Unwind/Rewind Math Questions

[drbitboy]

Trying to do 3rd order ramps in a PLCs is asking for abuse.

https://deltamotion.com/peter/wxMaxima/Seg1234567.html
And that is only ONE case.

Meh. Both of those are mostly bookkeeping. The only trick is figuring out the value of unity I e. the value of 1.

 

[drbitboy]

And I don't have to dodge anything

 

[Blue92]

Which is why I provided the graphical approach to the same solution. Was that lost on you too?

I understand the graphs. I guess Im having trouble seeing how I can compress all of this into a function block in Codesys.

 

[Peter Nachtwey]

Meh. Both of those are mostly bookkeeping. The only trick is figuring out the value of unity I e. the value of 1.

That is easy for you to say. You are not writing the code in ladder.

 

[drbitboy]

That is easy for you to say. You are not writing the code in ladder.

Meh, this is one case, so it will not be hard, OP says codesys, so probably ST, which will be trivial. Ladder would be much more aesthetic (or maybe anaesthetic lol).

 

[Blue92]

It will be ladder.

 

[MaxK]

What device(s) is used to accelerate / decelerate?
How is the acceleration currently going (curve/ramp...)?
How is the braking currently going?

 

[drbitboy]

Part IIIa

For the constant acceleration case, i.e. not jerk-limited:

​

We know the following a priori parameters

• D_CORE, unwind
• k', ratio of RPM to rate of change of diameter (empirically determined; negative)
  
• 400RPM, unwind speed at start of deceleration
• Δτ, duration of deceleration (chosen)

From those we calculate the diameter parameter, D₀, at which to start the deceleration:

• D₀ = D_CORE - ((400RPM÷k') × Δτ ÷ 2

Then the ladder logic, once it detects the the measured unwind diameter is less than D₀, and each time it measures the unwind diameter after that, makes the following calculations:

1. f = √((D_CORE - diameter) ÷ (400RPM÷k'))
2. v = (400RPM÷k') × f
3. RPM = v × k'

Where

f is a transformed time parameter
v is the target rate of change of the diameter
RPM is the output target unwind motor speed to achieve v
​

That is the simple part. The messy part will be handling edge cases (e.g. what if measured diameter is less than D_CORE, which would pass a negative operand to the square root instruction), measurement noise, etc.

 

[MaxK]

A center driven unwind feeding to a center driven rewind.
The unwind/rewind shafts only spin at a maximum 400 rpm, accel/decel rates will be operator inputs, but assume pretty slow.

How is this machine arranged?

 

[MaxK]

@Blue92, are my questions unclear? What exactly?

@drbitboy you solving the problem. What is the design of the machine for which the problem is being solved?

 

[drbitboy]

@Blue92, are my questions unclear? What exactly?

@drbitboy you solving the problem. What is the design of the machine for which the problem is being solved?

The original problem was understanding the math. I tried to fix that by moving from calculus domain to the graphical+algebra domain; the resulting formulae are the same, and are ultimately what the PLC has to implement.


@Maxk: did you read Post #1?

 

[MaxK]

@Maxk: did you read Post #1?

I read it 10 times or more, but didn’t understand – it’s very convenient to hide my-own stupidity and inattention behind the language barrier )))

A center driven unwind feeding to a center driven rewind.

"center driven" - [FONT=&quot]does this mean that the material is being wound/ unwound on/off the shaft?[/FONT]

[FONT=&quot]If this is true, [/FONT]

The unwind/rewind shafts only spin at a maximum 400 rpm, accel/decel rates will be operator inputs, but assume pretty slow.

[FONT=&quot]then there are two rolls of different diameters, interconnected by material, while the rotation speed of their shafts is the same[/FONT]

 

[drbitboy]

"center driven" - [FONT=&quot]does this mean that the material is being wound/ unwound on/off the shaft?[/FONT]

See e.g. here.


I am pretty sure that "X-driven" refers to the motive force that rotates each shaft.

Center-driven force is a torque applied at the center of the winder shaft i.e. a motor (or motor-driven gearbox) shaft in-line with the axis of the unwind and rewind shafts.

The alternatives are a surface-driven shaft i.e. a moving surface pressing against the material rolled around the shaft, and I suppose a nip (material coming off the shaft between a driven roller and an idler roller) would be another option for pulling material from the unwind side.

A center-driven unwind controls the rotation speed (RPM) of the shaft directly; a surface-driven unwind controls the material speed (e.g. feet per minute) directly.

 

[drbitboy]

[FONT=&quot]then there are two rolls of different diameters, interconnected by material, while the rotation speed of their shafts is the same[/FONT]

Not quite.


The two rolls are running at different speeds, driven by different motors. The ratio of their speed is the inverse of the ratio of the current diameters of the material leaving or loading the winders, because there are no dancers i.e. no accumulation of the material between the winders, so one linear foot of material coming off the unwind will mean one foot of material loading onto the rewind.

 

[MaxK]

Not quite.


The two rolls are running at different speeds, driven by different motors. The ratio of their speed is the inverse of the ratio of the current diameters of the material leaving or loading the winders, because there are no dancers i.e. no accumulation of the material between the winders, so one linear foot of material coming off the unwind will mean one foot of material loading onto the rewind.

I DO understand that this is how it should be, but Blue92 writes

“The unwind/rewind shafts only spin at a maximum 400 rpm, “

That is why I am asking Blue92 to give a clear description of his machine. And how it works now.