Imaging RX02 disks for simh (was: Re: MU-BASIC V2 and RT-11 V03B distribution disks.)

From the SIMH pdp11 RX driver (pdp11_rx.c) the disk size is computed as follows, and the byte offset into the file is computed by CALC_DA(trk,sec) given the PHYSICAL track (0..76) and sector (1..26) addresses used in accessing the controller. So the SIMH disk image should be 77*26*128 = 256,256 bytes for an RX01 format. RX02 format is same number of tracks and sectors, but has 256 byte sectors. So if you image an RX disk using logical operating system 512B blocks there are 494 of them (76*26*128/512 = 494) numbered 0..493. Track 0 is skipped in the filesystem (block 0 is at track=1 sector=1) for legacy compatibility reasons (with IBM). However, since the SIMH file is in physical track/sector order, if you read the disk image using logical device blocks, you have to know how the driver interleaves logical blocks onto physical track/sectors, as you must de-interleave to build the SIMH file. Or else you must run a program on the PDP-11 side that reads the RX drive as physical tracks and sectors, not using file system access commands. It's not pretty, but if you think about it enough it is the only way for SIMH to simulate the RX/RY devices and be operating system agnostic. Don From PDP11/pdp11_rx.c: #define RX_NUMTR 77 /* tracks/disk */ #define RX_NUMSC 26 /* sectors/track */ #define RX_NUMBY 128 /* bytes/sector */ #define RX_SIZE (RX_NUMTR * RX_NUMSC * RX_NUMBY) /* bytes/disk */ #define CALC_DA(t,s) (((t) * RX_NUMSC) + ((s) - 1)) * RX_NUMBY