GPS Satellites In View

[jimh]

Recently I became interested in some calculations for the elevation angle to GPS satellites. The essential dimensions of this problem are quite basic:

--the radius (r) of the Earth, which is taken as 6371-kilometers, and

--the altitude (a) of the GPS satellite orbit above the Earth, which is taken as 20,220-kilometers.

Ignoring any sort of refraction of the radio waves and considering the observer's height to be zero, we can diagram the problem as shown below:

Roughly to scale

newGPSHorizon.jpg (9.45 KiB) Viewed 3151 times

We can find angle θ from basic right-triangle trigonometry:

cosθ = r / (r+a)

Solving this we find the angle (θ) to be 76.13-degrees.

We can also solve for the distance (d) to the GPS satellite, which we find to be 25,816 kilometers.

These two calculations give us some useful information. The angular measurement from the observer's location will define the range of satellite orbit footprints that will be in view if we assume that the observer has a completely unobstructed field of view, able to see down to the geometric horizon. We can use this angle to predict what satellites will be in view.

If we assume the observer is standing at latitude 55-North, this means he is 35-degrees from the pole. Could a GPS satellite on the other side of the Earth be visible? Since the GPS satellites are in inclined orbits, the highest latitude they can reach is 55-degrees-North (or South). Since our observer can see a satellite on the horizon that is 76-degrees away, we can calculate how high a satellite on the opposite side of the Earth (that is on the observer's meridian) must be to be visible to him. We add 76-degrees to the observer's latitude, 55-degrees. This is 131-degrees. The latitude on the opposite side of the pole on the same meridian will be the supplementary angle (180 - α) or 49-degrees-N latitude. Since the satellites reach 55-degrees-N latitude, this means an observer at the relatively high latitude of 55-degrees-N will be able to see a GPS satellite rise to the North from the opposite side of the Earth.

The distance measurement shows us how much longer the path is to satellites that are on or near the horizon. A satellite that is directly overhead is 20,220-kilometers away. A satellite on the horizon is 25,816-kilometers away, or 5,596-kilometers farther away. Signals from orbiting satellites must travel through the layer of atmosphere known as the ionosphere, which exists roughly between altitudes of 50-kilometers to perhaps 1,000-kilometers. A signal from a satellite overhead makes the shortest possible route through the ionosphere, about 1000-kilometers. A signal arriving from on the horizon will travel a greater distance through the ionosphere, about 3,700-kilometers, or almost four-times greater. When signals transit through the ionosphere they are affected; their speed of propagation varies. Thus signals of low elevation will be subject to much greater influences from the ionosphere than signals coming from overhead. For this reason, a GPS receiver may employ an elevation angle mask, that is, it will ignore the signals from low-elevation satellites and not use them in a position solution. A typical elevation angle mask might be 15-degrees; the GPS receiver will not use signals whose look angle is lower than 15-degrees.

Roughly to scale

105angle.jpg (18.8 KiB) Viewed 3119 times

If we apply a 15-degree elevation angle mask, we can recalculate the angular distance that a GPS satellite will be in view (with at least 15-degree elevation or look angle). Solving this problem requires more advanced trigonometry: solving for a non-right-angle triangle with one angle of 105-degrees (90 + 15) and sides of r and r+a (as above). For that solution I had to turn to an on-line calculator at cossincalc.com for help. The beautiful algorithm there solved the trigonometry for me, finding the new angle (θ) to be 61.6-degrees. This new, smaller, angle of view means that a GPS satellite on the opposite side of the Earth would have to reach latitude 63.4-degrees-N to be in view. Since GPS satellites never get that far north, they would not be in view to our observer at 55-degrees-N latitude, if we assume the receiver is using a 15-degree elevation angle mask.

[jimh]

The lengthening of the path through the ionosphere when the satellite elevation is low is also a function of the ionosphere's height. First we solve for the path length at the horizon if the ionosphere has an altitude of 1,000-kilometers. A drawing explains the trigonometry:

Roughly to scale for 1,000-km ionosphere height

ionosphereDwg.jpg (16.26 KiB) Viewed 3114 times

The Earth radius (r) is 6,371-kilometers. The ionosphere height is assumed to be 1,000-kilometers. We solve the right triangle for the path distance (d), and find it is 3,707-kilometers. The shortest path is overhead, or 1,000-kilometers. Thus the horizon path is 3.7-times longer through the ionosphere.

This ratio is not fixed, as it depends on the actual ionosphere height. Let's assume on a certain day the ionosphere height was 500-kilometers. Now we solve for (d) and find it to be 2,573-kilometers. But the shortest path is now only through 500-kilometers of ionosphere, so the horizon path is 5.15-times longer through the ionosphere.

Because position fix accuracy is very dependent on precise timing, any change in radio wave propagation speed in the ionosphere introduces errors, and the longer the distance through the ionosphere (and the rest of the atmosphere), the more chance for variation. This helps explain why use of very low elevation satellite signals in a GPS position solution is often intentionally rejected by the GPS receiver by use of a look angle mask function.

For more on how the ionosphere (and atmosphere) affect GPS signals, see this excellent article on that topic:

The Ionospheric Effect
https://www.e-education.psu.edu/geog862/node/1715

[jimh]

Now that we know the included angle of the great circle distance between an observer and a GPS satellite that is at 15-degree elevation look-angle, we can find the distance away the satellite's ground track will be. Simple trigonometry will again solve this using the relationship of a circle's circumference to its radius as

Circumference = 2 × π × radius

Then we just multiply by the portion of the circumference of the great circle we are traversing, 61.6-degrees in this case, to find the distance of the great circle arc:

Distance of great circle arc = 2 × π × 6371-kilometer × (61.6/360)
Distance of great circle arc = 6849.6-kilometers