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Best fit alignment, MMC and positional tolerance.

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.
  • 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.
  • We are currently trying to get an answer from our customer about what feature -F- actually is, so no I'm not 100% sure. We have made the assumption that the design intent is to prevent eccentricity between the hole and R5.5, to keep wall thickness and prevent mismatch on final assembly. So we are going with 4x R5.5 until told otherwise.

    If -F- was the whole profile, then it would not constrain any further DOF after -A-.
  • If the teeth of the spline are constraining the rotation about "A", then anything you do with a best fit for the holes invalidates that rotation about "A" and the "F" is useless in that case as the teeth would control rotation about "A", and "A" controls the other 2 rotational DOF as well as 2 axial DOF (for a total of 5).
  • It's possible. Generally these drawings will say "Orientation of spline is important / unimportant to feature". This drawing doesn't say either.
  • 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.
  • Neil, I tried ACNO2D, this command seems to be used to compare actuals to nominals for 2D curves.

    It's not possible to add a diameter evaluation to POSITNXY so I added an evaluation just before.


    • #2.5
      commented
      Today, 02:50 PM
      Neil, I tried ACNO2D, this command seems to be used to compare actuals to nominals for 2D curves.

      It's not possible to add a diameter evaluation to POSITNXY so I added an evaluation just before.


    My mistake, I think you need BFIT2D, not ACNO2D (see attachment). For the POSITNXY ,you will need to report the diameter of the holes first.

    Attached Files
  • Neil,

    you need to use the evaluation symbol POSMMC, not POSITN, to take advance of MMC bonus
  • In PC-DMIS, you can use an iterative alignment to do this.

    However: a technique that works in ANY CMM software, regardless how primitive: make 4 different nominally equivalent alignments, each clocked to different hole (in Quindos, just use a DO loop for this). Now make 4 theoretical clocking points out in space (within same DO loop in Quindos). Now make a point cloud out of those theoretical points, construct a point at centroid of that cloud and clock to that. This splits the difference evenly.