Nadir Angle and Elevation Angle

[jimh]

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 20-degrees elevation above the horizon?

[Note: Updated January 2025: the original analysis used the orbital height of a GPS satellite. A GALILEO satellite is in a slightly higher orbit, 23,222-km compared to 20,180-km. I have re-calculated to use the GALILEO orbit height, and also to use an elevation angle of 20-degrees.)

A sketch shows the relationships, not drawn to scale:

Fig. 1. Geometry of the Nadir Angle

nadirAngle.png (110.58 KiB) Viewed 19484 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, 20-degrees, plus 90-degrees; C = 110-degrees
--we know length of side c; it's the satellite altitude 23222 plus earth radius 6271 km = 29593-km

We can find the other angles and lengths using a relationship known as the law of sines for a triangle. That law says

a/sinA = b/sinB = c/sinC

Solving the the remaining angles and sides gives

A (angular displacement from satellite ground point) = 58.3-degrees
B (the nadir angle) = 11.7-degrees
a (path length) = 26,802-km

Angle B is the desired calculation, the nadir angle which means the angular distance off of boresight to the Earth below. This means the signal from a transmitter that sees the satellite at an elevation above the horizon of 20-degrees will arrive at the satellite at a nadir angle of 11.7-degrees. This is of interest because the receiving antenna on the satellite must provide coverage out to at least 11.7-degrees to be useful for transmitters that see the satellite only 20-degrees above their horizon.

As a check we see if the three angles sum to 180:

A + B + C = 180
58.3 + 11.7 + 110 = 180

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

a = 26,802-km

The most direct path (when the satellite is directly overhead) is the satellite orbit altitude, 23222 km. The path length at 20-degrees elevation is about 3,600-km longer, so path loss with be greater.

All angular and length calculations checked with http://www.calculator.net/triangle-calculator.html

With the availability of a calculator, solving the trangle for various solution is made much easier. The focus in this discussion is on the nadir angle, and finding the width of the main lobe of the 406-MHz receiver antenna on a GALILEO satellite so that its gain pattern covers signals which will arrive at an angle that is the farest from the nadir 0-degree angle. With a bit on analysis, one can deduce that this situation occurs when the satellite elevation angle from the observer (in this case a person with a 406-MHz distress beacon) is at 0-degrees, that is, just at the observers geometric horizon.

In terms of nomenclature of the triangle in Figure 1 above, angle C is 90-degrees when the elevation angle is 0-degrees. To find the nadir angle B, we solve using inputs

b = 6371 km (Earth radius)
C = 90-degrees (required condition)
c = 29593 km

We then get the following calculated values
B = 12.4 degrees (nadir angle)
a = 28,899 km (path length)
A = 77.6 degrees (angular distance to ground point)

[jimh]

The view of the Earth from the satellite in which it is available at 20-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.

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 36900 times

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.

Calculating variable h

The height h can be calculated with right-triangle geometry; we use the right-triangle with hypotenuse r and angle θ, which will be A (from the diagram in Figure 1), or in this case 58.3-degrees. Because the sum of angles must be 180, the other angle must then be

h1 = 180 - 90 - 58.3 = 31.7-degree.

The length of the side opposite that angle h1 is then equal to

sin(31.7) x 6371 = 3348-km

The length of the side opposite the 31.7-degree angle h1 is equal to (r-h), or in other words h + h1 =r. This means h = r - h1:

h = 6371 - 3348 = 3023-km

This gives the area of the spherical cap as

Area = 2πrh
Area = 2π(6371 x 3023)
Area = 121,011,215 km2

Thus for a medium earth orbit satellite at altitude 23222 km, the area on the Earth from which it will be in view with an elevation angle of 20-degrees or more is about 121n-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 20-degree elevation mask) is 121,011,215 km2, or about 24-percent of the Earth's surface.

Alternate method

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 = 121,020,026 km2

This value is only 0.007-percent different from the value calculated using the other method, and that is probably due to some rounding of the factors employed in the first calculation.

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 given in a (linked) Wikipedia article. An interesting on-line calculator is also available to calculate these solutions.

[jimh]

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.

[jimh]

Revisiting this article after seven years, I thought it would be interesting to calculate the required elevation angle for an observer at the Earth surface in order for the nadir angle of arrival at the GALILEO satellite to not be greater than 12-degrees. (The 12-degrees figure comes from the plot of the GALILEO SARSAT 406-MHz antenna response in a related discussion.)

Using the handy triangle calculated at

https://www.calculator.net/triangle-calculator.html

makes this quite simple. Using the identifiers as shown above, we enter the values for the triangle as

c = 29593
B = 12
b = 6371

There are two possible solutions, but one of them is not valid in our case because it requires the path to the satellite to pass through the Earth. Discarding that solution we get

a = 27,292-km (path length)
b = 6,371-km (given)
c = 23,222-km (given)

A = 63.0-degrees (angular distance to ground point)
B = 13-degree (given)
C = 105-degrees (from which we get the elevation angle calculated below)

We then solve for the elevation angle as C - 90 = 15-degrees, and a path length of 27,292-km. Now we can go back to the article discussing path length and signal loss to see how this fits into the path signal margin. (Hint: the fade margin works out to be quite impressive.)

[jimh]

Another interesting result of solving the triangle problem illustrated above in Figure 1 (as recently revised to use an elevation angle of 20-degrees above the geometric horizon) is that the angle A tells us the angular distance between the observer's location and the satellite's ground point. In the particular situation illustrated in Figure 1, angle A is 58.3-degrees. Using a great-circle on which both the satellite ground point and the observer are located, we can compute the distance along the Earth surface. Using the Earth circumference as 24,900-miles, the distance is

Distance between observer and ground point = (58.3/360)×24900 = 4,032-miles

A further thought about the elevation angle
When I worked out the elevation angle on Earth to produce a signal arrival at the satellite of no more than 12-degrees off nadir (which is suggested by the antenna pattern on the GALILEO FOC satellites), I found a 15-degree elevation angle was necessary. (The triangle solution as calculated is linked here as the second solution to the problem; the first solution does not work because the satellite would be below the observer's horizon on Earth.) If the observer were at sea there is generally no surrounding terrain that would obstruct the view of a satellite at lower elevation angles, so certainly an elevation angle of even less than 15-degrees would still be useful. But my initial thinking about being useful at really low elevation angles was clouded because my perspective is from a maritime use of the SARSAT system.

SARSAT is certainly very useful for marine distress alerting and locating, but since it is a global system, it also should be useful for locations on land. When in a distress situation on land, there is no guarantee the view of the sky will be such that very low elevation angles to a satellite will be clear of obstructions. With that in mind, a 15-degree elevation angle seems to be more realistic and more likely to occur in actual use of the system. Perhaps the designers of the 406-MHz SARSAT antenna had that in mind when they configured the antenna gain and pattern as they did.