How thick is one sheet of paper problem !

[Handsonic]

Sorry, I'm new in this community, Scienceforums.net looks interesting so I JOINED !

How to solve this !? I'm not looking for an answer I just want to know how to solve it, formulas or etc? what do I need to do, sorry if my English is bad I'm still an ESLC Student ! in 11th Grade ! correct me if I spell something wrong !

Thanks xD

 

 

By the way, I may be in the wrong site ! Because this is Science ! not Geometry ! I apologize If I cannot post Homework about other subjects !

Edited April 6, 2015 by Handsonic

[studiot]

Hello and welcome, handsonic.

 

No it is fine to post here any sort of technical question.

 

But surely this is beyond 11th grade?

 

What you are asking is the length of part of a spiral.

Can you see that the paper forms a spiral round the inner tube with total length equal to the number of sheets times the length of a sheet?

 

Can you tell us what sort of approximate methods you can use to find and answer?

 

PS Is this really 11th grade homework or a challenge of some sort?

Edited April 6, 2015 by studiot

[Handsonic]

Hello and welcome, handsonic.

No it is fine to post here any sort of technical question.

But surely this is beyond 11th grade?

What you are asking is the length of part of a spiral.

Can you see that the paper forms a spiral round the inner tube with total length equal to the number of sheets times the length of a sheet?

Can you tell us what sort of approximate methods you can use to find and answer?

PS Is this really 11th grade homework or a challenge of some sort?

Thanks for replying !

Q1: Can you see that the paper forms a spiral round the inner tube with total length equal to the number of sheets times the length of a sheet?

Yes, Of course If the paper it is 3 inches times 3 inches and they are 500 pieces of paper, it would have a total length of 1,500 inches? (Correct me in If I am wrong)

 

Q2: Can you tell us what sort of approximate methods you can use to find and answer?

It is very complicated to me try to explain in English but ill try, Okay so I was trying to divide the area of the circle (the paper one without the paper board roll) by the numbers of sheet of paper, because in this problem they only provide the numbers of paper the size (sorry if I'm using a wrong term to mean 3 inches times 3 inches) and the diameter of the toilet paper, I was using area formula circles to try to figure out (total area of the circle where the paper covers only it is apparently 43.19689899), But any idea how can I get the thick (if there a higher level word for "Thick" please let me know) How can I divide 500 sheets of paper in area ? They are 2 different types of measures? How to? It is obvious that the area of a 3*3 Square its 9. Any hint or something? also I was wondering if exist a formula for this spiral? Also I know that the area of the inner roll (lets use some vocabulary from you "inner") It is 7.06858, I was trying to add the length of the paper to the perimeter of the inner roll Example, the perimeter is 9.42477796 (pi times d) which means that to complete a spiral with the paper we are going to need, 3.14 pieces of paper? (the perimeter divided by the length of the sheet of paper?), I think I am getting too much useless information to this, I was looking (in google ) for formulas to know the thickness of a paper, but I couldn't find something to figure out, I found many calculators to know different things about spiral, length, etc.

Q3: PS Is this really 11th grade homework or a challenge of some sort?

It is a homework, but it is a challenge too. My Geometry teacher always do similar hard questions like this, but this is very hard to solve. I suppose I am lacking some type of information or formula to solve this, that is the main reason that I am asking to try to learn something also this is not a graded homework. (In addiction he did not provided any type of hints or tips to solve this, just this information and the question)

PS: Thanks to reminding me to use "th" at the end of "11" thanks, I don't know how did I miss that.

Edited April 6, 2015 by Handsonic

[studiot]

If r is the radius of the inner tube then its circumference is [math]2\pi \left( r \right)[/math]
After the first time we have wrapped the paper, of thickness t, once round it completely the new circumference is [math]2\pi \left( {r + t} \right)[/math]
So the second time we wrap the paper completely round its length is [math]2\pi \left( r \right) + 2\pi \left( {r + t} \right)[/math]
The third time we wrap around the circumference has grown again to [math]2\pi \left( {r + 2t} \right)[/math]
So the length of paper is [math]2\pi \left( r \right) + 2\pi \left( {r + t} \right) + 2\pi \left( {r + 2t} \right)[/math]

So if we wrap around n times the circumference is edit [math]2\pi \left( {r + \left( {n - 1} \right)t} \right)[/math]
If you develop this it will give you a formula for the length of the paper in terms of the thickness and the number of whole wraps around.

 

Sorry had to edit the LaTex

 

This may be an 11th grade sum, I don't know your system.

Edited April 7, 2015 by studiot

[Handsonic]

If r is the radius of the inner tube then its circumference is [math]2\pi (r)[/math]

After the first time we have wrapped the paper, of thickness t, once round it completely the new circumference is [math]2\pi (r + t)[/math]

So the second time we wrap the paper completely round its length is [math]2\pi (r) + 2\pi (r + t)[/math]

The third time we wrap around the circumference has grown again to [math]2\pi (r + 2t)[/math]

So the length of paper is [math]2\pi (r) + 2\pi (r + t) + 2\pi (r + 2t)[/math]

 

So if we wrap around n times the circumference is [math]2\pi r(n - 1)t[/math]

 

If you develop this it will give you a formula for the length of the paper in terms of the thickness and the number of whole wraps around.

 

Hrm, So after I finish I will know how many wraps It has, which is proportional to the thick of the paper warped in an area of the circle right?

So it will be like, in the outer roll of paper is 2.5, after I got the number of rolls that the paper has, I just have the divide the numbers of the rolls that the paper has into 2.5 inches..? So that will be the thick?

Sorry, I took long to reply, I was reading and thinking about what you said, I was meditating what I was going to post, so I wont sound stupid because I know I'm just a High School student and you are really brilliant people, So If I say something stupid ! You may think that you are loosing your time, but I really appreciate the time that you are spending on trying to help me on this !

Also I joined to this community, I don't think I will be able to respond or solve many (like 99.9%) of the questions that are posted here, because I'm lacking some knowledge that you already know, But I will be reading yours! (not ours sorry EDIT !) concepts and ideas, because I love to learn ! xD

PS: Sorry if I post "xD" or emoticons like this, apparently this is a very serious forum ! But I cannot contain !

Edited April 7, 2015 by Handsonic

[studiot]

I should perhaps note that the shape is not a true spiral.

 

The first time the paper is wrapped round it is wrapped round a perfect circle since it touches the circular inner tube all the way round.

Where is come back to its start point the paper suddenly increases its radius by exactly one thickness of paper and again transits round in a perfect circle.

and so on.

 

We can, of course, form a second equation connecting the number of wraps or circles (n) and the paper thickness (t)

since each thickness is stacked side by side between the inner and outer radii of the paper.

 

We thus have two equations connecting n and t and can therefore solve them.

 

Please note the area of the paper is not relevent to this exercise.

It is the lenght that matters.

 

Of course the number of sheets times the sheet length still equals the total length.

[Handsonic]

 

I should perhaps note that the shape is not a true spiral.

 

The first time the paper is wrapped round it is wrapped round a perfect circle since it touches the circular inner tube all the way round.

Where is come back to its start point the paper suddenly increases its radius by exactly one thickness of paper and again transits round in a perfect circle.

and so on.

 

We can, of course, form a second equation connecting the number of wraps or circles (n) and the paper thickness (t)

since each thickness is stacked side by side between the inner and outer radii of the paper.

 

We thus have two equations connecting n and t and can therefore solve them.

 

Please note the area of the paper is not relevent to this exercise.

It is the lenght that matters.

 

Of course the number of sheets times the sheet length still equals the total length.

 

 

Working on it, also now that you say the area is not relevant, I can see the problem in a different point of view.

Edited April 7, 2015 by Handsonic

[studiot]

This question has interested others, but I can't post a full solution until after your deadline.

 

I made the thickness 0.00627 inches and the number of wraps 159.48, but of course my arithmetic is notedly fickle.

 

I can however help you further to develop a solution, though I can see you are a thinking sort of student who like to do for himself where possible.

 

So ask as many question as you like.

How are you getting on putting my post#4 formulae into an equation?

 

You will need to know what the sum to n terms for the natural or counting numbers is

 

1 + 2 + 3 + 4 +...............n = 1/2n(n+1)

 

So the sum of the first six terms is 1+2+3+4+5+6 = 0.5(6)(7) = 21

 

Do you understand summing series?

 

You will need to make a slight adjustment to the above to allow for the fact that we are starting at zero not 1

 

Can you do this?

Ask if you need help here.

Edited April 7, 2015 by studiot

[imatfaal]

could you not also use a cross sectional area calc

 

CSArea = Length * thickness

 

Also

 

CSArea = Area outer circle - Area inner circle

 

ie

 

Length * thickness = Area Outer Circle - Area Inner Circle

 

[latex]

 

(250*3)*thickness= \pi \cdot r^2_{outer} - \pi \cdot r^2_{inner}

[/latex]

 

edit

 

just noticed I have 250 for number of sheets - it should be 500 from OP

Edited April 7, 2015 by imatfaal
typo

[studiot]

I said others were interested.

It is a good problem, admitting of many methods of attack.

 

How did your numbers work out imatfaal?

Edited April 7, 2015 by studiot

[imatfaal]

I said others were interested.

It is a good problem, admitting of many methods of attack.

 

How did your numbers work out imatfaal?

 

Just over 8 thousandths of an inch (I think this might be the first time I have ever done a calc in thousandths of an inch)

 

http://www.wolframalpha.com/input/?i=%28500*3%29*t%3D%28pi*2.5^2%29-%28pi*1.5^2%29

[Greg H.]

could you not also use a cross sectional area calc

 

CSArea = Length * thickness

 

Also

 

CSArea = Area outer circle - Area inner circle

 

ie

 

Length * thickness = Area Outer Circle - Area Inner Circle

 

[latex](500*3)*thickness= \pi \cdot r^2_{outer} - \pi \cdot r^2_{inner}[/latex]

 

edit

 

just noticed I have 250 for number of sheets - it should be 500 from OP

Building on imatfaal's work, we can rewrite the equation in terms of thickness:

[latex](500*3)*thickness= \pi \cdot r^2_{outer} - \pi \cdot r^2_{inner}[/latex]

[latex](1500)*thickness= \pi \cdot (r^2_{outer} - r^2_{inner})[/latex]

[latex]thickness= \frac {\pi \cdot (r^2_{outer} - r^2_{inner})}{1500}[/latex]

 

I think.

Edit - plugging my formula in terms of thickness into wolfram, gives a similar answer to imfataal's.

 

http://www.wolframalpha.com/input/?i=t%3D((pi*(2.5^2+-+1.5^2))%2F1500)

 

As a side note, this is reasonably close to the thickness this guy got on his roll.

Edited April 7, 2015 by Greg H.

[studiot]

OK I now have 8thou after arithmetic correction by my series method, which is really also quite simple.

 

Since imatfaal has posted a valid method (I like it +1) here is mine.

 

 

Edited April 7, 2015 by studiot

[pavelcherepan]

I was thinking along the same lines with imatfaal, but I went with volume calculations.

 

[latex]V_{roll}= \pi r_{outer}^2*h - \pi r_{inner}^2*h = 3.14*2.5^2*3-3.14*1.5^2*3 = 37.68[/latex]

 

[latex]V_{sheets}=500*3*3*x = 4500x[/latex]

 

[latex]x = \frac{37.68}{4500} = 0.008373 \,in[/latex]

 

 

[imatfaal]

[latex]750=\int_{0}^{\theta}\sqrt{(1.5+(t/2 \pi)\theta)^2+(t/2 \pi)^2}d\theta[/latex]

 

[latex]2.5-1.5=\frac{\theta*t}{2 \pi}[/latex]

 

There we are two equations and two unknowns. This is the arc length of an archimedean spiral

[studiot]

I noted that the actual curve is not a true spiral or any regular curve.

[imatfaal]

I noted that the actual curve is not a true spiral or any regular curve.

 

I saw that - but also felt that it would approximate an Archimedean Spiral quite closely. I was looking for a third completely seperate method of modelling the answer

[michel123456]

As noted above, this is not a spiral, and the spiral approach is seemingly too complicated. Most probably intended.

It is a simple volume problem.

imagine simply cut the whole roll radially with a knife and open it, you obtain an open "cake" that has a specific volume. if you divide the volume of this "open cake" by the volume of a single sheet you obtain the number of sheets. If instead of the volume of a single sheet you are given the number of sheets then you can calculate the thickness. And since, as noted above, the width of the "cake" is the same as the width of a sheet, then it becomes even more simple, it is a surface problem. As solved above.

[studiot]

Unfortunately you cannot roll the paper round a cylinder without distortion if it has any thickness whatsoever.

In order to force it to lie 'flat' (contiguous) one side must be longer than the other.

[michel123456]

Unfortunately you cannot roll the paper round a cylinder without distortion if it has any thickness whatsoever.

In order to force it to lie 'flat' (contiguous) one side must be longer than the other.

So what?

The shape of the paper bears no importance. It could be crumpled all around.

Edited April 8, 2015 by michel123456

[studiot]

Because if part of it is stretched, or more difficult- compressed, then its volume and surface area changes.

Edited April 8, 2015 by studiot

[Greg H.]

Unfortunately you cannot roll the paper round a cylinder without distortion if it has any thickness whatsoever.

In order to force it to lie 'flat' (contiguous) one side must be longer than the other.

 

You could also simplypull the sheets apart one at a time and stack them up. Each sheet is a standard size (according to the packaging), so you end up with a rectangular solid, and the volume of each sheet becomes

 

[latex]

V_{sheet} = \frac{V_{solid}}{sheetcount}

[/latex]

[studiot]

The volume of the stack will not match the volume between the two cylinders exactly.

[Greg H.]

Because if part of it is stretched, or more difficult- compressed, then its volume and surface area changes.

 

Then to get an accurate value for the actual thickness of each sheet, we have to eliminate the roll - otherwise we risk corrupting our result.

 

Also, you would have to compress the sheets before you do your thickness measure anyway - toilet tissue is manufactured to be fluffly, not flat.

[studiot]

I think you misunderstand what I mean.

 

In tech drawing and differential geometry we can open out the surface of a cylinder (map it) to a plane because it has zero thickness.

 

As soon as we have a rectangular parallelepiped of substance the thickness becomes important when we bend it in any way.

 

I could add a 'transition curve' to the junction of each wrap in my series solution and also calculate for the centreline or mean radius but that would add complications of higher order, small in overall effect but large in difficulty.