Monthly Archives: January 2019

Lunar Eclipse of 1/20/2019

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It was very cold on lunar eclipse night here in the Chicago area, but also very clear, so this counts as good luck.

I set up my Orion Starmax 127 on its equatorial mount right in the driveway, but with no clocking – just the knobs. I used my Nikon D5100 camera body mounted for prime focus photography. The moon is just slightly too large for this ( especially the “supermoon” ! ) but I wanted a large image.

I adjusted the exposure time and ISO sensitivity as it approached totality, so that the sequence I’m showing here was at the “Hi 1” ISO setting with a shutter speed of 1/25 of a second.

Here is the first image of the sequence

The bright rim was just a very faint glimmer to the naked eye, and I’m not sure why it seemed to persist for so long.

The images are 2464 x 1632 pixels and are reduced to 493 x 397 in MS Paint.  I further edited them in MS Paint to position them consistently for the animation. I made the image size 540 x 420 and filled in the boundary ( after positioning ) using the “dropper” ( i.e. Color Picker ) to match the dark sky. This worked pretty well.

There was one frame where I clipped the bright rim, and this created a jarring discontinuity. So, I ( rather crudely ) filled it in with a facsimile. It’s easy to see as it goes by, but I think it makes the viewing easier.

The animation runs from 10:48 PM to 11:00 PM ( CST, ) according to the camera timestamps.

Animated gif of onset of totality

An interesting automorphic sudoku

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I came up with this in an Aha Moment, actually while lying in bed, as I was contemplating whether two 3×3 squares of a sudoku could be identical. The answer came quickly to me when I thought that if these squares were diagonally juxtaposed, they would be compatible if their elements were translated diagonally by one, along the same diagonal, of course, and these new ( identical ) squares were added to form a 2X2 sudoku.

Then, in a flash, I saw that this operation would allow THREE identical squares along a diagonal, with the diagonal translation performed twice, and with a third time producing the original square. That was it.

So the scheme is :

A B C
C A B
B C A

where A, B, and C are the 3×3 grids :

1 2 3   9 7 8   6 5 4
4 5 6   2 3 1   8 9 7
7 8 9   5 6 4   3 1 2

and the solution grid is:

1 2 3   9 7 8   6 5 4
4 5 6   2 3 1   8 9 7
7 8 9   5 6 4   3 1 2

6 5 4   1 2 3   9 7 8
8 9 7   4 5 6   2 3 1
3 1 2   7 8 9   5 6 4

9 7 8   6 5 4   1 2 3
2 3 1   8 9 7   4 5 6
5 6 4   3 1 2   7 8 9

So that’s the long and short of it, but it remains to specify the automorphism. Indeed, this whole question was perplexing to me, as I hadn’t been thinking in those terms.

Well, in the first place, we may specify the cyclic permutations of the “stacks” and “layers” ( i.e the columns and rows of the 3×3 squares ) .

Each of these, in our case, may be replicated, or reversed, by a permutation of the digits.

But also due to our particular configuration, these permutations may be replicated by permutations among the columns and rows within the stacks and layers.

So, at this point, I can only wonder how this particular case fits into the enumeration specified in the Wikipedia article.

So, it’s a mind bender, of this I feel sure.

We may also note that, as specified above, the downward rotation by one, among the layers, followed by, or “times”, the leftward rotation by one, among the stacks, “equals unity”, i.e. does not change the layout.

This is a consequence of the triplication of the three unique squares, and not usually to be found among other automorphisms.

 

… all presented for your consideration!