XRS

If you have a roll of paper 33 microns thick and wind it onto a core of 85mm diameter, what will be the diameter of the roll when there is 200m on it?

For extra marks show your workings

HankScorpio

Roughly 302mm inc core although I expect real world it would be a bit more.

CrisPDuk

Fcuked if I care

Our company has a separate department for that sort of thing

HankScorpio

3289 wraps is where you end up

mit

iTrader: (10)

How'd you get that? i make it about 570mm inc core?

ChefDude

so you start with Pi x 85mm for the length of one rotation, but each rotation/wrap is increased by 66 microns
85mm + 2 x (33 microns x (200m / (Pi x (85mm+[66microns extra for each new revolution complicated maths bit])))



do i get a prize?

HankScorpio

What CD said.

Circumference at 0 turns is 85mm x pi = 267.2427207 length of paper
Circumference at 1 turns is (85mm + 0.066) x pi = 267.4500658 length of paper
So after 2 turns, you've added 534.6927865 length of paper.

Extrapolating up to 2000000 mm of paper (thank you excel) and you get diameter of 217.008mm + your original 85mm.

I'm not convinced you should use the 33 microns absolutely as your diameter addition but that would require more thought than I would be prepared to give it.

Would be interested in the real/correct answer and method of calculation though.

Timwinner

iTrader: (3)

8.

AndyC_772

iTrader: (2)

Assume 1m width, Volume of paper = 33x10^-6 * 200 * 1 = 6.6x10^-3 m^3

Therefore, 6.6x10^-3 = PI*(r^2) - PI*(85x10^-3)^2

2.101x10^-3 + 7.225x10^-3= r^2

r = 96.6mm

Therefore, the diameter of the roll is 193mm.

speedking

I reckon:

Let the number of wraps = n

Circumference of first wrap = pi x d = pi x 85 = 267.0354mm

Each subsequent wrap the diameter increases by 2t = 2 x 0.033mm = 0.066mm.

The circumference increases by pi x .066 = 0.207345mm

For n wraps, the total length = n x 267.0354 + (the sum of 0 to n-1) x 0.207345mm

The sum of the first n integers is 0.5n(n+1)

So length = 267.0354n + 0.5(n-1)n x 0.207345 = 267.0354n + 0.103673n² - 0.103673n = 266.9317n + 0.103673n²

0.103673n² + 266.9317n – 200000 = 0

Solve the quadratic, n = 606.4, say 606.

The total diameter = 85 + 606 x .033 x 2 = 125mm.

We can check by saying that the average diameter = 105mm.

The average circumference = pi x 105 = 329.867229mm

Multiply this by 606 wraps = 199899.5mm.

QED

Depending how accurate you want it you must make allowance for the difference between the centreline distance and the inner circumference used above

speedking

HS you need a more up to date excel with the modern value for PI

HankScorpio

Ooops, that was 85.066, displaced by 1 turn so ammended.

Still 302 tho

WRX_Dazza

doesn't it depend where you take the measurement though !!!

think about it.

TelBoy

Do any of these solutions take account of the slightly increasing diameter of the tube with each subsequent layer of paper? You can't assume the diameter is a constant 85mm as the paper is wound on. It's a small difference, but it's not insignificant.

HankScorpio

Damn again, that's what happens when you're on the dole and only use excel once a month.

Yours is wrong though as C is different every time.


Using the amended 200k mm which I should have been all along - 125mm total diameter at 697 turns and 200m long length.

mit

iTrader: (10)

Good god!!!!!

HankScorpio

Yes.
Except corvid

TelBoy

Right ok, i've read speedking's solution more thoroughly. Just checking.

stara

weirdly, the extra length wound is approx .207mm greater than the previous wrap for every wrap. so wrap 100 is 20.07mm longer than wrap 1. wrap 750 would be 155mm longer than wrap 1. . ........ just look at speedkings answer

jaytc2003

iTrader: (1)

thats where you lost me

BOB.T

Is it 42?

speedking

Your method is good, unfortunately the execution is slightly awry

Ap = area of edge of paper = 200000 x 0.033mm = 6600mm²
Ac = area of core = pi x 85² / 4 = 5674.5mm²
Total area, A = 12274.5mm²
d = sqrt(4A/pi) = 125.0135mm

number of wraps = (125.0135-85)/0.066 = 606.2655

all as per my earlier answer

decepticon

surely all this pi and integers etc (though very clever) is completely un necessary for such a simple question?
quite simply:
area of original tube is 42.5x42.5x3.142=5674.5mm2
total area of paper is 200000mm x 0.033=6600mm2
add the 2 figures together = 12274.5mm2
12274.5/3.142=3907.09
square root of 3907.09 = 62.506757mm radius
125.01351mm diameter of roll
easy peasy

AndyC_772

iTrader: (2)

D'uh.... given the cold I'm currently suffering with it's a minor miracle I can even count to five, never mind algebra...

XRS

Well, it looks like that kept some of you amused on a wintry day

According to this website Calculator for Outside Diameter of Roll of Material (yes, there is a web site dedicated to this issue) the answer is 85.49cm. Unfortunately it doesn't give the workings.

Thanks for the input guys

HankScorpio

No, it doesn't.



Which would be, hhmmmm, what I said! (Once I had my m -> mm conversion correct )

speedking

You're welcome. With the correct values the website says d=125.014mm. I hope you're not responsible for specifying the storage space for 1000 such rolls, you will be a long way out in your assessment

@deception. Depends how accurate you want the answer. The area method is reducing the thickness of the outer layer. Not an issue in this example, but given say 4 wraps of something 100mm thick, a significant error could occur

XRS

I admit defeat I input the size of the core incorrectly

Don't worry, no lives were lost nor space under-specified, I just sent the web page to my brother who wanted the information in the first place.

Now to get back to retirement.........