Report on vintage "Programmer Electronnic Control" alias "RAF Tornado Computer"
Erik Baigar
erik at baigar.de
Thu Mar 15 02:28:12 CDT 2007
Hi Roger!
> >> Then 12 bit models were the 902, 102C, ARCH 105 and Minim or 12/12.
>
> I doubt you will find much apart from the names in timelines.
You are right - only very limited hits on google.
> normal use. Every extra Kg in the Avionics adds a lot to the overall
> takeoff weight. For a helicopter, I think the ratio was somewhere
> around six. Then there's the power, every extra watt needs bigger
> alternators/generators and a bigger engine to drive them, and more
> fuel to carry, which means more engine power and so on.
Never thought about this in that way. But sounds
absolutely logic.
> In 920 assembler, instructions did not have mnemonics, you used their
> decimal operation code values. To indicate 'B Line modification' in
> the assembler you wrote a / before the operation code. The assembler
> simply added 16 to the opcode, so /8 = 24, but you don't have the B-Line
> bit on your machine - its always on.
OK, I understand this now. What looked a typical list
file and of did you specify the operands those days?
Lot of assembly was done by hand, right?
> > The Programmer Electronic Control (Let's call it PEC in the future)
> > stores PC+1 (as your 18bit) to the specified location into
> > core. The new address is taken from the position in memory
> > where the index register points to. So in my case this is
> > a perfect table jump.
>
> Ah, so you do this ?
> 0 word with address of routine
> 11 link word
Yes, quite right. The 11 gets as operand the address
where to store the program counter for later returning.
Since the unit is 12bit and has 8192 words of memory
it uses two consecutive words for storing the return
address or loading the destination. And a bottleneck
again si, that the return address has to be stored
in the first 128 words of core. I think this is a
severe drawback of this machine: 4 bits for identifying
the instruction and only 7 for the operand. So the
0 and 11 are restricted to the "zero-page"...
> > Right. But here again, the bit 11 is the sign bit and
> > thus the shift does not affect the bit 11 of the
> > destination. I.e. bit 0 of A is moved to bit 10 of Q
> > in shifting right for example.
> I see. The Q register was always a bit weird.
Yes. From today's point of view this seems
strange. But considering the sign/value notation
it makes sense somehow.
More information later...
thanks,
Erik.
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