What am i doing wrong???
Datum reference error
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What are the datums A, B and C and their relations to each other? The error message explains it quite well, I think  "Secondary is neither parallel nor perpendicular to primary, so tertiary is not allowed". It could be an error of the print, or it could be an error in your interpretation of the datums.AndersI
SW support  Hexagon Metrology Nordic AB
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Measure vectorpoints at the locations in your screenshot.
Use iterative alignment and use the points for the alignment.PCDMIS CAD++ 2o19 R1 SP11
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This is a common issue with most of our MBD parts and below is my solution.
For this example lets assume that the Datum A points have a Z+ish vector, Datum B points have a Y+ish vector, and Datum C point(s) have a X+ish vector and lets assume you are preprogramming offline.
After my DCC alignment, I retake all my datum target points then I construct a point for every datum target point (except Datum C if there is only 1 Datum C target point). Your offset will be opposite the nominal of the axis you are concerned about so lets look at the 3 Datum A points. Say Datum A is used to level the Z axis and lets say that in the Z axis A1=1.0000, A2=1.050, and A3=1.010. My offset points for those 3 Datum points in the Z axis will be +1.0000, +1.050, and +1.010 respectively. The Z nominal for those constructed points, if done correctly, should be 0 as should be the constructed plane and the plane should have a 0,0,1 vector. The individual constructed points will retain the vector of the points they are dependent on.
Now do the same thing for the Datum B points in the Y axis and construct a line if there are 2 points and a plane if their are 3 points. The Datum C point I leave alone because I have not run into a part yet that has more than 1 Datum C point so no offset point is needed.
So sum it up quickly, you are going to construct a plane with a Z 0 / 0,0,1 vector nominal and do the same for Y using Datum B. A quick note on the constructed line for Datum B, I don't worry about offsetting the points in Z. You wont have a perfect 1 (or1),0,0 vector but that doesn't matter as long as the nominal J vector is 0.
So now you have a plane and a line that are perpendicular to each other while still being dependent on the actual datum target points so if the part varies from the nominal at those points, it will be reflected in your GD&T dimensions. Now to get your deviations in relation to the part alignment instead of the alignment Xact measure will construct based on your constructed datums, select "use current alignment" in the dropdown in the edit window for the GD&T dimension.Remembering my beautiful wife Taz who's life was taken from me on 6132020. I love you and I miss you.
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AndersI How do you mean? Offset points from the origin?Last edited by Amachineinsp; 09062019, 09:57 AM.
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