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Alignment with Datum Targets and Geometry


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I have a cylindrical part, laying along the X axis. The drawing calls for 4 Datum A target points, 2 at each end. Imagine the part sitting in 2 90° v-blocks. Datum B is the end face. And Datum C is a flat in the Z plane on the top.

I created a Geometry Best Fit of the 4 Datum A target points, with full translation and rotation. Next, I created a normal secondary alignment. I changed the Alignment reference to the previous GBF alignment and added Datum B plane for the X origin and the Datum C line for the planar rotation. Is this correct?

Screenshot 2020-11-09 090438.jpgScreenshot 2020-11-09 085412.jpgScreenshot 2020-11-09 085436.jpg

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I would fit cylinder to 4 points with locked radius. Drawing should contain base dimension for that. But i don't know if 4 points will not have issues with cylinder.

Or you can make 2 circles with locked radius and construct 3d line from them.
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I can't understand the logic of adding datum target to this like simple geometry.
Datum targets are added for complicated contour or geometry that can't be work with standard practice.
I think that four points for designing Fixture.
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I can see the dilemma. Datum A is supposed to constrain Y,Z location, but with 4 points arranged like that, how can it? The diameter/center point of the part could be anything and still fit on those 4 points. Maybe you could forget it's a cylinder and think of it as an irregular shape but that doesn't change the inability to establish Y,Z nominal. I maybe wrong, I don't know.
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Absolutely. When I looked at Y14.5, it said 2 sets of 3 equally spaced points at each end of the cylinder should be used.
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Check out paragraph 4.24.10 Also, datum targets on standard geometry could be used to identify where measure or fixture the part.
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If the Datum targets represent the contact (tangent) points of the part resting in v-blocks, why not Construct
a theoretical cylinder from those four points and use that for the spatial rotation in your alignment?

Are the Y & Z coordinates for the datum points .4808? I want to try something out.
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Create the 4 nominal datum points at .4808 Z & X (whatever in Y) and construct two
symmetries. Construct a 3d line from the symmetries = X+ (in your example) Spatial
or Planar Rotation?
150_e000be4961425d73b380fee2980f6502.png
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Not to sound ungrateful but nobody answered my original question about how one might create an alignment using datum targets and geometry.

Let's forget about 4 target datums on a cylinder and make it a multi-level plane-like surface with 4 datum targets at different heights.
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I don't know if I'd let it tumble is all 6 axis. I would suggest only having Translation in the Y, Z, and Rotation in the Y, Z active.

Have you tested it to see if the results you are getting make sense?

I've just started using Geometry Best Fit, and I actually have a lot of questions myself regarding it.

Typically the experience I have had with a 3d Best Fit though would be in the Base Alignment, and looped at least 3-5 times. Your case is unique though.
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