In the IBM PC 5150 documentation, it is stated that the floppy drive has 40 tracks, 8 sectors/track and 512 bytes/sectors, for a total of 163 840 bytes of data. It makes sense, 40×8×512 = 163 840 bytes. Ok. It is also said that a floppy drive has 45 cylinders...
The IBM 5-1/4″ Diskette Drive is a single sided, double density, 40 track unit. The Diskette Drive has a formatted capacity of 163,840
(page 2-110 from the IBM 5150 Technical Reference, 1st edition)
1 Head, 45 cylinders, 8 sectors/TRK, 512 bytes/sector,MFM.
(page 2-104 from the IBM 5150 Technical Reference, 1st edition)
Both manuals are available from https://www.pcjs.org/documents/manuals/ibm/.
How is it possible? How can the number of cylinders be different from the number of tracks? What happens if I want to access cylinder 42 when there are only 40 tracks?
I thought cylinders were tracks on top of the others, there should be the same number of tracks as cylinders, shouldn't there? Or are there 45×8×512 = 184 320 bytes, but 163 840 usable and some tracks are disabled? I suppose this is not the case because the address mode would have been THS for Track/Head/Sector instead of CHS.