How many satellites?

Fusion_power · 2013-07-21, 13:41

I'm posting this strictly for fun. There is no homework involved.

Given a sphere of Radius = R, and satellites orbiting at altitude = A, how many satellites would be required to ensure that at least one satellite is always visible from any point on the sphere?

Note that the number of satellites would increase for lower orbits because the coverage area would shrink.

Look at it as a problem involving the earth with radius 6,371 km, circumference 40,075 km, and orbital altitudes starting at 500 km and going up to 35,000 km.

I have a set of solutions and a formula that seems to work and will post it later.

Uncwilly · 2013-07-21, 13:56

Does your solution take into consideration the polar regions? They don't get good coverage by most sat systems.

Fusion_power · 2013-07-21, 16:31

Iridium has excellent coverage of polar regions with 66 satellites in 6 shells that are only a few degrees inclined from circumpolar.

Yes, my solution takes into account polar regions and the orbital velocity of the satellites. The only hint I will give is that orbits less than the radius R require more satellites.

c10ck3r · 2013-07-21, 23:13

5?

Fusion_power · 2013-07-23, 03:55

The closer the satellite orbits, the more satellites are required to provide 100% coverage. For an earth sized object of radius 6370 km, a 500 km high orbit would require a minimum of 36 satellites. Take the orbit out to 35,000 km (geosynchronous orbit for Earth) and a 6 satellite solution would work, 3 circumpolar, and 3 equatorial. I hate to even think of the problems that would have to be overcome to enable radio communications. This would include atmosphere absorption, doppler shifts, communications hand-offs as one satellite goes over the horizon, and dozens of others.

Anyone want to speculate on why it takes at least 6 satellites vs 2, 3, 4, or 5?

It is also worth noting that mountains and other surface features would interfere with line of sight coverage. To accomodate them, a custom solution like the 66 Iridium satellites would be required. Even with 66 satellites, there are some places and times that are not within line of sight to an iridium satellite.

science_man_88 · 2013-07-23, 12:09

Quote:

Originally Posted by Fusion_power

The closer the satellite orbits, the more satellites are required to provide 100% coverage. For an earth sized object of radius 6370 km, a 500 km high orbit would require a minimum of 36 satellites. Take the orbit out to 35,000 km (geosynchronous orbit for Earth) and a 6 satellite solution would work, 3 circumpolar, and 3 equatorial. I hate to even think of the problems that would have to be overcome to enable radio communications. This would include atmosphere absorption, doppler shifts, communications hand-offs as one satellite goes over the horizon, and dozens of others.

Anyone want to speculate on why it takes at least 6 satellites vs 2, 3, 4, or 5?

It is also worth noting that mountains and other surface features would interfere with line of sight coverage. To accomodate them, a custom solution like the 66 Iridium satellites would be required. Even with 66 satellites, there are some places and times that are not within line of sight to an iridium satellite.

Quote:

earth with radius 6,371 km

so b = 6371 for the radius of the cone of view so 2r=d is the diameter of a cone C=Pi*d = roughly 3 satellites needed around 2 major circles for a total of 6 ?

TheMawn · 2013-07-24, 05:38

This more or less involves a bunch of tangents. Instead of a spherical planet, if we started with a circle, we would have two concentric circles: The inner one is the planet and the outer is the orbit of the satellites.

Draw a point on the outer circle for the first satellite and then draw two lines through that point which are tangent to the inner circle (on on each side of the point) which represent line of sight. Repeat until you have full coverage.

This is basically the process of approximating the circle by a shape which encloses the circle. Think of a circle inside a hexagon, or a pentagon, or a square, or even triangle. In the limit, two satellites infinitely far from each other would suffice (does this mean a sufficiently long line would somehow enclose a circle? :P).

Some quick sketches at my desk tell me that a triangle with vertices 2R away from the center will enclose a circle of radius R. A square with vertices 2^1/2R away from the center will enclose a circle of radius R.

I did a bit more crunching with the notion of approximating a circle by a larger polygon. A bit of trig, a bit of sine law and poof, a formula giving the ratio of altitude to radius, depending on the number of satellites. Could be rearranged and implicitly solved for n, but to hell with that, lol. Just because I can do it doesn't mean I want to.

I've hit a bit of a wall at this point, though. I don't know that I can easily extend this to three dimensions.

LaurV · 2013-07-24, 06:02

I did not take the paper yet, no time for it, but I am trying hard in my mind and I honestly still don't get it why 4 satellites (like the vertexes of a tetrahedron, for example) won't cover all the sphere, especially when the sphere is about 6 (units) and the orbits are over 35 (units). The refraction on atmosphere and the fact that the sphere is not a perfect sphere actually works in our favor, if we place the tetrahedron conveniently. Now, it may be a trick of how the tetrahedron is moving, which I am missing, but I consider that it is not moving, the points are fixed compared with the earth (I think it called geo-stationary, or so).

sdbardwick · 2013-07-24, 06:55

Quote:

Originally Posted by LaurV

[...] but I consider that it is not moving, the points are fixed compared with the earth (I think it called geo-stationary, or so).

Geostationary orbit requires the satellite to be directly above the equator at about 36 000 kilometers ASL. However, that is too close to allow full coverage of a hemisphere; "At latitudes above about 81°, geostationary satellites are below the horizon and cannot be seen at all." [From Wikipedia, but it matches the result I remember from high-school physics back in the stone-age]

WMHalsdorf · 2013-07-24, 14:20

The best I can think of is 5 with 3 in polar orbit and 2 in equatorial orbit. None are in geostationary orbit because the earth rotates with respect to the orbits of the polar satellites the blind spots are perpendicular to the plane of the of the polar satellites thus requiring the 2 equatorial satellites.

Mini-Geek · 2013-07-24, 17:48

If A were sufficiently large and it was sufficient for a receiver some short distance above every point on the sphere to be able to see any satellite, rather than from every point on the sphere, you'd only need two satellites.

2013-07-21, 13:41	#1
Fusion_power Aug 2003 Snicker, AL 7×137 Posts	How many satellites? I'm posting this strictly for fun. There is no homework involved. Given a sphere of Radius = R, and satellites orbiting at altitude = A, how many satellites would be required to ensure that at least one satellite is always visible from any point on the sphere? Note that the number of satellites would increase for lower orbits because the coverage area would shrink. Look at it as a problem involving the earth with radius 6,371 km, circumference 40,075 km, and orbital altitudes starting at 500 km and going up to 35,000 km. I have a set of solutions and a formula that seems to work and will post it later. Last fiddled with by Fusion_power on 2013-07-21 at 13:42

2013-07-24, 05:38	#7
TheMawn May 2013 East. Always East. 6BF₁₆ Posts	This more or less involves a bunch of tangents. Instead of a spherical planet, if we started with a circle, we would have two concentric circles: The inner one is the planet and the outer is the orbit of the satellites. Draw a point on the outer circle for the first satellite and then draw two lines through that point which are tangent to the inner circle (on on each side of the point) which represent line of sight. Repeat until you have full coverage. This is basically the process of approximating the circle by a shape which encloses the circle. Think of a circle inside a hexagon, or a pentagon, or a square, or even triangle. In the limit, two satellites infinitely far from each other would suffice (does this mean a sufficiently long line would somehow enclose a circle? :P). Some quick sketches at my desk tell me that a triangle with vertices 2R away from the center will enclose a circle of radius R. A square with vertices 2^1/2R away from the center will enclose a circle of radius R. I did a bit more crunching with the notion of approximating a circle by a larger polygon. A bit of trig, a bit of sine law and poof, a formula giving the ratio of altitude to radius, depending on the number of satellites. Could be rearranged and implicitly solved for n, but to hell with that, lol. Just because I can do it doesn't mean I want to. I've hit a bit of a wall at this point, though. I don't know that I can easily extend this to three dimensions. Attached Thumbnails Last fiddled with by TheMawn on 2013-07-24 at 05:43

2013-07-24, 17:48	#11
Mini-Geek Account Deleted "Tim Sorbera" Aug 2006 San Antonio, TX USA 17·251 Posts	If A were sufficiently large and it was sufficient for a receiver some short distance above every point on the sphere to be able to see any satellite, rather than from every point on the sphere, you'd only need two satellites. Last fiddled with by Mini-Geek on 2013-07-24 at 17:48

2013-07-21, 13:56	#2
Uncwilly 6809 > 6502 """"""""""""""""""" Aug 2003 101×103 Posts 9784₁₀ Posts	Does your solution take into consideration the polar regions? They don't get good coverage by most sat systems.

2013-07-21, 16:31	#3
Fusion_power Aug 2003 Snicker, AL 7·137 Posts	Iridium has excellent coverage of polar regions with 66 satellites in 6 shells that are only a few degrees inclined from circumpolar. Yes, my solution takes into account polar regions and the orbital velocity of the satellites. The only hint I will give is that orbits less than the radius R require more satellites.

2013-07-23, 03:55	#5
Fusion_power Aug 2003 Snicker, AL 7×137 Posts	The closer the satellite orbits, the more satellites are required to provide 100% coverage. For an earth sized object of radius 6370 km, a 500 km high orbit would require a minimum of 36 satellites. Take the orbit out to 35,000 km (geosynchronous orbit for Earth) and a 6 satellite solution would work, 3 circumpolar, and 3 equatorial. I hate to even think of the problems that would have to be overcome to enable radio communications. This would include atmosphere absorption, doppler shifts, communications hand-offs as one satellite goes over the horizon, and dozens of others. Anyone want to speculate on why it takes at least 6 satellites vs 2, 3, 4, or 5? It is also worth noting that mountains and other surface features would interfere with line of sight coverage. To accomodate them, a custom solution like the 66 Iridium satellites would be required. Even with 66 satellites, there are some places and times that are not within line of sight to an iridium satellite.

2013-07-24, 06:02	#8
LaurV Romulan Interpreter Jun 2011 Thailand 2·5·31² Posts	I did not take the paper yet, no time for it, but I am trying hard in my mind and I honestly still don't get it why 4 satellites (like the vertexes of a tetrahedron, for example) won't cover all the sphere, especially when the sphere is about 6 (units) and the orbits are over 35 (units). The refraction on atmosphere and the fact that the sphere is not a perfect sphere actually works in our favor, if we place the tetrahedron conveniently. Now, it may be a trick of how the tetrahedron is moving, which I am missing, but I consider that it is not moving, the points are fixed compared with the earth (I think it called geo-stationary, or so).

2013-07-24, 14:20	#10
WMHalsdorf Feb 2005 Bristol, CT 3³×19 Posts	The best I can think of is 5 with 3 in polar orbit and 2 in equatorial orbit. None are in geostationary orbit because the earth rotates with respect to the orbits of the polar satellites the blind spots are perpendicular to the plane of the of the polar satellites thus requiring the 2 equatorial satellites.

2013-07-21, 23:13	#4
c10ck3r Aug 2010 Kansas 547 Posts	5?