Re: [CAD_CAM_EDM_DRO] EMC Encoders

Ethan Vos wrote:

> Thanks for the info.
>
> I made an error in my calculations and I actually only need 10,000
> steps/sec.
>
> Since I can't buy a 233 anymore, I'll have to use a 700+ so the 10,000 is
> doable.
>
> The next question is with the encoders. The steppers that I have are 200
> steps/rev with 1000 steps/rev on the encoder. I understand from the EMC
> documentation that the numbers need to be the same. What do I do?

No, they don't have to be the same. The problem is, that if the encoder
resolution
is higher than the stepper resolution, there are encoder positions that can't
be reached. This will cause dithering. The solution to the dithering is to
set
a deadband equal to one stepper step. This should allow the system to settle
at any step position and not want to move finer than that.

> I'm replacing a Parker Hannifin AT6400 controller which uses a Motorola
> 68000 series processor. Any idea what the clock speed would be on that and
> how much faster the 700+ P3 would be?

A real 68K, such as an MC68000L8 is 8 MHz. I think there was a 10 MHz
version.
The 68010 was good to 10 and 16, I think. The 68020 ran at 20 and 25 or so.
There were later 68030, 68040, etc. These early ones were true 16-bit CPUs,
with no
floating point, maybe even no integer multiply, no cache, etc.

A 700 MHz Pentium has floating point on chip, 32-bit integer arithmetic,
32-bit
addressing, cache, pipelining, out-of-order execution, etc. So, raw clock
speed
doesn't tell the whole story. So, the 700 MHz Pentium is at least 100 times
faster
when you throw floating point into the mix. Most likely, the Parker Hannifin
doesn't
use floating point, but uses a bunch of fixed point conversions to make things

faster.

Is the Parker Hannifin really their own system, or is it another maker's
system
(Allen-Bradley, Siemens, etc.) relabeled?

Jon