Page 217 - Machinery Component Maintenance

Machinery Alignment 199

At OB, -(.007) ‘4 ”) - (q) -.0683 - .0055 = -.0738

=
~
say lower OB .074 in.

Horizontal
(3
At IB, (.014) - N f -
i*O?)S
= .0316 N - .0145 S
.0316
.0145
S
N
=
-
= .0171 N: say move IB .017 in. north
f (y)S
At OB, (.014) r:’”)N ~
= .137 N - .0145 S
= .1225 N
say move OB .122 in. or 123 in. north

As you can see, the values found this way are close to those found ear-
lier. The main problem people have with applying these formulas is
choosing between plus and minus for the terms. The easiest way, in our
opinion, is to visualize the “as found” conditions, and this will point the
way that movement must proceed to go to zero misalignment. For exam-
ple, our bottom face distance is wide-therefore we need to lower the
feet (pivoting at plane A) which we denote with a minus sign. The ma-
chine element to be adjusted is higher at plane A-so we need to lower it
some more, which takes another minus sign. For the horizontal, our
north face distance is wider, so we need to move the feet north (again
pivoting at plane A). The machine element to be adjusted is north at plane
A, so we need to move it south. Call north plus or minus, so long as you
call south the opposite sign. Not really hard, but a lot of people have
trouble with the concept, which is why we prefer to concentrate on
graphical methods , where direction of movement becomes more obvi-
ous. We will get into this shortly, but first let’s do a reverse-indicator
problem mathematically.
For our reverse-indicator example, we will use the setup shown earlier
as Figure 5-6. Also, we must now refer to the appropriate data sheet,
Figure 5-18. Finally, we resort to some triangles, Figures 5-19 and 5-20,
to assist us in visualizing the situation.
(PXI conrinued on page 202)