Page 80 - Rotating Machinery Pratical Solutions to Unbalance and Misalignment

Rotating Machinery: Practical Solutions

B ▼ ▼
▼
A ▼ A ▼ C
R

B
C
▼
▼
Figure 5-6. Vector Addition

As an illustration, assume 0 degrees is at the 3 o’clock posi-
tion and that the degrees are laid off in the counterclockwise di-
rection. Weight A is 11.5 ounces @ 63 degrees; weight B is 7.0
ounces @ 160 degrees; and weight C is 8.5 ounces @ 0 degrees.
When the three vectors that represent the weights are placed
tail to head, and the resultant R drawn from the tail of the first
vector to the head of the last vector, the measured length of the
resultant is the equivalent weight. The angle formed by the result-
ant is the equivalent angle where the weight is to be applied. In
this instance, the equivalent weight is 14.8 ounces @ 60 degrees.
Once again the same results could have been obtained using
trigonometric functions. First, we can find all the vertical distances
and add them together and then the horizontal components can be
found and summed together.

Y = A sin α + B sin β + C sin γ + … (5.13)

X = A cos α + B cos β + C cos γ + … (5.14)
Therefore:

Y = 11.5 sin(63) + 7 sin(160) + 8.5 sin(0) or Y= 11.5 × .89100 + 7
× .342020 + 8.5 × 0.0 = 12.6407

and
X = 11.5 cos(63) + 7 × cos(160) + 8.5 × cos(0) or X = 11.5 × .45399
+ 7 × (–.93969) + 8.5 × 1.000 = 7.14303