[Ja...] Posted November 26, 2019 Share Posted November 26, 2019 Calypso can only compute datum shift/displacement in the case of features in the same workplane. Let's say I have a cylinder with normal axis on Z+ and a primary datum cylinder referenced at MMB with normal axis on X+. I do not get granted datum mobility. What if I create a theoretical cylinder about X+ with the actual data pulled from the Z+ cylinder, perform a best fit to the datum cylinder using best fit bore pattern with MMB of references, create an alignment from the best fit, and report the position of the Z+ cylinder back to that alignment? Is this "correct"? I can't seem to justify it either way. Link to comment Share on other sites More sharing options...
[Br...] Posted November 26, 2019 Share Posted November 26, 2019 If you pull data from cylinder that is really perpendicular to a cylinder you want to load it into, you are going to get some jacked up results. I'm not sure how you would do this anyway. I'm assuming this is a position characteristic with an MMB modifier. The toleranced feature is located to the DRF by basic dimensions. Its from this DRF that it is allowed to shift. I just don't see how you could recreate it in a completely different orientation and get a legitimate result. You might be able to do a geometry best fit with these features allowing only the degrees of freedom to best fit that the datum with MMB applied to it can constrain. Then you could find out the total amount it shifted from the datum feature, then decide by looking at the size of the datum feature, if that amount of shift that it moved would be acceptable or not. Link to comment Share on other sites More sharing options...
[Ja...] Posted November 26, 2019 Author Share Posted November 26, 2019 Brett, The reason your results will be off when pulling data is because, for example, how do you define the positional component in X of a cylinder whose normal axis lies along X? You do not, as it is infinite. However, a cylinder normal along Z will have a defined X component. To take that same X component and use it to define the X component of an Xnormal cylinder is funny, but again the component is not going to be considered going forward. It's just like if you define a cylinder along Z, Calypso will still assign it a "Z value". Is that the one particular location in Z that you will find that cylinder? Not at all. Assume a cylinder with axis along Z. Primary cylindrical datum (@MMB) with axis along X. Translational datum shift does not make sense along X (datum not defined along X) nor does it make sense along Z (feature not defined along Z). Translational datum shift does make sense along Y, but Calypso cannot compute this. Assume we ignore rotation and everything is dead straight. If the datum is .005 from MMC size, then we have .0025 potential motion in either of the two directions in which it can move, in this case we'd like to see Y. We create a theoretical cylinder from the feature of interest, but we force the axis along X. So the x component of position is wonky, but will not matter. Actually, same for Z. We captured a cylinder with the correct Y location. Now, we can do a best fit of bore pattern with primary datum being allowed to translate (in Y only). We then make an alignment from that bore pattern origin (which has shifted the allowed amount) and then reference the original feature in question back to that alignment, after the shift has been performed. It's a funny process, but an attempt to circumvent basic limitations in Calypso in simple translational cases to one datum only. Does this explanation make the process any more clear, or exaggerate any errors in thought? Link to comment Share on other sites More sharing options...
[Br...] Posted November 26, 2019 Share Posted November 26, 2019 Please sign in to view this quote. What I meant was if you Recall feature points into a feature that was measured perpendicular to the feature you are loading the data into, the sigma values will be through the roof and it will not be useful at all. I haven't tested it, but I cant see how this could work. Please sign in to view this quote. Maybe. If you are just going to create a theoretical feature to represent the datum feature projected onto the same work-plane as the toleranced feature whereby you use some formulas to fill in the actuals in X,Y, and the diameter. I think that might get you close. But if you try to recall feature points, it just wont work. Link to comment Share on other sites More sharing options...
[An...] Posted November 27, 2019 Share Posted November 27, 2019 See attached.Contribution_27_11_2019_b.pdf Link to comment Share on other sites More sharing options...
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