Nadir Angle and Elevation Angle
Posted: 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:
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)
a=25,226
b=6371
c=26551
Angles (degrees)
A=71.17
B=13.83
C=95
Checked with http://www.calculator.net/triangle-calculator.html
A sketch shows the relationships, drawn roughly to scale:
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)
a=25,226
b=6371
c=26551
Angles (degrees)
A=71.17
B=13.83
C=95
Checked with http://www.calculator.net/triangle-calculator.html