Elastic belt stretch, lead screw twist, and far end encoders

I finally got around to doing the math on the differences between
timing belt stretch, and shaft twist.
On the Gates website, there is a technical document that has a chart showing
elastic belt stretch, vs unit loading. This is the stretch that is
temporary, based on load, not the long term stretch that may occur during
use.
The document is at:
http://www.gates.com/facts/documents/Gf000289.pdf
Using this data, I made a few assumptions for calculation purposes:
Assume a 0.200"/rev frictionless ball screw.
Assume a 1:1 motor pulley to lead screw pulley ratio, using a 1" wide Gates
belt.
Assume a 6" center distance of the pulleys.
Assume 500lb mill table force.
Assume a 1" pitch diameter ball screw

From here, we need to get to the force at the timing belt. Since we
have zero friction in the ball screw, with a 5:1 ratio from the 0.2" lead,
and a 1/2" radius on the lead screw, the force on the belt to obtain 500 lb
table force is:
Torque on screw = Force * 1/2" radius
Table Force = 500lb=1/2Force * 5 (lead screw multiplier)
Force = 200 lb Since this is at 1/2" radius, that is 100inlb torque from
the motor.

From the Gates data, the EA value for a 1" belt at 200lb/in width
is about 37,000 (lb/in width/unit strain)
The technical data also shows the relationship between this value, and the
elastic belt stretch:
EA = (TxL)/(E1xW) where T= belt tension (lb), L= span length (in), E1=
elongation of span length (in), and W=belt width (in)
Doing the math, we get: 37,000=(200x6)/(E1x1) or E1= 0.0324" span stretch
Since this is belt stretch, it has to be divided by screw pitch get to table
movement error, so: 0.0324" * 0.2=0.0065"
This means that at 100inlb force on the motor, the 1" wide drive
belt has stretched elastically by 32 thousandths, causing the table to move
6.5 thousndths less than the desired amount. Of course, in a servo system,
the encoders will account for this, and keep the motor driving until the
proper position is obtained.

Now, lets look at the same error, but due to lead screw twist
instead of belt stretch, same assumptions, but assume a 36 inch length of
screw between the pulley, and the ball nut/table:
From physics, we know that the angle of twist of a steel shaft, in
Radians, is equal to (TxL)/(J*G), where
T=Torque (inlb), L=length (in), J= (pi*diameter^4)/32, and G=E/(2(1+v))
where E=Steel's modulus of elasticity, 30X10^6 psi, and v=poisons ratio, for
steel 0.333.
Plugging in all of the numbers, and using 100inlb torque, and
converting radians to degrees: (you can do the math on this one :-) we get
0.187 degrees of twist.
This amount also has to be converted into table motion: so since 360
deg=0.2", with a ratio we see that 0.187 degrees = 0.0001" table movement
error.
From this whole math lesson we learned the following:
With the CNC described above, the belt stretch causes an error of 0.0065",
and the screw twist 0.0001" or simply put, the belt error is 65 times the
size of the lead screw twist, and ball screws are not frictionless so
therefore, the belt stretch, being upstream of the friction, will be even
greater than this.
Since servo systems deal with this phenomenon every day, and do not
have major stability problems, or routinely go into damaging resonances, we
can conclude that putting an encoder on the far end of a leadscrew will have
no detrimental consequences. Also, since the ball nut can be either right
next to the pulley, or at the opposite end of the screw, placing an encoder
at the pulley end vs the opposite end will cause exactly the same amount of
error in the part, since we use the whole table. In other words, the shaft
twist is present as an error in all cases: When the encoder is at the pulley
end, the table will experience error from shaft twist when the nut is far
from the encoder, and vice versa when the encoder and table are at opposite
ends.
Comments?

Brad Heuver