Re: Star Wars mathbox math

Thanks very much. Amazing how one paragraph can bring so much enlightenment.

Funny I 'knew' that 'Road Accumulator' was a typo but thought the R had been subbed for an L as a bad duplicate. Was thinking 'Store' or 'Write' or 'Save' just not e for o to make 'Read'

Remaining divider tests now pass after replacing the LS374 at 4L in the dividend latch. On to the AVG intensity problem!

-DrG

On Sep 26, 2013, at 6:21 PM, "Neil Ward" <Neil@wardclan.f9.co.uk> wrote:

> Hi,
>
> From the man who designed the mathbox is an explanation of how 4000 hex represents 1.000, so 1.000 x 1.000 = 1.000
>
> http://www.jmargolin.com/uvmath/uvmath.htm#What Is
>
> I've looked at the mathbox myself, it's clever stuff:
>
> http://www.wardclan.f9.co.uk/Star%20Wars%20Mathbox.html
>
> Hope that helps.
>
>> PPS. Setting bit 8 of the instruction PROMs loads the accumulator, but what does it mean to set bit 9 and 'Road' the accumulator?
>
> It's a typo!! Should be "Read Accumulator" If you look at the schematics 'Instruction Strobe Gates' you can see that IP9 produces 'READACC' signal, which enables Accumulator outputs and writes to Math RAM.
>
> Good luck fixing your board!
>
>
>
> Regards,
> Neil
>
>
> ----- Original Message ----- From: "Douglas Gauck" <douglas@gauck.com>
> To: <vectorlist@vectorlist.org>
> Sent: Thursday, September 26, 2013 6:40 AM
> Subject: VECTOR: Star Wars mathbox math
>
>
>> Working on two SW main boards that complain about bad RAM at 5F and 5H even when replaced with known good chips. I think it's elsewhere as they also fail four of the six matrix tests (15, 16, 18 and 20), all the divider tests, and tests A, B, C and D. I would love to know how the math works as shown in the SP-225 Troubleshooting Guide for example:
>>
>> test 17 (passes) performs (5555-0000) x 4000 = 5555
>>
>> test 18 (fails) performs (0000-5555) x 0000 = 5555 if you go by the test or (0000-5555) x C000 if you go by the first diagram on page 12
>>
>> test 19 (passes) performs (2AAA-0000) x 4000 = 2AAA
>>
>> test 20 (fails) performs (0000-2AAA) x C000 = 2AAA
>>
>> test C performs ACC = (1B2C-0000) x 4000 + (196A-0000) x 4000 whose result should be 3496
>>
>> test D performs ACC = (2696-1B2C) x 4000 whose result should be 0B6A
>>
>> I understand binary and hex but neither is helping me understand how any of the above operations could calculate the way the guide says they do.
>>
>> Can anyone explain this?
>>
>> -Douglas
>>
>> PS. seems my Accumulator seems to hold onto the first value it gets, which is why 17 and 19 pass but 18 and 20 fail with 0000, A with 0001, C with 196A and D with 2696 which is interesting. Test 15 and 16 fail with 1B2C which I can't explain and B ends with 0000.
>>
>> PPS. Setting bit 8 of the instruction PROMs loads the accumulator, but what does it mean to set bit 9 and 'Road' the accumulator?
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>
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Received on Fri Sep 27 21:23:50 2013