Nadir Angle and Elevation Angle

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Nadir Angle and Elevation Angle

Postby jimh » Mon Dec 18, 2017 11:20 am

What is the angle of an incoming signal to a SARSAT MEO satellite (known as the nadir angle and measured in degrees from the satellite to the point on earth directly below) if the transmitter's view of the satellite is at 5-degrees elevation above the horizon?

A sketch shows the relationships, drawn roughly to scale:

Satellite in medium earth orbit in view with 5-degree elevation
earthSatelliteDwg5degree.jpg (49.42 KiB) Viewed 14967 times

The blue scalene triangle ABC can be solved as follows:
--we know the length of side b; it's the Earth radius; b = 6371 km
--we know angle C; it is the elevation angle, 5-degrees, plus 90-degrees; C = 95 degrees
--we know length of side c; it's the Earth radius plus the satellite altitude; 6371 + 20180 = 26551 km

We can find angle B from a relationship known as the law of sines for a triangle. That law says

a/sinA = b/sinB = c/sinC

Solving for B we can use our known values, b, c, and C:

B = arcSin [ (b*sinC)/c ]

B = 13.83-degrees

This is the desired calculation, the nadir angle. This means the signal from a transmitter that sees the satellite at an elevation above the horizon of only 5-degrees will arrive at the satellite at a nadir angle of 13.83-degrees. This is of interest because the receiving antenna on the satellite must provide coverage out to at least 13.83-degrees to be useful for transmitters that see the satellite only 5-degrees above their horizon.

Now we can find A, from the rule of triangles that the three angles must sum to 180:

A = 180 - B - C
A = 71.17-degrees

We can also find the path length, a, to the satellite, again

a = 25,226-km

The most direct path (when the satellite is directly overhead) is the satellite altitude, 20,180 km. The path length at 5-degrees elevation is more that 5,000 km longer, so path loss with be greater. It is non-trivial to compute the exact path loss because the signal traverses a lot more distance in the atmosphere of Earth. The effect of the atmosphere on radio signal attenuation is much more complex to calculate than the effect of distance on attenuation of radio signals in space. Because much of the added distance from the low-elevation path will be through the atmosphere, the effect of the atmosphere will be be greater than on the overhead path.

Angle A also tells us the distance to the ground point (GP) of the satellite from a location where the satellite is visible with elevation angle of 5-degrees. We compute from:

Earth circumference in 360 degrees = 40,075 km
Earth arc distance in 71.17 degrees = 7922.6 km

Here is a summary of triangle ABC
Length of sides (km)
Angles (degrees)

Checked with

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Re: Coverage area on Earth of MEO Satellite

Postby jimh » Tue Dec 19, 2017 9:56 am

The view of the Earth from the satellite in which it is available at 5-degree elevation or higher projects as a circular area on a sphere. If we assume the ground point (GP) of the satellite is at the north pole of the sphere, this area is known as a spherical cap. The area of the spherical cap is defined by

Area = 2πrh

where r is the radius of the sphere and h is the height from the circle's center to the pole. A hemisphere is a special case of spherical cap where h = r.

Spherical cap, from a drawing by Jhmadden; used under terms of the Creative Commons Attribution Share Alike 4.0
Spherical_cap_diagramSmall.png (32.66 KiB) Viewed 14929 times

In our case r is the Earth radius, 6371-km. The angle to the cap's outer rim, θ, is taken from our earlier calculation (above) of the coverage of a MEO satellite (shown there as angle A), and is 71.17-degrees. The height h is to be calculated from the geometry (see below). The height in our example is 4314.7-km. This gives the area of the spherical cap as

Area = 2πrh
Area = 2π(6371 x 4314.7)
Area = 172,718,190 km2

Thus for a medium earth orbit satellite at altitude 20180 km, the area on the Earth from which it will be in view with an elevation angle of 5-degrees or more is about 172.7-million km2. We compare with the total area of the Earth, as follows:

The area of a sphere is given by

Area = 4πr2

For Earth with radius 6371 km, the total surface area is then about 510,064,472-km2. The coverage area of the MEO satellite (assuming 5-degree elevation mask) is 172,718,190 km2, or about one-third of the Earth's surface.

The height h can be calculated with right-triangle geometry; we use the right-triangle with hypotenuse r and angle θ of 71.17-degrees. The other angle must then be 18.83-degrees. The side opposite the 18.83-degree angle is equal to (r-h). We compute its length from (r-h) = 6371 x sin(18.83), or 2056.3, and subtracting that from r (6371) gives the height, h = 4314.7 km.

The area of the spherical cap can also be calculated from

Area = 2πr2(1−cosθ)

which avoids having to find h. This method gives the area as

Area = 172717696 km2

To explain this derivation, note that the value of r must be

r = h + (rcosθ)

In that case, then we can say

h = r - (rcosθ)
h = r(1-cosθ)

Then we substitute that value for h in our original equation for area:

Area = 2πrh

which yields

Area = 2πr2(1−cosθ)

More properties of a spherical cap and how to calculate its dimension are give in a (linked) Wikipedia article. An interesting on-line calculator is also available to calculate these solutions.

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Re: Nadir Angle and Elevation Angle

Postby jimh » Sun Dec 24, 2017 6:25 am

Reworking the problem for a SARSAT satellite in low earth orbit (LEO) and allowing for a minimum elevation angle of 15-degrees (a more practical situation) we find the following:

For a satellite altitude of 850 km, Earth radius 6371 km, and elevation angle 15-degrees, our ABC triangle has

C = 90 + 15 = 105
c = 7221
b = 6371

Solving for B

B = arcSin [(6371sinC)/7221] = 58.454

Solving for A

A = 180 - B - C = 16.545

Solving for area of the spherical cap area

Area = 2πb2(1-cosA) = 10,559,951.6 km2

The arc length from observer to GP is then

Distance = 40,075 km × (16.545/360) = 1841.8 km

The percent of Earth surface covered is

Coverage = 10,559,951.6 km2/510,064,472 km2 = 0.02 or 2-percent

The LEO satellite covers much less area, however the path length a is much shorter:

a = [(cSinA/sinC)] = 7221sin16.454/sin105 = 2117.5 km

The path length to an LEO is much less, less than one-tenth the path length to a MEO satellite. This mean much less signal attenuation. Also note the nadir angle for the satellite antenna will be much greater, 58.454 degrees in this example. This requires the satellite receiver to have much wider antenna pattern to receive the signal coming up from a distress beacon transmitter.

On the other hand, one MEO satellite can cover about one-third of the Earth's surface, while one LEO satellite only covers about two-percent. The SARSAT system gets a lot more coverage from one MEO than it does from one LEO satellite. Since both MEO and LEO satellites are in motion with respect to the distress beacon, they can both develop a position solution using Doppler shift (Frequency of Arrival) and Time of Arrival techniques.