Gentlemen,
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Thank you for your contributions, it looks like we finally found real answer.
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I have found the last peace of it:
<gaps.png>
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In this text we see that there is four gaps that accompany each member in PDS dataset.
Also, i presume, that the same gaps are present at in directory block.
As Silvio Losa pointed out, there is 1729 cells in 3390 track.
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55996bytes / 1729cells = 32bytes in one cell.
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Now lets calculate gaps for one member only. As i said earlier, there is going to be two occurrences of those gaps,
one for member and other for directory block(?).
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15.75cells + 7cells + 7.75cells + 6.75cells = 37.25 or 38 cells
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38cells * 32byte = 1216bytes(for one set of gaps)
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But in PDS with only one member we have two sets.
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1216 * 2 = 2432(which is almost exactly the number of free space left in our experiment)
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Lets proof our observations with second example.
When there was 14 members in PDS, free space was equal to 231 lines.
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231 lines * 80bytes = 18480 bytes of free space
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For 14 members in PDS we should have 15 sets of gaps.
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1216bytes * 15 = 18240
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BINGO!!
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Thanks to Harold Grovesteen we now know that in emulated CKD drives all those gaps just collected at the track end.
Hope you found this investigation entertaining! :)
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Best wishes,
Andre
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