## Tuesday, January 9, 2018

### Lesson 8-5: Areas of Triangles (Day 85)

Lesson 8-5 of the U of Chicago text is called "Areas of Triangles." In the modern Third Edition, areas of triangles appear in Lesson 8-4.

This is what I wrote two years ago about today's lesson:

`Actually, I don't have much to say about today's geometry lesson at all, since Lesson 8-5 of the U of Chicago text is on the areas of triangles. You already know that the area of a triangle is half the product of its base and height. I did point out that this is proved for all three cases -- when the altitude is inside, outside, or aside of the triangle. This roughly corresponds to the acute, obtuse, and right triangles. But notice that if the altitude is drawn from the largest angle of a triangle, the altitude is always inside the triangle even if that angle is right or obtuse.`
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`One of the included questions gives the area of a quadrilateral with perpendicular diagonals as half the product of those diagonals. It applies to all quadrilaterals whose diagonals are perpendicular -- which includes kites (and rhombuses and squares, which are kites under the inclusive definition). In some texts this is a highlighted formula, but in the U of Chicago it is buried in an exercise.`

And so let me take advantage of the short lesson to squeeze in some extra topics. I suppose I can't avoid the big news story of the day -- rain and mudslides here in California. It's typical for mudslides to occur right around the same areas that have been burnt by wildfires. Therefore some Santa Barbara county schools are once again closed. It will be tough for them to make it to 180 days this year!

Last year, there were also rainstorms that occurred right when we returned from winter break. (Recall the post about the "California Snow Day"?) I wrote about how outdoor recess and lunch were cancelled, and that P.E. teachers played a movie indoors instead. Well, I found out that today during the rain, the students participated in indoor yoga.

Today there is a Google Doodle featuring a scientist -- Har Gobind Khorana, an Indian-American biochemist. He earned a Nobel Prize for helping to discover the structure of DNA and RNA. Yes -- all those A's, C's, G's, and T's (and U's in RNA) go back to Khorana.

I've already written about how I tried to teach DNA and RNA last year to my eighth graders. The lesson had two major problems. First, I used my Bruin Corps member -- a molecular bio major -- to teach the lesson, when I really should have developed science lessons without Bruin Corps. Second, I taught the lesson to eighth graders, when I should have followed the LAUSD Pacing Plan and gave the lesson to seventh grade instead.

Earlier I posted a song based on a DNA sequence. This wasn't for Mocha, but for the Google Fischinger player instead. I could rewrite the DNA song for Mocha -- but then again, I've already written so many coding music posts lately that I don't wish to post more code so soon.

In those DNA songs, we changed T to F, since the musical scale has F, but not T. Then again, we might recall that the old Bohlen-Pierce scale "s-delta" had both "t" and "u," so we could borrow T and U from that scale -- while keeping A, C, and G -- to make a real DNA/RNA song.

By the way, I also looked into the Atari 16-bit music to figure out the pattern. No, again we're not coding any songs today -- I don't have an Atari emulator, and Mocha can't handle the program listed at the above link:

https://www.atariarchives.org/c3ba/page045.php

Anyway, we see that in the chart for both 8- and 16-bit, higher numbers correspond to lower octaves and vice versa. This suggests that these values are raw "Degrees" which we don't need to subtract from Bridge 261 or any other number to find the ratio.

But notice that the highest B in 8-bit (B6) has a value of 15. So doubling this to 30 ought to give the next B an octave (2/1) lower (B5) -- but 30 corresponds to C6, not B3. Also, the highest B in 16-bit (B9) is 50, while doubling this to 100 produces C9, not B8.

It appears that there actually is a "bridge" that we must take into consideration -- except that this bridge is negative, not positive. For 8-bit, it's Bridge -1, while for 16-bit, it's Bridge -7. We can prove these bridges by starting with a low note with an odd value, such as C3 (243) in 8-bit. This note is actually 244 degrees away from Bridge -1. Meanwhile, C4 (which I assume is Middle C) is 121, or 122 degrees away from Bridge -1. So the calculated ratio is 244/122 = 2/1, a true octave. Likewise, for 16-bit we use -7 as the bridge -- C1 is 27357 (27364 from Bridge -7) and C2 is 13675 (13682 from Bridge -7), and 27364/13682 = 2/1, a true octave. As long as the lower note is odd (hence an even degree from Bridge -1 or -7), then all octaves should be 2/1.

In 8-bit, starting from 1 degree from the bridge and doubling produces notes 0, 1, 3, 7, 15, 31, 63, then finally 127. Apparently, these notes are all B's, so we can color these all as "white B" and then determine the rest of the note names using just ratios as we did with Mocha.

In 16-bit, starting with 1 degree from the bridge and doubling produces notes -6, -5, -3, -1, 9, 25, 57, 121, 249, 505, 1017, 2041, 4089, 8185, and 16377. According to the chart these notes are all A's. We could call these all "white A," but notice that the difference between 16-bit white A's and 8-bit white B's are not a just 9/8. This is significant because Atari apparently allows users to play harmony -- either two 16-bit voices or one 16-bit and one 8-bit voice.

I calculated the difference between 16-bit A and 8-bit B to be a just 9/8 plus about 21.5 cents -- or approximately one syntonic comma. Hence we could call the 16-bit A "yellow A" in order to show the relationship between 16-bit and 8-bit notes. ("Yellow" is still the only otonal color allowed, since all computer music is based on utonal EDL's.)

Okay, that's enough on computer music. Here are the worksheets: