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- From: lupe@ukw.uucp (Lupe Christoph)
- Newsgroups: comp.sys.sun.admin
- Subject: Re: why does an 6000 ft exabyte tape give 2.3Gb capacity?
- Message-ID: <1993Jan2.182125.20541@ukw.uucp>
- Date: 2 Jan 93 18:21:25 GMT
- References: <C06Bpu.2IJ@ie.utoronto.ca>
- Sender: news@stasys.sta.sub.org
- Organization: cic
- Lines: 41
-
- andy@ie.utoronto.ca (Andy Sun) writes:
-
- >Hi Everyone,
-
- >I know I am missing something, but I don't know what it is and will
- >appreciate your help.
-
- >I know for a fact (from the dump(8) man page) that a 2.3Gbyte 8mm
- >Exabyte tape has a length of 6,000 feet and a density of 54,000 bits/inch.
- >What I don't understand is WHY 6,000 feet with 54,000 bits/inch would give
- >2.3Gbyte?
-
- >54000 bits x 12 inch x 1 byte x 1 kbyte x 1 Mbyte x 1 Gbyte x 6000 feet
- > ---- ------- ------ ---------- ----------- ----------
- > inch 1 foot 8 bits 1024 bytes 1024 kbytes 1024 Mbyte
-
- >= 0.45Gbyte
-
- >Where does the extra factor of 5 come from?
-
- You are probably referring to this example:
-
- 2.3-GByte 8mm tape: dump dsb 54000 6000 126
-
- This computes like this:
-
- 6000 ft x 12 in/ft = 72000 in
- 54000 bpi is 54000 *byte* per inch, as dump assumes a 9-track tape
- when the 'c' is missing.
- 54000 byte/in * 72000 in = 3888000000 byte, or 3.62 GB
-
- Where did the remaining 1.32 GB go to?
-
- Into the "inter record gaps". Dump assumes that the drive will write
- an IRG after each block of data. At a block length of 126*512, or 63 kB,
- this accounts for about 2200 pseudopods, excuse me, pseudo-ft.
- --
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