[To...] Posted April 8, 2021 Share Posted April 8, 2021 Looking for suggestions on how to do a base alignment or secondary alignments for this part. The picture shows an arc segment part. Datum A is the top Plane. Datum B is called out in the notes "DATUM B IS CONSTRUCTED BY CREATING A CYLINDER OF RADIUS R10.000 (BASIC) OF PERFECT FORM CONTACTING DATUM TARGETS B1/B2." The 2 datum targets are shown on the sketch and are controlled by basic dimensions equally spaced from Datum C which is the small hole in the center.Lat.jpg Link to comment Share on other sites More sharing options...
[Jo...] Posted April 8, 2021 Share Posted April 8, 2021 If you made a Theo circle centered at Datum C location, large enough diameter, the intersection with the Radius at B1 and B2 would produce a rotation always perpendicular and Bisecting. Link to comment Share on other sites More sharing options...
[Da...] Posted April 8, 2021 Share Posted April 8, 2021 That is an interesting problem. My base/start alignment I would probably do: Spatial: Datum A Planar: Symmetry between the end planes X-Zero: Datum C y-Zero: Datum C Z Zero: Datum A If the ends are rough sawn, I would maybe use a line from datum C to the center of the outer arc to do the planar rotation. Either way, I would use that base alignment to measure datum B. I would try measuring it as two small clusters of vertical scans at the basic dimensions, the width of the datum targets, and constrain the cylinder size. I can't say I have ever tried that though. Either way, I would then recall datum B and C into a line and use that as a planar rotation in a secondary alignment. Then loop until it repeats and the distances from the datum targets to C is correct. My instinct is that unless your form is really good, it is going to be a real bear, and require a lot of looping. Link to comment Share on other sites More sharing options...
[Mi...] Posted April 8, 2021 Share Posted April 8, 2021 Is the nominal outer radius of the part R10.000 also? Link to comment Share on other sites More sharing options...
[To...] Posted April 8, 2021 Author Share Posted April 8, 2021 Please sign in to view this quote. yes Link to comment Share on other sites More sharing options...
[Jo...] Posted April 8, 2021 Share Posted April 8, 2021 Maybe I need to expound on my method. 1. Level with Datum A 2. Measure location of Datum C 3. Measure the R10.00 radius constrain the radius 4. Create a Theo circle at Datum C with the appropriate fixed radius to intersect the R10.00 at B1 and B2 5. Construct line from B1 to B2 for rotation. Align A,B,C but it will execute in the order shown because of dependencies. Loop Link to comment Share on other sites More sharing options...
[Mi...] Posted April 8, 2021 Share Posted April 8, 2021 Please sign in to view this quote. I'm not sure, but I would start with what John said and make a theoretical circle concentric to C. Figure out the diameter that would give you the basic distance you're given. I used 5 in the example and got 5.531, but some trigonometry or CAD should get you the actual number. Intersect this with the actual OD and report the actual distance. This won't come out to your basic distance (unless the part is exactly at nominal), so you'll need to make some sort of loop that iterates the size of the theoretical circle. That part I'm not sure how to do - maybe you or someone else here can figure it out. Once you get to intersections the correct distance apart, I'd make new alignment. X- and Y-origin at a symmetry point between the two intersections. Datum C for planar rotation. The distance from this alignment to datum B can be calculated from the basic dimensions you have. Again, I assumed 5" and got 9.682, but law of cosines/CAD should get you the actual number. Then just make a theoretical point offset this distance from your new alignment, and that's BCapture1.PNGCapture2.PNG Link to comment Share on other sites More sharing options...
[Da...] Posted April 8, 2021 Share Posted April 8, 2021 Since the basic lengths are equally spaced from the circle -C-, Then circle -C- is essentially datum -B- . You have basic dimensions in both directions ( diam and equi distant from -C- ). I am assuming that the centerline through -C- is also the centerline of the 10 mm. ? Link to comment Share on other sites More sharing options...
[Mi...] Posted April 8, 2021 Share Posted April 8, 2021 Please sign in to view this quote. The distance from -B- to -C- depends on the actual outer radius, not just the basic dimensions. If there's a position with respect to -B-, he needs -B-. Link to comment Share on other sites More sharing options...
[An...] Posted April 8, 2021 Share Posted April 8, 2021 Please sign in to view this quote. Bump! Link to comment Share on other sites More sharing options...
[Jo...] Posted April 8, 2021 Share Posted April 8, 2021 Please sign in to view this quote. Tom, what features are using the "basic" dimensions ? or is this a profile requirement ? Link to comment Share on other sites More sharing options...
[Mi...] Posted April 8, 2021 Share Posted April 8, 2021 I found the trig solution - no looping needed. The radius of the theoretical circle "x" can just be calculated from the actual outer radius, actual distance to -C- and basic distance between points: x=sqrt(B^2+R^2-2*B*R*cos(0.5*acos(1-A^2/(2*R^2))))Capture.PNG Link to comment Share on other sites More sharing options...
[To...] Posted April 8, 2021 Author Share Posted April 8, 2021 Please sign in to view this quote. Ironically, there are only a couple of positions to ABC and a surface profile A and a couple of surface profiles with no DRF. Seems like there could have been a bunch more profiles but they used standard dimensioning for 80% of the part. smh. really smh. Link to comment Share on other sites More sharing options...
[Jo...] Posted April 8, 2021 Share Posted April 8, 2021 Please sign in to view this quote. I have one part in particular that has basic dimensions to a feature that are completely negated by the DRF . It is apparent what they intended to control but failed in their execution. Link to comment Share on other sites More sharing options...
[Da...] Posted April 8, 2021 Share Posted April 8, 2021 Measure the 20.0 Circle. Assign an evaluation, ote, lsq, whatever.... Calypso reports one diameter size, of a perfect circle, with only one location. Let's say it's 20.25, ok ? Now measure the small circle...same thing right ? One diameter, one location . Now let's create a line from circle ones location to circle 2's location. Now the criteria for location B1 is that it is at 10(basic AND inside the part) radius and some other value from the line we created. B2 is identical except the opposite direction, but must be the same distance. What was just described was a chord. Since the cord must be bisected by the line we created it is perpendicular to that line. You can change the distance from the line to whatever you want, but it still perpendicular. Since the radius is basic for the location to both B locations, and it's being used to rotate the alignment ONLY, then the line B1,B2 is irrelevant. It is simply the perpendicular to the line we created. That being said, there must be some other measurement or angle called out that you are supposed to be reporting . That should be a piece of cake from there on out. Link to comment Share on other sites More sharing options...
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