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Rule1 not required

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aniiben,

Your sides would be perpendicular to within[ ]1[°] as per your note. Your maximum size would be 38.2mm plus 46.2mm[×]sin(1[°]).

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JHG
 
The maximum envelope would seem to be 38.4 mm -- this comes from the maximum local size of 38.2 plus the form error of 0.2 on the datum feature.
The minimum envelope would be 37.6 mm, derived in a similar way. (We usually don't speak about a minimum mating envelope for an external feature of size, but it makes for a good discussion question here).

John-Paul Belanger
Certified Sr. GD&T Professional
Geometric Learning Systems
 
I assume the question is not intended to take in consideration the default angle tolerances, but only shown geometrical callouts- flatness and parallelism.
 
greenimi,

I am assuming that angle tolerances matter. Rule[ ]#1 controls this.

--
JHG
 
Hi drawoh -- the corner angles have nothing to do with the question. And Rule #1 has nothing to do with angle tolerances.
The OP asked about the envelope of the 38 mm width dimension. Using terminology from Y14.5, it's a question about the unrelated actual mating envelope (UAME) rather than the related actual mating envelope (RAME).

John-Paul Belanger
Certified Sr. GD&T Professional
Geometric Learning Systems
 
I agree with J-P that rule 1 (or the lack thereof) has nothing to do with the angle tolerances. I suspect the angle tolerances applies with the now "infamous" unless otherwise specified.
 
J-P:

Sorry coming late to this thread. I was off for a long weekend.

I wrestle with the interrelationship between size and form when independency "I" is applied - is assume "I" is equal to "PERFECT FORM AT MMC NOT REQUIRED" . So, given your 38.4 value, which includes the flatness on Datum C, why did you ignore the parallelism with implied flatness on the other side?

Certified Sr. GD&T Professional
 
Mkscski -- I've created a hand sketch to show how I look at it (see attached). I didn't include the parallelism tolerance because I was assuming that the part is already at its maximum local size. If the right-hand surface of my sketch curved the other way (and made the parallelism additive), then it would fail the actual local size requirement.

John-Paul Belanger
Certified Sr. GD&T Professional
Geometric Learning Systems
 
 http://files.engineering.com/getfile.aspx?folder=14357fbb-6591-425b-a2df-6cf75dc991dd&file=MaxEnvelope.pdf
J-P:

Thanks. This was what I was thinking too. I am pretty sure ISO allows the GDT tol to be additive to the size when "I" is applied. And I assumed the Y14.5 did NOT have this interpretation.

Certified Sr. GD&T Professional
 
mkcski,
Could you, please, expand a little bit on how the ISO allows GDT to be additive and how the ASME (with "I" applied) doesn't?
I am trying to grasp the differences between these two concepts (ISO vs ASME) on an seemingly equivalent schemes.

Thank you
Gab
 
gabimot:

Give me a day or two to dig out and scan my ISO ref material - Alex Krulikowskis' "ISO Geometrical Tolerancing" guide pages 51 and 137, 144 if you have a copy.

Certified Sr. GD&T Professional
 
Thank you mkcski. You have the entire week... if you wish.
And will be a Christmas gift!
P.S. I do not have much ISO material (except this forum)and the search.
 
gabinot:

Your welcome.

Certified Sr. GD&T Professional
 
Thank you mkcski,
I will study (try to understand) your provided material and follow-up with additional questions (most likely) about the differences between ISO (default / Independency) and ASME ("forced" Independency I)
 
gabimot:

This topic is not one I have studied and examined in detail. Maybe you want to start another "independency" post.

Certified Sr. GD&T Professional
 
Let me try and I am sure if I will miss something or provide wrong information there are people/experts here who will set the record straight.

I do not know if ISO (with Independency principle, ISO 8015 stated) and ASME (with I symbol attached to 38±0.2 dimension) will have different extreme boundaries (OP question has been: what would be the maximum and minimum envelope that 38±0.2 dimension would never violate?)

In my opinion would be the same for both envelopes or extreme boundaries (38.4 and 37.6) Please, correct me if I am wrong. I am trying to learn from each and every discussion.

Also, the parallelism would not be added to the sum of flatness and maximum material size. I am not sure how can I explain why (would not be added), but I would say because the orientation (parallelism in our case) never locates only orients and the extreme boundaries are already determined by the flatness and the size tolerance. Am I correct? Any orientation error should only be toward the inside of the material in order to not be exceeding the limits of size.

What I know is: The tolerance zone is limited by two parallel planes a distance 0.2 apart and parallel to the datum plane C. The extracted (actual) surface shall be contained between two parallel planes 0,2 apart, which are parallel to datum plane C.

Hmm, any errors in my statements? Or better to ask any truth?

 
With ISO 8015 the linear measurements are all 2 opposing point measurements so the part could look like a banana or a wavy cut potato chip and still pass as long as every opposing point pair is within the tolerance. There really is no upper maximum on the envelope. ISO 8015 wants you to explicitly state each and every GD&T requirement rather than falling back on the envelope "crutch". I hate it.

----------------------------------------

The Help for this program was created in Windows Help format, which depends on a feature that isn't included in this version of Windows.
 
greenimi:

Look at the attachment J-P on 12 Dec 17. This clarifies the ASME boundaries very well. The ISO analysis is more expansive because "I" is the default. I am not up to snuff as to how the GDT tolerances interact with the size.

Certified Sr. GD&T Professional
 
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