Help on positional inspection method
Help on positional inspection method
(OP)
I have two identical holes on center line drilled on opposite ends of a part. One hole is datum A Ø.189/.195 and the other is the feature under consideration, also Ø.189/.195 with a position requirement of φ|Ø.002(M)|A(M)|. We are trying to verify conformance by using the tightest fitting gage pins in each hole, placing the pin in Datum A in the rollers of a concentricity gage with the indicator tip on the pin placed inserted into the feature under consideration. Using this method, I know we can add the bonus tolerance from the feature calculated by (Øgagepin-.189) to the positional tolerance of .002 and get up to .008. Can we also add the datum shift as well calculated by (Øgagepin[A]-.189) for a potential total positional tolerance of .014? My instinct says NO! But this has sparked a healthy debate and I wanted to get some feedback. We would prefer to check this with a functional gage or CMM, but that is just not in the cards right now.





RE: Help on positional inspection method
Think of it this way: If the datum were tagged to the other hole, and the position tolerance tagged to the opposite hole, the results would be the same. So bonus and shift in your specific example are kind of the same thing, but coming from different ends of the part.
John-Paul Belanger
Certified Sr. GD&T Professional
Geometric Learning Systems
RE: Help on positional inspection method
Certified Sr. GD&T Professional
RE: Help on positional inspection method
RE: Help on positional inspection method
Certified Sr. GD&T Professional
RE: Help on positional inspection method
Notice the table for that figure has "bonus" tolerance going downward, and the datum "shift" builds as the numbers go to the right. That shows that they essentially add together (again, for this special case where there is one feature, to one datum, and where the feature and the datum act in the same direction, inline with each other.)
John-Paul Belanger
Certified Sr. GD&T Professional
Geometric Learning Systems
RE: Help on positional inspection method
RE: Help on positional inspection method
John Acosta, GDTP Senior Level
Manufacturing Engineering Tech
RE: Help on positional inspection method
Imagine a part with total length 11.000 and diameter .194 holes in each end to a depth of 1.000. This would allow a rotational datum shift of about (0.005 / 1.000) = 5 mrad. If the axes of the holes were perfectly parallel, they could be offset by about .051 and still meet the requirement.
Make sure there aren't any simultaneous requirements preventing you from using all of the potential datum feature shift for this one feature.
pylfrm