[Prof. Templeton]

Cawker City, Kansas boosts the World's largest ball of Twine. The ball is 40 feet in circumference and contains 7,827,737 feet of twine. That's alot of twine. If the ball was a perfect shere and the twine uniform, what is the diameter of the twine in inches? Kudos for the simplest solution.

[Guest]

Its

.16 in.

or .0132 ft.

[Guest]

Circumferance = 2 * pi * R

R = 40ft / (2 * pi) = 6.366 ft

Volume of sphere = 4/3 * pi * R^3 = 1080.8 ft3

Volume of twine = pi * r^2 * h = pi * r^2 * 7,827,737 = 1080.8

r^2 = 1080.8/(7,827,737 * pi) = 4.39484 * 10^-5

r = 0.006629 ft = 0.08 in

d = 0.16 in, about 1/6"

Edited March 24, 2009 by voltage

[Guest]

Maybe I'm wrong but...

Let's assume the string is uniform in density and that there is no air between strands. Probably poor assumption, but required. The volume of the ball will be the same as the total volume of string. So

C=2(pi)R ~> R=76.39in.

V=(4/3)(pi)R^3=1867552 cu.in.

Treating the string as a very thin cylinder of equal volume:

V=(pi)r^2h

h=93932844in.(Length of string converted to in.)

r=0.07955229in.

D=0.15910458in.

[HoustonHokie]

Okay - starting with the basic information given:

A ball 40 foot in diameter is 40/(2*3.14159) = 6.37' in radius. The volume of a 6.37' radius sphere is (4/3)*(3.14159)*(6.37³) = 1080.76 cf. Now, take the volume and divide by the length of twine and you get 1080.76 / 7827737 * 144 = 0.0199 inch² for the area occupied by a cross-section of twine.

Here's where it gets tricky. The twine is presumably circular, so I'd be tempted to divide by pi/4 and take the square root to get the diameter of a circle with area = 0.0199 inch². But when laid side by side and packed in layers, I think each piece of twine will occupy a square area, rather than circular, so there will be tiny gaps between each consecutive strand of twine. There's probably some other optimal (hexagonal?) packing to be considered here, as well as what happens when you add a layer of twine, but I don't want to mess with it right now.

Assuming then that each strand of twine occupies a square cross-section, the diameter of the twine is sqrt(0.0199) = 0.141 inch, or about 9/64 inch.

[bonanova]

Cawker City, Kansas boosts the World's largest ball of Twine. The ball is 40 feet in circumference and contains 7,827,737 feet of twine. That's alot of twine. If the ball was a perfect shere and the twine uniform, what is the diameter of the twine in inches? Kudos for the simplest solution.

Adding that the string is of circular cross section which does not deform under compression,

V[ball] = 4pi/3 [40/2pi]³ = 1080... cu ft

hexagonal circular packing density [most efficient, so this sets upper bound for amount of string] = .906...

V[string] = 980... cu ft

A[string = V/L = pi Diam²/4 = .000125 ft.

Diam = .0126... ft = .151517 in

[Guest]

Adding that the string is of circular cross section which does not deform under compression,

V[ball] = 4pi/3 [40/2pi]³ = 1080... cu ft

hexagonal circular packing density [most efficient, so this sets upper bound for amount of string] = .906...

V[string] = 980... cu ft

A[string = V/L = pi Diam²/4 = .000125 ft.

Diam = .0126... ft = .151517 in

How can you justify using hexagonal circular packing density? The string does not behave like ordered particles. It wraps arouns itself in all directions. I agree that some packing factor should be used, but I don't know which one.

[Guest]

Cawker City, Kansas boosts the World's largest ball of Twine. The ball is 40 feet in circumference and contains 7,827,737 feet of twine. That's alot of twine. If the ball was a perfect shere and the twine uniform, what is the diameter of the twine in inches? Kudos for the simplest solution.

Simplest solution? Measure the end piece of twine that is sticking out of the ball....boom.

[Guest]

I get the question now

Edited March 25, 2009 by IDoNotExist

[Guest]

I don't know if this is the point, but it sounds like he is asking for the diameter of the twine and not the ball of twine, which makes Tickytock correct, but I am sure that he is asking for the diameter of the ball.

Based on the answers that are being thrown out there, it sounds like everyone else is figuring the diameter of the twine itself. I mean, if you know that a sphere has a 40ft circumference, the diameter is easy.

[bonanova]

How can you justify using hexagonal circular packing density?

The string does not behave like ordered particles.

It wraps arouns itself in all directions.

I agree that some packing factor should be used, but I don't know which one.

My reply says it sets an upper bound.

Would be interesting to estimate a lower bound.

Any ideas?

[Guest]

My reply says it sets an upper bound.

Would be interesting to estimate a lower bound.

Any ideas?

I'm not so sure that your original answer is an upper bound since you assumed:

the string is of circular cross section which does not deform under compression

I have some issues with your assumptions. First, the string in the ball is going to have much higher axial stress on it than the limp string started off originally. Therefore, it will elongate axially and shrink radially. To calculate the volume change of the string under tension, we would need to know it's poisson's ratio. However, that is still assuming no local interaction forces with neighboring strings, which would disrupt the simple . In reality, the neighboring strings will apply forces normal to the axis of the string (but not radial) that will lead to local deformation, meaning that the string is no longer circular. These deformations would be impossible to calculate without knowing the winding pattern of the ball. Both of these factors tend to shrink the volume of the string from it's original, un-stressed volume. Therefore, the string may be bigger in diameter than your upper bound.

[Guest]

I'm not so sure that your original answer is an upper bound since you assumed:

I have some issues with your assumptions. First, the string in the ball is going to have much higher axial stress on it than the limp string started off originally. Therefore, it will elongate axially and shrink radially. To calculate the volume change of the string under tension, we would need to know it's poisson's ratio. However, that is still assuming no local interaction forces with neighboring strings, which would disrupt the simple . In reality, the neighboring strings will apply forces normal to the axis of the string (but not radial) that will lead to local deformation, meaning that the string is no longer circular. These deformations would be impossible to calculate without knowing the winding pattern of the ball. Both of these factors tend to shrink the volume of the string from it's original, un-stressed volume. Therefore, the string may be bigger in diameter than your upper bound.

Them's big fancy city words. I still prefer my answer for a "simple" solution.

[bonanova]

I'm not so sure that your original answer is an upper bound since you assumed:

I have some issues with your assumptions.

First, the string in the ball is going to have much higher axial stress on it than the limp string started off originally.

Therefore, it will elongate axially and shrink radially.

To calculate the volume change of the string under tension, we would need to know it's poisson's ratio.

However, that is still assuming no local interaction forces with neighboring strings, which would disrupt the simple.

In reality, the neighboring strings will apply forces normal to the axis of the string (but not radial) that will lead to local deformation, meaning that the string is no longer circular.

These deformations would be impossible to calculate without knowing the winding pattern of the ball.

Both of these factors tend to shrink the volume of the string from it's original, un-stressed volume.

Therefore, the string may be bigger in diameter than your upper bound.

Each set of assumptions, of course, leads to a different result.

Supposed you assumed some reasonable values,

like .3 for Poisson ratio, 5 pounds force for tension,

and a decrease in packing density of say 50% from

hexagonal circular packing.

What would your bound be?

[Yoruichi-san]

I'm not so sure that your original answer is an upper bound since you assumed:

I have some issues with your assumptions. First, the string in the ball is going to have much higher axial stress on it than the limp string started off originally. Therefore, it will elongate axially and shrink radially. To calculate the volume change of the string under tension, we would need to know it's poisson's ratio. However, that is still assuming no local interaction forces with neighboring strings, which would disrupt the simple . In reality, the neighboring strings will apply forces normal to the axis of the string (but not radial) that will lead to local deformation, meaning that the string is no longer circular. These deformations would be impossible to calculate without knowing the winding pattern of the ball. Both of these factors tend to shrink the volume of the string from it's original, un-stressed volume. Therefore, the string may be bigger in diameter than your upper bound.

But the OP said to assume the ball was a perfect sphere and twine uniform...if you're stretching the twine along one side more than the other it's no longer uniform...

Edit: Spelling

Edited March 25, 2009 by Yoruichi-san

[Guest]

But the OP said to assume the ball was a perfect sphere and twine uniform...if you're stretching the twine along one side more than the other it's no longer uniform...

Yeah, but we're on to a whole different problem now...We're looking at bounds rather than the answer

Each set of assumptions, of course, leads to a different result.

Supposed you assumed some reasonable values,

like .3 for Poisson ratio, 5 pounds force for tension,

and a decrease in packing density of say 50% from

hexagonal circular packing.

What would your bound be?

You'll have to give me a minute on that, but I think that poisson should be lower (.2?) and tension should be higher 20 lbs? Working the calculations now...

Oops, also need a young's modulus...0.2E6 psi?

Edited March 25, 2009 by voltage

[Yoruichi-san]

Yeah, but we're on to a whole different problem now...We're looking at bounds rather than the answer

Actually...what I was trying to imply is that the only way to satisfy the 'twine uniformity' of the OP is to assume it's one solid mass of twine, i.e. assume the way it's deformed during stretching allows perfect packing, so that it's no longer circular in cross section...

[bonanova]

Actually...what I was trying to imply is that the only way to satisfy the 'twine uniformity' of the OP is to assume it's one solid mass of twine, i.e. assume the way it's deformed during stretching allows perfect packing, so that it's no longer circular in cross section...

Imagining Prof. T's expression as he reads the evolution of his simple puzzle. ...

[Guest]

hardly any change: 0.159.

When I increase T to 2000, I get 0.182

Edited March 25, 2009 by voltage

[Guest]

I'm not so sure that your original answer is an upper bound since you assumed:

I have some issues with your assumptions. First, the string in the ball is going to have much higher axial stress on it than the limp string started off originally. Therefore, it will elongate axially and shrink radially. To calculate the volume change of the string under tension, we would need to know it's poisson's ratio. However, that is still assuming no local interaction forces with neighboring strings, which would disrupt the simple . In reality, the neighboring strings will apply forces normal to the axis of the string (but not radial) that will lead to local deformation, meaning that the string is no longer circular. These deformations would be impossible to calculate without knowing the winding pattern of the ball. Both of these factors tend to shrink the volume of the string from it's original, un-stressed volume. Therefore, the string may be bigger in diameter than your upper bound.

You're completely ignoring local axial variations, such as knots and frays that periodically interrupt the twine's natural perfection. Also, because of the twine's curvature, the outer surface will undergo greater deformation. Have you taken that into account? And it goes without saying that the ball's oblateness may undergo tidal variations. PT, we urgently need to know the precise date and time the circumference measurement was taken!

[Yoruichi-san]

You're completely ignoring local axial variations, such as knots and frays that periodically interrupt the twine's natural perfection. Also, because of the twine's curvature, the outer surface will undergo greater deformation. Have you taken that into account? And it goes without saying that the ball's oblateness may undergo tidal variations. PT, we urgently need to know the precise date and time the circumference measurement was taken!

Good call...and don't neglect the effect of gravity...the weight of the twine on top will further compress the lower twine, causing it to be under higher stress...unless we assume the ball of twine is rolling at a constant rate...but then we could have to take into account the centripetal force...

Edit: Oh, yes, and don't forget material properties are temperature dependent...we need to know what the temperature was...and whether it was sunny or not (b/c the sunlight would head the ball of twin from the top and create a temperature gradient through the material)

Edited March 25, 2009 by Yoruichi-san

[Prof. Templeton]

I had initially intended for someone to convert the sphere into a cylinder and tell me how many smaller twine cylinders were in the bigger, which some have done. I now see the need to provide everyone with more accurate information so the exact diameter can be obtained. First let's examine the packing. The completed twine itself is composed of seven smaller yarns that are twisted about each other, these yarns are made up of between 11 and 14 individual fibers which are also twisted. Now the ball itself, contrary to the picture shown on the website, is enclosed in a hermetically sealed chamber that is conditioned to 21.5 degrees celcius and maintained at 20% relative humidity. All measurements were obtained at midnight on all hallows eve under the light of a waxing gibbous moon. But seriously, I appeciate everyones contribution and I don't mind when the original premise is expounded upon as long as it stays within the context of the OP. It gives people an outlet for all the ideas, concepts and formulae that are hexagonally packed inside their brains (as well as making a "hot topic", which just gives me pride).