I'm not sure I understand your example numbers, particularly the bonus
number of .277. I'm trying to recall what I used to know about geometric
tolerancing. If the actual diameter is greater than the lower limit (5.8
mm) doesn't the bonus position tolerance grow mm for mm until you get to the
max of 1.4? In other words, with a measured diameter of 6.123, should the
bonus be 6.123 - 5.8 = .323?

If I understand it correctly, if your data is set up like this in row 475:

H I J K
Nominal Tol +/- Actual Posit Tol
6 0.2 6.119 1


This formula should work:

=IF(J475<(H475-I475), K475,
IF(J475<(H475+I475),(K475+J475-H475+I475),(K475+2*I475)))



"ufo_pilot" wrote in message
...
Need a 'singe cell' (or best) formula to calculate the bonus tolerance for
a
true position on a circle for a dimensional check.
I have the formula in 6 cells, but can't get them shrunk enough to be able
to fit 3 cells or less.
using cells for: upper, lower, and total tolerance for diameter
(+/-0.2mm)
upper, lower spec. limits (6.2mm, 5.8mm)
actual measured diameter (6.123mm)
nominal diameter (6.00mm)
nominal true poisitional tolerance for circle
(1.00mm)
The formula will show that if the circle dia is smaller than the LSL
(5.8mm),
the circle will get 0.00 bonus added to the position tolerance of the
circle.
as the measured circle gets larger (LMC -least material condition) a bonus
is added to the circles positional tolerance. But a circle will never
receive
more than the positional tolerance (1.00mm) + total tolerance (0.4mm). If
the
dia. exceeds the USL(6.2mm),
the circle will max out at 1.4mm tolerance
the sample abobe would receive a bonus of 0.277mm for a total poistional
tolerance of 1.277mm