Dual File Structure CD for RT-11

Jerome Fine replies: Since RT-11 and ODS might not have had the same requirements, then I suppose it might still have been "rather fortunate" that RT-11 needs those same 64 blocks. As for the name of the "RT-11 file structure", I suppose that is what everyone else calls it as well. The overall details are simple for RT-11 Partition ZERO: Block 0 - Secondary boot block (the same as will ALL PDP-11 systems) Block 1 - Directory information Block 2,3,4,5 - Tertiary boot blocks Blocks 6 => 67 - 31 Directory Segments (2 blocks per directory segment) Unfortunately, RT-11 devices are a MAXIMUM of 32 MBytes or 64536 blocks. Thus, the same RT-11 file structure MUST be repeated every 65536 blocks and requires those 64 (actually 68 blocks, although as few as 8 blocks are sufficient since ONLY ONE directory segment is mandatory for each RT-11 file structure) blocks at the start of every RT-11 partition which starts at every multiple of 65536 blocks. Thus a CD can have about 20 RT-11 partitions - depending on how full the CD is with files. As I asked, I need to be able to make those 68 blocks available by some means. On the other hand, I doubt that it would work very well with a CD like the RT-11 Freeware CD which has over 21,000 files in over 900 directories. The only really useful situation would be with a limited number of (for RT-11) large files, probably each being on average about 500 blocks which is about 120 files per RT-11 file directory. Since all RT-11 binary distributions were about 5000 blocks or larger starting with V4.00 in 1980, a CD with RT-11 distributions would be ideal if there was more than 300 MBytes of distributions and the standard ISO9660 file structure was used along with the RT-11 files structure. In particular, starting with V5.00 of RT-11 up to V5.07 of RT-11, there were about 17 unique binary distributions for a total of over 100,000 blocks. That is only about 50 MBytes, but when the layered products are added along with the source code, I suspect the total would be more than 300 MBytes, although less than 600 MBytes that ONE CD can manage. What I am looking for is some method of leaving 64 blocks at the start of every RT-11 partition (every 65536 blocks or 16384 sectors). Note that in most cases, my choice for an RT-11 distribution would be an RL02 image which can hold a maximum of 20450 blocks. I would then TRUNCATE the unused blocks (usually all zeros) at the end and thus save about 14000 blocks for each binary distribution or DSK file even though the boot blocks would remain for an RL02 in blocks 0, 2, 3, 4 and 5 of the RT-11 file directory. With only a small number of RT-11 DSK distribution files, it would be relatively easy to set up the RT-11 file directory for each RT-11 partition for the small number of CDs that might be produced. It seems VERY unlikely. The first 8 blocks of the RT-11 file structure are very specific. On the other hand, if you don't restrict the operating system that the PC is running, when I used the CD that was produced, I was able to boot under Windows 98 SE. I am also sure that it will be possible to use the same CD and boot on a real DEC PDP-11. As I asked before, does anyone know if the source code for mkisofs.exe could be modified so that when the files are set up in the standard ISO9660 file structure, the first 64 blocks at the start of every RT-11 partition (every 65536 blocks) can be left empty? OR, a standalone program could be written to take a CD image and move the files and adjust the pointers so that the same result is produced. Sincerely yours, Jerome Fine -- If you attempted to send a reply and the original e-mail address has been discontinued due a high volume of junk e-mail, then the semi-permanent e-mail address can be obtained by replacing the four characters preceding the 'at' with the four digits of the current year.