Table 4-3 Case 2.7 - Datum A is axis, Datum B is axis. No free transforms, all quantities invariant. This DRF is valid if (A NOT= B) AND ((NOT(A // B))).
This validity condition implies an intersection (by lack of parallelism) that arrests the last free transform seen when A // B.
The definition of skew lines is such that they are not parallel.
Assume A and B are skew. Then, they are not coincident and not parallel so by Case 2.7 the DRF formed by the two is valid.
However, two skew lines should not arrest a remaining free transform as there would be no point in the coordinate system that is defined singularly at a point that would serve as the arresting point.
How would we "handle" a DRF consisting of two skew datum axes? I believe the validity conditions of Case 2.8 should be extended.
There are a few other cases in table 4-3 that would beg this same question.
You said Y14.5 1994. right ? The 1994 Gd&t standards book ? Thats what I am looking at and your "table" and "case" are not making sense so I asked for the page #'s ?
You said Y14.5 1994. right ? The 1994 Gd&t standards book ? Thats what I am looking at and your "table" and "case" are not making sense so I asked for the page #'s ?