[Er...] Posted February 22, 2023 Share Posted February 22, 2023 I'm curious how others would Define Datum B. in this situation. Thoughts? Link to comment Share on other sites More sharing options...
[Er...] Posted February 22, 2023 Share Posted February 22, 2023 Maybe... Symmetry of opposing cylinders-intersect at datum A-create a 3d line? Link to comment Share on other sites More sharing options...
[Ro...] Posted February 22, 2023 Share Posted February 22, 2023 Datum -B- would be a symmetry plane created from the outer tangents/apex of the outer cylinders. I would measure the outer cylinders as multiple level circles/curves and get the maximum outer points from each. Then I would create a plane for each side and recall the maximum points accordingly. Then I would create symmetry from those 2 planes. This will allow you to satisfy the perpendicular callouts. Link to comment Share on other sites More sharing options...
[Je...] Posted February 22, 2023 Share Posted February 22, 2023 Please sign in to view this quote. . Midplane of the two surfaces defined by the extension lines connected to the perpendicularity callout. . Link to comment Share on other sites More sharing options...
[To...] Posted February 23, 2023 Share Posted February 23, 2023 Please sign in to view this quote. On prismatic geometry, Maximum Point Construction will find the highest deviation point measured on the feature, which may not be at the outermost point in X. I think this would work if you scanned curves but I don't remember. I think I would create line segments at the nominal locations, (one for each circle path of each cylinder) of the B tangencies, recall the feature points from segments of the cylinders into the lines without modifying nominals. Constrain the lines to the normal vectors. Then, use Maximum Point Constructions to take highest point from each line to construct the planes. Note: I added Point1 feature (last slide) that was not necessary but it shows which point was created from the Max Point. If the part was thin, I would probably use circles. I'm curious if using a point instead of a line would work, i.e. recall points from cylinder segment or circle into a point defined at nominal apex point, considering we're only looking for the X value.Screenshot 2023-02-23 083333.jpgScreenshot 2023-02-23 083408.jpgScreenshot 2023-02-23 083630.jpg Link to comment Share on other sites More sharing options...
[Br...] Posted February 27, 2023 Share Posted February 27, 2023 Please sign in to view this quote. Couldn't you just recall all those curve points into a plane feature and use an outertangential evaluation? Link to comment Share on other sites More sharing options...
[To...] Posted February 27, 2023 Share Posted February 27, 2023 Please sign in to view this quote. When you take a view from 36,000 feet, things become more obvious. That's probably why you make the medium bucks...lol Link to comment Share on other sites More sharing options...
[Br...] Posted February 27, 2023 Share Posted February 27, 2023 Please sign in to view this quote. 🤣 Link to comment Share on other sites More sharing options...
[Er...] Posted February 27, 2023 Author Share Posted February 27, 2023 We ended up using outer tangential Cylinders with scans at 3 levels, and recalled them into a plane. Created them on each side and recalled them into a symmetry plane. This has proven to be very stable and repeatable. Thanks everyone! Link to comment Share on other sites More sharing options...
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