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ISO 1938 calculation example (plain limit gauges of linear size)

ISO 1938 calculation example (plain limit gauges of linear size)

hello Gentelmens,

with regards to ISO 1938...
I have it in front of my eyes and I am not sure if I am doing this correctly

I have a slot (to be checked with bar gauge type F) with envelope requirement, let's say 20mm +/- 0.2

Cause the size tolerance is not given as an ISO code, I am using first IT grade from Table7-11 with a tolerance interval lower than or equal to the tolerance interval in the same nominal size range

So 0.4mm is 400 micrometers, for size 20 it would be IT13 (T=330, z=36), table 10

Now, in the formula for the new state "z" is one of the factors.
There is also H - for the form limit values which are "half the values given in column 2xF in Table 6"...

So now I am going to Table 6.
For IT13 bar gauge:
size H - IT7
form and orientation 2xF - IT5

Going to ISO 286 IT grades table:
Size H for IT7 (20mm) = 21 micrometers
form and orientation 2xF for IT5 (20mm) = 9 micrometers (<- this shall be multiplied by 2?...)

can someone check if my values selection from tables is correct?

RE: ISO 1938 calculation example (plain limit gauges of linear size)

I, and probably more of us, would like to help you, but I'm afraid it won't be possible until you share some screenshots of the standard with us. I have access to quite a lot of ISO GPS standards, but 1938 is not amongst them.

RE: ISO 1938 calculation example (plain limit gauges of linear size)


It isn't clear to me what your actual goal is. Clarifying your question would probably be helpful.

I don't have ISO 1938 either, so I may not be much help.


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