Can we apply datum shift on datum feature itself?

To get to the extreme of this, it is possible for the features that are used as datum features to be such that there is no allowable datum shift available to FCFs that reference those datum features. Typical analysis is more interested in the maximum allowable datum shift but that is not guaranteed to occur in actual parts.

Datum shift as a term can only apply to the datum references in an FCF. It cannot apply to the feature itself - that's called feature tolerance.

SeasonLee,

Datum shift applies to the position tolerance for the 4X hole pattern. The datum feature references do not constrain all 6 degrees of freedom (A constrains u,v,z but w,x,y are unconstrained). In the paper gage evaluation, the part can be adjusted in w,x, and y to get the best result - this adjustment is datum shift. Y14.5 tends to describe datum shift in the context of MMB datum features, but the concept applies to open degrees of freedom as well.

The textbook says that the 4 holes fall within 0.6 tolerance zones after paper gage evaluation. This is true, but we can do better than that (it's hard to optimize down to the micron using paper gaging). The zones can fit in 0.537484 zones just from shifting in the x and y directions (at least this is what I get).

Can someone calculate the smallest zones that the 4 holes will fit in, if w rotation is applied as well?

Evan Janeshewski

Axymetrix Quality Engineering Inc.

axym said:

The zones can fit in 0.537484 zones just from shifting in the x and y directions (at least this is what I get).

Click to expand...

Evan,

May I ask you how did you calculate that number? As usually, I am trying to learn something from each and every discussion.

Did you use a software or you did it manually?

Thank you

I would tend to agree with 3DDave's analysis that this would fall within the realm of that feature's allowable MMC tolerance zone, not datum shift.

Evan,
Is that really datum shift or just as 3DDave noted just feature tolerance? The initial "rejection" based on each feature's "actual position" is based on location from the axis of the top left hole #1 as if it were a datum feature from which the rest of the holes are referenced, this isn't the case.

Consider the case if holes #1-#3 came such that they were coincident with their MMB boundary and constrained x/y translation and w rotation between those three holes - hole #4 could still independently shift within its own position/size tolerance while any features referenced by that datum feature pattern would not benefit from datum shift.

SeasonLee,

I used Excel to do the calculations.

chez311,

This is where I try to look at one thing at a time, and not be tempted to mix several things together or it gets too confusing.

Step 1 - evaluating the position tolerance on the 4X holes. The datum reference frame |A| is not fully constrained. So there is datum shift available when we're trying to get the 4 axes within the 4 tolerance zones. The shift is unlimited because the x,y, and w degrees of freedom are completely unconstrained. We can apply whatever shift gives us the best result for the 4 axes and their 4 zones.

Step 2 - evaluating the profile tolerances. The datum reference frame |A|B(M)| is not fully constrained either, and there is datum shift available. But the shift that we apply in this step has nothing to do with the shift that was applied in Step 1. It's a different FCF. This time, the shift is limited because the x, y, and w degrees of freedom are partly constrained (the 4 holes would have to fit over 4 virtual condition simulators). As 3DDave pointed out, there may be datum shift or there may be none - it depends on the condition of the as-produced holes. We can apply whatever shift gives us the best result for the 2 profile zones, within the constraint that the 4 holes have to fit over their 4 simulators. This shift is most likely different than the shift from Step 1.


Evan Janeshewski

Axymetrix Quality Engineering Inc.

I wonder what the datum shift 'lens' looks like. That is, for each amount of shift there is a corresponding allowable rotation positive and negative. Put onto a graph of delta-x vs delta-y, the amount of rotation positive and negative would create a closed surface above and below the x-y plane.

Four example, if the four holes in the datum feature were displaced radially outward from the nominal center to their virtual condition limits there would be no allowable shift or rotation so the lens would be a single point with no internal volume.

The largest lens should be when the holes are perfectly placed and at LMC.

****

The minimum solution is a circle that passes through 3 of the points and encompasses the fourth point; so it's just a circle fit. I have a manual spreadsheet that I'm playing with that has gotten close, but I'm too lazy to do a circle-fit this minute. Rotation of the point set makes no difference to the solution. I see that rotation of the original point set before conversion to paper-gage can make a difference; however, rotation of the paper-gage deltas does not make a difference.

Evan,

I get what you're driving at but isn't that sort of a loose (or at least very broad) definition of datum shift? Typically the way I think of it datum shift means that the boundary/material condition of the datum feature(s) allows them to shift in relation to the theoretical datum(s) and their simulator(s) - in this case the datum feature A doesn't shift in relation to the theoretical datum A and its simulator, it always makes full allowable (3 point) contact. Whats actually shifting* is the features in relation to their theoretically perfect true position.

I don't mean to be pedantic or fixate on semantics, it just seems to me that shift due to a datum feature's boundary/material condition is fundamentally different than shift due solely to unconstrained DOF dictated by a datum feature's inherent geometry.

*Edit: I want to clarify that I am of course talking about your Step 1 here. I fully agree your Step 2 is a textbook case of datum shift.

chez311,

It is a more broad definition, yes. Perhaps we need a more general term than "datum shift" (or datum feature shift/displacement, as Y14.5 uses). I don't think you're being pedantic - these are important details. But I would say that the shift from open degrees of freedom is fundamentally the same thing as shift from MMB datum features (and also the same as shift from a rocking/unstable datum feature). The only difference is that there are no constraints on the shifting when the DOF's are fully open.

Some interesting issues come up, that become almost metaphysical or philosophical. It would appear that datum feature A doesn't shift in relation to its simulator, but it does. It's just that the shifts in the x, y and w directions don't affect anything in this case - there is no absolute reference for these shifts. It's kind of like simulator A has infinite extent - like a gigantic surface plate with no edges. Now imagine that there were two 4-hole patterns, each with a position tolerance that references only datum feature A. Simultaneous requirements would apply. The shifts for each 4-hole pattern still have no absolute reference, but they both have to be the same. In other words, there is no specific place on the surface plate that the part needs to be put, but it must be at the same place for both patterns.

Evan Janeshewski

Axymetrix Quality Engineering Inc.

axym said:

But I would say that the shift from open degrees of freedom is fundamentally the same thing as shift from MMB datum features (and also the same as shift from a rocking/unstable datum feature). The only difference is that there are no constraints on the shifting when the DOF's are fully open.

Click to expand...

I agree that they are fundamentally similar in that they both allow movement but the lack of a an "absolute reference" as you noted or finite boundary would be for me where they diverge. If we apply your broad definition, a cylindrical MMB datum feature which comes in at its MMC size could constrain all available shift in x and y but since it still allows rotation about its axis would still allow "datum shift". So would the same feature specified at RMB. Unless all DOF are constrained in a particular DRF then per that definition datum shift would apply - this seems at the very least counter-intuitive.

axym said:

It would appear that datum feature A doesn't shift in relation to its simulator, but it does. It's just that the shifts in the x, y and w directions don't affect anything in this case - there is no absolute reference for these shifts.

Click to expand...

When I was referring to shift relative to A, I meant it in the terms of the fact that the datum feature is required to make maximum contact with its simulator - this is not the case with MMB datum features which is where the shift originates from. In fact one could easily imagine a situation where an MMB datum feature makes zero contact with its simulator.

Axym,
For everyones benefit, PLEASE, could you post your Excel calculations? Thank you very much.

Boredom drives me. Here are the macros I wrote:

Function circle_y(x1 As Double, x2 As Double, y2 As Double)
'Debug.Print "x1, x2, y2 = "; x1, x2, y2
circle_y = ((-x2 / y2) * x1 / 2) + ((y2 - ((-x2 / y2) * x2)) / 2)
'Debug.Print "circle_y = "; circle_y
End Function

Function distance(x1 As Double, y1 As Double, x2 As Double, y2 As Double) As Double
distance = Sqr((x2 - x1) ^ 2 + (y2 - y1) ^ 2)
End Function

Function radius(x0 As Double, y0 As Double, x1 As Double, y1 As Double, x2 As Double, y2 As Double)
Dim radius_pt1 As Double
Dim rotation_pt1 As Double
Dim theta_pt2 As Double
Dim radius_pt2 As Double
Dim newx2 As Double
Dim newy2 As Double
Debug.Print "********************************"
Debug.Print "dist 0-1 = "; distance(x0, y0, x1, y1)
Debug.Print "dist 1-2 = "; distance(x1, y1, x2, y2)
Debug.Print "dist 2-3 = "; distance(x2, y2, x0, y0)
x1 = x1 - x0
y1 = y1 - y0

x2 = x2 - x0
y2 = y2 - y0

rotation_pt1 = WorksheetFunction.Atan2(x1, y1)
Debug.Print "rotation = "; WorksheetFunction.Degrees(rotation_pt1)

radius_pt1 = distance(0, 0, x1, y1)
Debug.Print "radius_pt1 = "; radius_pt1

theta_pt2 = WorksheetFunction.Atan2(x2, y2)
radius_pt2 = distance(0, 0, x2, y2)
Debug.Print "theta_pt2 = "; WorksheetFunction.Degrees(theta_pt2)

newx2 = radius_pt2 * Cos(theta_pt2 - rotation_pt1)
newy2 = radius_pt2 * Sin(theta_pt2 - rotation_pt1)

Debug.Print "newx2, newy2 = "; newx2, newy2
radius = distance(0, 0, radius_pt1 / 2, circle_y(radius_pt1, newx2, newy2))

Debug.Print "circle_y = "; circle_y(radius_pt1, newx2, newy2)

Debug.Print "dist 0-1 = "; distance(0, 0, radius_pt1, 0)
Debug.Print "dist 1-2 = "; distance(radius_pt1, 0, newx2, newy2)
Debug.Print "dist 2-3 = "; distance(0, 0, newx2, newy2)
Debug.Print ""
Debug.Print "dist c-0 = "; distance(radius_pt1 / 2, circle_y(radius_pt1, newx2, newy2), 0, 0)
Debug.Print "dist c-1 = "; distance(radius_pt1 / 2, circle_y(radius_pt1, newx2, newy2), radius_pt1, 0)
Debug.Print "dist c-2 = "; distance(radius_pt1 / 2, circle_y(radius_pt1, newx2, newy2), newx2, newy2)

End Function

The main function is radius(), which takes 3 points as coordinate pairs. It translates the second and third points by the offset to the first so the first point is effectively at the origin. The second step is to take the angle and radius to the second point. It takes the distance and angle to the third point and uses the angle to the second point to rotate it to a new location.

The result is that one point is at the origin, the second point is on the x-axis, and the third point is moved to where it should be. Since the center of the circle will be on the vertical divider between the origin and the second point, one merely needs the intersection between that and the bisector of the segment from the origin to the third point. This is done with the function circle_y. As mentioned, the x coordinate is half-way to the second transformed point.

For this example I got 0.5374838499, limited to 10 places.

Obviously one can remove or comment out the debug.print statements. Thanks to C, numbering starts with 0.

axym said:

The zones can fit in 0.537484 zones just from shifting in the x and y directions (at least this is what I get).

Can someone calculate the smallest zones that the 4 holes will fit in, if w rotation is applied as well?

Click to expand...

I get an RFS actual value of diameter 0.51269299 using the constraint solver in SolveSpace. The number of holes in this example conveniently matches the number of unknowns.

Without rotation, I get an RFS actual value of diameter 0.53748385. This agrees with your result and 3DDave's.


pylfrm

Interesting - the minimum diameter appears to be from rotation of the DRF for the hole pattern such that the four delta points lie on a common circle in the paper gage technique.

I used the Excel goal seek on a sheet that transformed the measured values in rotation to find the minimum diameter. I'm a little surprised it worked because two of the points exchange influence over the bounding circle as the angle changes.

It got down to a diameter 0.5126929852 with a rotation angle of 0.0875888823930269 degrees.

As I mentioned earlier datum shift will only apply on the measured features, this is what I learned and understood in the past, so it confused me when I read the textbook, and the term "datum shift" used several times on the paper gauge evaluation. That's my personal thinking: Is it possible a new term will be created in the future by ASME Y14.5, just like MMB(vs MMC) added at 2009.

I am not quite sure how to calculate the shift no matter with Excel or SolveSpace, are there any easy ways to get these data?

Season

SeasonLee said:

I am not quite sure how to calculate the shift no matter with Excel or SolveSpace,.............

Click to expand...

I am with you Season.....

Glad to see I'm not the only one a little lost here.

3DDave/pylfrm,
At the risk of asking stupid questions, what exactly is this "minimum diameter" you guys are referring to? The code posted by 3DDave seems to just translate the input points and calculate the radius of the circle that passes through all three points at their basic locations, none of which involves the tolerances shown or any output that looks like the ~0.537 or ~0.513 stated elsewhere. Just a little confused as to how you guys are executing this.

The calculation was not for the shift but for the paper-gage method as described in the PDF example, which they seem to have over-simplified because they used a naive calculation that did not include allowable rotation of the part.

Essentially, to evaluate the location tolerance, one is sometimes interested in finding the best case fit of the measured data to a DRF. In this case the DRF is simply [A], so there's the option to shift and rotate the measured locations.

To deal with datum shift one needs information about the features that depend on the referenced datums as well as the datum features. As I wrote before, the general solution creates a 3D surface envelope, shaped something like a lens, that describes the limitation to translation and rotation and is dependent on the as-made datum features. To know if the required translation and rotation to accept a feature is available, one needs that feature. Without one, datum shift is meaningless.

The functions are so one can build any analysis one likes that depends on radii or diameters given X-Y coordinates. It won't handle trivial bad cases (such as all points or two points being coincident or three points colinear) but being in a spreadsheet one will observe the resulting error message(s).