I don't know of any being deployed long distance, though in principle they'd be really valuable for intercontinental backbones. Starlink fills a huge gap in existing infra, and there are places that won't see any sort of fiber, let alone fancy microstructured fiber, for the foreseeable future (or ever, obviously in the case of ships/aircraft). But the bandwidth isn't great. Each current sat does I think 20 Gbps, and though no doubt that'll increase over time that's literally orders of magnitude from this single cable alone. Having the sats support direct ground optical links for backbone usage might be interesting someday, but weather attenuation will never stop being a problem with that. Starlink is filling in the gaps for fiber infrastructure, not replacing it. They're complementary.
So I agree it would be great to see more advanced fiber deployed long distances and start to shrink latency for everyone, and interesting to know what technical obstacles remain if any (maybe a lot remain?). A 40% speed boost while still having massive bandwidth isn't nothing.
How do you splice a hollow optic fibre?
Each hop will add latency since signal needs regeneration. So it’s not clear to me a swarm of satellites is a real winner from a latency POV. Furthermore, given costs to put the constellation up there, it’s extremely expensive on a $/bit basis and not sure how it could compete against fiber.
The value of Starlink is providing service in areas lacking existing broadband infrastructure where the cost to provide service exceeds the cost of Starlink.
Might want to check with Pythagoras on that one..
The further you are from the other end, the less additional distance the satellite adds on.
Consider a right-angled triangle with base length d and height 550, corresponding to transmission from a base-station to a satellite. The hypotenuse has length sqrt(d^2 + 550^2), so the difference in length between the hypotenuse and base is sqrt(d^2 + 550^2) - d.
This has a maximum of 550 when d=0 (i.e., shooting straight up), and decreases as d increases: https://www.wolframalpha.com/input/?i=plot+sqrt%28d%5E2+%2B+...
Alternatively, consider the triangle inequality: the sum of the lengths of any two sides must be greater than or equal to the length of the remaining side. This directly implies that the difference in length between the hypotenuse and base is less than or equal the height [base + height >= hypotenuse implies height >= hypotenuse - base].
I think this is an over-simplification if we are chasing pedantics; There are cases where it will be more and others less so the slightly more precise wording might actually be "about 1100km."
To the larger picture: it seems we often lose that order of length on the ground due to existing network topologies and geographical limitations.
The hypotenuse is cos(angle)*base.
If you think about it at a minute if a sat is 500 miles up directly overhead that's the closest it ever will be, as it flies off the hypotenuse gets longer, not shorter.
So ideally you bounce off a sat overhead, (distance of 1100), any single hop will be longer, and to get across an ocean you'll likely need more than one hop.
Basically the sin(beam path) will will never be less than 550 and the length of the beam will never be less than 550.
Edit: Actually, this is always true: we are considering a right-angled triangle where the base is the horizontal distance from the ground station to point under the satellite, the vertical part is the 550 miles between the point under the satellite and the satellite, and the hypotenuse is the line joining the satellite and ground station.
> if a sat is 500 miles up directly overhead that's the closest it ever will be, as it flies off the hypotenuse gets longer
Yes: as the horizontal distance d increases, then the length of the hypotenuse (sqrt(d^2 + 550^2)) increases.
However, the difference between this and the horizontal distance (sqrt(d^2 + 550^2) - d) decreases.
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If the angle from the horizontal to the line between the satellite and base-station is theta, then:
sin(theta) = 550/hypotenuse => hypotenuse = 550/sin(theta)
tan(theta) = 550/base-length => base-length = 550/tan(theta)
difference in length = 550/sin(theta) - 550/tan(theta)
[which simplifies to 550 tan(theta/2)]
We are interested in angles between 0 degrees (horizontal - corresponding to the limiting case of infinite horizontal distance between the satellite and base station) and 90 degrees or pi/2 radians (straight up): https://www.wolframalpha.com/input/?i=plot+550%2Fsin%28x%29+...
This is always between 0 and 550. The triangle inequality holds: for a single hop from base-station to satellite, the increase in length is never more than 550.
But as you point out, there may also be multiple hops.
> So ideally you bounce off a sat overhead, (distance of 1100),
This is the shortest total ground-satellite-ground distance, but as you cover 0 horizontal distance it is the worst case: the difference between the ground-satellite-ground distance and the length of the direct ground-ground line is maximised.
1.) Think of the precision mirrors in the so often mentioned EUV-lithography equipment from ASML for latest generation chips from TSMC.
2.) Now imagine something like that on board of a satellite, maybe smaller.
3.) Have 2.) moveable with sufficient precision to bounce the rays from satellite to satellite in realtime, without having to regenerate them in any way for about 4 to 5 hops.
4.) problem solved by purely 'optical' mesh while signal is 'in orbit'.
kthxbaiiii!
More seriously, those mirrors are special because they use bragg reflectors to handle 13.5nm light. They're not special for their precision, nor their reflectance.
Setting that aside, the major problem with your proposal is that laser still have significant bream spreading. So the mirrors would need to be large enough to encompass a spread beam at every step, which adds weight and volume for both the mirror and the tracking mechanism. The tracking mechanism is particularly problematic because moving mass on a satellite affect the attitude, so you either need precision counterweights to null it out, or large reaction wheels.
Using MEMS mirrors instead would solve some of the mass issues, but MEMS mirrors have very limited tracking (typically limited to a single axis) which would probably render them impractical.
Far, far easier to just send and receive the signal at every step.
> Flatness is crucial. The mirrors are polished to a smoothness of less than one atom’s thickness. To put that in perspective, if the mirrors were the size of Germany, the tallest ‘mountain’ would be just 1 mm high.
edit: What I meant to say was rather something with that precision reflecting whichever wavelengths are used for laser communications. Which would be infrared, I guess? Or are we talking Maser?
edit: arrgh, forget it... one beam, reflected multiple times until 'end of the line', got it...(sigh)
But Starlink will never match the bandwidth and reliability that fiber can do, nor is it meant to. So it's not a replacement, just another awesome option.