Page 81 - Rotating Machinery Pratical Solutions to Unbalance and Misalignment

Field Balancing

Since the values of X and Y form a right triangle:

2
2
a + b = c 2 (5.15)
By rearranging Equation (5.15) and substituting the supplied data,
2 .5
2
c = ((12.6407) – (7.14303) ) = 14.585
The length c is the resultant, or the required balance weight.
In this case the weight is 14.6 ounces. Once again, the mathemati-
cal method is more accurate than the graphical method, but the
answers are very close.
To find the angle, several methods could be employed, but
we will use the arcsine (asin) function.

Degrees = asin (Y/R) (5.16)

Here, asin (Y/R) simply means the angle in degrees, whose sin =
(Y/R). Dividing the value of Y by R and looking in the sin table
for that value yields: asin (.8669) = 60.07 degrees.
Once again the mathematical solution is more accurate than
the graphical method, but not enough to warrant its use in most
cases. It is a good practice to work all problems using both meth-
ods to check your work.
The following illustration is provided to show that the planes
onto which the weight can be resolved do not have to be at right
angles to each other. Figure 5-7 represents a typical case of split-
ting the required correction weight into two equivalent weights.
The lines marked B, B’, C and C’ are not vectors, but lines
representing planes. Note that vectors must have arrow heads
indicating their direction of application.
To graphically resolve this problem, either the B plane or the
C plane can be moved parallel, forming either the B’ or the C’
planes. These planes represent the direction the equivalent
weights must act in, and therefore, must maintain the same rela-
tive angle.
Note that regardless of which plane was moved, it was