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The Dunant subsea cable

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Re: The Dunant subsea cable

#211

Post by tgtweak »

trollied wrote: post #72
tgtweak wrote: post #41

No mention of latency improvements? Seems crazy since oversea transit (tcp & single-channel) is usually latency (or loss) bound. I would expect it's better than going over public transit and legacy subsea fiber, but it would have been useful to see some comparison tests between POPs.

I'm not sure how easy it is to increase the speed of light in glass without some sort of new breakthrough.

Not travelling through 20 routers in the process tends to help. Again it would be good to get a tangible idea of how much better this is vs just stating the obvious about peak theoretical throughput.

Re: The Dunant subsea cable

#212

Post by ativzzz »

parliament32 wrote: post #166
tzs wrote: post #92

Earlier quoted context omitted.

It's probably cheaper than people would expect because the long run across the deep ocean is a lot more straightforward than most people would expect. 1. For the deep ocean parts of the route, cables and associated equipment (such as repeaters) are simply spooled out from the back of the cable laying ship, to settle on the ocean floor. 2. For shallow waters, the cable is buried. This is done by dragging a plow along…

I figure the hard engineering challenge is the repeaters. How do you build repeaters and power them, considering you can't really service or replace them ever over the lifespan of the cable (the deep ocean bits anyway)? A repeater every 80km is a whole lotta repeaters.

They are repairable: http://www.k-kcs.co.jp/english/solutionRepairingMethod.html

Re: The Dunant subsea cable

#213

Post by sliken »

jamessb wrote: post #181
ptudan wrote: post #175

Earlier quoted context omitted.

Meh, he said at least. There could be cases where you beam up then down nearly vertically (same city).

But the correct statement is "no more than" not "at least". 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:…

Er, no, "the difference in length between the hypotenuse and base is sqrt(d^2 + 550^2) - d".

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.

Re: The Dunant subsea cable

#215

Post by mmmBacon »

LargoLasskhyfv wrote: post #196
mmmBacon wrote: post #164

Earlier quoted context omitted.

Starlink satellites are in orbit 550km high. So any journey would add at least 1100km. Moreover not sure that a single satellite would be able to hit another one across transpacific distances and may need to go through multiple hops to get there. 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…

Are you sure about the necessary regeneration? Let me hand wave from the dark skies here for a moment: 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 realtim…

While an interesting idea, I think you’ve greatly understated the problem. First, lasers and coherent light beams diverge, light cannot stay perfectly collimated and it’s not really possible to collimate well over such long distances. So the receiver, >10,000km away, will “see” only a small cross-section of beam. The efficiency of this is defined by something called the overlap integral between the areas of the beam and the detector. Think of it like the amount of light from a flashlight that gets through a pinhole in a sheet of paper. This reduces the available signal power significantly. If you introduce mirrors you have the mirror loss plus the vignetting losses for each bounce. This is likely much worse.

Re: The Dunant subsea cable

#216

Post by jamessb »

sliken wrote: post #213
jamessb wrote: post #181

Earlier quoted context omitted.

But the correct statement is "no more than" not "at least". 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:…

Er, no, "the difference in length between the hypotenuse and base is sqrt(d^2 + 550^2) - d". 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 yo…

~~D'oh - yes, that formula is true only if the triangle is right-angled, which is true for only a single base length.~~

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.

-----------------------------------------------

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.

Re: The Dunant subsea cable

#217

Post by LargoLasskhyfv »

mmmBacon wrote: post #215
LargoLasskhyfv wrote: post #196

Earlier quoted context omitted.

Are you sure about the necessary regeneration? Let me hand wave from the dark skies here for a moment: 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 realtim…

While an interesting idea, I think you’ve greatly understated the problem. First, lasers and coherent light beams diverge, light cannot stay perfectly collimated and it’s not really possible to collimate well over such long distances. So the receiver, >10,000km away, will “see” only a small cross-section of beam. The efficiency of this is defined by something called the overlap integral between the areas of the beam…

But the reciever won't be be > 10,000km away in the configuration I mentioned. 4 to 5 'hops', remember?

edit: arrgh, forget it... one beam, reflected multiple times until 'end of the line', got it...(sigh)

Re: The Dunant subsea cable

#218

Post by anonuser123456 »

amit9gupta wrote: post #124
SloopJon wrote: post #4

Wow, talk about a barrier to entry. Google already has Curie from North America to South America and Equiano from Portugal to South Africa. They're also working on Hopper from North America to UK and Spain: https://cloud.google.com/blog/products/infrastructure/announ... I presume that the other trillion-dollar companies are getting in on the action too.

Welcome to the new form of Colonialism!!

Because Google owns the sea and no one else can lay cable...

Re: The Dunant subsea cable

#219

Post by [deleted] »

jxcl wrote: post #127
thomaslangston wrote: post #122

Earlier quoted context omitted.

More feasible would be transmitting neutrinos or some other signal that would not be blocked by the Earth.

> More feasible If they don't interact with the thousands of miles of earth between the source and the destination, they probably also won't interact with the receiver! :p Imagine the retransmission rates! https://en.wikipedia.org/wiki/Neutrino_detector

[deleted]

Re: The Dunant subsea cable

#220

Post by tinus_hn »

zenexer wrote: post #185
tinus_hn wrote: post #176

Earlier quoted context omitted.

In internetworking a Tier 1 carrier is a carrier that is so interconnected other parties pay to receive traffic from them.

T1 and Tier 1 are not the same thing.

That’s true. However at least in my memory back in the day people would call the 10 megabit directly connected university connections t1 lines, because of the tier 1 thing.
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