Tek-Tips is the largest IT community on the Internet today!

Members share and learn making Tek-Tips Forums the best source of peer-reviewed technical information on the Internet!

  • Congratulations JAE on being selected by the Eng-Tips community for having the most helpful posts in the forums last week. Way to Go!

Specifying Flatness of a Derived Median Plane -RFS 2

Status
Not open for further replies.

Kedu

Mechanical
Joined
May 9, 2017
Messages
193
Location
US
Parallelism between the surfaces shown in fig 5-8 / ASME Y14.5-2009 is controlled within:

a.) 0.11
b.) 0.08
c.) 0.02
d.) 0.04
e.) 0.15
f.) 0.07
g.) None of the above

Edit: maximum allowable parallelism between the two surfaces dimensioned with 16h11 (15.89 - 16.00)
 
chez311 - and I missed the link to the UAME in the median plane definition - live and learn. I do think that this is not well thought out in the standard, as most examples of such things show bent items of constant thickness without consideration to variability of thickness in a single part. I'm sure the blinders were on with sheet metal as the focus, a material form that is usually sufficiently uniform for any single part.
 
I will assume the following:
[ul] [li]The lower surface in the drawing is identified as datum feature A.
[/li][li]The upper surface in the drawing is identified as datum feature B.
[/li][li]The datum feature reference letters are marked on the actual part so it's possible to tell which surface is which.
[/li][li]A parallelism tolerance referencing datum feature A is applied to datum feature B.
[/li][li]A parallelism tolerance referencing datum feature B is applied to datum feature A.
[/li][li]The two dimensions not shown in the drawing are 21.00 and 52.56.
[/li][/ul]

And consider the following questions:
[ol 1][li]What is the maximum possible actual value of parallelism for datum feature B?
[/li] [li]What is the maximum possible actual value of parallelism for datum feature B assuming it is not larger than the actual value of parallelism for datum feature A?
[/li][/ol]

For question 1, consider a cross section consisting of a polygon with the following vertices:

[pre]
P1: ( 0.000, 0.000)
P2: (17.520, 0.000)
P3: (52.560, 0.110)
P4: (52.560, 16.000)
P5: ( 0.000, 16.000)
[/pre]​
Datum A is derived from the segment joining P2 and P3. Datum B is derived from the segment joining P4 and P5. I calculate parallelism of about 0.1650 for datum feature B, parallelism of 0.1100 for datum feature A, and DML flatness of about 0.0183.

For question 2, consider a cross section consisting of a polygon with the following vertices:

[pre]
P1: ( 0.000, 0.000)
P2: (17.520, 0.000)
P3: (52.560, 0.055)
P4: (52.560, 15.945)
P5: (17.520, 16.000)
P6: ( 0.000, 16.000)
[/pre]​
Datum A is derived from the segment joining P2 and P3. Datum B is derived from the segment joining P4 and P5. I calculate parallelism of about 0.1375 for datum features A and B, and DML flatness of 0.

I'm not saying these are the actual answers, just that 0.11 does not appear to be.


pylfrm
 
pylfrym,

Basically a teeter totter - I duplicated the values of the answers to the first one in GeoGebra and attached the file.
I'm not sure about the reason for the values; by moving the pivot one can generate larger parallelism amounts, but I guess it's keeping it to the 1/3 point of the overall length.

At the 1/3 position the effect of the change in thickness and resulting slope of the teeter totter multiplies the size tolerance by 1.5, which makes sense as the multiplicative inverse of 2/3.

GeoGebra is free at
To be more obvious in the example, change the X values for p3 and p4 to 60 and the slider limits (right click, properties) to 1-20, and either make it continuous or type in a desired value.


I think this is a png of the worksheet:
(how is image embedding supposed to work?? More specifically I have a javascript blocker and don't see which one might be interfering with URL/dragn and drop/browse of images. I uploaded and tried the URL from engineering.com; no go.)
 
I suspected that this might be more than 0.11 ;-)

And I was sure that if there is anyone who is going to prove this, it will be pylfrm.
 
3DDave said:
chez311 - and I missed the link to the UAME in the median plane definition - live and learn. I do think that this is not well thought out in the standard, as most examples

Looks like the answer is not very straight forward. I guess we cannot pick without reasonable doubt one answer versus the others.

I will go back a little bit to the definition (as suggested by 3DDave) of the derived median plane flatness – RFS- and I have noticed that 2009 standard states -- same fig specified by the OP in the original question -- Fig. 5-8 page 94
“The derived median plane of the feature’s ACTUAL LOCAL SIZE must lie between two parallel planes 0.04 apart ……..”

The new draft of Y14.5 removed ACTUAL LOCAL SIZE from the text.

New draft: "The derived median plane of the feature shall be within two parallel planes 0.04 apart ……."

I don’t think this is a change in definition, but more like a clarification and I don’t think that will change the outcome of the question posted here (whatever the answer is going to me be).

Am I correct?
 
greenimi - since the only way it makes sense is if it is based on actual local size, what do they do to describe how to get there?
 
3DDave,

Thanks for posting the GeoGebra file. That looks like a nice tool.

If P2 passes 1/3 of the overall length, the segment joining P1 and P2 becomes a candidate datum for datum feature A. The actual value of parallelism for datum feature B would then be 0 in my first example, or 0.055 in my second example.

The specific numbers (21.00 and 52.56) were just measured from the figure in the standard, and adjusted to be multiples of 0.03 for convenience.

The definition of "derived median plane" in the draft is the same as the one in ASME Y14.5-2009 which chez311 posted earlier. It's good that this definition does not involve actual local size, because that term is poorly defined in both the current standard and the draft.


Kedu,

Why did you ask this question?


pylfrm
 
pylfrm,

Why would the shorter length, P1-P2 become candidate over the longer P2-P3?

I'm trying to let GeoGebra grow on me. It's nice that it's parametric and it has a wide reach into mathematics. The interface is different enough, especially the dynamic independent scaling of the view, but I have had geometry vanish as soon as it's created and I have no idea what click/key caused it. What should I do, ask for a refund?
 
pylfrm said:
Why did you ask this question?

Because we have a design case (resembling with a “bread toaster” with auto-centering guides) and we are very much confused by the statement on the 2009 standard

“Where the straightness tolerance is used in conjunction with an orientation tolerance or position tolerance value, the specified straightness tolerance value shall not be greater than the specified orientation or position tolerance value”

“Feature control frame placement and arrangement as described in para. 5.4.1.2 apply, except the diameter symbol is not used, since the tolerance zone is noncylindrical. See Figs. 5-8 and 5-9.”



Therefore, we are “investigating” if the orientation tolerance (parallelism) between the two surfaces is already controlled by the flatness of derived median plane RFS shown on our design or a standalone parallelism is required.


I have to apologies as I haven’t realized the issue is more complex that I originally thought. I just hoped for a clear and straight answer. I guess that's not the case....
 
Kedu,

The way you designed the original question has very little to do with the statement given in para. 5.4.4.1 of Y14.5-2009:
"Where the straightness tolerance is used in conjunction with an orientation tolerance or position tolerance value, the specified straightness tolerance value shall not be greater than the specified orientation or position tolerance value.”

This statement applies to situations where a position and/or orientantion and straightness feature control frames are associated with a size dimension (for example see fig. 6-9 and imagine that additional straightness FCF is added). Per the statement the value of that additional straightness tolerance could not be greater than 0.2.

Unfortunately that statement is incorrect because a feature of size may have significant DML (or DMP) error and at the same time its axis (or center plane) may be perfectly oriented and located with respect to the specified datums. It is because in ASME feature's DML and axis (or DMP and center plane) are totally different animals.
 
Pmarc,
I might be out of sync here, but if the width (15.89-11.00) I am concern about, is located relative to A, B and C (let’s pretend) from the part’s main coordinate system (again, it is a controlled feature and its relative location is defined based on its functional requirements/ relative position to A, B and C) then its orientation is also controlled by the position (meaning the orientation of the center plane is implied as having a maximum as defined by the position), correct?

But if an orientation refinement is needed (let’s say for a functional reason) then this orientation (parallelism) value should be bigger than 0.04 (flatness) to not fail the statement posted above (the one you said is incorrect)?

And I do understand my mistake: there is no parallelism of two surfaces, but for this discussion (and help you provided) pretend there is parallelism of a surface relative to the datum plane derived from the other opposing surface.
 
Kedu said:
But if an orientation refinement is needed (let’s say for a functional reason) then this orientation (parallelism) value should be bigger than 0.04 (flatness) to not fail the statement posted above (the one you said is incorrect)?
Yes, that is correct.
 
3DDave,

ASME Y14.5.1M-1994 para. 4.3.2 (a) said:
The candidate datum set for a nominally flat primary datum feature is defined by the following procedure:
(1) Consider a plane P which is an external set of support for the datum feature. Let C be the set of contact points of the datum feature and P.
(2) Consider an arbitrary line L in P. Orthogonally project each point on the boundary of the datum feature onto L, giving the line segment L’. Consider regions of L’ that are within some fraction x of the endpoints of L’. That is, if the length of L’ is n, consider regions of L’ within a distance xn of the endpoints of L’. Unless otherwise specified on the drawing, the value of x shall be 1/3. If all of the orthogonal projections of the points in C are within either single region, then plane P is rejected as a valid datum plane.
(3) Do this for all lines in P. (Note that parallel lines will yield identical results.) If no line rejects P, then P is a candidate datum for the datum feature.

I'm really only thinking about this question in two dimensions, so there is only one line L to consider.

pylfrm
 
pylfrm,

I took the 14.5.1 explanation as meaning that the points could no solely be within the 1/3 end of either end, not that the zone of acceptable points stopped at the 1/3 points. So the acceptable flat could be 1/3 + eta, where eta = some arbitrarily small value. Otherwise, if the teeter point was at 50% there could be no candidate datum.

 
3DDave,

If we change P2 from (17.52, 0) to (18, 0) in the first example, a plane contacting P2 and P3 would still be a candidate datum, but a plane contacting P2 and P1 would become one as well. Would you agree?

The associated candidate actual values of parallelism would be about 0.1673 and 0. The actual value is the minimum candidate actual value, so it would be 0.


pylfrm
 
pylfrm,

Since there is no parallelism requirement in this example, it is not possible to determine which candidate should be chosen. It is possible that the larger value will be the result if the candidate is used to minimize some other requirement.

In any case, changing P2 to 52.56/2 results in a potential candidate resulting in a parallelism of about 0.22. If P2 is changed to (2/3)*52.56 minus a small amount, the P2-P3 may still be chosen to satisfy some other requirement and the resulting parallelism is even larger.

If a bolt was to be put through the right end 1/3 of the block, the candidate datum for a position tolerance would be that 1/3 and the resulting slope would be about 0.33.
 
3DDave,

I added parallelism requirements in my original list of assumptions so that actual values would be meaningful. A parallelism tolerance can't be part of a simultaneous requirement, so the choice of candidate datum for its evaluation would be completely independent of other requirements. Choosing a different candidate datum to minimize the actual value of a different tolerance would be allowed, but would not affect the parallelism.


pylfrm
 
pylfrm,

I am looking beyond inspection and into the evaluation of the installed condition. I think it is a reasonable to understand the potential outcomes. The rules concerning the simultaneous requirements offer a way to simplify inspection and increase the acceptance rate, but they interfere with evaluation of real part behavior. When the part is bolted down by one end, there is no option to reposition the part to reduce parallelism variation.

I think the effect is not noticed as a potential problem because most manufacturing operations provide form errors that are significantly smaller than the size variations. I suspect that most parts are seriously under-toleranced and have thought it would be great sport to ship unusable but completely compliant parts in order to make the point. I expect that some who caught on would whine under 'workmanship.' I have seen suppliers forced to do the opposite and bend over backwards to ensure parts meet significantly smaller variations than allowed by tolerance, where any deviation from that smaller range, though acceptable, would cause failure.
 
3DDave,

Agreed on all points.

3DDave said:
I suspect that most parts are seriously under-toleranced and have thought it would be great sport to ship unusable but completely compliant parts in order to make the point. I expect that some who caught on would whine under 'workmanship.'

The trick is to find a case where the unusable but compliant parts are substantially cheaper to produce. Then you have a much better defense when people complain.


pylfrm
 
Status
Not open for further replies.

Part and Inventory Search

Sponsor

Back
Top