Roll outer diameter calculation

[Artūrs]

Hello!

Please accept my apology if this is not the right category to post such a question.

Problem:
We are working in the field of manufacturing self-adhesive labels and recently I've been tasked with creating a formula for a roll outer diameter calculation.

Picture of the roll:


Known variables:
1. Strip layer thickness, µm
2. Roll (inner) core diameter, mm (strip starting point?)
3. Strip length, m (for strip ending point?). Calculated: label count * label height + distance between labels
4. Distance between layers = 0

I've made a simplified formula in Excel, however, it seems to be off by a margin of ~2%.
I would highly appreciate it if someone here could help me with this problem.

Thanks!

- Artūrs, the guy who did not pay the required attention to math at Uni.

 

[Dr.Peterson]

Hello!

Please accept my apology if this is not the right category to post such a question.

Problem:
We are working in the field of manufacturing self-adhesive labels and recently I've been tasked with creating a formula for a roll outer diameter calculation.

Picture of the roll:
View attachment 28405

Known variables:
1. Strip layer thickness, µm
2. Roll (inner) core diameter, mm (strip starting point?)
3. Strip length, m (for strip ending point?). Calculated: label count * label height + distance between labels
4. Distance between layers = 0

I've made a simplified formula in Excel, however, it seems to be off by a margin of ~2%.
I would highly appreciate it if someone here could help me with this problem.

Thanks!

- Artūrs, the guy who did not pay the required attention to math at Uni.

Please show us your formula. It may well be mathematically correct, or there may be a small correction we could suggest.

The hard part in doing this calculation in real life is that it is sensitive to the exact thickness, which is likely off a little due to the realities of how the material bends. When I have helped people with such calculations, I often recommend getting the thickness to use based on an actual roll (dividing the total thickness by the number of layers), or even to just multiply by a correction factor such as what you have found.

I would have put this under Geometry.

 

[Artūrs]

Please show us your formula. It may well be mathematically correct, or there may be a small correction we could suggest.

The hard part in doing this calculation in real life is that it is sensitive to the exact thickness, which is likely off a little due to the realities of how the material bends. When I have helped people with such calculations, I often recommend getting the thickness to use based on an actual roll (dividing the total thickness by the number of layers), or even to just multiply by a correction factor such as what you have found.

I would have put this under Geometry.

Thanks for your reply!

Here's an example with a formula (above the tables)
I've used the Outer diameter as a known variable to find out the length of the strip to find the number of labels in a roll.




Apparently, I've made a mistake with the Core diameter value - I did not consider the thickness of the actual carton core as a part of the overall core diameter, which made it 3 mm thicker and thus made the formula more precise:

The calculated number of labels on a roll: 6959
The actual number of labels on a roll: 7000

For smaller Outer diameter rolls the margin of error goes up to ~5%.
So a more precise alternative would be highly appreciated.

Considering the sensitiveness to the exact thickness made me think that aiming for precision might be unwise.
Perhaps, if there's no better alternative, I'll just add round up/down at the end.

 

[Dr.Peterson]

I'm trying to understand your formula and compare it to mine, but I'm missing some things.

I take it that C3 and C4 as the inputs, but the formula uses C15, whose meaning I can't figure out.

 

[Artūrs]

I'm trying to understand your formula and compare it to mine, but I'm missing some things.

I take it that C3 and C4 as the inputs, but the formula uses C15, whose meaning I can't figure out.

Here's the clarification
(C11 is not an input thought, in case of rolls it's always 0)

 

[Dr.Peterson]

I'm not sure what "core" diameter and "actual carton core" mean, but I suppose C15 is meant to be the radius of the actual inner layer of the strip, including a fixed 1.5 mm beyond the measured core.

Your formula seems to be what I would write as [imath]\frac{2\pi r+\pi(2r+nt)(n+1)}{h}[/imath], where r is the inner radius, t is the thickness, n is the number of layers, and h is the total distance taken up by one label. To put it differently, the total length is [imath]L = 2\pi r+\pi(2r+nt)(n-1)[/imath] which simplifies to [imath]L = \pi n(2r+(n-1)t)[/imath]. Where did you get your formula from?

Here is the formula I know, which comes from the fact that the cross-sectional area can be expressed as both [imath]\pi(R^2-r^2)[/imath] when rolled, and as [imath]Lt[/imath] when unrolled, where R is the outer radius and r is the inner radius:

[imath]L = \frac{\pi(R^2-r^2)}{t}[/imath]​

[As an aside, since [imath]R-r = nt[/imath], the total thickness of the n layers, this could also be written (if you knew the number of layers rather than the thickness) as

[imath]L = \pi(R+r)n[/imath]​

That's just the average circumference of a layer, times the number of layers.]

Taking [imath]R = 100[/imath], [imath]r = 41[/imath], and [imath]t = 0.134[/imath], this gives [imath]L = \frac{\pi(100^2-41^2)}{0.134} = 195036.6\text{ mm}[/imath], which means [imath]195036.6/28 = 6965.6\text{ labels}[/imath]. That's simpler and about as accurate in this example as yours.

I see that I can get something like your formula if I replace [imath]R[/imath] in mine with [imath](r + nt)[/imath]:

[imath]L = \frac{\pi((r + nt)^2-r^2)}{t} = \frac{\pi(2nrt + n^2t^2)}{t} = \pi n(2r + nt)[/imath]​

So yours is essentially equivalent to mine except that it accounts for one less layer for some reason.

The difference between my 6965.6 and your counted number of 7000 is probably accounted for by a tiny error in the thickness; if I round the number of layers to the whole number 440 and solve [imath]\frac{\pi 440(2(41) + 440t)}{28} = 7000[/imath] for the apparent actual thickness, I get 0.13589. A little air between layers could account for that.

 

[Artūrs]

I'm not sure what "core" diameter and "actual carton core" mean, but I suppose C15 is meant to be the radius of the actual inner layer of the strip, including a fixed 1.5 mm beyond the measured core.

Your formula seems to be what I would write as [imath]\frac{2\pi r+\pi(2r+nt)(n+1)}{h}[/imath], where r is the inner radius, t is the thickness, n is the number of layers, and h is the total distance taken up by one label. To put it differently, the total length is [imath]L = 2\pi r+\pi(2r+nt)(n-1)[/imath] which simplifies to [imath]L = \pi n(2r+(n-1)t)[/imath]. Where did you get your formula from?

Here is the formula I know, which comes from the fact that the cross-sectional area can be expressed as both [imath]\pi(R^2-r^2)[/imath] when rolled, and as [imath]Lt[/imath] when unrolled, where R is the outer radius and r is the inner radius:

[imath]L = \frac{\pi(R^2-r^2)}{t}[/imath]​

[As an aside, since [imath]R-r = nt[/imath], the total thickness of the n layers, this could also be written (if you knew the number of layers rather than the thickness) as

[imath]L = \pi(R+r)n[/imath]​

That's just the average circumference of a layer, times the number of layers.]

Taking [imath]R = 100[/imath], [imath]r = 41[/imath], and [imath]t = 0.134[/imath], this gives [imath]L = \frac{\pi(100^2-41^2)}{0.134} = 195036.6\text{ mm}[/imath], which means [imath]195036.6/28 = 6965.6\text{ labels}[/imath]. That's simpler and about as accurate in this example as yours.

I see that I can get something like your formula if I replace [imath]R[/imath] in mine with [imath](r + nt)[/imath]:

[imath]L = \frac{\pi((r + nt)^2-r^2)}{t} = \frac{\pi(2nrt + n^2t^2)}{t} = \pi n(2r + nt)[/imath]​

So yours is essentially equivalent to mine except that it accounts for one less layer for some reason.

The difference between my 6965.6 and your counted number of 7000 is probably accounted for by a tiny error in the thickness; if I round the number of layers to the whole number 440 and solve [imath]\frac{\pi 440(2(41) + 440t)}{28} = 7000[/imath] for the apparent actual thickness, I get 0.13589. A little air between layers could account for that.

Thank you for your help!
I will try to implement your suggested method and follow up if other questions arise.
Sorry for the unclear explanation.

 

[Artūrs]

Thank you for your help!
I will try to implement your suggested method and follow up if other questions arise.
Sorry for the unclear explanation.

Just a quick follow-up.
Everything works fine with an insignificant error margin. Thank you very much! Your help is highly appreciated!