brain teaser: paper mill problem

Quote: Originally posted by Magpie

lineal meters?

Quote: Originally posted by Fulmen

Volume core: pi * 0,05^2 = 0,00785m^2 Volume paper: 4000m * 0,15mm = 0,06m^2 Diameter roll: 2* root (0,06785/pi) = 0,880m Edit: Calculation error in last step. [Edited on 8-3-16 by Fulmen]

Quote: Originally posted by Magpie

Blogfast, I think you and I may have overthought this problem. The solution is very straightforward as shown by Fulmen.

Quote: Originally posted by Magpie

Somehow, magically, Fulmen did get the right answer even with the slipped decimal.

Quote: Originally posted by Fulmen

Quote: Originally posted by Magpie   Somehow, magically, Fulmen did get the right answer even with the slipped decimal. LOL. It's 2AM and I'm a bit tired. The miscalculation was in the second step, not the last.

Tired or not, your calc is correct. Congrats!

All this mathematical problem suppose that each layer of paper is fully adherent to the previous one --> no air gaps!


In the real world in such an event no sheet of paper could be taken away since adherence would be so strong that the roll wouldn't ne a roll anymore but be a full paper pipe ;-)

I did it by looking at the area on the end of the roll;

0.15*4,000,000=Pi*(x^2 - 50^2)

Where x is the radius of the tube.

This rearranges to;

Sqrt((600,000/Pi) + 2500) = x = 440mm

So the diameter is 880mm

It's the same as what Fulmen did, but he was talking about volume, then calculating area

@Blogfast, How do you do that nice lettering with sub/superscript and fancy Greek letters?

Quote: Originally posted by PHILOU Zrealone

All this mathematical problem suppose that each layer of paper is fully adherent to the previous one --> no air gaps!

Quote: Originally posted by Maker

It's the same as what Fulmen did, but he was talking about volume, then calculating area

Quote: Originally posted by PHILOU Zrealone

All this mathematical problem suppose that each layer of paper is fully adherent to the previous one --> no air gaps! In the real world in such an event no sheet of paper could be taken away since adherence would be so strong that the roll wouldn't ne a roll anymore but be a full paper pipe ;-)

Yes. But you might be surprised how tightly wound those paper rolls are! I worked in this industry for 11 years and this is an actual problem I was presented by the Finishing Room superintendent. The Finishing Room cuts the roll into rectangular sheets according to customer specifications.

After seeing Fulmen's elegant solution I am embarrassed to show my long-winded, over-thought, solution, but here it is for your enjoyment:

Solution:
Imagine the roll to be composed of an infinite number of thin shells (hollow cylinders) of thickness dx. Let x = the distance from the center of the roll to a shell.

The circumference of each shell = 2πx. Let A = the end area of the roll in m^2. Let the end area of each shell = dA.

Then dA = 2πxdx (equation 1)

Let L = the total length of the paper in m.
Then A = 0.15L/1000


Then dA = (0.15/1000)dL (equation 2)

Note: This "dA" is different than the one in equation 1 above but their integrals will be the same, ie, $$\int dA=A$$ in each case.

Equating equations 1 and 2 to eliminate $$\int dA$$

$$\int\frac {0.15}{1000}dL=\int 2πxdx$$

Let α =2π/(0.15/1000) = 41,888

Ie, the integral of dL = the integral of αxdx.

Integrating both sides to indefinite integrals

L = (αx^2)/2 + C (C is a constant of integration)

For the definite integrals:

L (0 to 4000) = (αx^2)/2 (0.05 to D/2)

4000 = (41,888/2)[(D/2)^2 – (0.05)^2] = 20,944[(D/2)^2 – 0.0025]

[(D/2)^2 – 0.0025] = 4000/20,944 = 0.1910

(D/2)^2 = 0.1885

D/2 = 0.4341; D = 0.868 m

@blogfast: Doing the above integral sign equations using LaTex just about drove me nuts!

Curious to see how well this method would work in a similar application I measured the ID and OD of a new roll of Scotch 33+ electrical tape. The package provided the thickness (0.177mm) and length (15.85m). Thank you 3M! My result was L = 15.7m. Not bad.


[Edited on 8-3-2016 by Magpie]

Quote: Originally posted by Magpie

@blogfast: Doing the above integral sign equations using LaTex just about drove me nuts! Curious to see how well this method would work in a similar application I measured the ID and OD of a new roll of Scotch 33+ electrical tape. The package provided the thickness (0.177mm) and length (15.85m). Thank you 3M! My result was L = 15.7m. Not bad.

Nice! So that's basically a third way...

Remember you can steal someone else's Latex, then modify it. Right click on any LaTex rendered formula and choose 'Latex'. A window opens that shows the 'code'. Cut 'n paste and modify to your heart's delight.

Most LaTex functions, here:

https://en.wikibooks.org/wiki/LaTeX/Mathematics

Proof reading? Go: https://www.mathjax.org/ and choose 'Live demo'. Cut and paste your roughs in there to see how it renders.


[Edited on 8-3-2016 by blogfast25]