Maths/trig help needed

[BJG145]

I'm thinking about hexagonal/microtonal keyboard layouts, and trying to work out the angle of the keys on the Lumatone.

On an Axis 64, the hexagons are flat...



...but on a Lumatone, they're slightly rotated.



The angle seems to be about 16 degrees clockwise, but I'm trying to work out exactly what it is. There's a lot of microtonal theory and history behind it...Bosanquet, Wilson, The Terpstra, 31 EDO etc...but I can't find that angle documented.

The idea is that it brings keys that are an octave apart in one particular layout into vertical alignment.



This is the pattern before rotation...



...and after.



How do I figure out what angle will achieve that...?

[pk.roberts]

It's been a while since I've done this stuff, but here's my guess.
A regular hexagon is made up of 60 degree equilateral triangles, so the distance between two opposite 'points' is twice the length of a 'side'.
I'm assuming the line you've drawn is the hypotenuse, so if we imagine a right angled triangle, with the adjacent side running parallel to the base of the hexagons, and describe the length of a side of the hexagon to '1 unit' the the lengths of the long side of the right angle triangle is 10.5 units and the length of the opposite side is 3 x 0.886 (the apothem of a hexagon with side of 1 unit) = 2.658 units. So the tangent of the required angle is 2.658/10.5 = 0.2531 giving an angle of 14.2033 degrees.
I think.
Better minds than mine will probably be along to correct me soon; but good puzzle!
..... and I had to learn the word 'apothem' - all I have to do now is work out a way to slip it into conversation.

[BJG145]

Wow, clever stuff. I just tried reversing that rotation on someone's diagram of the Lumatone to see if it would bring it back to horizontal, and it looks very close. I expect the difference is their drawing rather than your maths.



I've spent about 40 minutes trying to figure it out with ChatGPT via descriptions like this.

Imagine a hexagon H1 with the bottom edge horizontal. The height of this hexagon, the "distance across the flats" from top to bottom, is D. The centre of the hexagon is positioned at X1/Y1.

Now imagine a chain of hexagons starting with H1.

H2, H3 - leading diagonally up to the right
H4 - leading diagonally DOWN to the right, from the lower right edge of H3
H5, H6, H7 - leading diagonally up to the right
H8 - leading diagonally down to the right

What are the coordinates of H8?
What is the angle between H1 and H8?

It was making heavy weather of it, so I was just about to try and do it by hand. I can see that each step along the chain adds or subtracts D/2 to the height, depending whether you're following it up or down, so the Y coordinate for H8 will be:

Y1 + D/2 + D/2 - D/2 + D/2 + D/2 + D/2 - D/2 = Y1 + 3D/2

I was just starting to work out what the horizontal shift of each move is, in terms of D.

[BJG145]

I'm assuming the line you've drawn is the hypotenuse

...not sure what you mean by that. The line I've attempted to draw connects the midpoint of two of the hexagons that are related in a particular way within the pattern.

[N i g e l]

If you want to check your maths emperically, you could quickly draw up the pattern in somthing like MS WORD and measure it.

Insert, Shapes, Basic_shapes, hexagon & then cut'n'paste blocks as desired

[merlyn]

The triangle I see goes right through F#, G, G# and ends at the midpoint of the base of C. If we call the length of a side of a hexagon r, this distance is 7r. The other side is 2.6r.

Then use inverse tan:

tan^-1 (2.6/7) = 20.38 degrees

[BJG145]

Thanks for looking at it. I probably haven’t explained it well, but the key is that trail of eight hexagons I’ve highlighted in the straight layout. I’m trying to find the angle of rotation that will bring the one on the left into vertical alignment with the one on the right.

That’s the Lumatone way, also the Starr Labs Microzone I think. Bosanquet seems to be responsible; they’re an octave apart in that system. It’s about 14 or 15 degrees I think, but I’m not 100% convinced we have the exact answer yet…or if 14.2 is correct, I don’t understand the method.

[merlyn]

OK, I see what I've done. The adjacent side in the triangle I'm using is 11r.

So it's tan^-1 (2.6/11) = 13.28 degrees

[BJG145]

I don’t understand the “triangle”…? It’s the trail that I’m trying to use, the one I’ve highlighted. 14.2 is very close.

[merlyn]

This is the triangle:


I'll get it right this time. The length of the long red side is 10.5r.

tan^-1 (2.6/10.5) = 13.9 degrees

[BJG145]

Oh, I see…thanks!

Brilliant, much clearer now.

[BJG145]

...just trying to understand the discrepancy with @pk.roberts' result. He writes that: "3 x 0.886 = 2.658", but you use 2.6...is that it...?

It's all great stuff, just trying to catch up and understand it.

[BJG145]

...wait, ChatGPT says the apothem of a hexagon with side 1 is 0.866, not 0.886.

3 * 0.866 = 2.598

[merlyn]

There's more maths in calculating the short side of the triangle:

I didn't know it's called an apothem, but I did know it's sqrt(3)/2. Put that into a calculator and you'll see where the discrepancy is. pkroberts had everything else right apart from that numerical error of sqrt(3)/2 being 0.886, when it is 0.866.

[BJG145]

13.9 degrees looks spot on.



With the help of my AI linesman, I declare @merlyn the winner.



@pk.roberts was very close, and laid the foundations for this calculation, but: "There's many a slip twixt cup and lip".

I'm planning to build one of these microtonal monsters later this year, and your geometry is gold dust. Thanks all.

[merlyn]

Thanks, but I can't accept as I can't count to ten and a half.

[Hugh Robjohns]

Hmmm... I made it 15.2 degrees.

Working from this:



If the side of a hexagon has length 2, the apothem is Sqrt 3.

I counted 3 apothems vertically on the short side.
3x sqrt 3 = 5.196.

Counting along the base, I counted 3.5 sides plus 7 apothems.
Remember a side is length 2, so 3.5x2 =7 plus (7x sqrt 3) = 19.124

Inv tan (5.196/19.124) = 15.2 degrees.

...but I've not done trig in a long time and counting has never been a strength (I'm a "one, two, many" kind of guy). So e&oe..

[merlyn]

... Counting along the base, I counted 3.5 sides plus 7 apothems.

If you look at the long line it is either on a side which is r or it goes from a vertex to the centre which is also r. So crossing a hexagon on a side is r. Crossing vertex to vertex is 2r, plus 1/2r for the last bit. Following the long red side of the triangle from left to right we have:

r + r +2r + r + 2r + r + 2r + 1/2r

which adds up to 10.5.

I imagine due to eyes and optical illusions it doesn't look like the diameter of a hexagon from vertex to vertex is twice the length of a side, but it is.

[nathanscribe]

Has nobody noticed that the hexagons on the first two pictured devices are at different rotations to one another? The flat is horizontal on one, vertical on the other. This changes the relationship between staggered rows.

[BJG145]

Counting along the base, I counted 3.5 sides plus 7 apothems.

Those weren't apothems. The distance "across the flats" (concertina parlance, two apothems) is different from the distance "point to point". These are sometimes referred to as "short diagonals" and "long diagonals".

https://www.omnicalculator.com/math/hexagon

[Hugh Robjohns]

If you look at the long line it is either on a side which is r or it goes from a vertex to the centre which is also r.

Darn it — I hadn't spotted it was going through vertices. Doh! You're quite right.

Base of 10.5 x 2 = 21

So it's: Inv tan (5.196/21) = 13.9 degrees.

Apologies for my numbskull moment and thanks for the clarity. Should've gone to Specsavers....

[BJG145]

Has nobody noticed that the hexagons on the first two pictured devices are at different rotations to one another? The flat is horizontal on one, vertical on the other. This changes the relationship between staggered rows.

That's what we're trying to figure out...have figured out. Different controllers have hexagons with a flat base (C-Thru Axis), pointy base (Intuitive Instruments Exquis), and somewhere in between (Lumatone, Microzone).

Check out this beast...same deal.

https://www.starrlabs.com/product/microzone-u990/

*edit*

Currently offline. Their apothems blew up.

[OneWorld]

If you look at the long line it is either on a side which is r or it goes from a vertex to the centre which is also r.

Darn it — I hadn't spotted it was going through vertices. Doh! You're quite right.

Base of 10.5 x 2 = 21

So it's: Inv tan (5.196/21) = 13.9 degrees.

Apologies for my numbskull moment and thanks for the clarity. Should've gone to Specsavers....

Sounds like the numbskull error I would make. But, as the hedgehog said to the loo-brush "We all make mistakes sometimes" to some degree!

And on Specsavers, they told me they had a 2 for the price of one offer, I asked "Is that offer for those that have double-vision?"

[Moroccomoose]

I drew it in CAD and measured!

For horizontal flats the angle is 13.89deg
For vertical flats, the angle is 16.10deg

Stu.

[BJG145]

I drew it in CAD and measured!

For horizontal flats the angle is 13.89deg
For vertical flats, the angle is 16.10deg

Stu.

Great, thanks for confirming! I think it's not a coincidence that 13.9 and 16.1 add up to 30...