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Y14.5.1-1994

I have a question regarding Table 4-3 in Y14.5.1-1994.

If anyone else is familiar with the standard and wants to discuss or knows how I can get in touch with somebody I would appreciate it.
Parents
  • 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 think the answer is in fairly clear in sections 4.3.3, 4.3.4, & 4.3.5 on pages 15 & 16.

    Mr. Smartypants says there certainly IS A POINT in Case 2.7. Neutral face

    I suspect closer to your intent is an assertion that you can't imagine a scenario where such a datum scheme would be best practice or a logical representation of fit and function. None of which makes it mathematically invalid.

    I do not understand what you mean with regards to Case 2.8. Extended how? Why?



    Anywho, since this standard is rarely discussed and even less frequently understood, I will point out that I have written a bit about the next revision which is nearing publication. You can find that by looking for my posts about the HxGN Live Goonapoolooza that took place in SinCity a couple of months ago.

    HTH & ymmv


Reply
  • 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 think the answer is in fairly clear in sections 4.3.3, 4.3.4, & 4.3.5 on pages 15 & 16.

    Mr. Smartypants says there certainly IS A POINT in Case 2.7. Neutral face

    I suspect closer to your intent is an assertion that you can't imagine a scenario where such a datum scheme would be best practice or a logical representation of fit and function. None of which makes it mathematically invalid.

    I do not understand what you mean with regards to Case 2.8. Extended how? Why?



    Anywho, since this standard is rarely discussed and even less frequently understood, I will point out that I have written a bit about the next revision which is nearing publication. You can find that by looking for my posts about the HxGN Live Goonapoolooza that took place in SinCity a couple of months ago.

    HTH & ymmv


Children
  • 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.