Calculate the additive RPM required for tension control

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1. Nov 4, 2018

Stixoffire

In a web tension control system, where feed forward is to be used. A tension set point is desired lets say 20lbs, and a load cell system provides feed back. There will be dead weight on the load cell as the material with no pull sits on top of the load cell system. [Obviously I will need to compensate for that too].

The take-up reel diameter and speed are known values at all times of rotation - driven by a motor via a gearbox where both motor Hz / RPM and GB Ratio (25:1) are known values. For the moment assume constant speed 1750 RPM @ 60hz , diameter = 1 foot

The formula Torque = Tension (foot lbs) x Radius (in feet)
I have 20lbs on the load cell (don't know if I can call that foot lbs ) a roller with load cells under each side of the roller. 20 x .5 = Torque = 10'lbs
should I then Divide that by 25 (gearbox ratio) = .4

How can I calculate the additional RPM required to achieve the tension set point ?
Keep in mind there will also be a material on top of the load cell so dead weight would be a dead band and I will obviously need to offset by the dead weight ..

2. Nov 4, 2018

Staff: Mentor

Welcome to the PF.

Can you post a sketch or picture of the setup? That would help a lot. Just use the UPLOAD button in the lower right of the Edit window to attach a PDF or JPEG image to your post.

Also, what quantity are you feeding forward for the feedforward part of the control loop? How are you combining the feedforward and feedback signals to generate your motor control signal(s)? Or is the feedforward portion independent of the feedback control loop? (I have designed a DC-DC voltage converter with that characteristic, where there is feedforward from the input voltage level that helps the output voltage feedback to have better accuracy and line regulation...)

And I'm having a hard time understanding how a motor's RPM is converted into a tension in a rope. Is it via pulley friction?

EDIT / ADD -- Do you mean a tension control system for a roller-to-roller application like this?

http://www.cgb.com.au/images/Merobel Redex Andantex Web Tension Control Tension Page.jpg

Last edited: Nov 4, 2018
3. Nov 5, 2018

Stixoffire

I had difficulty in replying , I don't know what happened .. feed forward is being used - this will be my tension RPM value. PIDE Loop in PLC will control the motor.

It is a web not a rope - but basically the same.. So I know I can calculate Horsepower at x Diameter, n FPM, T tension and Build Up ratio at set point and also current value of Tension , reverse the formula and have an RPM (not sure if that is right or not - but I need to provide an RPM (actually it will be in Hertz ) for feed forward into the PIDE Loop

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4. Nov 5, 2018

5. Nov 5, 2018

Stixoffire

I have that document already and many more like it : See page 11 of 52 - you will note all of those WONDERFUL formulas for all of the methods Prior to ; and then on that page 11 ... a load cell provides feedback to a controlling module .. ..now what is the formula to provide that feedback in RPM? I would like to tell the author .. Yes Duh.. and the grand secret sauce is ... because every document I have come across is the same .. Load Cell oh it provides the feedback..

6. Nov 5, 2018

Tom.G

OK, to start, you can't. They are independent variables.
The force restraining the feed of the web determines the maximum tension. You can limit the torque the takeup reel supplies if you want to avoid rupturing the web, or control the torque if you want to maintain a constant tension on the takeup reel.

There are a few constraints you failed to mention.
Is the purpose to:
1. Maintain a given tension in the material on the takeup reel?
2. Avoid rupturing the web if there is an obstruction on the supply side?
3. Supply enough takeup torque to maintain a constant rate of web travel?
1. maybe with a torque limit?
4. Something else I may have missed?
Motor torque is directly proportional to motor currents. To limit the takeup torque you control the Current to the takeup motor. Note that when the current limit is reached that the motor voltage will drop, consequently reducing the motor speed*.

An alternate approach is to use a magnetic brake on the supply side to control the feed-side torque, or a magnetic slip clutch between motor and reel on the takeup side to limit or control the takeup torque.

If the motor is running at 25 times the takeup reel speed and you want the torque at the motor, yes, divide the reel torque by 25 (or multiply by 0.04). Torque ratio is the inverse of the speed ratio.

Cheers,
Tom

EDIT:
* For DC motors their speed is proportional to applied Voltage.
For AC synchronous motors, like you referenced at 1750RPM, their speed is proportional to the Frequency of the applied voltage.

Last edited: Nov 7, 2018 at 7:38 PM
7. Nov 5, 2018

Staff: Mentor

To be frank, your reply is a bit insulting, IMO. We are trying to help you given the sparse information you have posted.

Please post your mechanical drawing of your machine, including dimensions. Please post your feedback/gain equation for the control loop you have designed so far. And please clarify what you meant about your feedback and feedforward system -- I'm not understanding that part yet. Thank you.

8. Nov 6, 2018 at 4:34 AM

Stixoffire

I am not trying to be insulting - my apologies - I have just read about 100 different documents (maybe that is an under estimate) in the last 5 days and each one once it comes to load cell tension control - they all stop and declare the controller then controls the tension - no formulas of how to convert that, Which is why I am reaching out searching for an answer. I know there is probably some document somewhere that has the answer - after all this has been done in some fashion before.

9. Nov 6, 2018 at 10:11 AM

Dullard

Re-Read Tom.G's post. If tension is the target PV, then you have to think in those terms. The speed doesn't really matter - in simple terms: You want to 'speed up' when tension is low and slow down when it gets too high. The 'required speed' isn't (really) calculable because it's (almost entirely) a function of payoff/web friction/inertia. I don't mean to imply that math won't work - only that you'll probably never have the information that would be required to use it.

10. Nov 9, 2018 at 1:43 AM

Stixoffire

1: Maintain a given tension in the material on the takeup reel? YES.
2: It is a web, imagine a long roll of paper threaded down a machine..
3: Supply enough RPM's to the motor to maintain feed rate and also provide a constant tension on the take-up roll.

The feed rate is what is being provided by the other machine - I can not change the rate of feed from the other machine.

A VFD is being used to control the motor via the frequency reference.
Typically in these systems RPM is trimmed by an amount related to the load cell feedback ( at this point I am not sure if the values are guestimate percentage of RPM or a real life equation exists), either additional rpm or less rpm to control tension of the web.- I know that the material has a lot to do with the tension - but I am getting a feedback and I want that feedback to be maintained at x - so regardless of the material the feedback declares where I am at and I need to trim rpm one way or the other to achieve my set point.

I have my feed speed , I know my own speed (both commanded and via encoder feedback) , so I have that ratio of Feed/Takeup. I have the Tension in Lbs and Radius of the wind at all times in the process so I know Torque = Tension x Radius for process value PV and I also know what it is for Set Point - by simply placing the SP in for Tension in the formula.
I know the difference in torque required - but not the difference in rpm. I can get the actual motor torque that is being supplied via a feed back from the drive - so I do know current torque on the drive to the motor but I can't change the torque - the drive is in velocity mode - I can feed Frequency Reference.all I am allowed at present.

"Torque ratio is the inverse of the speed ratio" : this is helpful. I know the torque ratio and I also know the speed ratio Both the process value and also what the values should be at Set-point for given diameters.

Now having that information is there a way to utilize the inverse of torque ratio to quantify an RPM change - by taking the ratio TQratio (somehow inverting that and then speed ratio @ SP = feed rpm ratio . where I know what the feed rpm is and compute for takeup RPM ?

11. Nov 9, 2018 at 2:55 AM

Stixoffire

I really hate to fudge numbers as a guess of how much difference in RPM to trim.

I know my payoff speed, my take-up speed, my ratio's for both torque and rpm - what they should be at tension SP and also what they are at process value.
In a perfect world - where I follow at x speed , to vary that speed based on a web tension via a load cell - requires a trim rpm (or to be running the drive in torque mode). Since I am stuck with the drive in frequency mode - I need to trim rpm + or -.

Now the question with the data : knowing the torque ratio is inverse of rpm ratio and knowing what the values of those ratios are at for BOTH tension set point and tension process value, is there any way to perform some substitution / translation math ? So for example:

With actual values of : feed rate of 100 Feet PerMinute , Feed diameter (10 inches , R=5) , Take-up Diameter of 40 inches (R=20), Tension of 50 Lbs. , Tension SP of 40 lbs., I know RPMs will need to drop. I first need to calculate RPM at each location [ FPM / feed roller diameter ] and [ FPM / take-up diameter ]

Perhaps I could relate my torque ratio for SetPoint and ProcessVariable, calculate the variance between the two. Then using that variance as percent .

TQratio@SetPoint (known) / TQratio@ProcessVar(known) = ΔPercentage

ΔPercentage = RPMratio@SetPoint (known) / TargetRPMratio@ProcessVar(unknown)

TargetRPMratio@ProcessVar(unknown) = RPMratio@SetPoint (known) / ΔPercentage

TargetRPMratio@ProcessVar(Now known) = FeedRPM (known) / TargetRPM (unknown)

TargetRPM (unknown) = FeedRPM (known) / TargetRPMratio@ProcessVar(Now known)

12. Nov 9, 2018 at 3:09 AM

Tom.G

Ahh, enough information to come up with a speculative approach!

A few assumptions to get started:
1. The take-up motor is not grossly over-sized for the job.
2. You have, or can get, the detailed motor performance curves. From these you will determine the amount of slip at a specific torque. This will vary with speed.

Here is a first pass on a solution that may or may not feasible depending on your details; such as change in take-up reel size, motor characteristics... and probably a few other details.

Basic approach:
Use the variation of motor slip frequency versus torque.

The knowns are:
Web feed rate.
Gearing ratio between take-up reel and motor.
Take-up reel size (radius).
Desired web tension.
Measured web tension.
• Based on the take-up reel size, calculate the motor speed needed to gather the web at the known feed rate with zero tension.
• Using the motor curves and gearing ratio, do an intermediate calculation to add a smaller-than-needed additional torque to the take-up reel. This is just a safety factor to avoid a room full of webbing if/when things change a little.
• Compare Measured web tension with Desired web tension.
• Using the tension difference just calculated and the motor curves, find the slip frequency needed for the motor to produce the desired torque.
• Add (or subtract) this slip frequency to the currently command frequency sent to the VFD controller.
REMEMBER TO PUT IN FAIL-SAFES TO AVOID FULL SPEED OPERATION WHEN THE WEB BREAKS.
Maybe a dancer or optical sensor to avoid running without a web present.

For other than small changes, you will need to ramp the motor speed rather than use step changes. This is to avoid overshooting the tension and rupturing the web. If your VFD controller has ramping capability built-in it makes your programming easier.

You will find the control loop easier to stabilize if do NOT try to maintain tight control of the tension. There is probably a range of acceptable tension where you can use bang-bang control to keep tension in the central ±25% of the nominal range.

If you find the above impractical, change the VFD approach to a Torque motor and control that directly.

Cheers,
Tom

edit: looks like we were typing at the same time

13. Nov 10, 2018 at 1:30 AM

Tom.G

Just had a thought. The VFD drive for the take-up reel may already have torque control capability. Even if it doesn't, it may be more practical (cheaper) to just replace the VFD controller with one that does. That would save a bunch of programming and debugging!

Please let us know how this all works out.

Cheers,
Tom

14. Nov 10, 2018 at 8:09 AM

Stixoffire

It does indeed but my arms, hands and feet are tied.
I am in the the real world where the things you would do because it makes sense are not allowed because of non-sense.

So I trudged. and I have it working to some degree now ... but .. it could have been easier...

15. Nov 10, 2018 at 10:53 AM

Staff: Mentor

16. Nov 10, 2018 at 7:23 PM

Tom.G

Sounds like you are well on the road to success!
Any details you care to share about the easy/difficult aspects?
And if it's not confidential, what is the web material and dimension?

Tom