Practical Geography

175 soru

Soru 121Soru

During a chain survey along a baseline, a student uses a steel measuring tape that is shorter than its standard designated length. Which type of error will consistently accumulate in the total recorded distance?

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Cevap: Cumulative error

Cevap

Cumulative error
Using an instrument with a fixed structural defect, such as a tape shorter than standard length, causes the measurement discrepancy to accumulate continuously in one direction throughout the survey. This defines a cumulative (systematic) error.

Adım Adım Çözüm

1
Analyze the nature of the faulty instrument
The measuring tape is shorter than its standard length.
Using a shorter tape means every single chain length laid out on the ground will record a value higher than actual distance.
2
Determine the impact of the fault over multiple measurements
The error constantly adds up in the same direction throughout the fieldwork.
Errors that continuously build up in one direction are classified as cumulative or systematic errors.

Anahtar Kavram

Classification of Errors in Elementary Surveying
Soru 122Soru

Two agricultural processing centers in Osun State are separated by an actual ground distance of 7.5 km7.5\text{ km}. If the distance between these two centers on a regional planning map measures 15 cm15\text{ cm}, what is the Representative Fraction (R.F.) of the map?

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Cevap: 1:50,0001 : 50,000

Cevap

The Representative Fraction (R.F.) of the map is 1:50,0001 : 50,000.
To find the Representative Fraction, first convert 7.5 km7.5\text{ km} to centimeters (7.5×100,000=750,000 cm7.5 \times 100,000 = 750,000\text{ cm}). Setting up the ratio of map distance to ground distance gives 15750,000\frac{15}{750,000}, which reduces to 150,000\frac{1}{50,000}, yielding an R.F. of 1:50,0001 : 50,000.

Adım Adım Çözüm

1
Convert the ground distance from kilometers to centimeters.
7.5 km×100,000 cm/km=750,000 cm7.5\text{ km} \times 100,000\text{ cm/km} = 750,000\text{ cm}.
Representative Fraction requires both map distance and ground distance to be in identical units.
2
Formulate the ratio of map distance to ground distance.
Scale=15 cm750,000 cm\text{Scale} = \frac{15\text{ cm}}{750,000\text{ cm}}.
The formula for Representative Fraction is R.F.=Map DistanceGround Distance\text{R.F.} = \frac{\text{Map Distance}}{\text{Ground Distance}}.
3
Simplify the fraction to have a numerator of 1.
15÷15750,000÷15=150,000\frac{15 \div 15}{750,000 \div 15} = \frac{1}{50,000} or 1:50,0001 : 50,000.
An R.F. must always be expressed as a unit fraction (1:n1 : n).

Anahtar Kavram

Calculating Representative Fraction (R.F.) from map distance and ground distance
Soru 123Soru

During a prismatic compass survey along a open traverse line PQRSP-Q-R-S, a student records the following forward bearings (FB) and back bearings (BB):

- Line PQPQ: FB = 045045^\circ, BB = 225225^\circ
- Line QRQR: FB = 130130^\circ, BB = 308308^\circ
- Line RSRS: FB = 220220^\circ, BB = 042042^\circ

Based on these field observations, which of the survey stations is affected by local attraction?

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Cevap: Station R

Cevap

Station R is affected by local attraction.
Station R is the correct answer because it is the intersection station for both Line QR and Line RS. In magnetic compass surveying, when the difference between forward bearing and back bearing of a line is exactly 180180^\circ, both stations forming that line are free from local magnetic disturbance (as seen with Line PQ). Since Line QR (308130=178308^\circ - 130^\circ = 178^\circ) and Line RS (220042=178220^\circ - 042^\circ = 178^\circ) both display a 22^\circ error, the common station to both lines—Station R—must be the sole source of local attraction.

Adım Adım Çözüm

1
Calculate the difference between Forward Bearing (FB) and Back Bearing (BB) for Line PQ.
225045=180|225^\circ - 045^\circ| = 180^\circ
A difference of exactly 180180^\circ indicates that both Station P and Station Q are free from local magnetic attraction.
2
Calculate the difference between Forward Bearing (FB) and Back Bearing (BB) for Line QR.
308130=178|308^\circ - 130^\circ| = 178^\circ
The difference is 178178^\circ, showing a 22^\circ discrepancy. Either Station Q or Station R is affected by local attraction.
3
Calculate the difference between Forward Bearing (FB) and Back Bearing (BB) for Line RS.
220042=178|220^\circ - 042^\circ| = 178^\circ
The difference is 178178^\circ, showing a 22^\circ discrepancy. Either Station R or Station S is affected by local attraction.
4
Synthesize the results to isolate the affected station.
Station R is common to both erroneous lines (QR and RS), while Stations P and Q are verified correct from Line PQ.
Because Station R is present in both lines that exhibit the 22^\circ magnetic error, Station R is the station experiencing local attraction.

Anahtar Kavram

Local Attraction Detection in Prismatic Compass Surveying
Tahmini Süre:2m 0s
Soru 124Soru

On a topographical map extract drawn at a scale of 1:50,0001:50,000, a river valley features V-shaped contour lines that point consistently toward the south-west. Point X lies directly on the 450 m450\text{ m} contour line, while Point Y lies 5 km5\text{ km} further down the channel path on the 200 m200\text{ m} contour line. Based on these topographic observations, toward which direction does the river flow, and what is its average channel gradient between Point X and Point Y?

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Cevap: Toward the north-east at a gradient of 1 in 201\text{ in }20

Cevap

The river flows toward the north-east at an average channel gradient of 1 in 201\text{ in }20.
In topographic map interpretation, contour lines form 'V' shapes when crossing river valleys, with the apex of the 'V' pointing upstream toward higher elevation. Because the V-shapes point south-west toward the 450 m450\text{ m} contour, the river must flow downstream toward the lower 200 m200\text{ m} contour in the north-east direction. Furthermore, the gradient calculation requires dividing the Vertical Interval (450 m200 m=250 m450\text{ m} - 200\text{ m} = 250\text{ m}) by the Horizontal Equivalent (5 km=5,000 m5\text{ km} = 5,000\text{ m}), giving 2505,000=120\frac{250}{5,000} = \frac{1}{20} (or 1 in 201\text{ in }20).

Adım Adım Çözüm

1
Determine flow direction from V-shaped contour lines
The river flows toward the north-east.
V-shaped contour lines crossing a river valley always point apex-upstream (toward higher ground/source). Since the V's point south-west (toward 450 m450\text{ m}), the river flows down the elevation slope toward the north-east (toward 200 m200\text{ m}).
2
Calculate Vertical Interval (VI)
VI=250 m\text{VI} = 250\text{ m}
Difference in elevation between Point X (450 m450\text{ m}) and Point Y (200 m200\text{ m}) is 450 m200 m=250 m450\text{ m} - 200\text{ m} = 250\text{ m}.
3
Convert Horizontal Equivalent (HE) to consistent units
HE=5,000 m\text{HE} = 5,000\text{ m}
The horizontal distance along the river course is 5 km=5×1,000 m=5,000 m5\text{ km} = 5 \times 1,000\text{ m} = 5,000\text{ m}.
4
Compute stream gradient ratio
Gradient=250 m5,000 m=120\text{Gradient} = \frac{250\text{ m}}{5,000\text{ m}} = \frac{1}{20} (1 in 201\text{ in }20)
Gradient is expressed as VI/HE\text{VI} / \text{HE}, simplifying 2505000\frac{250}{5000} to 120\frac{1}{20}.

Anahtar Kavram

Valley Contour Interpretation and Topographic Stream Gradient Analysis
Soru 125Soru

During a chain survey along a baseline, a surveyor uses a 20 m20\text{ m} chain that is later discovered to be 0.10 m0.10\text{ m} too short due to bent links. If the recorded length of the baseline is 400 m400\text{ m}, what is the true ground distance of the baseline, and what category of error is produced by this tape defect?

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Cevap: 398.00 m398.00\text{ m}; cumulative error

Cevap

398.00 m398.00\text{ m}; cumulative error
The true ground distance is calculated using the formula True Distance=Recorded Distance×(L/L)\text{True Distance} = \text{Recorded Distance} \times (L'/L), where LL' is the actual chain length (19.90 m19.90\text{ m}) and LL is the nominal length (20.00 m20.00\text{ m}). Applying this yields 400×(19.90/20)=398.00 m400 \times (19.90 / 20) = 398.00\text{ m}. Furthermore, because the flaw in the instrument is constant and acts in one direction throughout the survey, the resulting error accumulates continuously, making it a cumulative error.

Adım Adım Çözüm

1
Determine the actual length of the chain
Nominal length L=20.00 mL = 20.00\text{ m}; actual length L=20.000.10=19.90 mL' = 20.00 - 0.10 = 19.90\text{ m}.
The chain is 0.10 m0.10\text{ m} too short, so each full chain layout measures only 19.90 m19.90\text{ m} of ground distance.
2
Calculate the true ground distance using the proportion formula
True Distance = Recorded Distance ×LL=400×19.9020.00=398.00 m\times \frac{L'}{L} = 400 \times \frac{19.90}{20.00} = 398.00\text{ m}.
When an instrument is too short, the measured distance is over-reported, so the true distance must be less than the recorded distance.
3
Classify the nature of the surveying error
Cumulative (systematic) error.
Errors arising from constant instrumental flaws (like an incorrect chain length) always act in the same direction and accumulate in magnitude as more chain lengths are measured.

Anahtar Kavram

Chain Surveying Distance Correction and Error Classification
Soru 126Soru

Match each elementary surveying accessory or instrument on the left with its precise operational function during field survey procedures on the right.

Soldaki öğeye tıklayın, sonra eşleşen sağdaki öğeye tıklayın

Öğeler

Optical square
Plumb bob
Clinometer
Trough compass

Eşleşmeler

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Cevap

Optical square matches setting out 9090^\circ offsets; Plumb bob matches centering equipment over ground pegs; Clinometer matches measuring vertical slope angles; Trough compass matches orienting a plane table to magnetic north.
Each instrument serves a distinct purpose in fieldwork setup and measurement: the optical square sets out 9090^\circ offsets; the plumb bob ensures precise vertical centering over ground pegs; the clinometer measures vertical slope angles for distance corrections; and the trough compass aligns plane tables to magnetic north.

Adım Adım Çözüm

1
Identify the primary function of the optical square
Optical squares utilize double-reflection optics to construct perpendicular lines (9090^\circ offsets) relative to a baseline.
It ensures offset distances measured to nearby features are exactly perpendicular to the main survey line.
2
Identify the primary function of the plumb bob
Plumb bobs suspend vertically under gravity to align an overhead instrument center with a peg on the ground.
Accurate centering eliminates index and displacement errors at station points.
3
Identify the primary function of the clinometer
Clinometers measure vertical slope inclination angles.
Knowing slope angles allows conversion of surface distances measured along slopes into true horizontal ground distances.
4
Identify the primary function of the trough compass
Trough compasses align the board edges of plane tables parallel to the magnetic north magnetic meridian.
Proper orientation ensures consistent directional accuracy across all station setups in plane table surveying.

Anahtar Kavram

Fieldwork functions of elementary surveying tools and accessories
Soru 127Soru

On a topographical map with a scale of 1:50,0001 : 50,000, a footpath connects a valley station at an elevation of 250 m250\text{ m} to a mountain peak at 650 m650\text{ m}. If the distance between the two points measured along the footpath on the map is 8 cm8\text{ cm}, what is the average gradient of the slope?

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Cevap: 1 in 101\text{ in }10

Cevap

The average gradient of the slope between the valley station and the mountain peak is 1 in 101\text{ in }10.
The correct answer is obtained by finding the difference in height between the two points (650 m250 m=400 m650\text{ m} - 250\text{ m} = 400\text{ m}) and dividing it by the ground horizontal distance (8 cm×500 m/cm=4,000 m8\text{ cm} \times 500\text{ m/cm} = 4,000\text{ m}). The resulting ratio 4004000\frac{400}{4000} simplifies to 1 in 101\text{ in }10.

Adım Adım Çözüm

1
Calculate the Vertical Interval (VI)
VI=650 m250 m=400 m\text{VI} = 650\text{ m} - 250\text{ m} = 400\text{ m}
Vertical interval is the difference in elevation between the highest and lowest points.
2
Calculate the Horizontal Equivalent (HE) using map scale
HE=8 cm×50,000=400,000 cm=4,000 m\text{HE} = 8\text{ cm} \times 50,000 = 400,000\text{ cm} = 4,000\text{ m}
Map scale of 1:50,0001 : 50,000 means 1 cm1\text{ cm} on map represents 50,000 cm50,000\text{ cm} (500 m500\text{ m}) on the ground.
3
Calculate the gradient ratio
Gradient=VIHE=400 m4,000 m=110\text{Gradient} = \frac{\text{VI}}{\text{HE}} = \frac{400\text{ m}}{4,000\text{ m}} = \frac{1}{10} or 1 in 101\text{ in }10
Gradient is expressed as the ratio of vertical rise to horizontal distance.

Anahtar Kavram

Topographic Gradient Calculation
Soru 128Soru

A drainage pattern where streams converge from surrounding elevated areas into a central inland depression or lake is best described as which of the following?

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Cevap: Centripetal drainage pattern

Cevap

Centripetal drainage pattern
Centripetal drainage pattern is defined by streams that flow inward from elevated margins toward a central depression, basin, or lake (such as Lake Chad).

Adım Adım Çözüm

1
Analyze the stream flow direction described in the stem.
The streams flow from surrounding high areas inward toward a central low point (depression or lake).
Identifying the flow direction relative to topographic relief determines the structural drainage pattern type.
2
Match the flow direction to standard geological drainage pattern definitions.
Inward converging flow characterizes the centripetal drainage pattern.
The term 'centripetal' denotes moving or tending toward a center point.

Anahtar Kavram

Classification of river drainage patterns based on structural control and terrain slope
Tahmini Süre:45s
Soru 129Soru

A geography student carrying out a chain survey along a designated baseline encounters a wide pond that obstructs direct distance measurement but allows clear line-of-sight vision across it. To determine the exact chainage distance across this obstacle without crossing the water, which of the following fieldwork procedures using elementary surveying instruments is correct?

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Cevap: Erect perpendicular offset lines at both edges of the obstacle using an optical square, chain a parallel line clear of the pond, and record the parallel segment length as equal to the obstructed distance.

Cevap

Erect perpendicular offset lines at both edges of the obstacle using an optical square, chain a parallel line clear of the pond, and record the parallel segment length as equal to the obstructed distance.
When chaining is obstructed by a body of water but vision remains clear, setting up right-angle offsets using an optical square allows the surveyor to construct a rectangular parallel line clear of the obstacle. Because opposite sides of a rectangle are equal, measuring the parallel line gives the exact distance across the pond.

Adım Adım Çözüm

1
Identify the type of obstacle in chain surveying.
The pond obstructs chaining (linear measurement) but does not obstruct vision (ranging poles are visible across it).
Choosing the correct geometric method depends on whether vision, chaining, or both are obstructed.
2
Apply geometric principles of rectangular offsets using appropriate surveying tools.
By setting off 9090^\circ angles at two points on the baseline using an optical square or cross-staff, equal perpendicular lines are measured to clear the pond.
An optical square ensures exact right angles (9090^\circ) to form a parallel line equal in length to the missing baseline segment.
3
Evaluate instrument functions to rule out incorrect procedures.
Clinometers, Abney levels, and prismatic compasses measure angles or bearings rather than linear baseline offsets across flat water obstacles.
Linear distance across obstacles in chain surveying requires geometric construction using offset instruments.

Anahtar Kavram

Obstacle Handling and Offset Procedures in Chain Surveying
Tahmini Süre:2m 0s
Soru 130Soru

A topographic map drawn at a scale of 1:25,0001 : 25,000 shows a hillside where point P on a ridge has an elevation of 580 m580\text{ m} and point Q near a river valley has an elevation of 430 m430\text{ m}. If the straight-line distance between P and Q measured on the map is 7.2 cm7.2\text{ cm}, what is the value of xx when the average gradient between the two points is expressed in the ratio form 1:x1 : x?

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Cevap: 12

Cevap

The numerical value of xx in the gradient ratio 1:x1 : x is 12.
To find the gradient ratio 1:x1 : x, first calculate the Vertical Interval (VI) by subtracting 430 m430\text{ m} from 580 m580\text{ m}, which gives 150 m150\text{ m}. Next, calculate the Horizontal Equivalent (HE) by multiplying the 7.2 cm7.2\text{ cm} map distance by the scale factor of 25,00025,000, giving 180,000 cm180,000\text{ cm} or 1,800 m1,800\text{ m}. Finally, divide VI by HE: 150 m1,800 m=112\frac{150\text{ m}}{1,800\text{ m}} = \frac{1}{12}. Expressed in the form 1:x1 : x, x=12x = 12.

Adım Adım Çözüm

1
Calculate the Vertical Interval (VI)
VI = 150 m
Subtract the lower elevation (430 m) from the higher elevation (580 m) to determine the vertical height difference.
2
Calculate the Horizontal Equivalent (HE) in meters
HE = 1,800 m
Multiply the map measurement of 7.2 cm by 25,000 to obtain ground distance in cm (180,000 cm), then divide by 100 to convert to meters.
3
Compute the gradient ratio and solve for x
Gradient = 1 / 12, giving x = 12
Divide the Vertical Interval by the Horizontal Equivalent: 150 m / 1,800 m = 1 / 12.

Anahtar Kavram

Slope and Gradient Calculation from Topographic Maps
Tahmini Süre:1m 30s
Soru 131Soru

In quantitative cartography, proportional circles are constructed such that the surface area of each circle is directly proportional to the statistical quantity being represented. On a demographic map of West Africa, a city with a population of 4,000,0004,000,000 is drawn using a proportional circle with a radius of 3.0 cm3.0\text{ cm}. Which of the following is the correct radius required to represent a neighboring metropolitan city with a population of 16,000,00016,000,000 on the same map?

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Cevap: 6.0 cm6.0\text{ cm}

Cevap

The correct radius required to represent the city is 6.0 cm6.0\text{ cm}.
The area of a proportional circle represents the statistical quantity (APA \propto P). Because circle area is calculated as πr2\pi r^2, the radius rr must vary with the square root of the population (rPr \propto \sqrt{P}). The population ratio between the two cities is 16,000,0004,000,000=4\frac{16,000,000}{4,000,000} = 4. Taking the square root gives 4=2\sqrt{4} = 2. Multiplying the initial radius of 3.0 cm3.0\text{ cm} by 22 yields the accurate radius of 6.0 cm6.0\text{ cm}.

Adım Adım Çözüm

1
Identify the relationship between statistical data and proportional circle dimensions.
The area of a circle A=πr2A = \pi r^2 is proportional to the statistical quantity PP. Therefore, rPr \propto \sqrt{P}.
Cartographic principles require symbol areas, not linear radii, to reflect statistical magnitudes.
2
Set up the ratio between the radii and the square root of populations for both cities.
r2r1=P2P1\frac{r_2}{r_1} = \sqrt{\frac{P_2}{P_1}}
Using ratios eliminates the proportionality constant kk and simplifies calculation.
3
Substitute given values into the ratio formula.
\frac{r_2}{3.0} = \sqrt{\frac{16,000,000}{4,000,000}} = \sqrt{4} = 2
Simplifying the population fraction yields a square root factor of 2.
4
Calculate the unknown radius r2r_2.
r_2 = 3.0 \times 2 = 6.0\text{ cm}
Multiplying the baseline radius by the scale factor yields the required circle radius.

Anahtar Kavram

Square root scaling rule for proportional circle radii in statistical maps
Soru 132Soru

In satellite remote sensing, sensors are categorized based on their energy source. Which type of sensor generates its own electromagnetic energy to illuminate a target and measure the backscattered signal?

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Cevap: Active sensor

Cevap

An active sensor is a remote sensing instrument that emits its own energy source to illuminate a surface and detect the reflected radiation.
The correct option is the active sensor because active remote sensing instruments transmit artificial pulses of electromagnetic radiation toward a target on Earth and record the echo or reflected signal.

Adım Adım Çözüm

1
Identify the energy mechanism in remote sensing systems.
Sensors are classified into two major operational modes: active systems (self-illuminating) and passive systems (solar radiation dependent).
Remote sensing classification depends on whether the system relies on internal or external energy emission.
2
Evaluate the definition of self-emitting systems.
Sensors such as Radar (Radio Detection and Ranging) and LiDAR (Light Detection and Ranging) send out pulses of electromagnetic energy and record the signal returned.
By generating their own energy, active sensors can operate day or night and penetrate cloud cover.

Anahtar Kavram

Active vs. Passive Remote Sensing Systems
Tahmini Süre:45s
Soru 133Soru

On a topographical map, a river flows along a valley floor where contour lines form distinct V-shapes. If the apexes of these V-shaped contours point toward the northeast as elevation increases from 200 m200\text{ m} to 500 m500\text{ m}, in which direction is the river flowing?

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Cevap: Southwest

Cevap

The river flows toward the southwest.
Contour lines crossing a river valley bend into a 'V' shape pointing upstream toward higher elevation. Because the V-shapes point northeast toward the 500 m500\text{ m} contour, the stream originates from the northeast and flows downhill toward the lower 200 m200\text{ m} elevation in the southwest.

Adım Adım Çözüm

1
Identify the rule governing V-shaped contour lines in river valleys.
V-shaped contour lines always point upstream toward higher elevation (the source of the stream).
Water flows downhill due to gravity, from higher elevation to lower elevation.
2
Determine the orientation of higher and lower elevations based on the given contour values.
Higher ground (500 m500\text{ m}) is in the northeast, and lower ground (200 m200\text{ m}) is in the southwest.
The prompt states that the apexes point northeast as elevation increases to 500 m500\text{ m}.
3
Deduce the flow direction of the river.
The river flows from northeast (500 m500\text{ m}) down toward southwest (200 m200\text{ m}).
River flow is always from higher contour elevation to lower contour elevation.

Anahtar Kavram

V-shaped contour interpretation and river flow direction
Tahmini Süre:1m 0s
Soru 134Soru

Match each Geographic Information System (GIS) analytical operation on the left with its defining function on the right.

Soldaki öğeye tıklayın, sonra eşleşen sağdaki öğeye tıklayın

Öğeler

Georeferencing
Buffering
Overlay Analysis
Spatial Query

Eşleşmeler

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Cevap

Georeferencing corresponds to assigning real-world spatial coordinates; Buffering corresponds to creating a zone or boundary perimeter of specified distance; Overlay Analysis corresponds to combining multiple thematic spatial layers; Spatial Query corresponds to filtering and selecting geographic features based on location.
Each GIS operation uniquely matches its defining procedure: Georeferencing grounds unreferenced spatial data using geographic coordinates; Buffering measures outward distance corridors around points, lines, or polygons; Overlay Analysis merges thematic map layers to examine spatial interactions; and Spatial Query retrieves features according to topological or geometric criteria.

Adım Adım Çözüm

1
Identify the primary function of Georeferencing
Georeferencing matches with assigning real-world spatial coordinates to an unreferenced map or imagery dataset.
Georeferencing ties digital pixel or vector data to actual geographic coordinate systems on Earth's surface.
2
Identify the primary function of Buffering
Buffering matches with creating a zone or boundary perimeter of specified distance around geographic features.
Buffering creates proximity zones (such as a 100-meter buffer around a river) for spatial distance analysis.
3
Identify the primary function of Overlay Analysis
Overlay Analysis matches with combining multiple thematic spatial layers to evaluate overlapping geographical relationships.
Overlay operations integrate disparate map layers (e.g., land use, slope, and soil type) into a unified analysis layer.
4
Identify the primary function of Spatial Query
Spatial Query matches with filtering and selecting geographic features based on location or spatial relationships.
Spatial querying allows users to select features meeting spatial conditions such as 'within 5 km of a highway'.

Anahtar Kavram

Core GIS Analytical Operations and Functions
Tahmini Süre:1m 0s
Soru 135Soru

Which elementary surveying instrument is specifically used by a surveyor to set out perpendicular offset lines at right angles (9090^\circ) from a baseline?

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Cevap: Cross-staff

Cevap

Cross-staff
The cross-staff is specifically constructed with sight lines fixed at right angles (9090^\circ), enabling a surveyor to quickly and accurately erect perpendicular offset lines from a main baseline.

Adım Adım Çözüm

1
Identify the specific function required in the question stem.
The requirement is an instrument used to establish perpendicular (9090^\circ) offset lines from a survey baseline.
Different surveying tools are designed for specific geometric tasks in the field.
2
Match the requirement with the correct elementary surveying tool.
The cross-staff contains fixed sighting slots set at 9090^\circ angles to easily lay out perpendicular lines.
Understanding fundamental instrument designs allows accurate identification of their field functions.

Anahtar Kavram

Functions of Elementary Surveying Instruments
Soru 136Soru

Arrange the following sequential stages of optical remote sensing data acquisition and image generation in their correct chronological order, from energy origin to final spatial raster output:

Öğeleri doğru sıraya koymak için sürükleyin

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Cevap

The correct sequence of stages in optical remote sensing data acquisition is: (1) Emission and propagation of electromagnetic radiation through the atmosphere, (2) Interaction and spectral reflection by surface features, (3) Detection and recording by satellite sensors, (4) Telemetric transmission of raw data to ground receiving stations, and (5) Pre-processing into calibrated raster imagery.
The optical remote sensing data workflow follows a physical sequence dictated by electromagnetic energy transfer. Energy emanates from a source and travels through the atmosphere to Earth's surface, interacts with targets, reflects back up to orbit where spaceborne sensors detect it, gets transmitted via radio telemetry to ground receiving stations, and is finally corrected into calibrated GIS raster datasets.

Adım Adım Çözüm

1
Identify the initial energy emission and atmospheric transmission
Solar radiation propagates through atmospheric gases toward Earth's surface.
Passive remote sensing relies on an external illumination source whose radiation must pass through the atmospheric window.
2
Determine surface target interaction
Radiation is selectively reflected based on the target's physical and chemical properties.
Spectral reflectance signatures are established when surface materials reflect unique proportions of specific wavelengths.
3
Identify signal capture at orbit
The satellite sensor array detects and quantizes the incoming reflected radiance into digital values.
Sensors on spaceborne platforms convert photons into electronic signals representing brightness levels.
4
Locate data transmission down to Earth facilities
Raw digital data streams are beamed via radio telemetry to ground receiving stations.
Spaceborne platforms must transmit captured observations to ground station receivers for storage and processing.
5
Trace ground calibration and raster rendering
Ground processing centers execute radiometric and geometric corrections to yield spatial raster maps.
Distortions caused by Earth rotation, topographic relief, and atmospheric interference must be corrected before GIS spatial analysis.

Anahtar Kavram

Remote Sensing Process and Data Acquisition Workflow
Soru 137Soru

A surveyor conducting a prismatic compass traverse along stations PP, QQ, RR, and SS records the following observed bearings:

- Line PQP-Q: Forward Bearing =14200= 142^\circ 00', Back Bearing =32200= 322^\circ 00'
- Line QRQ-R: Forward Bearing =07530= 075^\circ 30', Back Bearing =25800= 258^\circ 00'
- Line RSR-S: Forward Bearing =19000= 190^\circ 00'

If the local magnetic declination in the survey area is 530 W5^\circ 30'\text{ W}, what is the true geographic forward bearing of line RSR-S?

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Cevap: 18200182^\circ 00'

Cevap

The true geographic forward bearing of line R-S is 18200182^\circ 00'.
The correct answer is 18200182^\circ 00'. Evaluating line P-Q shows a difference of exactly 18000180^\circ 00' between its forward bearing (14200142^\circ 00') and back bearing (32200322^\circ 00'), confirming that stations P and Q are free from local attraction. Consequently, the forward bearing of line Q-R (07530075^\circ 30') measured at station Q is accurate, making its theoretical back bearing at station R equal to 07530+18000=25530075^\circ 30' + 180^\circ 00' = 255^\circ 30'. Comparing this with the observed back bearing at station R (25800258^\circ 00') identifies a local attraction error of +230+2^\circ 30' at station R. Applying this correction to the observed forward bearing of line R-S (19000190^\circ 00') gives a corrected magnetic bearing of 18730187^\circ 30'. Finally, subtracting the 530 W5^\circ 30'\text{ W} magnetic declination (True Bearing=Magnetic BearingWest Declination\text{True Bearing} = \text{Magnetic Bearing} - \text{West Declination}) yields 18200182^\circ 00'.

Adım Adım Çözüm

1
Identify stations free from local attraction
Difference for line P-Q =3220014200=18000= 322^\circ 00' - 142^\circ 00' = 180^\circ 00'. Both station P and station Q are free from local attraction.
When the difference between the forward bearing and back bearing of a line is exactly 180180^\circ, neither instrument station suffers from local attraction.
2
Determine local attraction at station R
Since station Q is correct, the true magnetic forward bearing of Q-R is 07530075^\circ 30'. The correct back bearing at station R should be 07530+18000=25530075^\circ 30' + 180^\circ 00' = 255^\circ 30'. The observed back bearing at station R is 25800258^\circ 00'. Local attraction at station R =2580025530=+230= 258^\circ 00' - 255^\circ 30' = +2^\circ 30'.
Any bearing taken FROM station R reads 2302^\circ 30' too high due to local magnetic interference.
3
Calculate corrected magnetic forward bearing of line R-S
Corrected Magnetic FB =19000230=18730= 190^\circ 00' - 2^\circ 30' = 187^\circ 30'.
Subtracting the positive local attraction error adjusts the reading to the correct magnetic orientation.
4
Convert magnetic bearing to true geographic bearing
True Geographic Bearing =Magnetic BearingWest Declination=18730530=18200= \text{Magnetic Bearing} - \text{West Declination} = 187^\circ 30' - 5^\circ 30' = 182^\circ 00'.
When magnetic declination is West, True Bearing equals Magnetic Bearing minus Declination (TB=MBW\text{TB} = \text{MB} - \text{W}).

Anahtar Kavram

Correction of Prismatic Compass Bearings for Local Attraction and Magnetic Declination
Tahmini Süre:2m 0s
Soru 138Soru

Match each slope measurement scenario on the left with its corresponding calculated gradient on the right.

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Öğeler

Vertical Interval =80 m= 80\text{ m}, Horizontal Equivalent =1.6 km= 1.6\text{ km}
Vertical Interval =150 m= 150\text{ m}, Horizontal Equivalent =600 m= 600\text{ m}
Vertical Interval =50 m= 50\text{ m}, Horizontal Equivalent =2.5 km= 2.5\text{ km}
Vertical Interval =200 m= 200\text{ m}, Horizontal Equivalent =1 km= 1\text{ km}

Eşleşmeler

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Cevap

Vertical Interval = 80 m, HE = 1.6 km matches 1 in 20 (5%); VI = 150 m, HE = 600 m matches 1 in 4 (25%); VI = 50 m, HE = 2.5 km matches 1 in 50 (2%); VI = 200 m, HE = 1 km matches 1 in 5 (20%).
Each scenario is correctly matched by converting all horizontal distances into meters and calculating the simplified ratio VI / HE, expressed as both a ratio (1 in N) and a percentage.

Adım Adım Çözüm

1
Convert all Horizontal Equivalent (HE) distances given in kilometers to meters so that units match the Vertical Interval (VI).
1.6 km=1,600 m1.6\text{ km} = 1,600\text{ m}, 2.5 km=2,500 m2.5\text{ km} = 2,500\text{ m}, and 1 km=1,000 m1\text{ km} = 1,000\text{ m}.
Both Vertical Interval and Horizontal Equivalent must be in the same units of measurement before calculating the ratio.
2
Apply the gradient formula Gradient=Vertical Interval (VI)Horizontal Equivalent (HE)\text{Gradient} = \frac{\text{Vertical Interval (VI)}}{\text{Horizontal Equivalent (HE)}} to each scenario.
Scenario 1: 801600=120\frac{80}{1600} = \frac{1}{20}; Scenario 2: 150600=14\frac{150}{600} = \frac{1}{4}; Scenario 3: 502500=150\frac{50}{2500} = \frac{1}{50}; Scenario 4: 2001000=15\frac{200}{1000} = \frac{1}{5}.
Dividing vertical rise by horizontal distance yields the simplified slope fraction.
3
Express each simplified fraction in standard ratio (1 in N1 \text{ in } N) and percentage formats.
120=1 in 20 (5%)\frac{1}{20} = 1\text{ in } 20\ (5\%), 14=1 in 4 (25%)\frac{1}{4} = 1\text{ in } 4\ (25\%), 150=1 in 50 (2%)\frac{1}{50} = 1\text{ in } 50\ (2\%), 15=1 in 5 (20%)\frac{1}{5} = 1\text{ in } 5\ (20\%).
This allows matching the calculated values directly to the corresponding choices on the right.

Anahtar Kavram

Slope and gradient calculation using Vertical Interval (VI) and Horizontal Equivalent (HE) with unit unification.
Soru 139Soru

A topographical map extract drawn to a scale of 1:100,0001 : 100,000 shows a footbridge over a stream at an elevation of 250 m250\text{ m} and a forest observation tower situated on a hilltop at an elevation contour of 450 m450\text{ m}. If the straight-line distance between the bridge and the tower measured on the map is 4 cm4\text{ cm}, what is the average gradient of the slope between these two locations?

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Cevap: 1 in 201\text{ in } 20

Cevap

The gradient between the two points is 1 in 201\text{ in } 20.
The gradient of a slope is computed as the Vertical Interval (VI) divided by the Horizontal Equivalent (HE). The vertical interval is 450 m250 m=200 m450\text{ m} - 250\text{ m} = 200\text{ m}. Using the map scale of 1:100,0001 : 100,000, a map distance of 4 cm4\text{ cm} equals 400,000 cm400,000\text{ cm} or 4,000 m4,000\text{ m} on the ground. Dividing 200 m200\text{ m} by 4,000 m4,000\text{ m} simplifies to 120\frac{1}{20}, expressed as 1 in 201\text{ in } 20.

Adım Adım Çözüm

1
Calculate the Vertical Interval (VI)
VI=450 m250 m=200 m\text{VI} = 450\text{ m} - 250\text{ m} = 200\text{ m}
Vertical interval is the difference in elevation between the two points.
2
Calculate the Horizontal Equivalent (HE) in meters
HE=4 cm×100,000=400,000 cm=4,000 m\text{HE} = 4\text{ cm} \times 100,000 = 400,000\text{ cm} = 4,000\text{ m}
Multiply map distance by the scale factor to get real-world distance, then convert centimeters to meters.
3
Calculate the slope gradient
Gradient=VIHE=200 m4,000 m=120\text{Gradient} = \frac{\text{VI}}{\text{HE}} = \frac{200\text{ m}}{4,000\text{ m}} = \frac{1}{20}
Gradient is expressed as the ratio of Vertical Interval to Horizontal Equivalent in identical measurement units.

Anahtar Kavram

Slope gradient determination using Vertical Interval (VI) and Horizontal Equivalent (HE)
Tahmini Süre:1m 30s
Soru 140Soru

A mining concession covers an area of 40 cm240\text{ cm}^2 on a topographical map drawn to a scale of 1:50,0001 : 50,000. If this map is reduced to a scale of 1:200,0001 : 200,000, what is the area of the concession on the reduced map in cm2\text{cm}^2?

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Cevap: 2.5

Cevap

The area of the mining concession on the reduced map is 2.5 cm22.5\text{ cm}^2.
When a map is reduced from a scale of 1:50,0001 : 50,000 to 1:200,0001 : 200,000, the linear scale is reduced by a factor of 44 (since 200,000/50,000=4200,000 / 50,000 = 4). Because area changes proportionally to the square of the linear change, the area scale factor is (1/4)2=1/16(1/4)^2 = 1/16. Multiplying the original area of 40 cm240\text{ cm}^2 by 1/161/16 gives 2.5 cm22.5\text{ cm}^2.

Adım Adım Çözüm

1
Calculate the linear reduction factor.
Linear scale factor k=50,000200,000=14=0.25k = \frac{50,000}{200,000} = \frac{1}{4} = 0.25
The linear change ratio is determined by comparing the old scale denominator to the new scale denominator.
2
Calculate the area scale factor.
Area factor k2=(0.25)2=116=0.0625k^2 = (0.25)^2 = \frac{1}{16} = 0.0625
Area scale varies with the square of the linear scale factor.
3
Determine the area on the reduced map.
New Map Area =40 cm2×0.0625=2.5 cm2= 40\text{ cm}^2 \times 0.0625 = 2.5\text{ cm}^2
Applying the area reduction factor to the original map area yields the new map area.

Anahtar Kavram

Relationship between linear scale factor and area scale factor during map reduction
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