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The Dunant subsea cable

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Re: The Dunant subsea cable

#181
post #175

Earlier quoted context omitted.

>> Starlink satellites are in orbit 550km high. So any journey would add at least 1100km Might want to check with Pythagoras on that one..

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: 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].

Re: The Dunant subsea cable

#182
post #136
post #127

Earlier quoted context omitted.

> 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

And you'd also have to ignore all the insane amounts of noise coming from regular neutrinos wizzing about in the universe.

If you've built a reliable detector, you've already built something that can intercept them. You just need to make a shroud around your detector and a tube facing your transmitter out of the same material.

Re: The Dunant subsea cable

#183

Earlier quoted context omitted.

How deep would this bore tunnel be at the centre most point if it was perfectly straight? Would it go into the mantle?

d=r(1/cos(s/(2r))-1)=3958(1/cos(6906/7916)-1)~=2198 mi; Yes, into the lower mantle with only 692 mi to go to the outer core. What I would love the internetz to explain is how to justify the h8 for HFT, yet the luv for Musk since he is the one in the driver's seat at the moment for this stuff with StarLink and the Boring company.

And tell me how one would service any problems that arise either by tectonic movements or breaching an alien/breakaway civilization hollow earth chamber?

Re: The Dunant subsea cable

#185

Earlier quoted context omitted.

T1 maximum speed 1.544 Mbps T3 is about ~45 Mbps Reference: https://en.wikipedia.org/wiki/T-carrier

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.

Re: The Dunant subsea cable

#186

Old timer story - years ago, Demon Internet (my old employer) was beginning to enter super-growth and wanted to really set itself up for trans-atlantic connections. So they decided to buy a T1 - 45Mbps link across the atlantic. Now it turns out that BT had only ever resold fractions of T1s - they had never actually had anyone want a whole one. And as such their sales commissions did not cap out. So, we rang up, a sal…

T1 maximum speed 1.544 Mbps T3 is about ~45 Mbps Reference: https://en.wikipedia.org/wiki/T-carrier

And it was probably an E3 (Uk/Europe is E1/E3, US/JP is T1/T3)

Re: The Dunant subsea cable

#187

Earlier quoted context omitted.

> Assuming the underwater cable itself is 10 times as expensive as regular cable, its about $150 million for 9000 km. Still sounds really inexpensive when I consider it contains a large number of repeaters and is meant to stay at the bottom of the ocean. Edit: Forgot to write, I haven't run the numbers myself but I enjoyed your reasoning here, you put a smile on my face : > At 5 knots, it would take about 1000 hours…

Fortunately you only need repeaters every 80 km or so, so you'd only need a bit over a hundred repeaters across the 9000 km span. Repeaters aren't terrible expensive, so they only add a few million to the total cost.

Checked your profile now, I belive it :-)

Re: The Dunant subsea cable

#188
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:…

Are all base stations directly underneath a satellite?

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.

Re: The Dunant subsea cable

#189

This cable is not just to be used by Google right? Or am I misunderstanding something? Fundamentally, infrastructure should be publicly owned and then rented by companies to use it, in this case it seems like Google physically owns the cables and infrastructure which would be a massive waste.

Feel free to convince your government and fellow citizens to use tax money to pay for such infrastructure. Google laying down their own cable isn't stopping anybody from doing so.

Re: The Dunant subsea cable

#190

Earlier quoted context omitted.

> Assuming the underwater cable itself is 10 times as expensive as regular cable, its about $150 million for 9000 km. Still sounds really inexpensive when I consider it contains a large number of repeaters and is meant to stay at the bottom of the ocean. Edit: Forgot to write, I haven't run the numbers myself but I enjoyed your reasoning here, you put a smile on my face : > At 5 knots, it would take about 1000 hours…

Fortunately you only need repeaters every 80 km or so, so you'd only need a bit over a hundred repeaters across the 9000 km span. Repeaters aren't terrible expensive, so they only add a few million to the total cost.

And how are potential repeater unit failures accounted for?
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