cross-posted from: https://lemmit.online/post/5025831
This is an automated archive made by the Lemmit Bot.
The original was posted on /r/ukrainianconflict by /u/Positive_Detective56 on 2025-01-27 15:23:10+00:00.
cross-posted from: https://lemmit.online/post/5025831
This is an automated archive made by the Lemmit Bot.
The original was posted on /r/ukrainianconflict by /u/Positive_Detective56 on 2025-01-27 15:23:10+00:00.
Messy napkin maths to explain this to myself, because it is wild to me that 41 km of anything fits in that. I will use the top picture and assume that the guy’s hand is about the same size as mine; my hands are reasonably large, so this should err on the safe side, and his hands are at a similar distance from the camera to the spool. That gives me a very approximate scale of 2 pixels per millimetre.
Diameter of the spool case: 365 px = 183 mm = 0.183 m Radius of the main section of the case: 91 mm = 0.091 m Length of the main section of the case: 477 px = 239 mm = 0.239 m
Volume of a cylinder = pi r r l v1 = pi * 0.091 * 0.091 * 0.239 v1 = 0.00622 m^3
I will assume that the shaft of the spool is the same diameter as the narrow part of the case
Diameter of the spool shaft: 55 px = 28 mm Radius of the spool shaft: 14 mm = 0.014 v2 = pi * 0.014 * 0.014 * 0.239 v2 = 0.00015 m^3
Subtract shaft from first volume = 0.00607 m^3
I’ll assume that the cable has the same volume, as if the case had zero-thickness walls and the cable filled the case perfectly. Since the cable has the same volume, dividing that volume by 41,000 m should give us the cross-sectional area.
0.00607/41000 = 1.48e-7 = 0.000000148 m^2
Square root of that to get the length and height of that cross section = 0.000345 m = 0.345 mm
This number can effectively serve as the diameter of the cable, since it’ll have a circular cross section. And… yeah, I can find 0.25 mm diameter optical fibre easily, so the numbers check out.
I work in telecommunication and see fiber quite regularly.
It really is thin, and fine is a better word. In general what you see is the coating of the fiber, and that already is very fine. The fiber itself is nearly invisible even when you hold a strand between your fingers against the light.
A single of these nearly invisible fibers can transport a GBit/s into your home. “Any sufficiently advanced technology is indistinguishable from magic” - yep, Arthur C. Clarke was right.
Funnily enough I have also worked with them, but all of mine were short lengths so I have absolutely no sense of scale for what kilometres of it looks like rolled up. Or had, I suppose, until now. I just used them to run light into awkward spots for sensors, though, so I wasn’t exactly maxxing out the potential applications
By a neat coincidence I also play a bit of guitar, and a quarter millimetre is the diameter of the thinnest string on most of mine. For any other guitarists reading, a 10 gauge string is 0.254 mm
0.2mm is also a common layer height in 3D printing.
Having a hard time imagining commercial grade optic wire a quarter of a mm think surviving things like a mildly stiff breeze against some bushes or tree branches.
I suppose unless you’re at the very end of the length of it, it can probably account for that to some degree by just letting more fibre out. This paper (which I have only skimmed, so sorry if I’ve got this wrong) seems to say that fibres even thinner than that — 0.15 mm — can handle a force equivalent to a couple of kilograms hanging off of them
Damn, fam did the math. Props.