Math Is Fun Forum

Muddyshoes

Member

Registered: 2009-02-10

Posts: 1

How much cardboard do I have?

Help me, Obi-Math-Kinobi.. you're my only hope...

I was hoping that you good folk could help me with a conundrum.

I have a roll of corrugated cardboard, used for construction.  It's used by taping it down over the floor during construction to help protect the floor from scratches and dirt.  I bought an unknown  quantity of it and am trying to figure out how much I have...

The cardboard itself is exactly 1/8 inch thick.  The diameter of the roll is 18 inches.  There is a core of 2 inches.  The width of the cardboard roll is 30 inches.

I'm trying to figure out:

1. How long the roll is.
2. What is the total square footage of the cardboard in the roll.

Any help will be greatly appreciated.  Unfortunately, I am not a math guru and will be happy to nod at your explanation of formulae and functions, but honestly, I'll probably have no clue how you worked your magic.

Thank you in advance!

Ron

mathsmypassion

Member

Registered: 2008-12-01

Posts: 33

Re: How much cardboard do I have?

I am going to offer you an aproximation of the amount of cardboard you have. This is the best I can do.
If the core is 2 inches in diameter than the length of the roll is aproximately:

L = pi × (2 × n + (n-1)×n/16) inches
where n is the number of times the cardboard is rolled (count the circles inside)
For 128 circles (and this the theoretical number of circles) the length would be around 3996.1 inches = 333 feet.

The total area of the roll :
L × 30 inches²  or 3996.1 × 30 inches ² = 832.52 feet ² if it is too hard to count the circles.
I would take 5% away of this number, just in case.
I would say you have almost 790.9 feet² of cardboard.

Last edited by mathsmypassion (2009-02-10 11:42:42)

janetpc2008

Member

Registered: 2009-02-03

Posts: 7

Re: How much cardboard do I have?

mathsmypassion - I think you have made several mistakes.

First, if the roll diameter is 18inches, with a 2inch core, then the radius is 9inches, with 1inch not used, leaving 8 inches to be wrapped. With 1/8inch per wrap, that gives 64 wraps (for your N) - NOT 128.

Second, the formula should be worked out as:
each wrap is 2 pi R
the first wrap is approximately the radius 1and1/8inches.
the next wrap is then the radius 1and2/8inches
and so on, giving the summation
1+1/8 + 1+2/8 + 1+3/8+ .... + 1+N/8
which equals
N*1 +1/8+2/8+3/8+....+N/8
=N+1/8*(N*(N+1))/2  -----------you used N*(N-1)
=N+N*(N+1)/16

Now put the summation with the circumference formula and we have the total length as:

2 pi *{ N + N*(N+1)/16}
=pi * {2N + N*(N+1)/8}

As you can see, my answer differs from yours mainly because I have 1/8 and you had 1/16.

So I would say that the roll is theoretically (using N=64) 2036inches long, or 170feet

With 2 and a half foot width of the roll, then the square footage is 425.

As you say, it will be smaller than this theoretical limit because the cardboard does not wrap infinitely close together.

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: How much cardboard do I have?

The model for our path will be:

sigma(t) = ( L(t) * cos(t) , L(t) * sin(t) )

That is, we go around and around a circle, the distance from the center will be given by L(t), which will increase as t does.  Now it might actually be more accurate to assume that L(t) is a step function of sorts, but it should be close enough to assume everything happens continuously.  That is, we shall take L(t) to be a line.

L(0) = 2.
L(2pi) = 2.25
L(t) = 0.25 / (2pi) t + 2

So our path will be given by:

sigma(t) = ( (0.25 / (2pi) t + 2) * cos(t) , (0.25 / (2pi) t + 2) * sin(t) )

To find the length of this, all we must do is compute the arc length. The major question is how many times do we wrap around?  I'm not sure if we should stop at a 9" radius or allow one more revolution after that.  If it's stopping right at 9", we would go from t = 0 to t = 56pi

Anyone want to compute this?  My mathematica license expired.

---

"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

mathsmypassion

Member

Registered: 2008-12-01

Posts: 33

Re: How much cardboard do I have?

I missed in my calculation one thing and you, janetpc2008, are right. So, I would say that muddyshoes has an estimation now.

JurassicPlank3

Guest

Re: How much cardboard do I have?

Christ on a bike, I wouldn't trust them as quantity surveyors.

Consider a 1'' width of the roll, the volume is PI*(R² - r²) = 1*3.142*(9² - 1²) =251.36 inch³

a square inch of carboard is 1*1*0.125 = 0.125 inches³

251.36/0.125 = 2010.88

2010.88/12

So the roll is approximately 167.5 feet long.

The width of the roll is 30 inches or 2'-6''

2'-6'' x 167'-6'' ~ 420 ft²

It is highly unlikely that the cardboard is exactly 1/8'' thick so you should probably allow +/- 5% either way. 400ft² ~ 440ft² using the lower figure for calculations.

Simplicity rules; ask any engineer!

mathsmypassion

Member

Registered: 2008-12-01

Posts: 33

Re: How much cardboard do I have?

Good to have you with us, JurassicPlank3. I believe muddyshoes couldn't imagine (when he asked the question) how much trouble he would cause with his problem. I consider that an engineer could do a better job. Thank you for your input.

janetpc2008

Member

Registered: 2009-02-03

Posts: 7

Re: How much cardboard do I have?

Christ on a bike, I wouldn't trust them as quantity surveyors.

Excuse me JurassicPlank3? My answer is the same as yours, just approached in a different way. I'm not sure what your problem is....