4 ms·

Latency in this case is dominated by time-of-flight to the satellite. At 100 Mbps with a 40 ms latency, there are about 2 megabits in the air between the groun

Latency in this case is dominated by time-of-flight to the satellite. At 100 Mbps with a 40 ms latency, there are about 2 megabits in the air between the ground station and CPE.

Denvercoder9 5y ago
> Latency in this case is dominated by time-of-flight to the satellite. Is it though? Starlink orbits at 550 km, time-of-flight from ground to satellite to ground would be only 3.7ms, twice that makes ~20% of the roundtrip latency.
- Robotbeat 5y ago
  550km is the closest approach. Usually it will be at least sqrt(2) times that (ie at least a slant angle of 45 degrees), maybe 2 times that. Plus the latency from the Gateway to the actual server. And the Gateways can have even greater slant angle to the satellite than the mobile terminals.
  - Denvercoder9 5y ago
    I don't think it'll usually be at a slant angle of at least 45°. The beta requires a field of view of 100° after tilting. I can't find the maximum tilt angle, but SpaceX has authorization to transmit only 25° degrees above the horizon, so the maximum slant angle is 65°.
    - Robotbeat 5y ago
      But think what that means in terms of SOLID angle, not linear angle.
      - Denvercoder9 5y ago
        Orbital planes are so close together that it doesn't make much difference. However, now that I've actually calculated it, your sqrt(2) factor seems to be about right for the average distance -- there's too few satellites per plane in the current phase. In this phase Starlink uses 72 orbital planes, with 22 satellites per plane, so 1440 satellites in total (they're almost there). It orbits at 550km above Earth's surface, so the orbit has radius 6921km, which gives an orbital length of 43486km. Separation between orbital planes varies depending on your latitude, but assume the worst case, where it is 43486km / 72 / 2 = 302km¹. Thus, the nearest orbital plane is at most 302km / 2 = 151km away from the orbital plane directly overhead. However, since the planes process, on average the nearest orbital plane is only half that, or 76km away from the plane overhead. Satellites within each plane have a separation of 43486km / 22 = 1976km. Thus, there's always a satellite at most 1976km / 2 = 988km away¹ from any point in each orbital plane, and on average there's a satellite half that away, or 494km. Adding all this together, the nearest satellite is on average √(550^2 + 76^2 + 494^2) = 743 km away (at the worst latitude). [EDIT: Actually, that's improper averaging, the correct average is obtained with ∫√(550^2 + x^2 + y^2) dx dy / ∫ dx dy on x=0..151, y=0..988, which yields 777km]. The original plan used 24 planes with 66 satellites, which reduces average distance to 617km. At more favorable latitudes the difference with the current design would be even larger. [EDIT: This should be 635km.] ¹ This is distance on the surface of the orbital sphere, straight-line distance is a bit less. It probably doesn't make much difference.
        thenewwazoo 5y ago
        This is a great comment. Thank you for writing it.
- bcrl 5y ago
  There are more than 2 roundtrips. Any MAC that has to perform time division multiplexing on a shared uplink has to poll all base stations over time to figure out which ones have data to transmit, and how much data is queued. Once the satellite knows how much data the ground station has to transmit, it assigns sufficient timeslots, transmits the assignment and then waits for data to come back. This is very similar to PON networks where upstream is shared, but the difference is sub-1ms latency vs 4-5ms latency. Sadly, this does have unfortunate latency implications for how long web pages take to load and render. Streaming video should, however, work swimmingly.
  - spookthesunset 5y ago
    It’s interesting they’d do TDMA over some flavor of code division multiple access (CDMA). From what I read it has to do with the fact the antennas are high gain directional antennas and not omnidirectional ones like on cell phones. With cell phones you are kinda walking around in a soup of cell signals all sharing the same spectrum at once... you and hundreds of other people are broadcasting in the same frequencies at the same time and they all tell each other apart because they all use a different “language”; the Wikipedia CDMA article does an excellent job explaining this. I would think that as more satellites get launched they could use WCDMA and signal from your station could be seen by multiple satellites in orbit much like a cell phone can reach multiple towers. Writing it out... I bet TDMA is required because the FCC would never grant a block of spectrum where hundreds of thousands of ground stations were using low gain, somewhat omnidirectional antennas to reach a constellation of satellites in space....
    - bcrl 5y ago
      It's even more complicated than that. Thanks to MIMO antenna arrays, signals from multiple ground stations can be received and decoded at the same time (MU-MIMO). The advances in radio MACs over the past 20 years is seriously impressive compared to what was considered high tech in the 1990s, and it's all a result of Moore's law making it cheaper to do more math in the same size and power envelope as older semiconductors.

Latency in this case is dominated by time-of-flight to the satellite. At 100 Mbps with a 40 ms latency, there are about 2 megabits in the air between the ground station and CPE.

Denvercoder9 5y ago

> Latency in this case is dominated by time-of-flight to the satellite. Is it though? Starlink orbits at 550 km, time-of-flight from ground to satellite to ground would be only 3.7ms, twice that makes ~20% of the roundtrip latency.

Robotbeat 5y ago
550km is the closest approach. Usually it will be at least sqrt(2) times that (ie at least a slant angle of 45 degrees), maybe 2 times that. Plus the latency from the Gateway to the actual server. And the Gateways can have even greater slant angle to the satellite than the mobile terminals.
- Denvercoder9 5y ago
  I don't think it'll usually be at a slant angle of at least 45°. The beta requires a field of view of 100° after tilting. I can't find the maximum tilt angle, but SpaceX has authorization to transmit only 25° degrees above the horizon, so the maximum slant angle is 65°.
  - Robotbeat 5y ago
    But think what that means in terms of SOLID angle, not linear angle.
    - Denvercoder9 5y ago
      Orbital planes are so close together that it doesn't make much difference. However, now that I've actually calculated it, your sqrt(2) factor seems to be about right for the average distance -- there's too few satellites per plane in the current phase. In this phase Starlink uses 72 orbital planes, with 22 satellites per plane, so 1440 satellites in total (they're almost there). It orbits at 550km above Earth's surface, so the orbit has radius 6921km, which gives an orbital length of 43486km. Separation between orbital planes varies depending on your latitude, but assume the worst case, where it is 43486km / 72 / 2 = 302km¹. Thus, the nearest orbital plane is at most 302km / 2 = 151km away from the orbital plane directly overhead. However, since the planes process, on average the nearest orbital plane is only half that, or 76km away from the plane overhead. Satellites within each plane have a separation of 43486km / 22 = 1976km. Thus, there's always a satellite at most 1976km / 2 = 988km away¹ from any point in each orbital plane, and on average there's a satellite half that away, or 494km. Adding all this together, the nearest satellite is on average √(550^2 + 76^2 + 494^2) = 743 km away (at the worst latitude). [EDIT: Actually, that's improper averaging, the correct average is obtained with ∫√(550^2 + x^2 + y^2) dx dy / ∫ dx dy on x=0..151, y=0..988, which yields 777km]. The original plan used 24 planes with 66 satellites, which reduces average distance to 617km. At more favorable latitudes the difference with the current design would be even larger. [EDIT: This should be 635km.] ¹ This is distance on the surface of the orbital sphere, straight-line distance is a bit less. It probably doesn't make much difference.
      - thenewwazoo 5y ago
        This is a great comment. Thank you for writing it.
bcrl 5y ago
There are more than 2 roundtrips. Any MAC that has to perform time division multiplexing on a shared uplink has to poll all base stations over time to figure out which ones have data to transmit, and how much data is queued. Once the satellite knows how much data the ground station has to transmit, it assigns sufficient timeslots, transmits the assignment and then waits for data to come back. This is very similar to PON networks where upstream is shared, but the difference is sub-1ms latency vs 4-5ms latency. Sadly, this does have unfortunate latency implications for how long web pages take to load and render. Streaming video should, however, work swimmingly.
- spookthesunset 5y ago
  It’s interesting they’d do TDMA over some flavor of code division multiple access (CDMA). From what I read it has to do with the fact the antennas are high gain directional antennas and not omnidirectional ones like on cell phones. With cell phones you are kinda walking around in a soup of cell signals all sharing the same spectrum at once... you and hundreds of other people are broadcasting in the same frequencies at the same time and they all tell each other apart because they all use a different “language”; the Wikipedia CDMA article does an excellent job explaining this. I would think that as more satellites get launched they could use WCDMA and signal from your station could be seen by multiple satellites in orbit much like a cell phone can reach multiple towers. Writing it out... I bet TDMA is required because the FCC would never grant a block of spectrum where hundreds of thousands of ground stations were using low gain, somewhat omnidirectional antennas to reach a constellation of satellites in space....
  - bcrl 5y ago
    It's even more complicated than that. Thanks to MIMO antenna arrays, signals from multiple ground stations can be received and decoded at the same time (MU-MIMO). The advances in radio MACs over the past 20 years is seriously impressive compared to what was considered high tech in the 1990s, and it's all a result of Moore's law making it cheaper to do more math in the same size and power envelope as older semiconductors.