One track per 19,000 focus movies
Posted: Tue Aug 15, 2006 6:31 am
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 pre-launch 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 pre-launch 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.