Semi-Daily Journal

You really need to internalize the cosine theorem of spherical trigonometry. Or buy an astrolabe.

What is the cosine theorem of spherical trigonometry? I remember trigonometry proper, but all I remember of spherical trigonometry is that Lt. Bush is inept at it in C.S. Forester's novels.

You really need to just pick up your blanket and move into the shade.

That's considered impolite when you are one of the people up on stage *presiding* over the graduation...

http://www.irbs.com/lists/navigation/0210/0168.html

Brad, you should be in Michigan. It was in the 50's all day, and it only sprinkled just a little. The main graduation in April was a beautiful clear day, high in the 60's. I attended law school graduation at the beginning of May (as a guest), and I was perfectly comfortable sitting in a summer weight wool suit (in the sun - my friend graduating, sitting in the shade, wore a sweater under her robes, and was comfortable).

So if you want comfort, come to Michigan. Talk to Hal - he was here. The weather here is *so good* that the University of Michigan closed for weather only once in the 20th century :)

Ever thought you were too smart for your own good?

Spherical trig is the business of solving triangles inscribed on a sphere. Both the "sides" (which are arcs of great circles) and the angles of such a tringle are measured in angular units.

Here is the triangle of interest to you: its three vertices are the pole, your zenith, and the sun. All this is done on a notional sphere concentric with the earth and "up there in the sky". Now the angle at the pole between the sides "pole-you" and "pole-sun" is the hour angle - basically the time difference between where you are and wherever it is the sun is overhead. And those two sides are known too; pole-you is your co-latitude and pole-sun is the sun's declination. The number you want is the length of side you-sun, which is nothing less than the altitude of the sun in your local system. Since you have a side-angle-side triangle there should be a result to calculate the missing side, and there is; the (spherical) cosine theorem.

cos(a) = cos(b)cos(c) + sin(b)sin(c)cos(A)

a is the side you want, a and b are the known sides, and A is the known angle opposite a.

Unlike Lt. Bush, Hornblower was an expert at spherical trig, and resorted to it to calm his mind, as in that lost tale "Chancellor Hornblower and the Liquidity Trap"

Well, there's Apollo's time, and then there's Mussolini's clock (the one that keeps the trains on time). To make an accurate calculation, you would have needed local solar time.

The simplest method is to count sunwidths between the sun's current (apparent) position and the cutoff, and measure the time for each sunwidth to be moved. Then start by linearly extrapolating the time remaining until cutoff, but constantly refine the extrapolation in a predictor-corrector way. Watch the skies.

Or, if you don't want to look at the sun but you can look at the shadows, try the same trick with head shadow widths (a proxy for the sun) and the cutoff's shadow. While it introduces an extra level of non-linearity, and more importantly a possible source of noise, the dynamic corrections will still cater for that eventually unless the noise swamps things.

Two points:-

- The allegory can be transferred to economic questions, in this sense (or did you already spot tha?). We do NOT always have to get full econometrically based modelling but only something with empirical corrections from the observables (we can often get away with hidden variables even if we hid them ourselves).

- People whose lives depended on doing this sort of thing in real time, like long distance ocean fliers, used to prepare ready reckoner tables in advance and design courses that would themselves bring out information. For instance they could aim off so as to be sure of reaching a moving line through their destination to a known side of that destination, then proceed along it until dead reckoning told them to start doing a square search or they found a beacon signal. Effectively, they were designing out noise the way a well designed experiment does, then revealing as much as practical of what was hidden.

Faster and probably not as accurate 'old sailors' method.

Hold your arm towards the horizon, with your wrist bent, count the number of hand lenghts between the horizon and the sun, each hand length is approximately one hour of remaining sunshine.

Works fairly well for me!