"BASIC" Dimension tolerance in First Article Inspection Report

[UchidaDS]

I have a part showing BAISC dimension, with profile, flatness and parallelism control.
The default tolerance in this drawing is +/-.005"
see attached. If required showing the measurement (diameters and height) in the First Article Inspection Report, what is consider out-of-spec,
Diameter: 3.798"
Diameter: 5.004"
Height: .630"

 

[pmarc]

Well, to me those values should not be reported at all. Or in other words, they can be reported, but as stand-alone numbers they mean nothing.
On this print there is only one dimension that requires to be reported in terms of number - and that this the size of datum feature B, dia. 5.543 +/-.005 (default tolerance).
For other features, inspection should rather try to check whether the actual surfaces of the part lie within boundaries defined by particular geometric tolerances (flatness, parallelism and profile).

 

[CheckerHater]

One thing to note from the very beginning: DEFAULT TOLERANCE NEVER APPLIES TO BASIC DIMENSION! (I cannot over-emphasize that )

Then point-by point:

Diameter 3.798 should measure 3.796 to 3.800

Creator of the drawing probably wants diameter 5.004 measure 5.000 to 5.008 but it isn't clear on the drawing

Height .630 should measure .626 to .634

BUT, drawing also applies certain orientational / locational requirements, that for some reason are not part of your Report

 

[pmarc]

CH,
Even if we just for a moment assume that reporting the numbers only is OK (which is not, like I already said), the values you mentioned should be different, don't you think?

Dia. 3.798 should fall within 3.794-3.802.
Dia. 5.004 should fall within 4.996-5.012.

 

[UchidaDS]

I understand the "DEFAULT TOLERANCE NEVER APPLIES TO BASIC DIMENSION", the BASIC is established by the tolerances in feature control frame.
The uncertain I have is the height (.630) and the diameter (5.004).
Where is the range getting from? If it is from the flatness, isn't it suppose to the for
Height .630 should measure .629 to .631 which is contribute by the flatness.

Thanks for the help!

 

[UchidaDS]

My understanding is:
It is unilateral tolerance and the PROFILE of .004 should divided by two = .002. And this BASIC dimension is established by PROFILE tolerance (.004) at that view.
Hence: Diameter 3.798 should measure 3.796 to 3.800.

 

[pmarc]

UchidaDS,
The upper surface of the part can lie within .626-.634 bandwidth which center is located at basic .630 from datum plane A. But in the same time the difference in readings can't be greater than .002, which comes from parallelism tolerance value. Or saying it differently, the parallelism tolerance zone of .002 width is able to float within larger profile tolerance zone of .008 width.

As for bottom surface of the part, flatness is verified as a totally independent callout. That is, it has no impact on the tolerance values applied to the upper surface of the part. The bottom surface will be fine if all its points lie within .002 tolerance bandwidth.

Regarding diameter 3.798, profile tolerance does not define unilateral, but bilateral tolerance zone. Draw a circle of 3.798 diameter, then draw two more circles concentric with the first one. One with RADIAL offset .002 outside of dia. 3.798 circle, and second with RADIAL offset .002 inside the dia. 3.798. Check the diameter values. They will not be 3.796 and 3.800, but 3.794 and 3.802.

 

[axym]

Hi All,

As usual, I agree with pmarc on all counts. Reporting measured values for basic dimensions is not a best practice. These values may tell you something about the part, but they often give misleading indications about the feature's conformance to the geometric tolerance.

UchidaDS,

The basic dimensions are theoretically exact and are not affected by tolerances in the feature control frames.

The possible height range for the part is determined by the Profile tolerance zone of 0.008, equally distributed around the basic height of .630. The Flatness tolerance on datum feature A does not affect this, because the datum is established from the extremities of the datum feature.

Evan Janeshewski

Axymetrix Quality Engineering Inc.

http://www.axymetrix.ca

 

[waqasmalik]

Agreed with pmarc

 

[UchidaDS]

Thank you guys!!
This help a lot!
I understand that BASIC dimension are theoretically exact; one of the reason for this post is I am looking for " tolerance numbers" due to the "First Article Inspection Report".
totally agree with pmarc

 

[pmarc]

I also thank you for agreeing with me.
I somewhat understand the need to provide numbers in this case - this is what most of inspection report templates require. Unfortunately at the same time this is not what print (geometric definition of the part) demands from inspection. I really hope that future standard Y14.45 on Measurement Data Reporting (which I heard is going to be released next year) will provide necessary requirements on how to report this kind of stuff. Time will tell...

 

[CheckerHater]

pmarc,

I have to agree that my numbers represented the bad case of Fridays.

I was surprised though that you found M<-->N tolerance zone adequate to determine value of [5.004] dimension
Are we talking about"point-to point" measurement taken on the very edge, or you believe the tolerance applies to the entire vertical cylindrical body?

 

[pmarc]

The M<-->N tolerance zone is not adequate to determine value of [5.004] dimension - that is correct.

 

[UchidaDS]

Eh?
Why is M<-->N tolerance zone inadequate to determine value of [5.004] dimension?

 

[CheckerHater]

See picture

 

[UchidaDS]

So for [5.004] dimension, if needed to report, the tolerance should be the default tolerance +/-.005".
In other word, the diameter 5.004" should not have BASIC; OR the "M" should extend further, that way
5.004+.008 = 5.012 MAX and
5.004-.008 = 4.996 MIN
Am I on the right track?

Not sure how you get 5.000 and 5.008, please help.

 

[CheckerHater]

I was wrong on 5.000-5.008, that's the part pmarc got right.

Now, you still need basic 5.004 as part of your M to N profile.

One way is to move your M point "one step down", so profile will apply to vertical cylindrical surface as well.

Then diameter of said surface will be 4.996-5.012 as pmarc noted.

The other way will be to create separate toleranced dimension for that surface, but it will appear as duplicate and may be confusing.