Magpie
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brain teaser: paper mill problem
Paper coming off a paper machine is wound into a roll on a cardboard core of 10cm diameter. The paper caliper (thickness) is 0.15mm. What diameter
roll of paper is needed to contain 4000 lineal meters?
The single most important condition for a successful synthesis is good mixing - Nicodem
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blogfast25
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You mean SI meters?
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Fulmen
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Volume core: pi * 0,05^2 = 0,00785m^2
Volume paper: 4000m * 0,15mm = 0,6m^2
Diameter roll: 2* root (0,60785/pi) = 0,880m
Edit: Calculation error in second step.
Editedit: Corrected my previous correction.
[Edited on 8-3-16 by Fulmen]
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blogfast25
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I'm getting 0.335 m.
Reasoning to follow.
[Edited on 8-3-2016 by blogfast25]
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Magpie
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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] |
Fulmen, did you slip a decimal point in the area of the paper, ie, should it be 0.6m^2.
-------------------------------------
Somehow, magically, Fulmen did get the right answer even with the slipped decimal.
Blogfast, I think you and I may have overthought this problem. The solution is very straightforward as shown by Fulmen.
[Edited on 8-3-2016 by Magpie]
The single most important condition for a successful synthesis is good mixing - Nicodem
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DraconicAcid
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I get 49.8 cm radius, so 98.4 cm diameter.
Please remember: "Filtrate" is not a verb.
Write up your lab reports the way your instructor wants them, not the way your ex-instructor wants them.
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blogfast25
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| 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.
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I don't see how it can be done that way.
My reasoning:
Radius for n winds:
$$R_n=R_0+n\tau$$
with $$\tau=0.15\:\mathrm{mm}, R_0=0.05\:\mathrm{m}$$
Length of n th wind:
$$L_n=2\pi(R_0+n\tau)$$
Total length of n winds:
$$L=\Sigma_0^n[2\pi(R_0+n\tau)]$$
$$L=\frac{\pi}{\tau}\big[(R_0+n\tau)^2-R_0^2\big]$$
Reworked:
$$\pi\tau n^2+2\pi R_0n-L=0$$
$$L=4000\:\mathrm{m}$$
The positive root is:
$$n=2600$$
$$R_{2600}=0.5+0.00015 \times 2600=0.44\mathrm{m}$$
or a diameter of 0.88 m.
Need to run some checks now...
[Edited on 8-3-2016 by blogfast25]
[Edited on 8-3-2016 by blogfast25]
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Fulmen
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LOL. It's 2AM and I'm a bit tired. The miscalculation was in the second step, not the last.
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blogfast25
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Tired or not, your calc is correct. Congrats! 
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DraconicAcid
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Oh, I see where I went wrong. My reasoning was the same as Fulmen's- the cross sectional area of the paper is 4000 m x 0.15 mm, which is 6000 cm^2.
Wrapped around a cylinder of radius 5 cm, this gives a total radius of 44.8 cm (to which I randomly added 5 cm for the radius of the core, although
that was already taken care of).
Please remember: "Filtrate" is not a verb.
Write up your lab reports the way your instructor wants them, not the way your ex-instructor wants them.
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Fulmen
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Well, Blogfast figured it out too, even though he went the long route. It's interesting to see how one can solve the same problem in entirely
different ways, I wonder if there are more solutions to this.
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PHILOU Zrealone
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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 ;-)
PH Z (PHILOU Zrealone)
"Physic is all what never works; Chemistry is all what stinks and explodes!"-"Life that deadly disease, sexually transmitted."(W.Allen)
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Maker
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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?
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blogfast25
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Yawn. It's a BRAIN TEASER!
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Fulmen
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*sigh*
As I've said, it as 2AM and I was barely awake. I'm not correcting my corrections to my first correction, I fear I'll only introduce more errors :-)
We're not banging rocks together here. We know how to put a man back together.
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Magpie
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| 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]
The single most important condition for a successful synthesis is good mixing - Nicodem
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blogfast25
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| 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.
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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]
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zed
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Hmmm. How about the air between the layers? Do you remove it via Vacuum and/or compression? Might be quite a bit of volume. Oh, I see the guys
thought of that.
Me, I'm a weak math type of guy. I would calculate Wt/Meter, and then weigh out 4000 lineal meters worth.
Phewww. Glad I didn't try to twist my head around that one.
I'm going to reward myself with an ice-cream cone.
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