Tropospheric time delay is caused by two effects: angular bending of the radiowaves, which increases the path length over what it would be in free space; and a decrease in the propogation velocity. Both effects result from a change in the index of refraction as a function of altitude, N(h), of the troposphere, which varies with temperature, pressure and humidity. However, the index gradually decreases approximately linearly from mean sea level up to 1 km in altitude and exponentially thereafter.
N(h) is essentially independent of frequency up to 30 GHz. Of interest to the GPS user is the range associated with the tropospheric time delay. Because path length is also related to elevation angle, a correction must be made to compensate for the increased oblique travel through the troposphere.
The range error can be defined by:
$$R(h,\theta) = f(\theta) \times \Delta R(h)$$
Where
\(R(h,\theta)\) = Total range error (metres)
\(\Delta R(h) \) = Range error as a function of altitude (metres)
h = Altitude above mean sea level (metres)
and
\(f(\theta)\) = Range error factor (mapping function) as a function of elevation angle.
$$f(\theta) = \begin{cases}
{1 \over{sin \theta + 0.00143 / (tan \theta + 0.0455) }}, & \text{for } \theta \lt 90^\circ \\
1, & \text{for }\theta = 90^\circ
\end{cases}$$
and
\(\theta\) = Elevation angle between GPS receiver and satellite.
\(\Delta R(h) = \Delta R_1(h) + \Delta R_2(h) + \Delta R_3(h)\) metres.
where
\(\Delta R_1(h)\) = Range error for altitude 0 km \(\le h \le\) 1 km
\(\Delta R_2(h)\) = Range error for altitude 1 km \(\le h \le\) 9 km
\(\Delta R_3(h)\) = Range error for altitude 9 km \(\le h \le h_{sv}\)
\(h_{sv}\) = Altitude of satellite in km
Note: All altitudes are referenced to mean sea level.
For: 0 km \(\le h_u \le\) 1 km
$$\begin{align}
\Delta R_1(h) &= \int_{h=h_u}^{h=1km} (N_s + h \Delta N) dh = {(N_s h + 0.5 \Delta N h^2 )}_{h_u}^{1km} \times 10^{-3} \text{metres}\\
\Delta R_2(h) &= 1430 \times 10^{-3} \text{metres}\\
\Delta R_3(h) &= 732 \times 10^{-3} \text{metres}
\end{align}$$
where
\(h_u\) = Altitude of GPS receiver (km)
\(N_s\) = Surface refractivity index at mean sea level
\(\Delta N = -7.32 exp (0.005577 N_s) \)
and
$$\Delta R(h) = [ N_s(1- h_u) + 0.5 \Delta N(1 - h_u^2) + 1430 + 732 ] \times 10^{-3} $$
Note: The global mean value of \(N_s\) at mean sea level is 324.8.
For: 1 km \( \lt h_u \le \) 9 km
$$\begin{align}
\Delta R_1(h) & = 0 \\
\Delta R_2(h) & = \int_{h=h_u}^{h=9km} N_1 exp [-C(h - 1)] dh \\
& = \left({-8N_1 \over{\ln ({N_1 \over{105}} ) }} exp \left[0.125(1 - h) \ln \left({N_1\over{105}} \right) \right] \right) _{h_u}^{9km} \times 10^{-3} \\
\Delta R_3(h) & = 732 \times 10^{-3}
\end{align}$$
where
\(N_1 = \) Refractivity index at h = 1 km
\(N_1 = N_s + \Delta N \)
\(C = \frac18 \ln \left( {N_1 \over {105 }} \right) \)
and
$$\Delta R(h) = \left[ \left({-8N_1 \over{\ln ({N_1 \over{105}} ) }} exp \left[0.125(1 - h) \ln \left({N_1\over{105}} \right) \right] \right) _{h_u}^{9km} + 732 \right] \times 10^{-3} \\
$$
For: 9 km \( \lt h_u \le h_{sv}\) $$\begin{align} \Delta R_1(h) & = 0 \\ \Delta R_2(h) & = 0 \\ \Delta R_3(h) & = \int_{h=h_u}^{h=h_{sv}} 105 exp [-0.1424(h - 9)] dh \\ & = {\left({-105 \over{0.1424}} exp \left[-0.1424(h - 9) \right] \right)}_{h_u}^{h_{sv}} \times 10^{-3} \\ \\ \Delta R(h) & = {\left({-105 \over{0.1424}} exp \left[-0.1424(20186.8km - h_u) \right] \right)}_{h_u}^{h_{sv}} \times 10^{-3} \end{align}$$
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