Musical Idea 38: Maximally Staggered Entry

06.28.2014 § Leave a comment

Some time ago, my lovely girlfriend Lauren and I were thinking about getting a hexagonal mirror. But how to mount it, we wondered: rotated such that sides were parallel to the floor, with points pointing at the walls; or the other way around, with points pointing up and down, and sides parallel to the edges of the walls? —Or how about neither? We could hang it rotated halfway between these: a 30 degree offset from either parallelism. Halfway seemed like the most natural next thing to experiment with, but now this had opened a whole ‘nother can of worms! What about 15, or 20 degree tilts? Or 12? Which would be the most beautiful?

We never did end up getting that mirror. However, some time after that I found myself shopping for the best malachite box for Lauren as a gift. They were generally available only in rectangular and hexagonal versions. Frustrated with this mundanity, I began to imagine pentagonal boxes, and heptagonal ones… and then pentagonal ones with heptagonal lids. As soon as I began to wonder what the most beautiful orientation of the sides of the lid to the sides of the base would be, I was reminded of our hexagonal mirror rotation dilemma. I realized that this was the same problem, only with the ante upped from 4:6 to 5:7! (If I’d’ve stuck with rectangles and hexagons here, I’d’ve had exactly the same problem — the four sides to my wall transferred to the four sides of the box, and the six sides to my mirror transferred to the six sides of the lid.) So here I was, trying to figure out what the most beautiful relative rotation of these two regular polygons should be: probably as asymmetrical as possible — with no shared vertices between the two, and no parallel sides between them, either — but from there, how to define perfection? What factors would make an offset rotation most compelling?

There’s a trick I heard of used in Ewe music called “staggered entry,” whereby the pulse streams which constitute their polyrhythms do not begin sameltimeously. This way, beats in the two streams never coincide. This is a fantastic feature, to me. That’s because it corrects for the most undesirable features that standard polyrhythms exhibit:

Standard polyrhythms exhibit mirror symmetry. Their second halves are merely their first halves, reversed. Whenever you choose to align two different regular pulses with each other such that a beat in one coincides with a beat in the other, as a consequence, on the other side of the rhythm, one interval will be centered either on a beat or another interval (depending on whether your streams pulse an even or odd number of beats per repetition), and either side of this centering and coincision will be chiral.

However, if you stagger the streams’ entries, this mirror is shattered. The symmetry is broken, and the number of distinct durations in the final output — that is, with both streams considered together — is doubled. Without the symmetry, it becomes significantly more difficult to perceive the polyrhythmic relationship, which is beautiful. It’s still just as much simply there, and I appreciate this increased subtlety.

Charles Ives’s Universe Symphony certainly has its time and space. Indeed, when you stack dozens of regular pulse streams atop each other, all perfectly centered, and make the contour of the Farey sequence your aesthetic focus, it’s pretty awesome. But in general, if you’re going to set quintuplets against septuplets, in my book, you might as well stagger them.

Which brings us back to the mirror and box. These are all the same problem. Just imagine one repetition of a polyrhythm as the circumference of a circle, and the regular pulse streams as regular polygons inscribed in it. And realize that this works for pitch as well, if you treat the circumference as one loop around your pitch class equivalence, and the vertices as defining pitches in a scale or tuning system. 

 

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Now we must start to find a generalized solution for any given combination of regular polygons.

The first thing to notice is that we don’t have to look through the entire 360 degrees of a circle to find our most beautiful staggerings. Only a tiny sliver of this range is relevant. If you begin with the two polygons coinciding at a point and rotate them apart, soon enough a different pair of points will come into coincision. It’ll be the same thing we started with, just with the entire thing rotated a bit — but since all we care about is the relative rotations of these two polygons, we may as well be back where we started. The only rotations which matter are within this range.

In fact, the only rotations which matter are within half of that range, because the two halves of it are symmetrical: anything you can do on one side of it can be achieved on the other side (with a flipping or rotating of the entire thing thrown in).

Rotating by exactly half of this range leads to comparable amounts of undesirable symmetry. If you have one polygon with an odd number of sides and the other with an even number of sides, the result will be that one side of the even one will have a point of the odd one centered in it, and the opposite side of the even one will have a side of the odd one centered in it (as opposed to one point of the even one centered in a side of the odd one opposite the coincision). And if you have two polygons with odd numbers of sides, the result will be that one point of each will be centered in one side of the other (as opposed to one side of each centered on each other opposite the coincision). And if you have two polygons with even numbers of sides — well, you should revisit what you’re doing, because it’s pointless to do this for polygons whose sides aren’t reduced to lowest terms — you might as well factor out any 2’s or 3’s or whatever they share, or you’ll be wasting your time looking at redundancies around the circle.

But let’s pause a moment and look at this: what exactly is this range, within half of which lay all our interesting possibilities? It’s simply the positive difference in side lengths of the two polygons, divided by the positive difference in number of sides. Why? Well, it’s intuitive if you look at examples. Rotating a square out of coincision with a triangle, clearly the next moment it comes into coincision is the difference between the triangle’s side length and the square’s side length; nothing further needs to be adjusted, and that’s because their side lengths are only one apart. However, if you tried to rotate a pentagon out of coincision with a triangle by the difference in their sides, clearly that would also lead to a coincision, but one coincision would have already happened elsewhere on the circle halfway along that way. And so on for triangles with heptagons and other higher ‘gons.

The most interesting rotations will probably be the purest, most simple ones. So let’s look first at rotating by one-third of this range, by one-fourth, and by one-fifth. Once we get to fifths, we can also look at rotation by two-fifths, since that’s still less than one half, and thus interesting (that is, for example, we didn’t care about two-thirds since it’s on the other side of one-half and thus equivalent to one-third, and three-fourths is similarly the same as one-fourth [and two-fourths is, of course, just one-half again]). Then one-sixth. Then one-seventh, two-sevenths, and three-sevenths.

 

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As an illustrative example, let’s take a triangle and a heptagon, and stagger them by one-fifth. 

  • A triangle, or 3-gon, has 360/3 = 120° to a side and a heptagon, or 7-gon, has 360/7 = ~51.43°.
  • The positive difference in side length is ~68.57°.
  • The positive difference in number of sides is 7 – 3 = 4.
  • ~68.57° / 4 = ~17.14°.
  • One-fifth of ~17.14° is ~3.43°.

 Let’s anchor the triangle with one point pointing up, and rotate the heptagon by this ~3.43°:

1 In the first column we see the degrees of the circle at which either polygon intersects it.

In the second column are the arc lengths cut by these vertices. You can see clearly enough that there are a bunch equal to ~51.43°, which are from the sides of the heptagon unbroken by the triangle. The other arc lengths are the three heptagon sides that get broken by the three points of the triangle, into differently sized pieces each time, but these sets of two always sum to ~51.43° (~13.71° + ~37.71°; ~30.86° + ~20.57°; 48° + ~3.43°). Similarly, you will find consecutive arc lengths that sum to 120°; every side of the triangle is broken by heptagon points, into either three or four parts.

But we are interested less in absolutes and more in relatives here — degrees of a circle won’t help us directly when transferring these findings to pitch and rhythm; so let’s look at these arcs in terms of each other. If you reduce them into common terms, their simplest proportions to each other are in the third column. When looking at proportion rather than absolute degrees, one side of the triangle is represented by 35, and one side of the heptagon is represented by 15. Naturally they are still in the ratio 7:3 with each other. You can still see the summation groupings by 35 (1+15+15+4, 11+15+9, 6+15+14) and 15 (15, 15, 4+11, 15, 9+6, 15, 14+1).

The ordering of these proportions is plain enough to explain thusly — you see the same basic effect of intervals getting split at a gradually progressing point within themselves from interval to interval that you see with traditional polyrhythms — it’s just that it’s asymmetrical, except for the stagger which causes that asymmetry, which is symmetrical. But what’s more interesting to look at it are the proportions themselves. To get a better look at them, let’s sort them: the fourth column.

A natural next thing to note at this point is their differences when sorted this way, which is our fifth column, Delta. At the beginning, it looks like there’s a pattern to these sorted deltas, an alternating 2 and 3 — but then it gets interrupted by the ceiling, the max, which is simply the length of the side of the polygon with shorter sides.

However, we can “fix” this by eliminating these “maxes”: what we have to do is return to the unsorted list of intervals and begin merging the non-canonical maxes with their neighbors. Begin with the smallest resultant interval and merge it with its neighboring max (big ones tend to live next to little ones). In this case, that will be the 1 & and its 15. Together, they create an interval which is proportionally 16. You will notice that 16 falls into the pattern of deltas, since it is 2 above 14. If you merge the next 15 with the 4, you get 19 — again falling into place in this pattern, being 3 above 16. Keep going until you’ve eliminated all the maxes, and this makes our sixth column, Fixed. You can still see the summation groupings by 35 here, but the groupings by 15 are no longer apparent. That is because all of the points we deleted in order to merge intervals and conceal the ceiling we were hitting with the heptagon’s shorter sides came from the heptagon. We deleted exactly enough points from the heptagon until it contributed as many points to the resultant rhythm as the polygon with fewer points did — that is, we deleted 4 to bring the 7 down to 3 like the triangle. In the end, the heptagon leaves its mark only ever-so-slightly; this part of the rhythm or tuning is defined by but will likely not be felt any longer as “7-ish”. The ratios of durations between the points it still contributes to the rhythm are 2, 2, and 3, which do sum to 7, but since these remaining points always get distributed as evenly as possible this way — that is, they wouldn’t have been deleted to create spaces of 1, 3, and 4 — it makes it tough to perceive their difference, especially as your n-gons increase in number of sides.

Sorting them, you’ll end up with 11, 14, 16, 19, 21, 24. That’s our seventh column, FixSort.

Looking at the deltas to these FixSort intervals, we see the pattern is made consistent: 3, 2, 3, 2, 3. That’s our eighth column, FxSrtDlt.

 

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Something that — perhaps only because I’m not a mathematician — is striking to me is that when you start sketching and testing these out, it turns out that no matter which two polygons you choose, for each rotation (in terms of the positive difference in their side length over the positive difference in their number of sides), the sorted proportions of the resultant “fixed” intervals they produce fall on the same sequence.

I know that’s a ton to attempt unpack all at once, so I’ll retrace from scratch:

Basically what I’m saying is, I could pick any two polygons — like a triangle and square, or pentagon and enneagon — instead, and as long as I stagger them by one-fifth (in the same manner as we did the triangle and heptagon), their FixSort column will come out the same: 11, 14, 16, 19, 21, 24. To be clear, it’s not just their deltas (that is, this won’t be transposed up or down to like 12, 15, 17, 20, 22, 25) — it’s those absolute numbers. However, it is possible to take different subsets from that continuum. For example, if there had been less or even no merges, it could start as early as 4, 7, 9, 11… and it can go up higher and higher if you choose really high number of sided polygons.

Let’s show you, for a pentagon and hendecagon, now, also rotated by one-fifth:
2

As you can see, even though we chose a completely different set of two shapes, because we staggered them by the same rotation (one-fifth), the proportions between the resulting arcs fall on the same sequence. The only difference is that with the pentagon and hendecagon we started at 16, its sixth number, because the first six had to be deleted when merging with maxes (including the trivial min), while with the triangle and heptagon we started with 11, its fourth number, because only the first the four had to be deleted. Basically if I haven’t already said this , you have to delete down until the same number of points are contributed from each n-gon, in other words, from the one with the smaller number of sides (It’ll make something like an mos out of the larger n-gon, maximally even). Also, with the latter, we go up to 39, its 15th number, whereas for the former we only made it up to its 9th — add the numbers of sidesa nd subtract 1 to get this number.

So what this comes down to is that for each pair of n-gons, all I have to do is give a diagram of variables a, b, c, d, e, f … representing the escalating sizes of sides, and then you fill them in with the sequence for the particular stagger.

Here are the ratio sequences: 3

What you might notice is that multiples of the denominator are always missing from the sequences. That is, for a rotation by 1/2, all multiples of 2 are missing — it’s just the list of odd numbers. For 1/3, it’s the list of all numbers which aren’t multiples of 3.

Comparing the sequence for 1/5 with the one for 2/5, you can see that they are the two possible different patterns for picking 2 numbers out of every five: for 1/5 it is the pattern which picks the numbers one away from multiples of 5 (e.g. 4 & 6, 9 & 11), and for 2/5 it is the pattern which picks the numbers two away from multiples of 5 (e.g. 3 & 7, 8 & 12). The pattern is the same for 1/7, 2/7, and 3/7: for 1/7 it is the numbers one away from multiples of seven, for 2/7 two away from them, and 3/7 three away.

 

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As we’ve seen, we can fix a set of arc proportions for any pair of staggered n-gons by removing its maxes, reducing the set to only contain a sub-sequence of the sequence for the chosen stagger. However, in any real system of pitches or durations derived from these arcs, we will care not only about intervals from one point to the next point, but also from each point to every other point. For instance, if we had a 7/4 rhythm going X—X—X—— we would understand that our rhythm contains not only the intervals 2 and 3, but also 4 (2+2) and 5 (2+3) (and not 1 or 6). In other words, now we need to take a look at the sums of consecutive arcs.

When polygons are staggered by proper fractions like we’ve been doing, consecutive arcs will sum to one of two things.

The first thing should be unsurprising: further up numbers in the current stagger’s sequence. This sequence is cyclically embedded.

The second thing is: multiples of the denominator of the stagger. Yes, these are numbers which are strictly forbidden, as discussed in the previous section, from being in the stagger’s sequence — the sequences are in fact defined by their members’ amount of difference from these multiples. Arcs sum to such multiples when you’ve chosen consecutive arcs that connect two vertices of a single polygon, however many sides of that polygon the total arc may encompass. For example, look back to our pentagon vs. hendecagon example. Suppose we want to know about the interval from ~197.67° to ~263.13°. If you sum its arc proportions 14 + 11 + 25 you get 50. 50 is a multiple of 5, all multiples of which are missing from the sequence for the 1/5 stagger. We get 50 exactly because these consecutive arcs connect vertices on the hendecagon spanning two of its sides, and each side of the hendecagon covers an arc of proportion 25. (You can always determine the proportion of a staggered n-gon’s side by multiplying the n of the other n-gon by the denominator of the stagger. For the pentagon against the hendecagon, then, one side’s arc would be 11 x 5 = 55.)

Because the latter of these two types of consecutive arc sums are non-canonical for the current stagger’s sequence, you might be tempted to try to get rid of them by deleting further intervals.

Once the initial maxes are eliminated, however, selecting which intervals to delete becomes a less obvious problem. One thing we’d like to preserve is the continuity of the proportions — that is, so far the proportions, while the smallest may be much smaller than the largest, other proportions bridge them continuously — so we should try eliminating the smallest ones first. However, there is no good generalizable strategy for doing  this. Starting with the smallest arc and working your way up, you can try to merge all adjacent arcs which do not connect vertices on the same n-gon (meaning that they were probably merged in the original max-eliminating phase and thus now represent three arcs from the original). Or you can try to do the opposite, creating another round of maxes and then eliminating those in a similar way as before. But regularly you will leave “stranded” arcs, which were not upgraded; generally this is because you’re now grouping arcs in sets of three, and while the number of arcs after fixing maxes is necessarily even (twice the number of sides of the n-gon with less sides) unless it is divisible by six it is not also divisible by three.

The most you can ever hope to do, anyway, is delete any redundancies in those sums. Suppose you had a pentagon staggered against a hexagon: deleting any two non-adjacent pentagon vertices would leave only two arcs, one of proportion of two of its sides and the other of proportion of three of its sides; while deleting any any two adjacent plus one other of the hexagon’s vertices would leave three arcs, of proportions of one, two, and three of its sides, respectively. Thus, these arcs will at least all be different from each other, as is the way of the arcs canonical to the stagger’s sequence. However, they will still always be higher multiples of the stagger’s denominator and thus outside your sequence. This may be undesirable, but they are a necessary part of this system and its style. They must be there for the sequence to be generated in the first place, so you should embrace them within your harmony or rhythm when you use them.

Here are some suggestions for doing so.

Switch between musical phrases where the n-gon sides are hammered through and phrases which  avoid them entirely. In superparticular staggers, pairs of consecutive arcs alternate between summing to a side of one n-gon and the other n-gon, so “whole note” scales on these will be almost like moment of symmetry scales: maximally even subsets of / deletions from equal tempered tunings.

You could play a scale which involves skipping to the next next pitch around the circle which is on the other n-gon. By choosing the next next one you avoid moving chromatically, in order anyway — you may eventually touch every note available after going around enough times.

Or you could embrace the presence of these multiples by choosing a stagger which accommodates them. For instance the 2/7ths stagger creates a pattern of +4, +3, and the multiples of 7 will split that 4 into two 2’s, thus maintaining the maximal evenness of the sequence.

Finally perhaps it makes more sense for the musical applications of these staggerings to be more macro than micro. For rhythms, that is, play these intervals not like sixteenths but maybe even as structures of numbers of bars of one thing or another — big enough that your audience can really hear the 9, 12, 16, 19, 23 -ness of the music and figure out how these numbers relate on that alternating-delta stagger sequence. If you overlay musical patternings which correspond to the constituent n-gons’ sides then you may actually be able to communicate the beauty of n-gon staggering through music.

Basically I have to admit that I don’t know if we can hear meta-intervals. What I mean is: I know that if you took one of these stagger sequences and just made them a sequence of pitches rather than intervals, or hashed marks at those addresses in an array of cells creating a rhythm rather than using them as intervals, in other words overlapping the intervals from the same starting point rather than laying them end-to-end, you’d definitely get relatively nice sounding things. Laying them end-to-end as we are doing, even in this very particular way, it is not clear if that particularly way will contribute to any niceness of the sound of a musical system, or if it will simply be heard as yet another musical system with irregular intervals. However, that is what this system is about: maximizing interval variety using a method which has a chance of producing something beautiful.

 

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In an earlier section we eliminated the need to draw a different diagram for each stagger of a particular pair of two n-gons, reducing that to a single diagram labeled with a, b, c, d, e, etc. which only needs to be substituted with the numbers from the sequence for your given stagger.

Now we will eliminate the need to even draw different diagrams for each pair of n-gons! As it turns out, there are also patterns for pairs of n-gons depending on how different their numbers of sides are. That is, a triangle and a square fall in the category of (n):(n+1) staggers, or superparticulars, along with the square and pentagon, and octagon and enneagon. They all share similar patterns of a, b, c, d, e … etc. placements around the circle. The same goes for the (n):(n+2) staggers within each other, and the (n):(n+3)’s. Observe the diagrams included here to see the patterns at work.

 

(n):(n+1).

  1. A single max is next to the smallest arc and gets merged in. Draw a line of symmetry across the circle dividing it in half, the same number of arcs on each side; the line should touch the counterclockwise side of the merged max.
  2. The next clockwise arc is the smallest, and its mirror image is the next smallest.
  3. Two further clockwise from the smallest is the third smallest, and its mirror image is the next smallest.
  4. Eventually you will cross the line of symmetry with the two-clockwise move; that’s okay, just keep going. Finally you will end with the arc counterclockwise from the merged max, which is the second largest, and whose mirror image is the merged max which is largest.

 

(n):(n+2).

  1. The final merged max is the largest arc. Across from it is the penultimate merged max, the second largest arc. Draw a diameter of symmetry across the circle which places these two arcs as centers of their respective halves.
  2. The arc one counterclockwise from the largest arc is the smallest, and its mirror image across the diameter of symmetry is the next smallest.
  3. The arc two counterclockwise from the second largest arc is the next smallest, and its mirror image is the next smallest.
  4. Continue this process of alternating between largest and second largest arc, and incrementing the number of arcs away you choose. Eventually you will choose the largest arc but have incremented enough away from it that you choose the next largest arc as the next smallest, and the mirror image will be the next smallest: the largest.

 

For (n):(n+3), I haven’t quite gotten it, but it appears that it involves drawing a diameter of symmetry and matching next smallest arcs across it as with :(n+1) and :(n+2), and selecting the next set by going around the circle a number of arcs from the previous chosen arc (for square vs. heptagon, 2; for pentagon vs. octagon, 6; for octagon vs. hendecagon, 10… but I cannot figure out why for the heptagon vs. decagon it begins as one thing and then changes to another: starts 4 then when it runs out of those that work, switches to 3). Also I can’t tell why it starts where it starts.

Ultimately what these patterns reveal is that the symmetry I sought to avoid can be further and further obscured, but never escaped. Whatever pattern two staggered n-gons in a ring produce in one half, one way or another get undone in reverse in the next half.
Regarding staggering more than two n-gons together: it’s way too much for me to explore here. I started staggering a pentagon against a random established triangle against square staggering, but it was like every degree or couple of degrees there was a new way that two arcs were the same or two vertices intersected. Since I could not find a triangle against square staggering which is particularly beautiful, I don’t even know that the I one I chose was the most appropriate one to be exploring, so I gave up for now.

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