Datum Shift Explanation and Advantages

[kakalee1]

Hi engineers,

I have a question regarding about specifying datum feature at RMB. What are the advantages of having a datum shift since we are not getting any more or less bonus tolerance? Wouldn't it complicate the inspection process since the part can now move (shift) around?

There is an example from the book that I can't wrap my head around it, and I hope someone here can help me out. On the attachment, for option b and c, I don't understand what the y mean by "to ensure that datum precedence is not violated..." and how they choose the 0.4 and 0.2 for each option.

Any help is appreciated.

 

[Belanger]

First, just focus on your opening paragraph.... What are the advantages? One advantage is that the inspection process could simply use a physical, fixed-size, functional gage. Yes, the part can shift, but at least we can instantly see if the part passes or not!

Second: if the part in its intended function can shift around (at least before being subsequently bolted down), then our tolerancing should imitate that, so that we can get as much flexibility in the tolerance as possible.

You say there is no bonus tolerance, which is technically correct, but there could be a resulting "datum shift." Although this doesn't make the tolerance zone itself get larger, it does allow the zone(s) to move around, thus giving more tolerance in a way.

As for the picture you give, each case has this "datum shift," but depending on the datums referenced in the callout where there is an asterisk, we might have different values for the datum shift. For each grouping of datums referenced at the end of the feature control frame, we can only take the datum shift from the callout that shares those same datum references (minus the D at the end).
In other words, if the asterisked callout only references D, then the part is not held against A for that tolerance, therefore the two feature control frames already attached to D are not carried over to the asterisked callout. The numbers they give actually represent the fixed-gage that could be used, but that will affect how much looseness a real part feels when placed on that gage.

Perhaps a little tricky, I realize, but this is certainly one of the slightly more advanced topics in GD&T!

John-Paul Belanger
Certified Sr. GD&T Professional
Geometric Learning Systems

 

[CheckerHater]

Now, when it comes to second part of your question "Wouldn't it complicate the inspection process since the part can now move (shift) around?"

Belanger already explained that datum shift has advantages when functional gauge is used. With CMM - not so much. CMM systems (and CMM people) don't like the idea of datum feature able to move. So datum shift is often ignored even if it may result in theoretically good parts being rejected.

 

[drawoh]

kakalee1,

I think Dingy is the one who pointed out that if your datum is a feature of size at MMC, you can fabricate a fixture that will locate on it. If your datum is RFS, you are going to have to figure out where it is. RFS is convenient for CMMs. MMC is convenient to mechanical fixtures.

Never forget that your part has to work. Presumably, your FOS datum picks up a feature when your part is installed. At MMC, your part may fit exactly, allowing zero movement. Your bolt holes have to be positioned accurately enough that your bolts go through. If your feature is not at MMC, you have some wiggle room to position your sufficiently accurate bolt pattern over the mount holes.

If your part must be located precisely, an FOS is the wrong locating point, and the wrong datum.



--
JHG

 

[kakalee1]

To all,

Thanks so much for all the response. They are all really helpful.
The Q.A department at my company always checks the parts with CMM machines, and the part is always stationary during the inspection process. That's why I really didn't understand the idea behind datum shift.

I have a follow up question. Let's use the picture from my first post as example. Assume that I am making a functional gage for that piece to inspect the 3.5 dia hole. Now that the part is allow to shift around both B and D datum, how would I establish datum reference frame? Should I still use the datum reference frame established from the gage?
Since the part is allowed to shift, is it possible that it shifts too much that the 3.5 dia hole is out of the 0.3 diametrical tolerance?

 

[Belanger]

You said it right: The datum reference frame is established from the gage, which isn't moving. However, the part can shift on that gage; it can shift until it sits in the datum reference frame in such a way that it meets the tolerances tied to that datum reference frame.

If there is not any shift (looseness), then the part's tolerances appear to be tighter because you don't have the ability to shift it around.

John-Paul Belanger
Certified Sr. GD&T Professional
Geometric Learning Systems

 

[pmarc]

Since the part is allowed to shift, is it possible that it shifts too much that the 3.5 dia hole is out of the 0.3 diametrical tolerance?

Yes, it is possible, but as long as you find at least one position of the part in the gage in which axis of the 3.5 dia hole is within 0.3 dia position tolerance zone, the feature is good.

 

[3DDave]

Datum shift is intended to represent allowable movement relative to the mating part - so it doesn't matter if the part being inspected is held motionless or not.

It seems to me that the third case 'c' in the picture attached to the original post is incorrect.

The explanation below the figure fails to note that perpendicularity constrains the diametral allowance for position.

I think the shape of the datum D simulator is not a diameter, but an obround that is 7.3 X 7.5

I'm a torn on the concept of whether the datum B simulator is in exactly the same location for validating datum feature D or is allowed to find a new location. If the former, there should be definition that identifies such sub-FCF groups (in this case A}B(m)) as a simultaneous requirement. If the latter, then the obround length of the datum D simulator should include the amount of possible difference between the two conditions.

 

[greenimi]

[Yes, it is possible, but as long as you find at least one position of the part in the gage in which axis of the 3.5 dia hole is within 0.3 dia position tolerance zone, the feature is good.]

Pmarc,
Your latest comment *IS IT* or *IS NOT* in the same line of thinking of the candidate datum set defined by Y14.5.1 (math standard)?

Somewhere Mark Foster (Applied Geometrics) said:
"I would contend that your choice of datum planes is only one of an infinite number of candidate datum planes (to use Y14.5.1 terminology) and that you have a "rocker." In instances where we have candidate datum planes, the standard allows us to "optimize" and choose *the* datum plane that best suits our specification/application. I could have a mobile coordinate system for any reason, such as a datum feature of size referenced at MMB, and find an infinite number of candidate locations of the datum reference frame where the measurements would be out of spec. The point of optimization is to find the place that works, not to demonstrate the plethora of places that don't work."

 

[Belanger]

To 3DDave... I would gently disagree with a couple of your statements about that graphic. The perpendicularity makes datum feature D appear to be even better than the position tolerance (in terms of orientation). But the MMB of datum feature D must account for the worst case, which is the additional 0.4.

And the datum feature simulator would indeed be round. The apparent tolerance carved out within different scenarios may be a different shape over the span of many parts, but the key is to focus on the datums being referenced for each option (a) through (c). However, feel free to elaborate.

John-Paul Belanger
Certified Sr. GD&T Professional
Geometric Learning Systems

 

[3DDave]

In the A|B(m) frame of reference the D feature can be closer or farther from B based on the diametral tolerance allowed D.

Relative to A, it cannot tilt more than the perpendicularity tolerance allows.

It's very similar to the case where the [28] basic dimension is replaced with a directly toleranced dimension, but retaining the perpendicularity. The set of possible geometries is an obround.

 

[Belanger]

Then we agree that the datum feature simulator is to be round? (Notice that the figure is focusing on simply calculating the MMB, not the tolerance area created by possible shift.)

John-Paul Belanger
Certified Sr. GD&T Professional
Geometric Learning Systems

 

[3DDave]

The simulator is a slot.

 

[Belanger]

Hi again Dave...
First, you should realize that the picture is from the standard (page 61).
Second, we are talking about the datum feature simulator for datum D, and there are three options being discussed in that figure.
Third, the text right below the figure clearly states in each option that the datum's MMB is a diameter, so any datum feature simulator would be built to simulate that diameter. The datum feature simulator doesn't simulate the shape of the tolerance zone on the 3.5 mm hole, which is which I think you're trying to visualize.

It's all explained on the lower left side of the graphic given in the OP, under 4.11.6.1 (a), (b), and (c), as well as other paragraphs that relate to the MMB concept.

John-Paul Belanger
Certified Sr. GD&T Professional
Geometric Learning Systems

 

[3DDave]

That it shows a diameter is what is wrong. The correct simulator for that situation is not circular.

I am referring to the shape taken by the 7mm diameter when it is allowed the movement afforded in the same DRF as 'c'.

The suggested interpretation does not match the set of allowable places 'D' could get to. If one took all the possible parts that met the requirements applied to D and transformed them to no longer include C, then it is an obround. Picture the view as if standing at the center of B; would D ever appear to be more than it's MMC and perpendicularity tolerance wide? Without a reference to C, there is nothing that fixes the direction of the view and so nothing that causes adding a location tolerance to the width.

It looks to me like an overly complicated case that was incorrectly analyzed. As I've mentioned before, one of the greater failings of previous versions of the standard is that they only show individual parts and how to inspect/interpret them and don't show the effects of tolerance schemes on the relation of features on multiple parts. I presume the same case applies here, where no mating part is shown to clarify the need for the 'c' case.

 

[pmarc]

Pmarc,
Your latest comment *IS IT* or *IS NOT* in the same line of thinking of the candidate datum set defined by Y14.5.1 (math standard)?

Short answer is - yes, it is in the same line of thinking. From inspection point of view, at least one position of the part in the gage in which considered feature meets its tolerance is sufficient to say that the feature is good.

------
3DDave,

In my opinion the standard is correct in (c), although I would really like to understand why you think it is otherwise. Unfortunately I am afraid I am not following this:

The suggested interpretation does not match the set of allowable places 'D' could get to. If one took all the possible parts that met the requirements applied to D and transformed them to no longer include C, then it is an obround. Picture the view as if standing at the center of B; would D ever appear to be more than it's MMC and perpendicularity tolerance wide? Without a reference to C, there is nothing that fixes the direction of the view and so nothing that causes adding a location tolerance to the width.

Assuming I am picturing it correctly, I disagree that when standing at the center of B and looking at D, the D would appear *just* as a 7.3 width. Material of feature D would indeed occupy max 7.3 width, but the width would additionally be allowed to float 0.1 left and 0.1 right from true position centered at the datum axis B, thus producing 7.5 wide virtual condition. And since the true position of feature D is fixed at basic 28 from datum axis B, this 7.5 width would actual be dia 7.5 cylinder.

 

[3DDave]

The reason it cannot be seen to float to the left or right of the axis is that there is no direction in A|B(m) to measure that offset from. That's what the original |C(m)] provides - a single direction to view the possible locations of D, however that alignment is not part of inspecting the location of the 3.5mm hole.

For every possible compliant solution to locating D, draw a line from the center of B to the center of D where their axes intersect A. Then drop a vertical from the top-center of cylinder D to A. The maximum distance between these two points from D projected onto A is the radial amount of perpendicularity. The maximum allowable perpendicularity is dia 0.2. That is all that can be added to the MMC diameter in a direction that is perpendicular to the line between the center of B and the center of D along the surface A.

Along that same line, the intersection of the D axis with the line can be closer or farther away by the amount of the position tolerance and at MMC at the nearest and farthest from B the axis of D must be perfectly perpendicular to A.

As a contrast.

Suppose the original limitation on D did not include a reference to C(m). This would mean that D could be anywhere in a 360 degree location around B and still be acceptable. To make a mating part that would align with A|B(m)|C(m) and avoid touching D would require a circular slot in order to clear all the possible locations for D. The annular width of the slot is 7.5mm

However, if a mating part was to align with A|B(m)|D(m) it would only have to clear the radial variation and the tangential tilt of D. This is the case no matter where D ends up around the circle.

The length of the slot is the same as the width of the annulus and the width of the slot is the MMC + perpendicularity at MMC, hence an oblong of 7.5mm X 7.3mm.

 

[greenimi]

3DDave and pmarc,
Just curious: what you are talking about/debating has anything to do with "tertiary datum problem"? (solution could be adding a translation modifier)

Discussed here:

http://www.eng-tips.com/viewthread.cfm?qid=363878

Thank you

 

[3DDave]

greenimi,

It's an interesting overlap, but this case concerns the shape of the datum simulator while the tertiary datum problem focuses on the location of the simulator. I'd have to think more to see if they are simple transformations of one to the other.

In both cases, the lack of supporting practical applications suggests that what is presented in the standard is either an edge case or just a solution to a non-existent problem.

 

[powerhound]

So here's a quickie gage I drew to inspect the c scenario. Imagine the best fit as a removable pin. I wasn't going to create an assembly and all. This should get the idea across. Will this work? If not, why not?

John Acosta, GDTP S-0731
Engineering Technician
Inventor 2013
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