Number of satellites needed for global 4-fold coverage as a function of altitude?

Question

I want to estimate how many satellites (and how many number of planes) are needed for achieving a 4-fold global coverage on earth (global coverage with at least 4 satellites in view in each location). To do this I basically compute the area of coverage for a single satellite (A_Cap), and I divide the total earth surface by the coverage of a single satellite. The equations I use are (based on Figure 1):

Acap = 2*pi*Re^2*(1-cosd(Phi))

Where Phi is the cap angle and is computed as:

Phi = acosd(Re/(Re+h)*cosd(Alpha))-Alpha;

Where Re is the radius of the Earth, h is the orbit altitude and Alpha is the minimum elevation angle for consider the satellite in view.

To compute the numbe rof satellites I need I do:

Nsv = N_fold*A_Earth/Acap

Where N_fold is the needed number of satellites in view at any location (in my case 4). Doing this I belive I'm overestimating the number of satellites (since I'm just multiplicing the coverage of 1-fold by N). Do you have any other idea how I could solve this problem?

Answer 1

Some notes on your approach, which I believe is a good one as a first order approximation.

Doing this I belive I'm overestimating the number of satellites

I would rather say it has to be an underestimate.

1. Circles do not perfectly tile. To achieve a 1-fold coverage, for instance, there must be some overlap, requiring $\frac{2\pi}{3\sqrt{3}} \approx 1.21$ times more circle area.

For higher order coverings, the cover efficiency will asymptotically approach $1$, but for a 4-fold covering it's not clear that you can do better than just layering four 1-fold coverings (and even if some more clever solution exists, it's only going to be marginally better). So 20% more satellites just from the static geometry of the problem seems reasonable.

2. Satellites are not static. Even if you find a geometric solution that covers the whole planet 4-fold, it's likely to instantly break when their required relative motion is taken into account. The coverage pattern has to change as a function of time, and you are likely going to need extra satellites to make sure the 4-fold coverage holds all the time instead of just a specific instant.

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For something that's clearly an overestimate, but will work, you could place satellites in longitude planes, tight enough to provide 4-fold coverage.

Some initial number juggling gives me an efficiency maximum where the planes are 1.55 single-satellite coverage radius apart, with the satellites in a plane spaced 0.36 at radii.

This should give a provable upper bound, but is inefficient because it gives unnecessary amounts of coverage near the poles.

Answer 2

Perhaps you are looking for a relationship in the form of (or wondering whether such a relationship exists):

Nmin(A_cap,n) = μ(n) * (A_Earth/A_cap)

A_cap: area of an instantaneous coverage by a single satellite, constant in time (circular orbits), modelled as a spherical cap.

Nmin: smallest number of satellites in a practical constellation that can provide seamless n-fold coverage.

μ(n): a fitting constant, independent of A_cap, function of n.

Fortunately, such approximate relationships exist, and μ(4)~7.2, whereas μ(1)~2.

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However, I don’t know of any mathematical derivation of μ(n), even for μ(1). Mostly, μ(n) is derived from testing different constellations constructed using heuristic reasonings, such as the so-called Walker constellations (Wiki).

• Case n=1

Let’s start with n=1 to get familiar with some published results.

Yuri Ulybyshev wrote a nice review in 2008 titled Satellite Design for continuous coverage: short historical survey

Figure 1, reproduced here for convenience, gives a plot for n=1 and Elevation =10° (that he called α).

As you have noted yourself, if you call Phi the Earth-centered half-cone angle of the spherical cap representing the individual coverage, then A_cap=2 π RE2 (1-cos(Phi)) So that (A_Earth/A_cap) = 2/(1-cos(Phi))

What I am claiming here is that the tight lower bound for N displayed in Figure 1 of Ulybyshev follows the trend of the relationship:

   Nmin= 4/(1-cos(Phi))

In other terms, μ(1) ~ 2.

Here is a check point so that we are on the same page with the detailed calculation:

H=1000Km (and El=10°) => Phi =21.6° => 2/(1-cos(Phi))= 28.4 => Nmin=56.8

This result (μ(1) ~ 2) was obtained independently by Beste in Design of satellite constellation for optimal continuous coverage. It is paywalled but Figure 3 is available (reproduced here, ψ is the half-cone angle that we called Phi).

• Case n=4,

Take GPS. Since we know H (20200 Km) we can compute their Phi (66.3°), assuming that their design Elevation is 10° (a reasonable assumption for Satnav). We also know GPS requires 24 satellites. From this we can make the informed guess that μ(4) ~ 7.2, assuming that the designers of GPS did optimize their constellation for minimum number of satellites.