > 2
C) n < 2
D) n > 72
At last, we have a true first semester Algebra I problem. The answers are set up to test student knowledge of two important ideas. First to solve -6n < -12, instead of multiplying by -6, we should divide by -6. This eliminates choices A) and D). The other is that whenever we divide both sides of an inequality, we must flip the direction of the inequality. Thus the answer is B).
Last week when I subbed in an eighth grade Algebra I class, dividing inequalities by negative numbers arose in the context of finding the domains of radical functions. What, for example, is the domain of the function f(n) = sqrt(-6n + 12)? A few students remembered to flip the direction of the inequality, while others didn't. This is something I must make sure my students remember when I teach my summer Algebra I course.
Question 8 of the SBAC Practice Exam is on trig ratios:
Consider this right triangle.
[In Triangle ABC, C is a right angle, BC = 20, AC = 21, AB = 29.]
Enter the ratio equivalent to sin B.
This Geometry question is also straightforward. The sine ratio is opposite divided by hypotenuse. The side opposite B is AC = 21, while the hypotenuse is 29. Therefore sin B = 21/29.
SBAC Practice Exam Question 7
Common Core Standard:
Solve linear equations and inequalities in one variable, including equations with coefficients represented by letters.
SBAC Practice Exam Question 8
Common Core Standard:
Understand that by similarity, side ratios in right triangles are properties of the angles in the triangle, leading to definitions of trigonometric ratios for acute angles.
Commentary: Solving simple inequalities such as this one is covered in Lesson 4-6 of the U of Chicago Algebra I text. And the sine ratio, meanwhile, is defined in Lesson 14-4 of the U of Chicago Geometry text.
Which inequality represents all possible solutions of -6n < -12?
A) n < 72
B) n > 2
C) n < 2
D) n > 72
At last, we have a true first semester Algebra I problem. The answers are set up to test student knowledge of two important ideas. First to solve -6n < -12, instead of multiplying by -6, we should divide by -6. This eliminates choices A) and D). The other is that whenever we divide both sides of an inequality, we must flip the direction of the inequality. Thus the answer is B).
Last week when I subbed in an eighth grade Algebra I class, dividing inequalities by negative numbers arose in the context of finding the domains of radical functions. What, for example, is the domain of the function f(n) = sqrt(-6n + 12)? A few students remembered to flip the direction of the inequality, while others didn't. This is something I must make sure my students remember when I teach my summer Algebra I course.
Question 8 of the SBAC Practice Exam is on trig ratios:
Consider this right triangle.
[In Triangle ABC, C is a right angle, BC = 20, AC = 21, AB = 29.]
Enter the ratio equivalent to sin B.
This Geometry question is also straightforward. The sine ratio is opposite divided by hypotenuse. The side opposite B is AC = 21, while the hypotenuse is 29. Therefore sin B = 21/29.
SBAC Practice Exam Question 7
Common Core Standard:
Solve linear equations and inequalities in one variable, including equations with coefficients represented by letters.
SBAC Practice Exam Question 8
Common Core Standard:
Understand that by similarity, side ratios in right triangles are properties of the angles in the triangle, leading to definitions of trigonometric ratios for acute angles.
Commentary: Solving simple inequalities such as this one is covered in Lesson 4-6 of the U of Chicago Algebra I text. And the sine ratio, meanwhile, is defined in Lesson 14-4 of the U of Chicago Geometry text.
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