Center plane ISO 5459 2

[cr7]

Hi guys,

looking at 5459:2011 I would like to be sure, would the DRF (center plane) be exactly the same in both callouts?

 

[pmarc]

cr7,

Unfortunately, there is no easy answer to this question.

According to ISO 5459:2011, when a common datum is established from multiple features the following applies:
"⎯ The associated features that establish a common datum are by default constrained in location and orientation to each other. If the modifier [DV] is placed in the tolerance frame after the letter(s) identifying a common datum, then the linear distance between the members of the collection of features that make up this common datum shall be considered variable.
⎯ The orientation and location constraints correspond to the intrinsic characteristics introduced by the collection of features and are specified by TEDs. The values of 0 mm, 0°, 90°, 180° and 270°, and equally divided linear or angular dimensions may be implicit TEDs and not indicated."

This would indicate that the two cases are not equivalent, but the problem is that 5459 assumes that the relationnship between the features used to establish a common datum is defined with TEDs. In your case it is not and the standard does not give an explicit example of what to do then.

 

[jassco]

Hi, Cr7:

The 2nd FCF does not make sense to me as datum features A and B don't have to be in parallel. I think the 1st one with derived medium plane is correct one to use.

Best regards,

Alex

 

[Tarator]

cr7,

I am assuming you want to use the option with A-B because you want to use Datum A and B in some other FCFs by themselves? For that option, it might be a good idea to call out the distance between the 2 datum features as BASIC dimension, and somehow define their relationship with respect to each other (such as location, parallelism, etc.). I am not sure about ISO, but in ASME, since the datum simulators are considered perfect in size, shape, location, you would use the datum translation symbol ( looks like a triangle |> ) to ensure you would hold the part tight as in the other option (Datum E).

Or you can have all 3 datums (A, B, and E). You will just need to define their relationships with each other.

 

[cr7]

The 2nd FCF does not make sense to me as datum features A and B don't have to be in parallel.

Datum features that will give datum E would also likewise not be parallel and still provide derivative (the same as A-B in this case). But I think I get what pmarc and Tarator pointed out about TEDs in common datums. I just can't imagine a use case when A-B with TED would be used. I just feel that my example with perpendicularity isn't good example.

 

[Eurotex]

Datum plane E would be obtained by collapsing two perfectly parallel planes around the feature.
Datum plane A-B would be obtained by placing two unrelated perfect planes on each datum feature and creating a midplane. That is my current understanding of how common datums are established, since each datum feature is "equally weighted".
If these interpretations are connect, these two datum reference frames are not the same (in general).
Agree?

 

[cr7]

, since each datum feature is "equally weighted"

Yes. And this would be the job for MinMax (Chebyshev) algorithm as I understand. And wouldn't that algorithm calculate exactly the same result for both, datums E and A-B? (And again, thanks to pmarc and Tarator I get that A-B is deficient without TED)

 

[Eurotex]

Rereading pmarc's reply makes me think my interpretation of the common datum was incorrect. Given his definitions, why would pmarc not think the cases are equivalent? I see two 90 degree TEDs linking the two planes. I think this means that datum plane A-B would be obtained by collapsing two perfectly parallel planes around the feature - meaning that both datum reference frames are identical.

cr7, I don't know anything about the MinMax (Chebyshev) algorithm.