Early floppy disks had significant spacing between blocks, e.g. a PC 5.25" disk had a theoretical capacity of about 500K but was formatted as 360K, nine blocks per track, with a substantial fraction of a block's worth of space between blocks.

What was the equivalent figure for early hard disks? For example, a common size for PC hard disks in the eighties was 20 megabytes. On a hard disk like that, how much of the space was used for spacing between blocks?

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    I don't think that can be easily answered, like quite some of your questions - The necessary spacing is a function of rotational speed of the disk (which differed a lot) and time the circuitry needed to get ready for the next sector (and move the data from the last sector out of the way) and the sync speed the circuitry could achieve and the amount of sync bytes needed for proper synchronisation. All of this could be varied a lot.
    – tofro
    Dec 11, 2017 at 22:39
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    minuszerodegrees.net/manuals/Seagate/… page 5, "track format as shipped" . Interesting document for anyone interested in early HDD technology, since this is one of the few PC compatible disk drives that are all-hardware. Dec 12, 2017 at 21:53

2 Answers 2


The difference between 360KB and 500KB is not entirely due to space between sectors. A lot of the difference is due to clock sync bits, sector identification markers, checksum/ECC bits, etc. Keep this in mind otherwise comparisons are meaningless.

Taking the DEC RP03 disk pack drive as an example of an "early hard disk", I took a look at the maintenance manual for the associated RP11 controller:

RP11-C disk pack drive controller maintenance manual

The RP03 drive uses RP02P disk packs (actually the same packs as the IBM 2316), and these packs are 'hard-sectored' via a metal disc with 20 sector notches (plus 1 index notch). Rather than having 20 small sectors, the RP03 ignores every other notch to end up with 10 sectors per track, allowing each sector to be twice as big. These packs are spun at 2400 rpm, which works out to 40 revolutions per second, or 25 milliseconds per revolution. Dividing by 10 sectors gives 2.5 milliseconds per sector, or 2500 microseconds.

Page 4-24 of the maintenance manual indicates that a continuous stream of 0 bits would be written as a 2.5 MHz signal while a continuous stream of 1 bits would be written as a 5.0 MHz signal. This means that the 'bit clock' is 2.5 MHz, or 0.4 microseconds per bit.

Page 3-10 reports that each sector consists of 165 37-bit words, which is 6105 bits. At 0.4 microseconds per bit, that's 2442 microseconds per sector. Note that the actual data capacity of each sector is 128 36-bit words (4608 bits) so a significant fraction of the theoretical capacity is occupied by parity/checksum bits, clock sync bits and other housekeeping (1497 bits out of 6105, or 598.8 microseconds).

2500 microseconds minus 2442 microseconds leaves 58 microseconds of spacing between each sector, or 2.32% of the track (58 divided by 2500).

This is a very small amount of overhead compared to the 598.8 microseconds used for error checking and other overhead (24%).


It's a feature or the storage scheme - and similar schemes will end in similar results.

The maximum you're asking for is defined by the bit clock. So take the time needed for one revolution and divede it by the clock uses. That's the theoretical capacity within the frameset of your question. Subtract the 'real' data and you'll get the overhed. And as a result the Ratio is about the same as with diskettes. Roughly 1/3rd.

Just, as @tofro already noted, it's useless, as it doesn't tell anything about the media or its real capacity, or what would be possible by adapting the drive system. Just compare Apples variable speed 3.5 drive to an identical fixed speed drive. Same can be reached with variable speed bitstreams. And then there's the media. The 500 KiB you mention is a very conservative seting anyway.

Playing with formating, gaps and alike makes only sense if one can't change the real parameters - which usually restricts this to nerdy games.

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