Unwind/Rewind Math Questions

[L D[AR2P#0.0]]

I may have missed it in the previous posts, but how/where do you set the required tension in the material?

 

[Puddle]

OP said in the first post there's no tension measurement on the machine. I would imagine this would be a much simpler task if there was.

I'm just watching and lurking as I'm also working on a re-roll project, just very slowly around production.

 

[MaxK]

The third-grader solution of the task imagined by me (the scheme is attached)

(the task is simplified as much as possible, to understand the approach to the solution, or I'm just too lazy to solve)

R is the radius of the roll of all material wound on the shaft
r0 - shafts radii
r1, r2 are the radii of rolls 1 and 2 at some point in time
w1, w2 are the angular velocities of rolls 1 and 2 at some point in time
h - material thickness
p is the density of the material and shafts
L - width of material and shafts
J - torque
E - rotational energy
F - friction force

J = (m * r^2) / 2 - solid cylinder

E = (J * w^2) / 2 = (m * r^2 * w^2) / 4 = (p*L*pi*r^2 * r^2 * w^2) / 4 = (p *L*pi* r^4 * w^2) / 4

Total rotational energy of the system

E = p*L*pi* (r1^4 * w1^2 + r2^4 * w2^2) / 4

w2 = w1 * r1 / r2

E = p*L*pi* (r1^4 * w1^2 + r2^4 * (w1 * r1 / r2)^2) / 4 = p*L*pi* (r1^4 * w1^2 + r2 ^2 * w1^2 * r1^2) / 4

r2^2 = R^2 + r0^2 – r1^2

E = p*L*pi* (r1^4 * w1^2 + (R^2 + r0^2 – r1^2) * w1^2 * r1^2) / 4

Find the radius of the roll 1 r1 at which it is necessary to apply a brake shoe to create a constant friction force F

E = F * l = F * pi * (R^2 – r1^2) / h

E=E

p*L*pi* (r1^4 * w1^2 + (R^2 + r0^2 – r1^2) * w1^2 * r1^2) / 4 = F * pi * (R^2 – r1 ^2) / h

p*L* (r1^4 * w1^2 + (R^2 + r0^2 – r1^2) * w1^2 * r1^2) / 4 – F * (R^2 – r1^2) / h = 0

(p*L*(R^2 + r0^2) * w1^2 / (4*F) + 1 / h) * r1^2 – R^2 / h = 0

r1 = R/ (h*p*L*(R^2 + r0^2) * w1^2 / (4*F) + 1)^0.5

At the moment when the roll 1 radius becomes = calculated r1, you need to turn off the drive and apply a brake shoe with friction force F


Linear increase in friction force from 0 to 2* F followed by a linear decrease to 0

l * h = pi * (R^2 – r1^2)
0.5 * l * h = pi * (R^2 – r1x^2)
0.5 * (R^2 – r1^2) = (R^2 – r1x^2)
r1x = (0.5 * (R^2 + r1^2))^0.5

If
Measured roll radius 1 > r1
Then

If
Measured roll radius 1 < r1x
Then
Target braking force = (Measured roll radius 1 - r1) / (r1x –r1) * 2* F
Else
Target braking force = (R - Measured roll radius 1) / (R -r1x) * 2* F
End if

End if

 

[drbitboy]

...but Blue92 writes

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

Ah, I see. Yes, that is an idiom (ambiguity?) of English.

It might be better written "the unwind/rewind shafts spin at a maximum [of] only 400 rpm" i.e. the "only" is a modifier to "maximum" that means "only this fast, which isn't very fast, and no more" i.e. the general sense of the whole sentence is that "even at maximum speed these shafts spin slowly."

 

[L D[AR2P#0.0]]

OP said in the first post there's no tension measurement on the machine.

Yes, but you still need to be able to set it.

 

[drbitboy]

Yes, but you still need to be able to set it.

I've been assuming one drive is set for speed control, the other is configured to be DRC* torque-limited at the DRC* equivalent speed ± a few percent, so the torque-limited drive never gets to its speed setpoint, can only apply so much torque to try to get there, and that torque provides the tension. Isn't that a common practice with web handling?

* DRC = Diameter-Ratio-Corrected

 

[MaxK]

I've been assuming one drive is set for speed control, the other is configured to be DRC* torque-limited at the DRC* equivalent speed ± a few percent, so the torque-limited drive never gets to its speed setpoint, can only apply so much torque to try to get there, and that torque provides the tension. Isn't that a common practice with web handling?

* DRC = Diameter-Ratio-Corrected

I think that the issue may be more interesting than it seems. (that is why I ask Blue92 for details about it machine)

I will explain in the picture #33

The winding cycle starts when there is nothing on the red shaft (roll 1). I.e. it is not too difficult for the red shaft drive to reach the set speed.

Further, in the process of winding, in order to maintain the angular velocity of the red shaft, some “force” (rather steady) must be applied. This “force” is quite possibly enough to maintain the required tension on the material (even without the effect of the blue shaft roll 2)

It is obvious that there is some point when the tension of the red shaft (roll1) is no longer enough to prevent the material from vibration.

strictly speaking an interesting task:
Calculate the "point" of the beginning of braking by the blue shaft (roll 2) providing
- prevention of vibrations of the material
- relative uniformity of material tension (winding thickness)

According to the algorithm:
-Up to a certain radius, the red shaft drive maintains a certain angular velocity
-after a certain radius, the blue shaft drive maintains some "braking torque"

 

[Blue92]

I've been assuming one drive is set for speed control, the other is configured to be DRC* torque-limited at the DRC* equivalent speed ± a few percent, so the torque-limited drive never gets to its speed setpoint, can only apply so much torque to try to get there, and that torque provides the tension. Isn't that a common practice with web handling?

* DRC = Diameter-Ratio-Corrected

Bingo. In the end, turns out we had a function block saved from a previous project that essentially serves the same purpose as @MaxK's formula above.