`e'min` vs `e'min'i0` behaves differently in `rot 1 $ arp "up"`

Questions and Answers
1

The rot gives different results when working on e'min compared to e'min'i0. I would expect that both chords would give equal results since i0 is the "zero-th" inversion, i.e., no inversion. Right?

rot 1 $ arp "up" $ n "e'min"

yields

(0>⅓)|n: 7.0n (g5)
(⅓>⅔)|n: 11.0n (b5)
(⅔>1)|n: 4.0n (e5)

while

rot 1 $ arp "up" $ n "e'min'i0"

yields

(0>⅓)|n: 11.0n (b5)
(⅓>⅔)|n: 4.0n (e5)
(⅔>1)|n: 7.0n (g5)

Is this behavior by intention?

2

i0 is not the 0th inversion - inversions work by adding extra is eg subsequent inversions would look like:

e'min e'min'i e'min e'min'ii e'min e'min'iii e'min e'min'iiii and so on

3

The reference describes it with the i2 notation. So I concluded that i0 is the "root position" of the chord.

Using the first and second inversions, both notations yields equivalent results using the rot. Using the i2 notation, the rot moves the events in the same way (first event moves to third position) compared to the ii notation where the "no-i" version, i.e., e'min moves to the other direction (as mentioned above).

rot 1 $ arp "up" $ n "e'min'i1" and rot 1 $ arp "up" $ n "e'min'i" yields

(0>⅓)|n: 16.0n (e6)
(⅓>⅔)|n: 7.0n (g5)
(⅔>1)|n: 11.0n (b5)

and rot 1 $ arp "up" $ n "e'min'i2" and rot 1 $ arp "up" $ n "e'min'ii" yield

(0>⅓)|n: 19.0n (g6)
(⅓>⅔)|n: 11.0n (b5)
(⅔>1)|n: 16.0n (e6)

So I was wondering whether this behavior has some intention.

4

Well, that's embarrassing - guess who wrote that part of the reference :stuck_out_tongue:

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I stand corrected, the syntax must have been updated and I forgot all about it. (looking at the date, I was actually holed up in bed isolating from my family with a bad case of covid, so no wonder I don't remember!)

I'm not sure why the rot changes between them ... the output without rot should be equal for both (I don't have tidal here, so testing on estuary - I get identical output without rot). If that is equal then the output from rot should also be identical. What do you see if you remove rot for both e'min and e'min'i0?

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5 
rot 1 $ arp "up" $ n "e'min'i0"

yields

(0>⅓)|n: 11.0n (b5)
(⅓>⅔)|n: 4.0n (e5)
(⅔>1)|n: 7.0n (g5)

without rot 1 it yields

(0>⅓)|n: 4.0n (e5)
(⅓>⅔)|n: 7.0n (g5)
(⅔>1)|n: 11.0n (b5)

So rot 1 shifts pattern one step to the right.

rot 1 $ arp "up" $ n "e'min"

yields

(0>⅓)|n: 7.0n (g5)
(⅓>⅔)|n: 11.0n (b5)
(⅔>1)|n: 4.0n (e5)

and without rot 1 it yields

(0>⅓)|n: 4.0n (e5)
(⅓>⅔)|n: 7.0n (g5)
(⅔>1)|n: 11.0n (b5)

So rot 1 shifts pattern one step to the left.

So both un-rotated patterns are equal, but rot 1 applied changes them.

6

This is a puzzle for sure, well beyond me though I'm afraid - I'll defer to people who are much better versed in the inner workings than I am :wink:

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