Desertfox,
As to your answer"if datum B is checked against
datum A for perpendicular then why make it a datum just givedatum B a tolerance for perpendicularity to datum A.Similar for datum C just give a tolerance for squareness to either", the reason why Datum B is referred to A is because that is the only available datum that is referred to. I am doing that cos it's only when all Datum A, B and C are firmly referred to(C related to A and B and B related to A) that the Imaginery Datum Reference Frame(DRF) is established.
Based on your answer.. say in the example of a rectanglar block, we can select the three(supposedly in theory mutually perpendicular) faces as datum features although the actual part is not mutually perpendicular. A set of imaginery mutually perpendicular planes A, B and C is only established by best fit in order of its relation(B related to A and C related to A and B)
I am asking this question cos there was a question that came up from my work. It was suggested by my peer that I need jus Plane, Line and Pt to establish my imaginery DRF. Anything more than that is "overconstraining" my part. My take is that Plane, Line and Pt is the minimum to constrain a part.I could specify Plane, Plane, Plane or any other combination so long as all the degree of freedom is taken care of. While in reality, three planar surfaces of an actual part is not perfectly mutually perpendicular,
an imaginery mutually perpendicular DRF can still be established from the actual part planar surfaces through best fit in order of A, B and C. Is my understanding correct?