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## Balancing Weights: Radius Changes & Splitting

Now that you have pinpointed that the original vibration problem was indeed unbalance, you need to arrive at a permanent solution with the proper correction weights and locations.

As a hypothetical example, let's take a look at trying to balance a rotor, where weights (washers) can be placed only at nine evenly-spaced bolt head positions at a radius of 16 inches. See the illustration below.

For this example, it is determined that the rotor can be balanced within acceptable tolerance with a temporary weight of 2.2 ounces placed at an angle of 250 degrees and at a radius of 14 inches. Strictly for convenience and expediency, this is the same radius where all trial (calibration) and corrective weights were placed.

Now we need to go back to a fundamental relationship where unbalance is defined by weight X (times) radius relationship. Thus, to figure the proper weight at a radius of 16 inches at the same angle, we simply do a proportional calculation, where

w(t) x r(t) = w(p) x r(p)

and w represents weight, r is radius, t is trial and p is permanent.

Plugging in the numbers,

w(p) = [w(t) x r(t)]/r(p) = [2.2 x 14]/16 = 1.925 ounces.

If a bolt head existed on this rotor at 250 degrees, we would simply pull out a combination of washers at a total weight of 1.925 ounces and we would be done. However, this is not the case, so we have to work out a weight splitting at the exiting bolts - one located at 240 degrees and one located at 280 degrees. This is accomplished with trigonometry and the "law of sines" for triangles or by a "vector splitting," as addressed below. First, let's take a look at the geometry one more time and concentrate on the balancing area on the rotor. See the figures below.

For vector splitting, we ideally would use polar graph paper for precision, but the procedure is to first draw a vector length at the 250 degree angle to represent the balance weight of 1.925 ounces. Next, we draw lines from the center at the angles where weights can be placed, in this case, at 240 and 280 degrees. After that, we complete a parallelogram with sides cutting through the end of the weight vector. Where these parallel lines pass through the angle lines then determine the proportional weight splitting. For this example, my original weight vector representing 1.925 ounces was drawn exactly at one inch. The length of the line at 240 degrees was measured at 25/32 inches, hence specifying 1.50 ounces. The line length at 280 degrees was measured at 9/32 inches, hence specifying 0.52 ounces.

Thus, we have arrived at a final clean solution, with washers weighing 1.50 ounces being placed at the bolt head at 240 degrees and washers weighing 0.52 ounces being placed at the bolt head at 280 degrees.

In concluding this discussion, it is interesting to note that even with all the sophisticated computer algorithms out there today doing most of the work, it is sometimes good to go back to basic principles and examples to better understand how radius changes and weight splitting can figure into your final balancing solution.

Dennis Shreve is Channel Support Engineer for the Channel Partner Sales organization with Commtest Inc. He has 40 years of experience in designing and developing electronics and software systems and leading projects for real-time industrial process monitoring and control applications. Over the past 21 years, he has specialized in predictive maintenance (PdM) technologies and vibration detection, analysis and correction methods for maintaining machinery health. www.commtest.com

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