Chapter 16, Problem 16.22P

To determine

Convert the geocentric coordinates obtained for station troy to geodetic coordinates using WGS84 ellipsoidal parameters.

The value of $D_r=4,792,248.634m$

The longitude for station troy== $-76˚08’1.77”\left(west\right)$

The latitude for station troy= $41˚23’16.4$

The geodetic height of station troy = $39.67m$

Concept used

= $\sqrt{X_{T }^2}+Y_T^2$ ....................(1)

Where X_T= X coordinate for station troy

Y_T= Y coordinate for station troy

The longitude for station troy (ƛ_T= 2 ${\tan }^{-1}\frac{D_T-X_T}{Y_T})$ ....................(2)

The latitude for station troy ƛ₀= ${\tan }^{-1}\frac{Z_t}{D_T\left(1-e^2\right)}$

Z_T= Z coordinate for station troy

e = eccentricity of the WGS84 reference ellipsoid

Determine the radius in the prime vertical (R_NP) = $\frac{a}{\sqrt{1}-e^{2}si{n}^2{\varphi }_0}$

Determine the improved value of the station troy(ƛ₀)_T= ${\tan }^{-1}\frac{Z_T+e^{2}RNF \sin {\varphi }_0}{D_T}$

Determine the geodetic height of station troy (h_T) = $\frac{D_T}{\cos {\varphi }_0}-R_{NP}$

Explanation:

Determine the value of

= $\sqrt{X_{T }^2}+Y_T^2$

X_T= X coordinate for station troy = $1,148,486.1133m$

Y_T= Y coordinate for station troy = $-4,652,593.5370m$

= $\sqrt{X_{T }^2}+Y_T^2$ = $\sqrt{(1,148,{486.1133}_{ }^2}+4,652,{593.5370}_{}^2)$ = $4,792,248.634m$

The longitude for station troy (ƛ_T= 2 ${\tan }^{-1}\frac{4,792,248.634-1,148,486.1133}{4,652,593.5370}$ )

=- $76˚08’1.77”$ (west)

Hence,the longitude for station troy ==-76°08'1.77" (west)

The latitude for station troy ƛ₀= ${\tan }^{-1}\frac{Z_t}{D_T\left(1-e^2\right)}$ = ${\tan }^{-1}\frac{4,194,{864.2659}_m}{4,792,248.634\left(1-0.0818190842\right)}$

= ${\tan }^{-1}\frac{4,194,864.265}{4,760,167.501}$ = $41˚23’16.4”\left(North\right)$

Determine the radius in the prime vertical (R_NP) = $\frac{a}{\sqrt{1}-e^{2}si{n}^2{\varphi }_0}$

   $\begin{array}{l}Substitutea=6,378,137\hfill \\ E=0.08181313084\hfill \\ {\varphi }_{0=}41˚23’16.4”\hfill \end{array}$

R_NP= $\frac{6,378,137}{\sqrt{1-\left({0.08181313084}^2\times si{n}^2\left(41˚23\text{’}16.4\right)\right)}}$ = $\frac{6,378,137}{\sqrt{1-0.002926}}=6,387,488.742m$

Determine the improved value of the station troy(ƛ₀)_T= ${\tan }^{-1}\frac{Z_T+e^{2}RNF \sin {\varphi }_0}{D_T}$

= $\begin{array}{l}{Z}_t=4,194,864.2659m\\ e=0.08181313084\\ D_T=4,792,248.634\\ {\varphi }_{0=}41˚23’16.4”\\ R_{NP}=6,387,488.742m\end{array}$

(∆₀)_T= ${\tan }^{-1}\frac{4,194,{864.2659}_m+4,194,{864.2659}_m{}^2\times 6,387,488.742\times \sin 41˚23\text{’}16.4\text{”}}{4,792,248.634}$

= $41˚23’16.4$

Determine the geodetic height of station troy (h_T) = $\frac{D_T}{\cos {\varphi }_0}--R_{NP}$

$\begin{array}{l}{D}_T=4,792,248.634\\ {\left({\varphi }_0\right)}_T=41˚23’16.4\\ R_{NP}=6,387,488.742m\\ hT=\frac{4,792,248.634}{\cos 41˚23\text{’}16.4}-6,387,488.742=39.67m\end{array}$

The geodetic height of station troy = $39.67m$

The value of $D_T=4,792,248.634m$

The longitude for station troy = $=-76˚08’1.77”\left(west\right)$

The latitude for station troy $=41˚23’16.4$