Per ASME Y14.5.1M-1994, the actual value of RFS position deviation is equal to the size difference between the RAME (using the appropriate DRF) and the UAME.
My statement above is intended to be as general as it sounds, but to explain in more detail I will sacrifice some generality and use the specific example of an external cylindrical feature.
First, definitions from ASME Y14.5.1M-1994 para. 1.5:
r(P) = the distance of a point P to true position, in the case that the datum reference frame is understood from context
rAM = actual mating size (radius)
t0 = a specific tolerance given on a drawing or part specification
I will also define the following:
actual_value = actual value of the feature's position deviation
size_UAME = diameter of the feature's unrelated actual mating envelope
size_RAME = diameter of the feature's related actual mating envelope (orientation and location constrained using the DRF of the position tolerance)
ASME Y14.5.1M-1994 does not use the terms 'UAME' and 'RAME', so the two types of AME are not always clearly distinguished. I will use the ASME Y14.5-2009 definitions of these terms for clarity. Unfortunately we are left to determine from context that
rAM refers to the UAME size in this case.
Per Table 5-1, the tolerance zone (volume in which material is prohibited) is defined as follows:
r(P) > b
To get the tolerance zone boundary, convert the inequality to an equation:
r(P) = b
For the actual value, we are interested in the smallest such boundary not violated by the feature. Note that this is equivalent to the definition of the RAME with orientation and location constrained. Substitute, accounting for the diameter vs. radius discrepancy:
size_RAME / 2 = b
Per Table 5-2, the size of the tolerance zone is as follows:
b = rAM + t0 / 2
Substitute as discussed above:
b = size_UAME / 2 + actual_value / 2
Combine the two equations, eliminating
b:
size_RAME / 2 = size_UAME / 2 + actual_value / 2
Multiply by 2:
size_RAME = size_UAME + actual_value
Rearrange:
actual_value = size_RAME - size_UAME
Additionally, if it were as you suggest (size difference between UAME and RAME) wouldn't that be zero in this case? Unless there is orientation error, which I do not think we were assuming any with this simplified 2D case, the UAME and RAME would be identical (99.544) - right?
I was referring to the RAME with both orientation and location constrained using the DRF of the position tolerance. I had forgotten that Y14.5-2009 says "constrained either in orientation or location or both", so I failed to specify.
Does this clear things up?
pylfrm