Inside Diameter Of A Torus

[Little Bang]

If I have a tube with diameter X what is the smallest torus I could produce?

[sanctus]

If the length of the tube is not specified I would say infinitely small ;-)

[modest]

If I have a tube with diameter X what is the smallest torus I could produce?

 

The volume of a torus is [math]V=2 \pi^2 R r^2[/math] where [math]R[/math] is the distance from the center of the tube to the center of the torus and [math]r[/math] is the radius of the tube. I think the smallest torus would be where [math]R \to r[/math] meaning the volume would be [math]V=2 \pi^2 r^3[/math] (where r is half of X given in the OP).

 

Intuitively, I think that would be right because the volume would be the same as a cylinder of height [math]2 \pi r[/math] and a cylinder's volume [math]\pi r^2 h[/math] which would give [math]V = \pi r^2 h = \pi r^2 (2 \pi r) = 2 \pi^2 r^3[/math].

 

This is assuming that you mean a torus with an inner radius of zero (ie the size of the hole in the center goes to zero) and an outer radius of [math]2r[/math].

 

The surface area would be [math]4 \pi^2 r^2[/math].

 

~modest

 

EDIT: I'm sorry, I just noticed that the tread's title is "inside diameter of a torus". Sanctus' answer suddenly makes sense to me—that the inner diameter is infinitely small.

[Little Bang]

Modest, I think you both gave basically the same answer but the detail you gave is what I was looking to find, Thank you.

[sanctus]

But little Bang, you do not need the math for it. Sorry, yesterday I was tired that is why I did not elaborate. Just imagine a tube, if you just bend it to make a torus, you will get a torus under the condition that the tube is long enough (ok for this "long enough" you would need math). So if you have a tube which is exactly long enough then you will have an inner radius of zero.