6 ms·

Yeah, it's hard to describe it as unforeseeable when it's happened to you before and is readily modelable.

• johnchristopher 13y ago
  The guardian states this phenomenon is difficult to model: http://www.theguardian.com/artanddesign/shortcuts/2013/sep/03/walkie-talkie-death-ray-buildings-heat http://www.theguardian.com/artanddesign/shortcuts/2013/sep/0...
  • stephencanon 13y ago
    Whatever the guardian says, it's a very basic optics problem, and not at all difficult to model. It's mostly a matter of whether or not anyone bothers to do it.
    • mikeash 13y ago
      In terms of difficulty, it's one step harder than modeling which direction something will fall if someone drops it off your building.
      • lucisferre 13y ago
        Well except you have to model the position of the sun for every hour and day of the year, but sure other than that just like it.
        • wmeredith 13y ago
          Or you could use a simple rule of thumb. Something like, "Let's not build skyscrapers that are giant parabolic reflectors aimed at things of value (people, cars, other buildings, etc.)" This seems like a really stupid mistake to make. TWICE.
          • jcampbell1 13y ago
            ...or you could put the concave side on the North. That also solves the problem.
            • bodhi 13y ago
              Then you build one in the southern hemisphere. Oops!
              • dsuth 13y ago
                In the southern hemisphere everything is upside down too, so the rays would reflect off into the atmosphere, I expect.
        • mikeash 13y ago
          I bet a modern calculating machine could run through all of the permutations in no more than a couple of weeks!
          • deleted 13y ago
            [deleted]
            • jlgreco 13y ago
              No way, it would be near instantaneous. Position of the sun in the sky is really trivial stuff. Hell, don't even bother, just use this: http://aa.usno.navy.mil/data/docs/AltAz.php http://aa.usno.navy.mil/data/docs/AltAz.php
              • mikeash 13y ago
                I gotta know, what did that comment say?
                • jlgreco 13y ago
                  Something about it taking a few hours. Wikipedia apparently has a pretty nice page on calculating the position of the Sun if anybody is interested: http://en.wikipedia.org/wiki/Position_of_the_Sun http://en.wikipedia.org/wiki/Position_of_the_Sun It's basically a lot of trig and a bunch of constants; pretty neat. Interestingly the information given on that page is only valid for a shorter timeframe than the information given on the USNO page.
        • 6d0debc071 13y ago
          That's not particularly computationally intensive.
        • ChuckMcM 13y ago
          And the position of the sun relative to the building is also a simple formula based on time of year. Given that this is the country that gave us freakin' Stone Henge you have to expect that someone remembered that there is a formula to predict exactly where the sun will be relative to a spot in England every day of the year. The clever ones actually used that to tell them what day it was since they couldn't be bothered to invent the wrist watch.
          • gonzo 13y ago
            upvoted for Stonehenge reference.
          • swombat 13y ago
            The position of the sun is very easy to predict, 90% of the time, in England: it's behind the clouds!
            • easytiger 13y ago
              * this summer being an exception!
              • swombat 13y ago
                Yes - apparently it'll be 2020 before we see a summer like this again, though.
        • frobozz 13y ago
          In which case, it is much simpler. To work out where an object will fall, you have to model the path of various objects of all shapes, masses and densities with wind of all strengths and directions (including shifting and gusting winds), and considering the original orientation of the object, and any spin placed on it at point of release (in three axes). Just working out the angles at which light from a distant ball of fire will hit a surface should be a piece of cake. You could probably do it pretty easily for every point (assuming a certain size of point) at that distance at which it isn't completely hidden, just in case the planet suddenly changes its orientation.
        • anigbrowl 13y ago
          So what? People have been doing that for centuries using almanacs. And with computers that sort of thing becomes stupidly simple. IIRC that functionality is even built in to Google Sketchup.
        • tripzilch 13y ago
          > Well except you have to model the position of the sun for every hour and day of the year That sounds exactly like some things xkcd has done in the "what if?" series. He's done all sorts of variations on similar calculations. And while xkcd is smart, he's not superhumanly so (his fame is due to the unique combination of creative thinking, humour, geekery and smarts). In other words, this is possible and a good engineer should be able to calculate this. Especially when you know that any concave reflective surface is going to focus the light somewhere. In addition you know that due to the movements of the sun during the day and the year, this focal point will in 3D to different positions. So yes, odds are that it's going to hit some spot, somewhere, where it's going to be able to do damage, at some point during the year. Then you get to make a simple risk calculation about expected strength of sunlight that time of the year. And then you will remember to realize that over the course of the lifetime of this building, climate always has a non-zero probability of powerful outliers[0], and the conclusion is: Simply don't build gigantic concave mirrors in residential areas. [0] I'm not 100% sure if this is the case for strength of sunlight, since it's mainly dependent on what angle the sunlight arrives, which is not really up to chance. Still, even in the winter when the sun is quite low, as long as the sky is clear, I can easily light an incense stick with an A4-size fresnel lens. And I guess a whole building-side has a "slightly" bigger margin of how narrow the point of focus needs to be.
  • jerf 13y ago
    It may be difficult to model precisely what will happen because of the little fiddly sub-millimeter differences in window positions mattering. It is not difficult to understanding that a generally concave surface facing the sun is very likely to have some sort of significant problem. To the extent that the exact result is hard to model, simple engineering prudence would suggest the "'Doctor, it hurts when I do this.' 'Don't do that then.'" response.
    • marvin 13y ago
      Yeah, this really seems...kind of obvious. A concave reflective surface will focus any light source, and in particular, the sun.
  • berkut 13y ago
    It's not - given an accurate geometry model and reflective window material, it'd be really easy to use a raytracer to test this: * Renderers like Maxwell handle caustics nicely, and you could get away with rough photon-mapping (instead of time-consuming but very accurate MLT) * Most renderers have very accurate physical sky light models which accurately model the intensity and position of the sun in the sky based on time of day and long/lat coords. It'd be easy to python script a run of renders for each minute in the day (10 seconds for a render) over the next 5 years and then run an analysis of the rendered image for intensity on the ground. Just set the ground material to not be lit by any diffuse GI rays or direct lighting, and it would literally just render out the reflected sun on the ground (or off other buildings if you wanted to do it REALLY accurately). It'd be time consuming and CPU intensive to do the rendering for all that: 2,630,880 renders for each minutes for 5 years, but you could probably work out an envelope you could skip (definitely the night time) after doing the first year. 7308 hours render time for the full 24-hour period for 5 years. But do-able, and very parallel over multiple machines.
    • sampo 13y ago
      You know, indirect lighting surely doesn't melt plastic, so you absolutely don't need Metropolis Light Transport here. And also you don't need to use photon maps. The computations to map the intensity of specular reflections from the building's windows would the equivalent to creating a photon map on a scene, not using the photon map to render a complete scene.
      • lutusp 13y ago
        > You know, indirect lighting surely doesn't melt plastic ... That depends on the light's intensity. If with the term "indirect" you include specular and diffuse reflection, then yes, in the right circumstances, the energy level can be spectacularly high. But computer simulations need to be done very carefully, because the issue isn't visible light reflection, it's infrared reflection. For the building in question, the windows are glazed with a coating meant specifically to pass visible light to the interior but reflect infrared radiation (both to preserve building heat in the winter and reject heat in the summer). What that means is that, as bad as the reflected visible glare seems to those passing by, the infrared energy levels are much higher, and easily explain the damage.
• mseebach 13y ago
  Everything is obvious once you know the answer.
  • SiVal 13y ago
    This is true, and we often forget it. But since they knew the answer after the Las Vegas building, the risk that this might happen should have been obvious.