# Remote Sensing, GIS, GPS and Photogrammetry

 Question 1
If the size of the ground area is $6 \mathrm{~km} \times 3 \mathrm{~km}$ and the corresponding photo size in the aerial photograph is $30 \mathrm{~cm} \times 15 \mathrm{~cm}$, then the scale of the photograph is 1 : ____ (in integer).
 A 15000 B 20000 C 25000 D 30000
GATE CE 2023 SET-2   Geomatics Engineering
Question 1 Explanation:
Scale of Photograph $(\mathrm{S})=\frac{\text { Photo distance }}{\text { Ground distance }}$

For Length scale :
\begin{aligned} S & =\frac{30 \mathrm{~cm}}{6 \mathrm{~cm}} \\ & =\frac{30 \mathrm{~cm}}{60 \times 10^{4} \mathrm{~cm}} \end{aligned}
$\Rightarrow \quad \mathrm{S}=\frac{1}{20000}$

\begin{aligned} S & =\frac{15 \mathrm{~cm}}{3 \mathrm{Km}} \\ & =\frac{15 \mathrm{~cm}}{30 \times 10^{4} \mathrm{~cm}} \end{aligned}
$S=\frac{1}{20000}$
$\therefore$ Scale of photograph, $S=1: 20000$
 Question 2
An aerial photograph is taken from a flight at a height of 3.5 km above mean sea level, using a camera of focal length 152 mm. If the average ground elevation is 460 m above mean sea level, then the scale of the photograph is
 A 1 : 20000 B 01:20 C 1 : 100000 D 1.986111111
GATE CE 2022 SET-1   Geometics Engineering
Question 2 Explanation:
\begin{aligned} H &=3.5Km=3500m\\ f&=152mm \\ h_{avg}&=460m \\ Scale&=\frac{f}{H-h_{avg}} \\ &= \frac{152 \times 10^{-3}}{3500-460}\\ &=\frac{1}{20000} \end{aligned}

 Question 3
A camera with a focal length of 20 cm fitted in an aircraft is used for taking vertical aerial photographs of a terrain. The average elevation of the terrain is 1200 m above mean sea level (MSL). What is the height above MSL at which an aircraft must fly in order to get the aerial photographs at a scale of 1:8000?
 A 2600 m B 2800 m C 3000 m D 3200 m
GATE CE 2019 SET-2   Geomatics Engineering
Question 3 Explanation: Given focal length = 20 cm
as we know scale of vertical photograph $=\frac{f}{H-h_{avg}}$
its given as 1 : 8000
Hence,
\begin{aligned} \frac{f}{H-h_{avg}} &=\frac{1}{8000} \\ \frac{20 cm}{(H-1200) \times 100cm}&=\frac{1}{8000} \\ \Rightarrow \;\;H &=2800m \end{aligned}
 Question 4
An aerial photograph of a terrain having an average elevation of 1400 m is taken at a scale of 1:7500. The focal length of the camera is 15 cm. The altitude of the flight above mean sea level (in m, up to one decimal place) is ______
 A 2680 B 2460 C 2525 D 3620
GATE CE 2018 SET-2   Geomatics Engineering
Question 4 Explanation:
\begin{aligned} h &=1400 \mathrm{m} \\ \text { Scale } &=1: 7500 \\ f &=15 \mathrm{cm} \\ \text { Scale } &=\frac{f}{H-h} \\ \Rightarrow\quad \frac{1}{7500} &=\frac{15 \times 10^{-2}}{H-1400} \\ \Rightarrow\quad H &=2525 \mathrm{m} \end{aligned}
 Question 5
A square area (on the surface of the earth) with side 100 m and uniform height, appears as 1 $cm^{2}$ on a vertical aerial photograph. The topographic map shows that a contour of 650 m passes through the area. If focal length of the camera lens is 150 mm, the height from which the aerial photograph was taken, is
 A 800 m B 1500 m C 2150 m D 3150 m
GATE CE 2018 SET-1   Geomatics Engineering
Question 5 Explanation:
\begin{aligned} A&=100 \times 100 \mathrm{m}^{2}\\ \text { Area on photo, }\\ a &=1 \mathrm{cm}^{2} \\ \text { Scale } \quad 1 \mathrm{cm} &=100 \mathrm{m} \\ f &=150 \mathrm{mm} \\ h &=650 \mathrm{m} \\ \text { Scale } \quad &=\frac{1}{100}=\frac{1}{100 \times 10^{2}}\\ &=\frac{1}{10000}=\frac{f}{H-h}\\ \Rightarrow \quad \frac{1}{10000}&=\frac{150 \times 10^{-3}}{H-650} \\ \Rightarrow \quad H&=2150 \mathrm{m} \end{aligned}

There are 5 questions to complete.