Hi at all. I have few questions on how the geometry is calculated. 1) I would like to understand. this table represents an 3740 IBM standard disk D: Drive Characteristics 1944: 128 Byte Record Capacity 243: Kilobyte Drive Capacity 64: 32 Byte Directory Entries 64: Checked Directory Entries 128: Records / Extent 8: Records / Block 26: Sectors / Track 2: Reserved Tracks Its DPT is: 1A00 03 07 00 F200 3F00 C000 1000 0200 Or: SPT = 001A = The number of 128 byte records per track. BSH = 03 = The block shift count. BLM = 07 = The block mask. EXM = 00 = The extent mask. DSM = 00F2 = Disk storage maximum (the largest block number). DRM = 003F = Directory maximum (the largest directory entry). DAB = 00C0 = Directory Allocation Block AL1: = 00 & AL0: = C0 [0C0H] CKS = 0010 = Directory check size. OFF = 0002 = Track offset (number of reserved tracks). I already know there are 26x128x77 = 256256 bytes But how does this table represent them? 1944 sectors x 128 bytes = 248832 bytes = 243x1024 is the capacity actually available for the files The system takes 26 sects/track x128 bytes/sect x2 sects= 6656 bytes So 256256 - 248832 = 768 bytes They are for the directory but how does this number come out? 2) I have instead this table of a BASF 6104 Floppy Disk Drive managed through a Shugart SA-1403 SASI Controller: B: Drive Characteristics 9216: 128 Byte Record Capacity 1152: Kilobyte Drive Capacity 64: 32 Byte Directory Entries 64: Checked Directory Entries 2048: Records / Extent 128: Records / Block 512: Sectors / Track 1: Reserved Tracks Its DPT is: 0002 07 7F 0F 4700 3F00 8000 1000 0100 SPT = 0200 = The number of 128 byte records per track. BSH = 07 = The block shift count. BLM = 7F = The block mask. EXM = 0F = The extent mask. DSM = 0047 = Disk storage maximum (the largest block number). DRM = 003F = Directory maximum (the largest directory entry). DAB = 0080 = Directory Allocation Block AL1: = 00 & AL0: = C0 [0C0H] CKS = 0010 = Directory check size. OFF = 0001 = Track offset (number of reserved tracks). Oddly, the controller seems to manage the drive as 32 sect/tracks x256 bytes/setc x77 trakcs x2 heads In this case I don't know exactly the parameters of the diskette 9216 sectors x 128 bytes = 1.179.648 bytes = 1.152x1024 is the capacity actually available for the files The system takes 512 setcs/track x32 bytes/sect x1 tracks = 16.384 bytes In case 1 the gross capacity is 256256 while in case 2 which is it? 3) what could be the correct parameters to insert in 22disk in order to transfer files between a type 2 disk and msdos? Thanks a lot at all who will wish to answer to me. Enrico

So in the case of IBM 3740 Standard Disk Support the 768 bytes are an unusable remnant due to the choice of the 1024byte block size. Yes, the second case disk is a backup disk !!! The machine has a SASI card that interfaces to an old Shugard SA-1403 SASI Controller that manages a 10MB winchester SA-1004 hard disk with 32 sectors of 256byte / track (with 256 tracks and 4 heads for a total of 32 * 256 * 256 * 4 = 8MB) and a disk drive with 32 256byte / track sectors yet (with 77 tracks and 2 heads for a total of 32 * 256 * 77 * 2 = 1.2MB) Physically the directory starts at 10000H and ends at 13FFFH since the first data on the disk image is stored starting at location 14000H. I sincerely thank you for your kind and perfect response. In fact from the specifications of the DPT I could not understand. Experience in this case counts a lot. THANK YOU -----Messaggio originale----- Da: cctalk [mailto:cctalk-bounces at classiccmp.org] Per conto di Patrick Finnegan via cctalk Inviato: mercoled? 10 febbraio 2021 21:52 A: Enrico email.it; General Discussion: On-Topic and Off-Topic Posts Oggetto: Re: 8 "disc geometry I've spent a fair bit of time staring at DPT's so we'll see if I can help.. On Wed, Feb 10, 2021 at 7:35 AM Enrico email.it via cctalk < cctalk at classiccmp.org> wrote: The disk is blocked into 1K blocks (8*128 bytes), and there are 243 blocks, starting after the reserved tracks (2). So there is 256256 total capacity -- 2*128*26 bytes on the reserved tracks, 243*8*128 bytes usable, and 768 bytes of stranded capacity, since you can't have a partial block. The directory starts on block 0 and occupies 2 blocks. For AL0 and AL1, the word is 0C000h, which has two 1 bits, marking the first two blocks as directory blocks. This matches up with DRM, which says there are 40h directory entries (32 bytes each), for a total of 2048 bytes. actually No, that 768 bytes is just unusable space. You have a total of 512*128 bytes per track - 64K. I assume the SASI controller is remapping things around, and the disk doesn't physically have tracks this big. Each block is 128*128 bytes, or 16k And there's a total of one reserved track (64K) plus 72 blocks (1152K), for a total disk size of 1245184 bytes (1216K). The directory is all in the first block (16KB), with a maximum of 64 entries. I'd guess that the floppy was meant to hold backup or install sets or something, and not lots of small files, as this sorta layout lets you minimize disk overhead if you do that. You will probably have some issues directly using the floppy disk on a PC with those parameters, since I suspect the controller is remapping things and they don't line up on track boundaries. If you can dump a raw image with eg imagedisk, you can probably get that to work with 22disk. I don't know enough about 22disk to help you, but its author is on the list and may be able to help. Pat