hexagon logo
Register Sign in
  • Replies 9 replies
  • Subscribers 15 subscribers
  • Views 263 views
  • Users 0 members are here
Options
Related

Best fit alignment, MMC and positional tolerance.

WM321
Hello.

I'm trying to tolerance the 4x Ø4.49 holes on this part to datums [A][F], and have the following questions.



1. How do I create a best fit rotation as required for datum -F- (through the 4x R5.5)?

I have went ahead and completed the rest of the program and used a rotation between the left and right R5.5 in place of the best fit, and I am having issues with the reporting of the position and MMC.

2. How do I properly add MMB to the positional tolerance?

I have used the MMCADD evaluation within POSITNXY, and evaluated the diameter directly before this. See evaluations suffixed _LEFT, MMCADD should say 0.094, not 0.090, and it is not affecting the positional tolerance calculation. Where am I going wrong?





Thanks.
Parents
  • Are you 100% sure about the "F" datum being all 4 of the R5.5? I think if it was all , then the "F" callout would be WITH the 5.5TYP callout.
    Are you 100% sure that the "F" datum IS the R5.5 and not the entire outer profile?
    IF the "F" is NOT all of the R5.5 and is only the one it is pointing to, then there is no best-fit for the holes as the center bore and the single R5.5 make a rotation line for the alignment.
  • in reply to Matthew D. Hoedeman
    A is the spline and constrains two rotational and two translational degrees of freedom. F is the entire outer surface profile and constrains the final rotational degree of freedom (around A). There is nothing fixing the final translational degree of freedom (along the axis of A) - unless F is tapered. It's been a while since I've used Quindos but from memory, you would have to create an alignment which sets datum A as the spatial control (Z+) and origins X & Y. You would then need to scan around F at multiple depths and perform a 2D, rotation only best-fit ( ACNO2D?). The position calls for MMC on the considered features, which you appear to have added correctly to the POSITNXY command. However, if I'm remembering correctly, I think the holes that are being reported need to have their diameter & it's associated tolerance included in the POSITNXY output - unless it has been output previously - in order for Quindos to be able to calculate the bonus.
Reply
  • in reply to Matthew D. Hoedeman
    A is the spline and constrains two rotational and two translational degrees of freedom. F is the entire outer surface profile and constrains the final rotational degree of freedom (around A). There is nothing fixing the final translational degree of freedom (along the axis of A) - unless F is tapered. It's been a while since I've used Quindos but from memory, you would have to create an alignment which sets datum A as the spatial control (Z+) and origins X & Y. You would then need to scan around F at multiple depths and perform a 2D, rotation only best-fit ( ACNO2D?). The position calls for MMC on the considered features, which you appear to have added correctly to the POSITNXY command. However, if I'm remembering correctly, I think the holes that are being reported need to have their diameter & it's associated tolerance included in the POSITNXY output - unless it has been output previously - in order for Quindos to be able to calculate the bonus.
Children
No Data
© 2024 Hexagon AB and/or its subsidiaries. Privacy Policy | Cloud Services Agreement