4 ms·

> Because the speed of light in a vacuum is 30% faster than in optical fiber Non-Australians, don't get too excited. Those satellites are at 340 miles, so that

> Because the speed of light in a vacuum is 30% faster than in optical fiber Non-Australians, don't get too excited. Those satellites are at 340 miles, so that adds 680 miles of latency (3.66 ms) plus two to three Starlink hops, which cancels out some of that "speed in a vacuum." The way to get almost c on earth is via direct microwave links.

cryptonector 6y ago
> The way to get almost c on earth is via direct microwave links. But line of sight. Short-haul.
- davedx 6y ago
  You'd be surprised how far HFT companies run their microwave link networks to get better latencies :) https://sniperinmahwah.wordpress.com/2014/09/22/hft-in-my-backyard-part-i/ https://sniperinmahwah.wordpress.com/2014/09/22/hft-in-my-ba...
  - ed_blackburn 6y ago
    From one space elevator to the next. Yeah?
- redis_mlc 6y ago
  FYI: both Canada and the US have had transcontinental RF links. The repeaters slow down the transmission by about 30%. However those were fairly low-bandwidth.
Retric 6y ago
The minimum latency is 3.66 ms, but depending on the angle to the satellites involved it may be adding significantly less than 680 miles to the journey. Aka if it’s 340 miles east and 340 miles up that’s 480 miles to the satellite, adding just 140 not 340 miles. Outside of HFT, most networks are far from the shortest great circle routes between you and the other end, which further completes the issue.
- redis_mlc 6y ago
  The round-trip is not 140 miles to a satellite at 340 miles. Regarding cable placement, in some cases it's circuitous. But sometimes not, expecially in 2020.
  - Retric 6y ago
    You misunderstood. (Ignoring the earth being a sphere.) A satellite directly overhead means you need to travel to that altitude. However, if it’s not overhead light is traveling the hypotenuse of a right triangle where X is the distance to a point underneath the satellite and Y is the altitude of the satellite. That distance is the square root( X^2 + y^2). From there you need to travel to a different base station. Assuming an ideal path where the satellite is directly between two locations that are 680 miles apart, that adds up to 2 * ( sqrt(340^2 + 340^2) ) ~= 961.7 miles vs 340 + 340 miles, or an added 281.7 miles not 680. In other words 1.414x the distance rather than simply adding 680. Clearly the earth is not a flat and your very unlikely to be in that situation, but assuming you can reach several satellites at the same time it is likely one of them will be roughly in the direction you want to go.

cryptonector 6y ago

> The way to get almost c on earth is via direct microwave links. But line of sight. Short-haul.

davedx 6y ago
You'd be surprised how far HFT companies run their microwave link networks to get better latencies :) https://sniperinmahwah.wordpress.com/2014/09/22/hft-in-my-backyard-part-i/ https://sniperinmahwah.wordpress.com/2014/09/22/hft-in-my-ba...
- ed_blackburn 6y ago
  From one space elevator to the next. Yeah?
redis_mlc 6y ago
FYI: both Canada and the US have had transcontinental RF links. The repeaters slow down the transmission by about 30%. However those were fairly low-bandwidth.

Retric 6y ago

The minimum latency is 3.66 ms, but depending on the angle to the satellites involved it may be adding significantly less than 680 miles to the journey. Aka if it’s 340 miles east and 340 miles up that’s 480 miles to the satellite, adding just 140 not 340 miles. Outside of HFT, most networks are far from the shortest great circle routes between you and the other end, which further completes the issue.

redis_mlc 6y ago
The round-trip is not 140 miles to a satellite at 340 miles. Regarding cable placement, in some cases it's circuitous. But sometimes not, expecially in 2020.
- Retric 6y ago
  You misunderstood. (Ignoring the earth being a sphere.) A satellite directly overhead means you need to travel to that altitude. However, if it’s not overhead light is traveling the hypotenuse of a right triangle where X is the distance to a point underneath the satellite and Y is the altitude of the satellite. That distance is the square root( X^2 + y^2). From there you need to travel to a different base station. Assuming an ideal path where the satellite is directly between two locations that are 680 miles apart, that adds up to 2 * ( sqrt(340^2 + 340^2) ) ~= 961.7 miles vs 340 + 340 miles, or an added 281.7 miles not 680. In other words 1.414x the distance rather than simply adding 680. Clearly the earth is not a flat and your very unlikely to be in that situation, but assuming you can reach several satellites at the same time it is likely one of them will be roughly in the direction you want to go.