Apologies for the formatting

My apologies for the lack of paragraph breaks in my previous post. I am attempting to fix them now. If I can’t do so quickly, I’ll post more info about the bug here.

Update: OK, it appears to be rendering correctly now. I’m going to use this post to experiment and see if I can discover the source of the bug.

Update: It seem to be just one of those transient things; now everything works.

Oh well, let’s make this an open thread. Suggested conversation topic: beautiful numerical coincidences. For example, there are just about \pi \times 10^7 seconds in a year.


20 thoughts on “Apologies for the formatting

  1. It’s closer to 10^{7.5}.

    (Assuming you’re reading this rather than my email: can you confirm/deny my cyclic flats question from last week?)

  2. Music theory numerology: 3/2 is really close to 2^{7/12}, and 5/4 is close to 2^{1/3}.

  3. As to the paragraph breaks, it can depend which of WordPress’ text editors you’re using. If you use the rich text editor it tends to strip out line breaks.

  4. How about :e^{\pi\sqrt{163}} - 744 = 640320^3 (okay, not exactly, but really close!).

    And this is a little silly, but sometimes you can fool students who rely too heavily on their TI graphing calculators to believe that sin(11) = -1, or that cos(355) = -1.

  5. I’ve heard that some mischievous engineers would claim that e^\pi-\pi = 20 was an identity to trick their colleagues into searching for nonexistent bugs in their floating point routines.

  6. According to Cartier in (I think) the Notices, Grothendieck has come to believe that the true value of the speed of light is exactly 300,000 km/s, and that the published value has been altered by the devil.

    Not that this one is all that beautiful, just convenient or amusing. Actually Wikipedia leaves this blog in the dust on this topic.

  7. 1 mile/1 kilometer is close to the golden ratio, so if you have a Fibonacci number of miles, the next Fibonacci number is roughly that distance in kilometers. Naturally, it also works with sums.

    Also, if you’re n feet above the ocean, then the horizon is about 1.23\sqrt{n} miles away. For meters and kilometers, the factor is 3.57. This works well until your altitude is a significant fraction of the radius of the earth.

    Incidentally, a brief calculation with this approximation puts to rest a rather laughable Bay Area legend, which claims that the view from Mount Diablo covers more land area than that from any other place except Kilimanjaro.

  8. Scott – shouldn’t the elevation of the surrounding land matter? A big mountain surrounded by other big mountains will have a much smaller view area.

  9. Yes, but there is no shortage of relatively isolated peaks that are much taller. Even if we limit ourselves to California peaks, Shasta admits much more visible land area.

  10. Scott, were you assuming that the radius of the earth is about 8000 miles? Geometric mean of n feet and 4000 miles is about .87n^{1/2} miles.

  11. @Scott –

    grasping at straws here, but what about a isolated mountain, with a distant rising hills, that allow you to see over the usual horizon? I think the argument for Mt Diablo is meant to rely on the fact that you can see over a large piece of the Central Valley, but then also see most of the Sierra — whereas if the Sierra weren’t there, the land would be below the horizon. No idea if this actually works, though.

    On the other hand, I remember once skiing at Kirkwood with Vaughan, and him pointing out Mt Diablo from the top of a ski lift. You certainly can see it from some unexpected places.

  12. I guess I should retract my assertion that the legend is laughable, since all of the viable candidates fall within about one order of magnitude. I still think some distant hills won’t overcome a factor of three from Shasta’s altitude advantage, especially since they aren’t all that distant. In particular, the line of sight from Kirkwood to Diablo is “only” about 120 miles long (grain of salt – I measured this by marking a piece of paper held up to Google maps).

    However, this brings up a couple questions about measurement of visible land area. For example, should you count only the patches of land that are visible, or take some kind of geodesic convex hull of the visible points? Also, should you count the actual surface area (here there be dragons), or the subtended spherical angle, viewed from the center of the earth?

  13. Scott C-

    Given the mountain formations around Shasta (and having been on top of both) I have trouble believing that you can see more surface area from Shasta than from Diablo. Diablo is about 38 hundred feet above sea level and Shasta 142 so your estimate would predict the Shasta view to be a bit less than twice as large. However if you take into account the reasonably sized mountain ranges near Shasta and the general topography of the two regions I would bet that your estimate falls apart to the point where Diablo wins out. But perhaps I’m wrong… I’d be interested to know if you have any proof.

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  14. Carl,

    I remember the view from Shasta being a bit anticlimactic, but I think it was because the most notable nearby feature was underfoot, and everything else was really far away. The view to the north and west is somewhat obscured by a mountain range, but it is relatively unobstructed to the east and south, and the stuff near the horizon contributes a large fraction of the viewable area.

    Also, I get a naive area factor of 3.67 rather than 2-epsilon, because area is proportional to the square of radius (in the zero curvature limit).

    I’m not sure what sort of proof you’re looking for, but the US Geological Survey provides a National Elevation Dataset with 1 arcsecond resolution (60GB download), and one could in theory compute approximate lines of sight from that. It might make a pretty cool Google map thing – I think the Web 2.0 crowd uses the word “mashup” to descibe what I have in mind…

  15. I actually started doing exactly this yesterday, before remembering I had better things to do. You can get lower resolution elevation models as well, which are probably good enough.

    I’d gotten as far as loading the data into Mathematica, and writing the function that returned an elevation when fed a latitude and longitude, but got bored before working out how to calculate lines of sight. If anyone wants it, I can give you what I wrote.

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