Your right, things are a little confusing.
SECTORS - Think of it as 224 pieces of pie.  It is, I believe, physical.
CELL - Also physical, but I think of them as little chunks of data, which may 
be your data or control data for the hardware.
TRACK BALANCE - How much room is left on the track if you were to write a 
single block.  Look up TRKBAL macro.

That extra calculation is for device control information, part of which I know 
is CRC, or at least that is what I was told.  All that stuff other than the 
COUNT-KEY-DATA areas are for the hardware and we mortals can't see it, but it 
is there.

Chris Blaicher
Technical Architect
Mainframe Development
P: 201-930-8234  |  M: 512-627-3803
E: [email protected]

Syncsort Incorporated
2 Blue Hill Plaza #1563
Pearl River, NY 10965
www.syncsort.com

Data quality leader Trillium Software is now a part of Syncsort.


-----Original Message-----
From: IBM Mainframe Discussion List [mailto:[email protected]] On Behalf 
Of R.S.
Sent: Tuesday, January 23, 2018 10:22 AM
To: [email protected]
Subject: Re: CKD details

Fortunately your explanation was clear enough to find out each "mini-record" 
has at least 10 cells (9 control plus a t least one for data).

However you used word "SECTORS". I used  "34-byte DATA CELL", following the 
3390 documentation (other names can be found elsewhere).
It seems the documentation use SECTOR for something else. The
SA22-1025-00 says the 3390 device has 224 sectors per track and 222 for 3380. 
So, the SECTOR  cannot be the same as DATA CELL.

I have no idea what is the SECTOR.


BTW: I also do not understand the formula. For me it should be just
Numberofcells=9+RoundUP(KL/34)
of course you provided right formula as it is documented, but I simply don't 
understand the purpose of KN and 6 and 232. i'm pretty sure it is not just a 
magic, but what is the rationale behind?

Last but not least: THANK YOU VERY MUCH for the explanations you gave me, I 
appreciate it.

Regards
--
Radoslaw Skorupka
Lodz, Poland







W dniu 2018-01-23 o 15:48, Christopher Y. Blaicher pisze:
> With all I put in the last post, I forgot to answer WHY 20 sectors at a 
> minimum for a record.  It is because at a minimum a record consists of a 
> COUNT and a DATA section. Each one takes a minimum of 10 sectors, so with 
> even a 1 byte record you need 20 sectors.  If you had a 1 byte key and 1 byte 
> of data you would need 30 sectors, 10 for each part.
>
> Chris Blaicher
> Technical Architect
> Mainframe Development
> P: 201-930-8234  |  M: 512-627-3803
> E: [email protected]
>
> Syncsort Incorporated
> 2 Blue Hill Plaza #1563
> Pearl River, NY 10965
> www.syncsort.com
>
> Data quality leader Trillium Software is now a part of Syncsort.
>
>
> -----Original Message-----
> From: IBM Mainframe Discussion List [mailto:[email protected]] On 
> Behalf Of Christopher Y. Blaicher
> Sent: Tuesday, January 23, 2018 8:30 AM
> To: [email protected]
> Subject: Re: CKD details
>
> See the following for how to calculate a sector 
> https://www.ibm.com/support/knowledgecenter/en/SSB27H_6.2.0/fa2mr_sectval.html
>
> Question 2: why 20 sectors at a minimum.  It's a long answer.
> We tend to think of a record as three unified parts, COUNT, KEY and DATA, but 
> really they are three mini-records (for lack of a better description). Each 
> part needs enough cells for the data, plus 9 cells of CRC and other control 
> information that we can never see.
>
> So, for a COUNT field it takes 10 cells; for a KEY field, if present, and a 
> DATA field the calculation is:
>
> SECTORS = 9 + (KL + (6 * KN) + 6) / 34)
> Where KN = (KL + 6) / 232
>        KL = Key length
> Change KL to DL and do the same calculation for the data area.
>
> Question 3: Are you running under VM and using a mini-disk?  VM formats their 
> R0 differently, I believe.
>
> Hope this helps.
>
> Chris Blaicher
> Technical Architect
> Mainframe Development
> P: 201-930-8234  |  M: 512-627-3803
> E: [email protected]
>
> Syncsort Incorporated
> 2 Blue Hill Plaza #1563
> Pearl River, NY 10965
> www.syncsort.com
>

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