In shaft alignment, the shaft of the driven machine is typically considered stationary,
while the other, the shaft of the driver, is movable.
The movable shaft is adjusted in relation to the stationary shaft to bring both shafts
into alignment and to satisfy machinery alignment tolerances.
In this video, we'll be alternating between 3D VIEW and GRAPHICAL VIEW.
In graphical view, the horizontal length of each square represents one inch, while the
vertical length of each square represents five one-thousandths of an inch.
While we'll only be working in the vertical plane, the same concepts also apply to the
horizontal plane.
Before we start, let's measure and indicate in our graph view the positions of the front
and back feet of the stationary machine, the center of the coupling, and the front and
back feet of the movable machine.
Beginning with our shafts in perfect alignment, lets introduce some offset misalignment by
raising the front and back feet of the movable machine equally by 20 thou.
The movable shaft centerline is now offset from the stationary shaft centerline by 20
thou at every point along its length.
Because we've added the same amount to both front and back feet, the shafts are parallel.
There is no angular misalignment.
So our offset misalignment is 20, and our angularity is 0.
In our next example, we'll induce both offset and angular misalignment by raising only the
back feet by 20 thou while leaving the front feet where they are.
Now, the movable shaft centerline is sitting 20 thou high at the back feet, while still
sitting at 0 thou of offset at the front feet.
So, what is our angularity?
In geometry, this concept is called SLOPE, or RISE OVER RUN.
Angularity, or slope, expresses the amount of rise or fall of the movable shaft over
each unit of measurement along the shaft's length.
Angularity is expressed as THOUSANDTHS PER INCH.
From the graph, we can see that the shaft rises 20 thou between the front and back feet
of the motor.
The distance between the front and back feet of the motor is 10 blocks across, or 10 inches.
Divide the rise of 20 thou by the run of 10 inches to get the angularity, which is 2 thou
per inch.
Now, what about the offset?
As you can see from the graph, the offset is now a different amount at every point along
the shaft centerlines.
You might ask, why is the offset expressed as a single number?
It is because there is a standard reference point for offset misalignment HERE, at the
plane of power transmission, typically at the center of the coupling.
The graph shows that the offset is 20 thou at the back feet, and zero thou at the front
feet.
We know that the offset changes by 2 thou for every inch we move along the shaft.
The coupling center is 4 inches from the front feet.
Over those four inches, the movable shaft will drop an additional 8 thou (4 inches x
2 thou).
Therefore the offset at the coupling center is -8 thou.
In this example, the movable shaft is sitting low at the coupling but high at the rear feet.
Since we induced the angular misalignment by adding 20 thou to the back feet, the only
way to correct the misalignment is by removing 20 thou to lower the rear feet.
At this point, we might hear an aligner ask, "I can see that the shaft is low at the coupling.
So why is the laser telling me to LOWER the feet, if the shaft is ALREADY low at the coupling?
It just doesn't make sense."
Careful examination of this graph provides the answer.
In another Concepts of Alignment video we explain how you measure the offset at the
coupling and the angularity.
We then show how to use the offset and angularity values to calculate the corrections necessary
at the feet of the moveable machine.
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