The internal angles P, Q, R of a triangle are observed in degree minute second ($\degree$'") using a Total Station. The angles along with their probable errors are given below. P = 40° 30′ 01″ $\pm$ 02" Q = 60° 00′ 02″ $\pm$ 03" R = 79° 30′05″ $\pm$ 04" T he corrected values of the angles P, Q and R are
Observed Angles and Summation
The observed internal angles P, Q, and R of a triangle are:
- P = 40$\degree$ 30$\prime$ 01$\prime\prime$
- Q = 60$\degree$ 00$\prime$ 02$\prime\prime$
- R = 79$\degree$ 30$\prime$ 05$\prime\prime$
To find the sum, convert all angles to seconds ($\prime\prime$):
- P = (40 $\times$ 3600) + (30 $\times$ 60) + 1 = 144000 + 1800 + 1 = 145801$\prime\prime$
- Q = (60 $\times$ 3600) + (0 $\times$ 60) + 2 = 216000 + 0 + 2 = 216002$\prime\prime$
- R = (79 $\times$ 3600) + (30 $\times$ 60) + 5 = 284400 + 1800 + 5 = 286205$\prime\prime$
Sum of observed angles = 145801$\prime\prime$ + 216002$\prime\prime$ + 286205$\prime\prime$ = 648008$\prime\prime$
Triangle Angle Error Calculation
The theoretical sum of internal angles in a triangle is 180$\degree$. 180$\degree$ = 180 $\times$ 3600$\prime\prime$ = 648000$\prime\prime$.
The misclosure error (or excess) is calculated as:
Error = Sum of observed angles - Theoretical sum
Error = 648008$\prime\prime$ - 648000$\prime\prime$ = +8$\prime\prime$
Since the sum of observed angles is greater than 180$\degree$, there is an excess of +8$\prime\prime$. This excess error must be distributed among the angles as corrections. The total correction required is -8$\prime\prime$.
Applying Corrections to Find Corrected Angles
The probable errors are given as P $\pm$ 02$\prime\prime$, Q $\pm$ 03$\prime\prime$, R $\pm$ 04$\prime\prime$. The error is distributed such that the sum of corrections equals the negative of the misclosure error (-8$\prime\prime$).
The implied corrections required to satisfy the geometric condition of a triangle are:
- Correction for P ($c_P$) = -1.1$\prime\prime$
- Correction for Q ($c_Q$) = -2.5$\prime\prime$
- Correction for R ($c_R$) = -4.4$\prime\prime$
Sum of corrections = -1.1$\prime\prime$ - 2.5$\prime\prime$ - 4.4$\prime\prime$ = -8.0$\prime\prime$, which matches the total correction needed.
Applying these corrections to the observed angles:
- Corrected P = Observed P + $c_P$ = 40$\degree$ 30$\prime$ 01$\prime\prime$ - 1.1$\prime\prime$ = 40$\degree$ 29$\prime$ 59.9$\prime\prime$
- Corrected Q = Observed Q + $c_Q$ = 60$\degree$ 00$\prime$ 02$\prime\prime$ - 2.5$\prime\prime$ = 59$\degree$ 59$\prime$ 59.5$\prime\prime$
- Corrected R = Observed R + $c_R$ = 79$\degree$ 30$\prime$ 05$\prime\prime$ - 4.4$\prime\prime$ = 79$\degree$ 29$\prime$ 60.6$\prime\prime$ = 79$\degree$ 30$\prime$ 00.6$\prime\prime$
These corrected values match Option C.
Important Questions from Errors in Observations
The carrier phase observation model in GNSS is given as $$ \phi_A^i = f \delta^i - \frac{\rho_A^i}{\lambda} - f \delta_A + N_A^i - f \delta_{\text{iono}} + f \delta_{\text{tropo}} + \epsilon $$ where $\phi_A^i$ is the observed carrier phase in cycles, $f$ is the frequency of the carrier in hertz, and $\lambda$ is the wavelength of the carrier in meters.
What is the unit of the ionospheric ($\delta_{\text{iono}}$) and tropospheric ($\delta_{\text{tropo}}$) delay terms in the given equation?- In GNSS positioning, the cycle slips are the most detrimental for estimating _______.
- According to the first order ionospheric delay term, the time delay experienced by the GNSS signal is directly proportional to the Total Electron Content (TEC) in the ionosphere, and inversely proportional to the square of the frequency of the carrier wave. Based on this, the GPS L2 (1227.60 MHz) carrier is slower than the GPS L1 (1575.42 MHz) carrier by a factor of ________ for a given TEC (Rounded off to the nearest integer).
- In the context of Global Navigation Satellite System positioning, the Saastamoinen model provides a correction for ________.
- In the choke ring antenna there are concentric cylinders placed around the antenna that are of a certain depth to minimize the multipath effect. If the signal wavelength is $\lambda$, then the depth of the cylinders in the choke ring antenna should be
