Re: Re: How does G02 & G03 work?

Ray,
You are right, I used the wrong example. I was doing it from memory and I
used the Bridgeport Absolute I, J values. (been translating BPT files to
AHHA)
Bridgeport does also use unsigned incremental I and J and when you run them
graphically in AHHA you will get some arcs that are right, some that are
wild and some that won't run. The ones that go wild just need a minus sign
somewhere and the ones that don't run I have to play with.
Darrell

----- Original Message -----
From: Ray Henry <rehenry@...>
To: <CAD_CAM_EDM_DRO@onelist.com>
Sent: Wednesday, February 09, 2000 9:17 AM
Subject: [CAD_CAM_EDM_DRO] Re: How does G02 & G03 work?


> From: Ray Henry <rehenry@...>
>
>
> Jon, Darrell, and other nc writers
>
> I confess to being a bit confused by the lines of nc code offered in the
> posts below and have mixed my comments in.
>
> My understanding is that i,j,k, and r are incremental when they are used
> with g02, g03 and are figured from the current tool position at the start
> of the block in which they are included. Fagor (the only book I've got
> handy) uses g06 if you want to enter absolute values for i,j,k, or r.
>
> I've dropped a graphic from my backplotter of both of the desired tool
> paths you speak of in
>
> http://www.linuxcnc.org/Dropbox/g02ij.gif
>
> (I used Will Shackelford's revised emccanon.cc with emc to plot this)
>
> >Message: 1 (digest 152)
> > Date: Tue, 8 Feb 2000 00:31:29 -0800
> > From: "Darrell" <dgehlsen@...>
> >Subject: Re: Re: How does G02 & G03 work?
> >
> >Jon,
> >g00 g90 x-2.0 y0.0
> >g01 x-.5
> >g02 x0.0 y-.5 I-.5 J-.5
>
> For me, the g02 block above does not run at all. You seem to be telling
> the control to set the arc center both back on x and negitive on y from
the
> current location.
>
> What you describe is wanting to slide the arc center negitive in y by the
> distance where you want the current arc move to end (y-.5). The j is a
> correct offset from the current y0 position but I should be 0 or left off.
>
> >y-2.0
>
> Here, since g02 is modal in emc, an error is generated because there is no
> i,j,k,or r value given on this line. To achieve your linear move using
> emc, the block would have to be written -- g01 y-2.0
>
> >On a FANUC or AHHA control makes two lines with a 90 deg tangent arc.
> >Substitute
> >g02 x0.0 y-.5 I.5 J.5
>
> Again by putting both i,j in as positive values an incremental control
> assumes you want the arc center ahead and positive rather than just ahead.
> Emc discovers that it can not arc from the existing location to the
desired
> location using that arc center. An nc block that will create such a
radius is
>
> g02 x0 y0.5 i.5 j0
>
> (but not if you have tool radius compensation on.)
>
> >and the arc starts at the end of the x move at a 90 degree angle going up
> >and around clockwise and ends at the beginning of the y move which is
again
> >at a 90 degree angle.
> >Draw it out and you will see what I am talking about. Both g02 lines are
> >valid clockwise moves on either FANUC or AHHA and I thought FANUC was as
> >close to a standard as I could find.
> >Bridgeport and Allen Bradley both use the unsigned version but
unfortunately
> >both are obsolete.
> >What code would you use if you wanted the second shape?
> >Darrell
> ----- Original Message -----
> From: Jon Elson <jmelson@...>
> To: <CAD_CAM_EDM_DRO@onelist.com>
> Sent: Monday, February 07, 2000 10:33 PM
> Subject: Re: [CAD_CAM_EDM_DRO] Re: How does G02 & G03 work?
>
>
> > From: Jon Elson <jmelson@...>
> >
> >
> >
> > Darrell wrote:
> >
> > > Not wanting to cause any confusion but betweentwo points you could
> > > have two
> > > arc centers for any given I, J combination. One example would be two
> > > lines
> > > at 90 degrees to each other forming a square corner. Now filett that
> > > corner
> > > with a .5" radius. Assuming that the corner was at X0Y0 and the X and
> > > Y
> > > lines extended in the negative direction and the arc was clockwise
> > > from X to
> > > Y you would have an I.5 and J.5. EMC would cut what you want but a
> > > FANUC
> > > would use the I+.5 J+.5 to cut an arc of 270 degrees with both ends of
> the
> > > arc 90
> <snipped>
> I suppose the moral of the story here is that there is no one true
> R(ecommended) S(tandard) 274 and you'll just have to make sure of what
your
> own control or cam program assumes about the nature of i,j,k, and r before
> you try to run someone elses nc code.
>
> BTW I've been thinking about writing some processors for emc that will fix
> nc differences and can be run from the tkemc program editor. Some are
> already working on this for mastercam. I'd need clear descriptions of
what
> code is generated by commonly used cam/?? systems. Need bobcad and vector
> using the closest post. ahha and flashcut. Any others. If you have such
> writeups you could attach to me, rehenry@..., I'd really appreciate it.
>
> Ray
>
>
>
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