6 ms·

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

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 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.

mrtnmcc 6y ago
>> Starlink satellites are in orbit 550km high. So any journey would add at least 1100km Might want to check with Pythagoras on that one..
- ptudan 6y ago
  Meh, he said at least. There could be cases where you beam up then down nearly vertically (same city).
  - function_seven 6y ago
    So, "at most" then, right? The further you are from the other end, the less additional distance the satellite adds on.
  - jamessb 6y ago
    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+550%5E2%29+-+d+from+d%3D0+to+d%3D1000 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].
    - imoverclocked 6y ago
      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.
      - jamessb 6y ago
        Yes, this is an oversimplification: the original statement seemed to be based on getting a fact about trigonometry backwards, and I was just trying to resolve the underlying confusion.
    - sliken 6y ago
      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.
      - jamessb 6y ago
        ~~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+-+550%2Ftan%28x%29+between+x%3D0+and+x%3Dpi%2F2 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.
LargoLasskhyfv 6y ago
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 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!
- morei 6y ago
  Those 'precision EUV mirrors' achieve a reflectance of about 70% i.e. they aborb ~30% of the EUV light that reaches them. :) 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.
  - LargoLasskhyfv 6y ago
    Hrm. Taken from https://www.asml.com/en/technology/lithography-principles/lenses-and-mirrors https://www.asml.com/en/technology/lithography-principles/le... : > 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?
- 6y ago
russdill 6y ago
1100km / c is 3.7ms. In free air, light speed is 50% faster. So long as the distance you are covering is more than 2200km, you'll overcome that. Of course, there's also the consideration that there can also be a lot of hops in terrestrial links and it's often very far from a straight line path.

mrtnmcc 6y ago

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

ptudan 6y ago
Meh, he said at least. There could be cases where you beam up then down nearly vertically (same city).
- function_seven 6y ago
  So, "at most" then, right? The further you are from the other end, the less additional distance the satellite adds on.
- jamessb 6y ago
  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+550%5E2%29+-+d+from+d%3D0+to+d%3D1000 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].
  - imoverclocked 6y ago
    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.
    - jamessb 6y ago
      Yes, this is an oversimplification: the original statement seemed to be based on getting a fact about trigonometry backwards, and I was just trying to resolve the underlying confusion.
  - sliken 6y ago
    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.
    - jamessb 6y ago
      ~~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+-+550%2Ftan%28x%29+between+x%3D0+and+x%3Dpi%2F2 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.

LargoLasskhyfv 6y ago

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 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!

morei 6y ago
Those 'precision EUV mirrors' achieve a reflectance of about 70% i.e. they aborb ~30% of the EUV light that reaches them. :) 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.
- LargoLasskhyfv 6y ago
  Hrm. Taken from https://www.asml.com/en/technology/lithography-principles/lenses-and-mirrors https://www.asml.com/en/technology/lithography-principles/le... : > 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?
6y ago

russdill 6y ago

1100km / c is 3.7ms. In free air, light speed is 50% faster. So long as the distance you are covering is more than 2200km, you'll overcome that. Of course, there's also the consideration that there can also be a lot of hops in terrestrial links and it's often very far from a straight line path.