Comments on: Answers to any of the following questions about digital storage? http://antiquerecordplayers.info/answers-to-any-of-the-following-questions-about-digital-storage.htm Sun, 20 May 2012 04:03:38 +0000 http://wordpress.org/?v=2.7.1 hourly 1 By: billrussell42 http://antiquerecordplayers.info/answers-to-any-of-the-following-questions-about-digital-storage.htm/comment-page-1#comment-1596 billrussell42 Fri, 18 Sep 2009 12:47:00 +0000 http://antiquerecordplayers.info/answers-to-any-of-the-following-questions-about-digital-storage.htm#comment-1596 1) At the end of the spiral, the diameter is 12 cm, or the circumference is 38 cm. At the other end, the beginning, the diameter is about 4 cm at a guess, so the circumference is 13 cm. Take the average of the two as 25 cm. In the distance between a radius of 2 cm and a radius of 6 cm, there is 4 cm. Divided by 1600 nm that is 25000 groves. Multiply that by average length of 25cm to get 625000 cm. I could get more exact numbers by reading online the red-book spec for CDs b) 44100 words/sec x 32bits/word x 1 byte/8 bits = 176400 bytes. but 44100 words/sec seems wrong, I'd have to check. . 1) At the end of the spiral, the diameter is 12 cm, or the circumference is 38 cm. At the other end, the beginning, the diameter is about 4 cm at a guess, so the circumference is 13 cm. Take the average of the two as 25 cm.

In the distance between a radius of 2 cm and a radius of 6 cm, there is 4 cm. Divided by 1600 nm that is 25000 groves. Multiply that by average length of 25cm to get 625000 cm.

I could get more exact numbers by reading online the red-book spec for CDs

b) 44100 words/sec x 32bits/word x 1 byte/8 bits = 176400 bytes.
but 44100 words/sec seems wrong, I’d have to check.

.

]]> By: jrrymiller http://antiquerecordplayers.info/answers-to-any-of-the-following-questions-about-digital-storage.htm/comment-page-1#comment-1597 jrrymiller Fri, 18 Sep 2009 12:47:00 +0000 http://antiquerecordplayers.info/answers-to-any-of-the-following-questions-about-digital-storage.htm#comment-1597 I will go for number 1. Consider each track as a circle. You know how many circles there are because you know how many nm have data tracks. Just find the length of the middle track (the average) and multiply that by the number of circles total. This should be a very good estimate of the length of the spiral. My sense is that this will be a very large number because there are many circles. I will go for number 1. Consider each track as a circle. You know how many circles there are because you know how many nm have data tracks. Just find the length of the middle track (the average) and multiply that by the number of circles total. This should be a very good estimate of the length of the spiral. My sense is that this will be a very large number because there are many circles.

]]> By: Max H http://antiquerecordplayers.info/answers-to-any-of-the-following-questions-about-digital-storage.htm/comment-page-1#comment-1598 Max H Fri, 18 Sep 2009 12:47:00 +0000 http://antiquerecordplayers.info/answers-to-any-of-the-following-questions-about-digital-storage.htm#comment-1598 2. 1/8 bytes/bit * 32 bits/word * 44100 words/second * 60 seconds/minute * 4.0 minutes 3. Channels and word size are redundant. 44100 bits/second * 60 seconds/minute * 80 minutes 4. check the internet, it has nice pictures 2. 1/8 bytes/bit * 32 bits/word * 44100 words/second * 60 seconds/minute * 4.0 minutes
3. Channels and word size are redundant.
44100 bits/second * 60 seconds/minute * 80 minutes
4. check the internet, it has nice pictures

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