For the past few months I’ve been constructing models of rock clouds, in order to try to find a companion rock of DA14 that would land on Chelyabinsk at 3:20 UT on 15 Feb 2013.
I’ve published snapshots of my simulation model from time to time, but it’s hard to convey what’s happening that way.
Today I wondered if I might be able to create a YouTube video. I found something called Debut which could record video from screen, and got it working, and was soon watching a short AVI movie on my computer. And then I tried uploading it to YouTube. This was very easy.
So now I’ve got a 3 minute video which shows the Earth passing through a dense cubical rock cloud, made up of 343 rocks, on 15 Feb 2012. The rocks start coming in from the bottom right hand corner. Some of the rocks are light grey in colour, to indicate that they’re below the Earth, and are a bit hard to see. The red ones are above the Earth in the z axis we’re looking down. After the rock cloud has passed, I zoom in to see the 150 red impact sites, first looking down the z axis, and then down the x axis, and then down the y-axis. It’s a little bit clunky – the earth was rotating far more smoothly than in the video clip.
The reason that I’m looking at this dense cloud of rocks is because I want to find one that will be thrown by this close approach into an orbit which will return to the Earth on 15 Feb 2013, and will land on Chelyabinsk on 3:20 that day. Needle in a haystack.
Anyway, I think the video speaks for itself. It’s pretty cool that it’s also recorded the mouse position, and mouse clicks.
All it really needed was some spacey music to go with it. A bit of Pink Floyd might have been an idea (but I don’t know if you’re permitted to do that).
P.S. BrianB will be pleased to hear that I’m now using NetBeans for Java development.
P.P.S. Here’s another nice impact. The rock cloud started out as a perfect cube, but as it approached the Earth its near side was accelerating faster towards the Earth than the far side, and its two sides were accelerating inwards towards each other, so that the rocks that didn’t hit the Earth, but skimmed over its surface, were drawn out into a ribbon, and swapped sides.
Frank Davis 
Fascinating. And impressive that you should have the knowledge, skill and patience to create theses. Would it be possible to make a computer model of the surface of a liquid in a rectangular container moving randomly in and about all three planes, and to more accurately assess the correct measurement of the surface at rest?
And a curse upon my typos!
A swimming pool? In principle it shouldn’t be too difficult. But in practice I’ve never managed to construct a simulation model of water. And I’ve had several tries at it.
I was thinking more of a tanker at sea… There could be potential for more accurately gauging the tanks without the guesswork involved at present (get the highest (possibly) and lowest (possibly) readings, and half the sum) However, if you have tried, without success, then I will accept its difficulties will be way beyond me.
I haven’t tried very hard to model liquids. It may be much easier than I think. People build models of atmospheres, and I don’t know how to do that either.
I played ”HOOKEY” from the fight for 3 days! Went to Cherokee Super Bingo and lost my arse1
The first thing that needs sorting out, however, are the words: meteor, meteorite, meteoroid and asteroid. Meteoroids and asteroids are objects in space. Meteoroids can be bits of asteroids or bits of comets. When they are burning through Earth’s atmosphere they are, for a few seconds, called meteors. If anything survives that fiery descent, the rocks found on the ground are called meteorites.
Fourthly, 2012DA14 was not going “too slow” for a related fragment to arrive at 17km/second: its radiant, relative velocity to the Earth before being accelerated was 12600mph. That is 5.6 km/ sec. If you add to that the freefall velocity of 11.2 km/sec (the corollary of escape velocity) you get 16.8 km/sec. Add to that the eastward rotation of the earth at 55degrees north at an Azimuth of 9 degrees south of east (0.2 km/sec) you arrive at precisely 17km/sec. This is the same calculation that Zuluaga and Ferrin (and now, NASA) must have done in reverse for their version of the reconstruction of the trajectory: I calculated the radial speed of their hypothesised orbits at the Earth’s position (r value/ radius from sun=1AU) on the day of impact (but without the Earth’s gravitational influence added) and ended up with 34.8 and 35.2 km/sec for the 2 posited orbits. That amounts to 5 and 5.4 km/sec relative to the Earth, respectively. Adding the freefall velocity and the eastward rotation you get 16.4 and 16.8km/sec. The difference between these posited orbits and the posited 2012DA14 fragment is that they invoke the head-on trajectory solution with little or no curvature as they are pulled into the gravity well. If it’s a bulls-eye hit the curvature is zero. The Zuluaga and Ferrin video shows the meteor coming in from about 3 degrees above the solar plane. The NASA video now shows the same.
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