Ukraine

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.