🚀 go-pugleaf

>> It doesn't have to mean that at all.  It probably means that the Mac is
>> slower because half of it's clock cycles are given up to the video hardware.
>> The effective 68000 clock rate is only 4Mhz.
>
>People seem to like to say "The effective 68000 clock rate is only x MHz"
>where the x is replaced by some number from 3 to 7 depending on the
>person's opinion of the mac.

But there's really nothing wrong with saying what the effective clock rate is;
it should have nothing to do with your opinion of the MAC.  What it has to
do with is how much of your program executes in ROM and how much in RAM.

The original statement above is definitely wrong, first of all.  The video
"steals" cycles from the CPU only for part of the time (roughly
proportional to the ratio of the time the CRT beam is writing on the screen
to the sum of that time plus the time the beam is doing horizontal and
vertical retraces).  Nowhere near "half" the cycles are given up to the
video hardware.  Furthermore, this only happens when the CPU tries to
access dynamic RAM while the display is accessing it, so when the CPU is
executing from ROM, the delays don't occur.

There's something else you have to consider, too, though, in comparing to
the IBM PC (which the referenced message later did).  The IBM PC has its
DMA controller programmed to periodically do a dummy DMA transfer to refresh
the dynamic RAM.  The length of the transfer is sufficient to scan all
the combinations of the [row or column, I forget which] addresses in the
memory array.  It does this fairly often, and takes a good bit of time.

On the Mac, the video RAM accessing does this for you, so you don't have
the dummy DMAs taking up time. (However, the IBM PC's refresh time takes
significantly less time than the Mac's video-accessing time).
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