One track per 19,000 focus movies
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One track per 19,000 focus movies
What is the probability that there is a track in any given movie? Or, put the other way, how many movies need to be searched to find one track, on average?
Taking the prelaunch estimate of 45 tracks in the whole array as a starting point, and that the whole array has an area of 1000 cm^2, there are an average of 45 tracks/10^3 cm^2. A focus movie covers an area of about 300 x 400 micrometers (if you believe the provided scale) which is 12,000 micrometer^2 = 1.2 x 10^5 micrometer^2.
One cm = 10^4 micrometer, so one cm^2 = 10^8 micrometer^2. Thus the array area in micrometer^2 is 1000 x 10^8 micrometer^2 = 10^11 micrometer^2
If there are 45 tracks on the whole array, that means there are 45 tracks per 10^11 micrometer^2 (Still with me? Good...)
That is, the ratio of tracks to area is 45 to 10^11 micrometer^2. This ratio can also be expressed as tracks per focus movie area. We set the ratios equal and solve for tracks:
tracks/area = 45 tracks/10^11 micrometer^2 = X tracks/focus movie area
Substitute the focus movie area 1.2 x 10^5 micrometer^2:
45 tracks/10^11 micrometer^2 = X tracks/1.2 x 10^5 micrometer^2
X = (45 x 1.2 x 10^5)/10^11 = (54 x 10^5)/10^11 = 54 x 10^6 = 5.4 x 10^5
That is, there are 0.000054 tracks per focus movie, on average. Putting it the other way, you will have to search, on average, 1/(5.4 x 10^5) = 1.9 x 10^4 = 19,000 movies to find one track.
Of course, the 45 tracks per array estimate might be off by a factor of ten either way. So there might be as many as one track per 2,000 movies, or as few as one per 200,000. Or there might be none...
Let me know if you found any mistakes in my calculation.
Taking the prelaunch estimate of 45 tracks in the whole array as a starting point, and that the whole array has an area of 1000 cm^2, there are an average of 45 tracks/10^3 cm^2. A focus movie covers an area of about 300 x 400 micrometers (if you believe the provided scale) which is 12,000 micrometer^2 = 1.2 x 10^5 micrometer^2.
One cm = 10^4 micrometer, so one cm^2 = 10^8 micrometer^2. Thus the array area in micrometer^2 is 1000 x 10^8 micrometer^2 = 10^11 micrometer^2
If there are 45 tracks on the whole array, that means there are 45 tracks per 10^11 micrometer^2 (Still with me? Good...)
That is, the ratio of tracks to area is 45 to 10^11 micrometer^2. This ratio can also be expressed as tracks per focus movie area. We set the ratios equal and solve for tracks:
tracks/area = 45 tracks/10^11 micrometer^2 = X tracks/focus movie area
Substitute the focus movie area 1.2 x 10^5 micrometer^2:
45 tracks/10^11 micrometer^2 = X tracks/1.2 x 10^5 micrometer^2
X = (45 x 1.2 x 10^5)/10^11 = (54 x 10^5)/10^11 = 54 x 10^6 = 5.4 x 10^5
That is, there are 0.000054 tracks per focus movie, on average. Putting it the other way, you will have to search, on average, 1/(5.4 x 10^5) = 1.9 x 10^4 = 19,000 movies to find one track.
Of course, the 45 tracks per array estimate might be off by a factor of ten either way. So there might be as many as one track per 2,000 movies, or as few as one per 200,000. Or there might be none...
Let me know if you found any mistakes in my calculation.

 DustMod
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This topic came up back in May also. You can compare notes with others here
http://stardustathome.ssl.berkeley.edu/ ... c.php?t=15
http://stardustathome.ssl.berkeley.edu/ ... c.php?t=15
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Your math is fine dstew, but a problem arises right from the beginning when you state
This would in fact result in significantly less than the official 1.6 million focus movies estimate that has been made public. http://www.planetary.org/programs/proje ... story.html states that "since each image will cover an area of 260 x 340 microns, and since each image will include a 10% overlap with its neighbor, the microscope will need to focus on 1.6 million different locations to cover the entire surface of the collector." 1.6 million/45=35,555 focus movies per particle, as stated some weeks ago in the thread mentioned above by DustBuster. Once again though, your calculations are perfect, but your initial estimates may have been off, that's all.dstew wrote: Taking the prelaunch estimate of 45 tracks in the whole array as a starting point, and that the whole array has an area of 1000 cm^2, there are an average of 45 tracks/10^3 cm^2. A focus movie covers an area of about 300 x 400 micrometers (if you believe the provided scale) which is 12,000 micrometer^2 = 1.2 x 10^5 micrometer^2.
The integral sec y dy
From zero to onesixth of pi
Is the log to base e
Of the square root of three.
Um...times the square root of the fourth power of i.
From zero to onesixth of pi
Is the log to base e
Of the square root of three.
Um...times the square root of the fourth power of i.
Thanks for your reply, Ferrum. Actually, the two calculations are subtly but really different. The 1.6 million focus movies are an oversampling of the array area by some amount, the 10% overlap mentioned. I am not sure if it means 10% on each edge, or 10% total area of overlap. In any case, since the movies overlap, assuming there are 45 tracks on the array, there will be more than 45 tracks in the 1.6 million movies, some movies sharing a track in the overlap area. So the number of movies per track will be somewhat less than 35,555, depending on how much overlap there is between the movies.[/quote]
Exactly dstew, great point...since it is probable that a few tracks are located near the edge of the movie, I had a discussion similiar to this on another (nonofficial) stardust board, in which I stated that although one could be the first to discover a particle on a particular movie, you may not be the first to in fact see the particle, since another movie with the same track (presuming that the track is on the edge of the movie) may exist, and someone may have noticed a track on that other movie first. Its an interesting question, since several people are convinced that they will be able to name particles, when in fact another individual may have discovered the same particle on another movie first, thus earning the naming rights.
The integral sec y dy
From zero to onesixth of pi
Is the log to base e
Of the square root of three.
Um...times the square root of the fourth power of i.
From zero to onesixth of pi
Is the log to base e
Of the square root of three.
Um...times the square root of the fourth power of i.
Ferrum wrote: This would in fact result in significantly less than the official 1.6 million focus movies estimate that has been made public.
Its already been posted by several dustmods that the project lucked into a camera that has a bigger ccd, and therefore there are 700,000 +/ movies.
Since this throws all the calculations out the window, time to start over. (or just divide by 2).
Correst me if I am wrong...PLEASE
This thing about 45 tracks total or what I read was a possible 45 something per viewing movie???
I really did not see anything that said the project would find 45 tracks???
I read they don't know and it could be ZERO or a Million???
Help here from a mod may clear this up???
Howie
I really did not see anything that said the project would find 45 tracks???
I read they don't know and it could be ZERO or a Million???
Help here from a mod may clear this up???
Howie
Better estimate
Thanks icebike for the info on the bigger ccd. My original estimate for the area of a focus movie was just a visual guess that the area was 300 x 400 micrometer^2. By measuring the 100 micron scale, and then the focus movie area, I calculate that the focus movie area is actually 360 x 480 = 1.72 x 10^5 micrometer^2. Substituting this more accurate focus movie area into the original calcualtion, one gets 7.74 x 10^5 tracks per movie, or 12,900 movies per track. That is more encouraging.