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

Thanks for looking at this Hugh.
I've added more comments into the text. Hope its not too confusing.

--- In CAD_CAM_EDM_DRO@y..., Hugh Currin <currinh@O...> wrote:
> 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.

I was making the assumption that the lead screw acted at a
lever/wedge. In other words, a triangle of 0.2" and 1" sides, with
the screw the hypotenuse. In this analogy, the screw force was
multiplied by the ratio of the sides. From your comments, I gather
that this was a false assumption of how screw force's work.
However, if it only takes 15.9 ftlbs to drive a 500lb load (plus
that pesky friction) it seems that we would all be using smaller
motors on our machines.

>
> > 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.

You're correct, in keeping things simple, I used a 1" pulley, the
same as the lead screw pitch diameter, probably smaller than that
most people use. Your 3 " pulley would have 1/3 the load, a better
design.

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"
>

So now we have two data points on the 1" belt:
0.0017" at 7lb load (10.6 inlb/1.5 inch pulley radius=7lb)
and 0.0324" at 200 lb load. The stretch is not linear, per the EA
curve.

> > 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
>

Anoter good reason to use large diameter pulleys! By increasing the
diameter, you reduce the impact of stretch by a factor of pi. Over a
3:1 bang for your buck, and less load and stretch as well.

> 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.

I agree with this. As I'm getting into this more, I think that the
far end encoders on the old Tape-O-matic will stay where they are.
Its not only easiest, but may even be the best design.
>
>
> 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

Looks like I may have to dig out a book on how screw threads work...
Are you sure that the 2pi is in the force calc, and not a power calc?
(I don't have any of my books nearby, but it just seems that 15inlb
won't really generate 500lb of thrust on a 0.2" lead screw.)
Brad Heuver