Pu’Er

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Now that winter has arrived here, it’s time for a warm cup of tea and a few in-house pictures of the experience, as threatened.

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Late last year I received two small Pu’er tea cakes as a gift. My conception of tea has changed over the years from tea bags over branded tins with generic names to loose tea from single tea gardens, and my expectation likewise from powder to beautifully rolled leaves, to be consumed as fresh as possible.

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So how would one dare to press tea leaves into bricks or cakes, and let them ripen? It helped a little that the cakes were nicely wrapped. So I took a quick course. Pu’er tea undergoes a special kind of fermentation that  can take many years. People buy raw Pu’er and let it mature like good wine. Alternatively, one can buy cooked Pu’er tea that has undergone a special procedure to accelerate aging. Prices vary considerably. Preparation is a story by itself. 

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One carefully breaks the cake into chunks. I placed about 10g of Ripe Pu’er into a steal tea infuser, and brew cup sized portions, using boiling waters. The first infusion steeps only 5 seconds to clean and loosen the leaves. Then I let the tea steep  for 30 seconds, increasing the time by 15 seconds for each subsequent infusion. This seems to do the job. 

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What you get is strong brew unlike anything else. It is very far away from the elegance of a Darjeeling or the floral delicacies of an Oolong. You get strong earthy notes, some fish, some mushroom, which I found, to my surprise, not unpleasant. Later infusions become more mellow and reveal complexity. The best: Near and far, I seem to be the only one around who likes it, so I can have it all for myself…

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Rationality 2

Last time I was asking about cyclic polygons with rational vertices and rational edge lengths, like the one below.

10 gon

It is easy to find all points on the unit circle with rational coordinates as (cs(t), sn(t)) with rational t, using

Formulas

Moreover, any pair of such points can be rotated into each other using a rational rotation R(t). Here t is again a rational parameter and not the angle of rotation. So we are interested in finding superrational rotations R(t) that are rational and displace points on the unit circle by a rational distance.

Surprisingly, the solution is quite simple: A rotation is superrational if and only if it is the square of a rational rotation.

Let’s first assume that u is rational so that R(u) is rational. The square R(u)² is then also rational. We can determine how it displaces points on the unit circle as follows: Suppose R(u) maps (1,0) into the point (c,s), which is rational. As R(Then  R(u)² maps (c,-s) into (c,s), and the distance between these two points is obviously 2s, R(u)² is indeed superrational.

Superrational

Now assume that R(v) is superrational, and write it as R(u)². We need to show that u is rational. Again let (c,s) be the image of (1,0) under R(u). As R(v) is superrational, we see that at least 2s and hence s must be rational, as above. Now we compute, using the matrix form of R(u), the image of (1,0) under R(u)² as (c²-s², 2cs). As R(v) is rational, 2cs must be rational. As we already know that s is rational, it follows that also c is rational. Thus R(u) is a rational rotation, and we are done.

This shows that superrational rotations form a group, namely the group of squares of all rational rotations. And this implies that a superrational cyclic polygon has necessarily all its diagonals rational…

What’s next? Probably some dreary landscapes from wintry Indiana or heart warming pictures of tea leaves. But of course, there are more questions: Can we also have superrational spherical polyhedra? I don’t know yet…

Rationality

Let’s do something simple today, something rational. We all know what a circle is, and most should know that there are points on the unit circle with rational coordinates, like (3/5, 4/5). This is because of the birational map called stereographic projection, known since antiquity:

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Take a point t on the real axis, connect it to the north pole (0,1) on the unit circle with a straight line, and find the second intersection of that line with the circle. This is a rational expression in t. So when t is rational, you get a point on the circle with rational coordinates. This is, of course, also a quick way to get (all) Pythagorean triples. But today we are going elsewhere. Now that we have many points with rational coordinates on a circle, we can make rational polygons, like the one below.

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This 9-gon is not only rational, but super-rational in the sense that all its edge lengths are rational numbers. Try it out. Even better: All the diagonals are rational as well. Is this a miracle? Are there others? No and yes, of course. Let’s get started:

Formulas

Using rational versions of the sine and cosine functions, we can write down rational rotation matrices. They will (for rational t) rotate any point with rational coordinates on a unit circle to another point with rational coordinates. What we are interested in are superrational rotations: Those that rotate a point to any other point. The example above suggests that there are many of those.

I will give the answer next time. For the moment, only a hint: The superrational rotations form a subgroup of the group of rotations. Which is it?

2008 Recap

Yes, that’s right. Let’s begin the year with a recap of not last year, but of 2008, the year 10 years ago.

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This year brought photographically two significant changes into my life: My move to full frame digital (and the ability to use a handful of SLR lenses I still had from film days), and the adjustment to the Indiana landscape.

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It is not that the Indiana landscape is featureless. It is more a assembly of countless insignificant features that tire the eyes, with occasional exceptions.

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Some are less obvious then others, but the only chance finding them is to look.

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Sometimes I am being asked why I bother carrying a heavy camera when there is nothing worth to photograph.

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Visiting some of the state parks has helped to open the eyes, like McCormicks Creek, Turkey Run, Shades, or Falls of the Ohio. This had been a good year.

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