Calculating diameter of a ball made of infinitely long but infinitely thin thread?

Question

This is slightly abstract question, with aim to determine what parameters are being missed in the hereby provided problem's description, and what are potential solutions to this problem based on various conditional assumptions.

Given: There is an infinitely long but infinitely thin threat, and it is coiled into spherical ball. Question: Would diameter $D$ of such spherical ball tend to shrink $D\to 0$, or would it remain as is $D=1$, or would it tend to expand $D\to\infty$? Please set examples of formulas that may determine the diameter of such thread ball under certain circumstances?

Ball of Thread 1 - Dense

Ball of Thread 2 - Sparse

Offtop: This question with these exactly conditions I've came across on Facebook, where a guy by claiming that he is a mathematician, has insisted that the only solution to this problem would be by L'Hospital's Rule applied to $V=πr*​2Хh$ where $r \rightarrow 0$ and $h\rightarrow\infty$, thus volume and hence diameter of the thread ball would too $\rightarrow0$. Since I could visualize in my mind that not necessarily at every circumstance this would be true, but couldn't support my predictions mathematically, I came here to seek your support and to collect some valuable counter-arguments that would clarify my doubts. Prior to asking for your help I've Googled hard to find any information on this or similar issues, but with no luck. And it is not easy to restore mathematical knowledge, once you lose practice. You guys with every argument mentioned in your replies and comments have contributed a lot to my understanding of the subject, specifically what essential parameters have been missing in the description, and how other solutions based on various assumptions may be calculated. Thank you a lot for your attention and your knowledge sharing!

Answer 1

Suppose the thickness of the thread is $t$ and its length is $l$. Then the volume of the thread is $\frac\pi4t^2l$. Thus, we need a radius of at least $r$, where $$ \frac{4\pi}3r^3=\frac\pi4t^2l $$ That is, the radius of the ball would need to be at least $$ r=\left(\frac3{16}t^2l\right)^{\!1/3} $$ Due to the fact that the thread would leave some enclosed space, the radius of the ball would need to be somewhat greater.

Thus, to get a meaningful, finite radius, the infinite length must be inversely proportional to the square of the infinitesimal thickness.

Answer 2

Presumably Volume = $h*\pi*r^2= \frac 43\pi*R^3$ where $h$ is the length of the string and $r$ is the radius or half the thickness of the string, and $R$ is the radius or half the diameter of the ball. So

$D/2 = R = \sqrt[3] {\frac 34*hr^2}$.

$=\sqrt[3]{\frac 34 \infty_1*(\frac 1{2\infty_2})^2}$

Which can be any value we like depending on what assumptions we make about how the "infinities" are related to each other.

If we assume the thread is such that it is always made of the same volume of material we get $h*r^2 = K$ for some constant $K$, we get $r^2 = \frac Kh$ and $R = D/2 = R = \sqrt[3] {\frac 34K}$ which is a constant no matter how long the thread is.

But if we assume the thread will get get $t$ times longer every second and will get $t$ times thinner every second we get $h = h_1*t$ and $r= r_2/t$ and $D/2=R=\sqrt[3]{\frac 34*h_1*(\frac{r_1}2)^2*\frac 1t} = 0$.

If we assume, for some utterly perverse reason, that the length of the wire is inversely proportional to the cube of the thickness of the wire, that is $h^3*r= L$ for some constant, then $D/2=R=\sqrt[3]{\frac 34*h*L}=\infty$.

We can make any other assumption we like to make any other value we like.