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A-B alignment using best fit question

oddam
Good morning, All,
I have a question concerning the use of an A-B print alignment. First, I am using pcdmis 2022.1, cad and print, (cannot post a section of the print, not authorized, but I did the alignment). I am measuring a tube in automotive XYZ, datum A is in the X axis, datum B is in the Z. I take four vector points for each datum, use the best fit 3D, vector min max, rotate & translate, highlight the features and click compute. This is my manual alignment, points fall on the cad, all looks good. I then copy the manual alignment features and paste it after the alignment, change the id's and input the dcc and moves needed. I then run the dcc features and it measures the part, features are on the cad, I make a new BF alignment just like the manual and dimension the dcc features using the t value, tolerance ± .05mm and the results report less than .06mm. For a non-machined tube and using four points, I think this is good, measure the other features and they all seem reasonable for deviation. I then looked closely at the cad; it looks like the features (all) are minorly shifted in the Y axis (different deviations, not all the same). This is my question, what locks the alignment in the Y axis on a BF 3d? I am questioning myself to make sure I am not making a mistake. If you see an error, please let me know, or you have another way to align, I am open to try. Thanks in advance.

Odda
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  • Best fit just does what it say's on the tin.. Try's to fit your points as close to nominal as it can it does not lock any one feature over another. You see where it say's Weight with a number 1 you can change the number to give a feature priority over another. I only use best fit with extreme reluctance you are better off working out a way to do a basic alignment locking down the 6 degrees of freedom.
  • Your setting all seem like they should work OK. If you are having problems with locking down one of the translational degrees of freedom, my guess is that you need to include more points along that vector. Without seeing the part, I'd say you need at least 6 points on datum A (like a cylinder) to constrain 4 degrees of freedom. If Datum A is a plane, then Datum B needs at least 6 points.

    Another possibility is that those 2 datums alone just don't work to constrain all 6 DOF.
  • in reply to Cris_C
    Both A & B are planes. I agree with your last statement, the two datums alone just don't work. After a review with management, we came to a solution of using the center line of the tube for the B and C datums, leveled to the A, this should work, especially since the centerline of the tube is the main feature. Thanks for the input, it all helps.
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