2314 Disk pack

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Kylie63
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2314 Disk pack

Post by Kylie63 »

In Digital Computer Fundamentals by Thomas C Bartee Qu 55 on page 392 asks-"Assuming that a disk in the IBM 2314 system rotates at a speed of 2400 RPM as stated and that we read from eight tracks simultaneously, how many bits per second can be read from one of these disks?"

Data sheet gives No. of surfaces-20,
Storage capacity-20 MB, Tracks-200,
Transitions/inch-4,400,
Data rate 25M bits/s,
Ave access time-60ms,
RPM-2400,
Detent-Mech,
Actuator-Hydr,
Disk coating min-90 u-in,
Head fly height min-80u-in.

Answer given in book is 3,600,000 bits/second. I just cannot reconcile these figures. Any ideas folks?

Many thanks

Kylie
William Collins
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Re: 2314 Disk pack

Post by William Collins »

No, nor me.

It gives a data-rate of 25M bits per second. I can't see a way from that number to 3.6M bps with the data given. But, it is hardware.

It is also very old hardware. I'd not be too concerned about it.
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mickeydusaor
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Re: 2314 Disk pack

Post by mickeydusaor »

a 2311 and 2314 go back to the early 1970's, so I do not know why you would even care about this type of information.......
Kylie63
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Re: 2314 Disk pack

Post by Kylie63 »

Hello to both of you and many, many thanks for responding!

In reading the book I am trying to be thorough, but probably also a bit OCD! There have been a good deal of mistakes so far in this book so I am not sure if I am being dumb in not getting the answer or Thomas Bartee has goofed again! But I do want to do all I can to ensure I have got the best understanding I can from this text-in spite of the fact, which you very correctly point out, that the things in question are entirely redundant!

Please excuse my ignorance here, but I assume you are both closely involved with IBM. So if people with powerful knowledge of IBM equipment believe the answers by the author are incorrect, that would be more than good enough for me. I have tried to contact Prof Bartee on Twitter, but receive no reply.

Once again, sincere thanks to the both of you for taking the time to respond!
enrico-sorichetti
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Re: 2314 Disk pack

Post by enrico-sorichetti »

cheers
enrico
When I tell somebody to RTFM or STFW I usually have the page open in another tab/window of my browser,
so that I am sure that the information requested can be reached with a very small effort 8-)
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Anuj Dhawan
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Re: 2314 Disk pack

Post by Anuj Dhawan »

Kylie63 wrote:In Digital Computer Fundamentals by Thomas C Bartee Qu 55 on page 392 asks-"Assuming that a disk in the IBM 2314 system rotates at a speed of 2400 RPM as stated and that we read from eight tracks simultaneously, how many bits per second can be read from one of these disks?"
I've not read that book and was looking for some "free chapters" on google books or online before replying but in vain. But I think it's tough to get the glimpse of the exact chapter as this disk usage and release goes back to 1965; as said by others too. But the point I'm trying to bring up is -- does the book not give you a formula too to derive the per second rate too? So that the number can be verified against the formula/logic as well.
Thanks,
Anuj

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Kylie63
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Re: 2314 Disk pack

Post by Kylie63 »

Many thanks for the links. The "History" link I had already found, but the manual I had not and am now digesting!

DCF does not actually set out the calculations you have to make, but the method is implied by the text. The intention of the subsequent questions at chapter's end are, I believe, designed to provoke thought by forcing the student to delve into the subject beyond what is offered directly by the text.

The way I think this question was intended to stimulate the student was to present a mode of operation for the 2314 which is not conventional and not as its normal operation is set out, thereby compelling the student to work through the fundamentals of Disk Pack operation. And this I have clearly not been able to reconcile.

The best I have achieved is 3,600,000 / 8 (indicated simultaneously read tracks) = 450,000
450,000 / 40 (Revolutions per second) = 11,250 bits per track

BUT the date sheet reflects bits per track as 7,294, so you see my difficulty!

Also, if the data sheet reports the pack will deliver 2.5M bits per second in reading 20 Disks together (as I understand it does), how can you possibly have a 3.6M bits per second output from reading just 8 tracks?

I do believe this is yet another error by the author, but I want to exhaust all avenues in view of the fact that he was a Harvard professor!
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