> On Mon, Sep 23, 2019 at 9:00 PM Jon Elson <el...@pico-systems.com> wrote: > If all arcs are exactly 90 degrees aligned to start on one > axis and end on the other, then they will look like that.
Ok, lets assume the origin of said orthogonal axes is the center of the desired arc or spiral. An arc from a start point on one axis will sweep through 90 degrees to a point on the other axis if and only if both the start and end points are the same distance from the origin. In this case, the center of such an arc written in G-code would have only an I or J coordinate and that coordinate would lie on the origin of those axis. An arc from a start point on one axis to an end point on another axis, where the distance from each point to the origin is different (as in a spiral approximation), will not sweep through 90 degrees if the center of the arc is written with only an I or J coordinate. Further, if the center only has an I or J coordinate relative to the start point, then the arc will not intersect the other axis at a 90 degree angle. As such, the arcs generated will not be tangent to the subsequent arc or any previous arc. However this lack of tangency is inherent to any method of trying to break up a spiral of constantly increasing radius using a finite number of arcs. I suspect the problem lies in the method for calculating CentDist on line 148. I'm not sure that method actually produces a I or J coordinate that is equidistant from both the start of the arc and end of the arc, regardless of decimal precision. On Mon, Sep 23, 2019 at 9:00 PM Jon Elson <el...@pico-systems.com> wrote: > On 09/23/2019 10:29 AM, Chris Kelley wrote: > > I plotted out that snippet of code in AutoCAD and the coded arcs don't > line > > up with the end points. > > > > I'm not great with C but looking through Jon's code I wonder if it has > > something to do with how it will never output an arc with both I and J > > coordinates. > If all arcs are exactly 90 degrees aligned to start on one > axis and end on the other, then they will look like that. > > I have no idea why Mach3 will run it, perhaps Mach has a much wider > > tolerance for arc-center-end-point mismatch or it is just converting the > > tiny arc moves to lines. > > > > > Mach does have relaxed tolerances on the arcs. > > Jon > > > _______________________________________________ > Emc-users mailing list > Emc-users@lists.sourceforge.net > https://lists.sourceforge.net/lists/listinfo/emc-users > _______________________________________________ Emc-users mailing list Emc-users@lists.sourceforge.net https://lists.sourceforge.net/lists/listinfo/emc-users
