Composite Tolerance

I am having trouble understanding composite tolerance, especially when it comes to a bolt hole pattern like this. I am wondering how you would dimension these true positions. Here's my idea using legacy dimensioning:

Level and translate to A, rotate to B, and report the position of 1 hole (any hole?...lets say the 12'O'Clock hole) and give it a .001 tolerance. Then, would I stay leveled to A&B, translate to the hole that I just reported on (12'O'Clock hole), and report the positions of the rest of the holes with a .0005 tolerance?

Thanks so much...I prefer to use legacy...

NinjaBadger over 1 year ago in reply to kingsld1

That's not correct.

The upper segment controls the orientation and location of the considered features with respect to the datums.

The lower segment is only controlling the orientation of the considered features with respect to the datums (and each other)
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kingsld1 over 1 year ago in reply to kingsld1

Respectfully disagree. In my 40 plus years of GD&T, I've never seen a composite position callout that repeated all of the datums in the lower frame. If the upper and lower DRF's are exactly the same, then how does the pattern float? The lower frame always drops at least the last Datum from the upper frame.
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Mike Rivard over 1 year ago in reply to kingsld1

Yes you're right, you usually lose the end datum on a composite
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Neil Challinor over 1 year ago in reply to kingsld1

NinjaBadger is correct, it is perfectly valid for the lower segment's DRF to contain all three datums, it's just the way each segment behaves that changes. There is an example in ASME Y14.5 2018 - fig 10-43 (page 222)
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kingsld1 over 1 year ago

Found example in 2009 version. Page 138

However, assuming I'm reading this correctly, then Ninja Badgers post should read "For the lower segment in, use the above alignment, then just best fit, 3D ROTATION only to the 6 holes,"

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NinjaBadger over 1 year ago

Found example in 2009 version. Page 138 {"alt":"Click image for larger version Name:\ttempsnip.png Views:\t0 Size:\t322.3 KB ID:\t522150","data-align":"none","data-attachmentid":"522150","data-size":"small"}

However, assuming I'm reading this correctly, then Ninja Badgers post should read "For the lower segment in, use the above alignment, then just best fit, 3D ROTATION only to the 6 holes,"

No, for the lower segment the orientation of the DRF is constrained by Datums, hence translation only. Technically you could do a 3D translation but it will make no difference as we're dealing with cylindrical tolerance zones.
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kingsld1 over 1 year ago

The Y14.5 example posted specifically says that the lower segment controls rotation only. The Y14.5 example also says that the upper segment controls both rotation and translation. The only significant difference between the y14.5 example and the original post is the tertiary datum. I'm not following why translation would be required.
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NinjaBadger over 1 year ago

It says it's constrained in rotation with respect to the datums.

In the ASME example the top segment controls all 6 DoF. That is the orientation and location of the tolerance zones to the datums (as defined by the boxed dimensions) .

The lower segment only constrains the orientation.
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kingsld1 over 1 year ago

Orientation = Rotation ??
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Lake Monster over 1 year ago

this right?

to the book example:
the upper tolerance zone can't rotate or translate.
the lower tolerance zone can't rotate, but is allowed to translate (within the confines of the upper)
the actual feature can rotate and translate as long as it is within the confines of both tolerances

if the lower was to AB the smaller tolerance could rotate with respect to C (but needs to stay within confines of upper)
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