Chapter 16, Problem 16.7P

To determine

Standard deviations for the adjusted quantities.

$\pm 8.5$

Given that:

Three horizontal angles observed around the horizon of the station A is $85˚07’15”,134˚26’48”and140˚26’15”$

Explanation:

Assuming the equal weighting or residuals for three angles as ʋ₁, ʋ₂, ʋ_3.

Writing the equations as

   $\begin{array}{l}\text{x=85°07'15"+v1}\mathrm{................................}\text{(1)}\\ \text{y=134°26'48"+v2}\mathrm{...............................}\text{(2)}\\ \text{x=140°26'15"+v3}\mathrm{...............................}\text{(3)}\end{array}$

The sum of the adjusted angles should be equal to 360°

   $x+y+z= 360˚$ ..................... (4)

Substituting equations (1), (2), (3) in equation (4) we get the equation as follows

   $85°07^{\prime }15"+v1+134°26"48"+v2+140°26\text{'}15"+v3=360°$

   $v_1+v_2+v_3=360˚-360˚0’18”$ .................. (5)

   $v_3=-18”–(v_1+v_2)\dots \dots \dots \dots \dots \dots \dots \dots \dots \dots$ (6)

From the equation of equation of least squares

   ${{\displaystyle \sum v}}^2=v_1{}^2+{\text{ v}}_2{}^2+v_3{}^2$ ........................ (7)

Substitute value of equation (6) in equation (7)

We get the equation

   ${{\displaystyle \sum v}}^2=v_1{}^2+{\text{ v}}_2{}^2+(-18”–{(v_1+v_{2)})}^2$ ................ (8)

Partial derivation of equation (8) with respect to ʋ₁we get

   $\begin{array}{l}\frac{\partial {\displaystyle \sum v^2}}{\partial v1}=0\\ \frac{\partial {\displaystyle \sum v^2}}{\partial v1}=2v_1+2(-18”–(v_1+v_{2)})\left(-1\right)\\ 2v_1+2(-18”–(v_1+v_{2)})=0\\ 2v_1+\text{ 36" +2}{v}_1+2v_{2)}=0\\ 4v1+2v2=-36"\mathrm{...........................}(9)\end{array}$

Partial derivation of equation (8) with respect to ʋ₂we get

   $\begin{array}{l}\frac{\partial {\displaystyle \sum v^2}}{\partial v2}=0\\ \frac{\partial {\displaystyle \sum v^2}}{\partial v2}=2v_2+2(-18”–(v_1+v_{2)})\left(-1\right)\\ 2v_2+2(-18”–(v_1+v_{2)})=0\\ 2v2+\text{ 36" +2}{v}_1+2v_{2)}=0\\ 4v2+2v1=-36"\mathrm{...........................}(10)\end{array}$

Solving equations (9) and (10)

   $\begin{array}{l}v1-v2=0\\ v1=v2\end{array}$

Substitute ʋ₁ for ʋ₂in equation (9)

   $\begin{array}{l}4v_1+2v_2 =-36”\hfill \\ 6v_2=-36”\hfill \\ v_2=6”\hfill \\ So,v_1=v_2=6”\hfill \end{array}$

Substitute the above values in equation (6)

We get

   $\begin{array}{l}{v}_3=-18”-\left(-6”-6”\right)\hfill \\ =-6”\hfill \end{array}$

Therefore, the probable value of three angles is -6"

Used equations:

   ${\sigma }_0=\pm \sqrt{\left(\frac{v_1^2+v_2^2+v_3^2}{m-n}\right)}$ ........................(11)

Where m = number of observations and

n = number of unknowns

Standard deviation for the adjusted quantities can be determined using

   $\sigma xi=\sigma { }_0{}_{ }\sqrt{\frac{2}{3}}$ ............. (12)

Answer:

Substitute $V_{1=}{V}_2=V_3=-6” ,m=3n=2$ in equation (11)

We get

   ${\sigma }_0=\pm$

   $\sqrt{\frac{{\left(6\right)}^{2 }+{\left(6\right)}^{2 }+{\left(6\right)}^{2 }}{3-2}}=\pm 10.4”$

Standard deviation ${\sigma }_0=\pm 10.4”$

Standard deviation for the adjusted quantities ${\sigma }_{xi}$ ) can be determined using

Substitute ${\sigma }_0=\pm 10.4”$ in equation (12)

We get

   $\sigma xi=\pm 10.4{"}_{ }\sqrt{\frac{2}{3}}=\pm 8.5"$

Conclusion:

The standard deviation for the adjusted quantities is $\pm 8.5"$