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

CAD/CAM Crew:

Comments below mixed in. :-)

At 10:58 AM 4/6/01 -0400, Brad wrote:

> I finally got around to doing the math on the differences between
>timing belt stretch, and shaft twist.

--snip--

> 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.

I think there is a 2*pi missing here. It should follow the power screw
equations which give (for no friction):

T = (L*W)/(2*pi) = torque in lead screw.


Here L is screw lead (.2"), W is table load (500 lb), pi is 3.14159.

With this equation the torque on the lead screw is 15.9 in-lb.

> 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"

I don't think you've accounted for the pulley diameter here. The equation
relating torque and belt tension involve the pulley radius as:

T = F*R

so the belt tension for say a 3" diam pulley will be 15.9/1.5 = 10.6 lb, in
this case. Then the belt stretch using the above information from Gates
will be:

37000 = (10.6*6)/(E1*1.0) or E1=.00172"

> 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.

This E1 relates back to lead screw motion again using the pulley radius.

E1 = theta*R

This gives a rotation of the lead screw (theta) of .00172/1.5 = .001146
radians. The table motion is related to lead screw angular motion with:

X = (L*theta)/(2*pi)

where theta is rotation of the lead screw in radians, L is again the screw
lead. The table motion X is then .000036" (0.036 thousands).

> 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.

Looks good but with the torque of 15.9 in-lb this calc gives a table motion
of .000016" (0.016 thousands).

>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.

If the encoder is placed on the free end of the lead screw the portion of
that lead screw between the encoder and nut carries no torque. With no
torque there should be no twisting of the shaft.


So, if this is right, the error from belt flex is about twice the error
from lead screw wind-up (at least in this case). But both errors are under
a tenth of a thousand and I'd think negligible.

I hope this adds to the analysis. If you see any problems with these
calculations please let me know. Thank you.

Hugh Currin
Klamath Falls, OR