Chapter 16, Problem 16.35P

To determine

Parameters of anerror ellipse

   $\begin{array}{l}Theparametersoftheerrorellipseare:\\ t=71˚55’36”\\ S_x=\pm 0.033ft\\ S_y=\pm 0.021ft\end{array}$

Given that

(A^TWA)^-1 = $\left[\begin{array}{cc}qxx& qxy\\ qxy& qyy\end{array}\right]$ = $\left[\begin{array}{cc}0.0004894& 0.0000890\\ 0.0000890& 0.0002457\end{array}\right]$

Standard deviation of unit weight is given as S₀= $1.45$

Determine the rotational angle t using the equation:

$\tan (2t)=\frac{2(0.0000890)}{0.00}$

We know that

   $\begin{array}{l}{q}_{xy}=0.0000890\hfill \\ q_{xx}=0.0004894\hfill \\ q_{yy}=0.0002457\hfill \end{array}$

$\tan (2t)=\frac{2(0.0000890)}{0.0002457-0.0004894}=-0.730406$

   $t=\frac{1}{2}\left[{\tan }^{-1}(-0.73046)+{180}^{\circ }\right]={71}^{\circ }55\text{'}36"$

Determine the values of semi major axis $S_m=S_0\sqrt{qxx{\sin }^{2}t+2qxy\cos \theta \sin \theta }+qyy{\cos }^2\theta$

We know that,

   $\begin{array}{l}\begin{array}{l}{q}_{xx}=0.0004894\hfill \\ q_{yy}=0.0002457\hfill \\ q_{xy}=0.0000890\hfill \end{array} \\ S_0=1.45\end{array}$

S_x=S₀ $\sqrt{0.0004894{\sin }^2({71}^{\circ }55\text{'}36")+2((0.0000890)\cos ({71}^{\circ }55\text{'}36")\sin ({71}^{\circ }55\text{'}36"))+(0.0002457{\cos }^2{71}^{\circ }55\text{'}36")}$

= $\pm 0.033ft$

Determine the values of semi minor axis S_m=S₀ $\sqrt{qxx{\sin }^{2}t-2qxy\cos \theta \sin \theta }+qyy{\cos }^2\theta$

We know that,

   $\begin{array}{l}{q}_{xx}=0.0004894\hfill \\ q_{yy}=0.0002457\hfill \\ q_{xy}=0.0000890\hfill \\ S_0=1.45\hfill \end{array}$

S_y=S₀ $\sqrt{0.0004894{\sin }^2({71}^{\circ }55\text{'}36")-2((0.0000890)\cos ({71}^{\circ }55\text{'}36")\sin ({71}^{\circ }55\text{'}36"))+(0.0002457{\cos }^2{71}^{\circ }55\text{'}36")}$

= ±0.021ft

Conclusion:

   $\begin{array}{l}Theparametersoftheerrorellipseare:\\ t=71˚55’36”\\ S_{x=}\pm 0.033ft\\ S_{y=}\pm 0.021ft\end{array}$