Your Products have been synced, click here to refresh
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.
I cannot find an answer in those sections. It says they must be basically oriented. My question pertains to skew axes. How would we then define basically oriented?
Are you talking about the point in "2.7"?
My intent was more of a mathematical curiousity.
I didn't mean 2.8, I meant 2.7.
I saw your post and will be waiting for the new standard.
© 2024 Hexagon AB and/or its subsidiaries. | Privacy Policy | Cloud Services Agreement |