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March 30, 2006
Inertia of Wound Rolls
Inertia compensation is a big deal at unwinds and winders. The torque demand of a unwind or winder comes from three sources: tension x radius, inertia x accel/decel, and torque system losses (from nip, bearings, and couplings).
A question on roll inertia came in recently:
>I'm a novice to web tension control applications but I would like to learn what the proper calculations are that must be made in order to find out the inertia of a center driven unwinder and the torque required to maintain a certain tension.
>For example, let's assume a roll of plastic film weighing 1000 lb. and having a 24" diameter. The Desired tension is 1.0 PLI (Pounds per linear inch). I would like to learn how to make this calculation correctly.
Here's the calculation:
Torque from inertia is I x alpha
I = rotational inertia of the roll
alpha = accel / decel rate in radians per second^2
I = pi x rho x w x (Rout^4-Rin^4) / 2 / g
rho is density in lbs/in^3
g is gravity
[NOTE: Two sharp blog readers pointed out an error in the original post of this equation that was missing the width term. It is now correct, but I want to give out a big thanks for their help.]
I have a spreadsheet where I can calculate all this.
The 1000-lb roll of film at 24" is likely 50" wide on a 6" OD core. This crunches out to an inertia of 200 in-lbs-s^2.
What is alpha?
alpha can be found from line speed and accel time.
If you accelerate this roll to 1000 fpm in 15 seconds, the accel alpha is about 1 radians per second-squared.
So the inertial torque is I x alpha or about 200 in-lbs. At 12-in radius, this is 16 lbs.
If you have a braked unwind and the brake applies 800 in-lbs or 48-lbs of tension at 12-in radius. During the accel, the tension would increase to 60 lbs due to the added inertial torque.
Clear? Let me know.
Advanced control (what really should be standard) has inertia compensation built into the tension control plan to adjust automatically for this natural source of tension upsets.
Posted by Tim Walker at March 30, 2006 10:22 AM
Comments
I do not understand the equation:
I = pi x rho x (Rout^4-Rin^4) / 2 / g
If rho=lb/in^3 and g=in/sec^2
then I=lb/sec^2, which is not the correct units.
Using:
I =(Wght/g)*(Ro^2+Ri^2)/2
then I=in-lb-sec^2, correct units
Posted by: Mark at December 4, 2006 08:42 PM
Mark is absolutely correct that this equation had a problem. In my initial post, I had left out the w (for width) in the inertia equation. It is correct now.
-tjw
Posted by: tjwalker at March 16, 2007 07:22 AM
