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► The standard parallels of a Lambert conical orthomorphic projection are 07° 40 minutes N and 38° 20 minutes N. The constant of the cone for this chart is:
A) 0.60
B) 0.39
C) 0.92
D) 0.42
Constant of cone = Sin(latitude of parallel of origin)
Latitude of parallel of origin = (07°40'N + 38°20'N) : 2 = 23°N
Constant of cone = Sin(23) = 0.39
► The constant of cone of a Lambert conformal conic chart is quoted as 0.3955. At what latitude on the chart is earth convergency correctly represented?
A) 23° 18 minutes
B) 66° 42 minutes
C) 68° 25 minutes
D) 21° 35 minutes
On a Lambert conformal conic chart, the earth convergency is correct at the parallel of origin:
Constant of cone = Sin(latitude of parallel of origin)
0.3955 = Sin(latitude of parallel of origin)
Sin-1(0.3955) = 23° 18 minutes
► A Mercator chart has a scale at the equator = 1 : 3 704 000. What is the scale at latitude 60°S?
A) 1 : 1 852 000
B) 1 : 7 408 000
C) 1 : 3 208 000
D) 1 : 185 200
Scale at B = Scale at A x (Cos(latB) : Cos(latA) )
Scale at (60°S) = 3 704 000 x (Cos(60°S) : Cos(0°N/S) )
Scale at 60°S = 1 : 1 852 000
► On a Direct Mercator chart at latitude 15°S, a certain length represents a distance of 120 NM on the earth. The same length on the chart will represent on the earth, at latitude 10°N, a distance of:
A) 122.3 NM
B) 117.7 NM
C) 124.2 NM
D) 118.2 NM
Distance at B = Distance at A x (Cos(latB) : Cos(latA) )
Distance at (10°N) = 120 x (Cos(10°N) : Cos(15°S) )
Distance at 10°N = 122.3 NM
► On a Lambert conformal chart the distance between two parallels of latitude having a difference of latitude = 2° , is measured to be 112 millimetres. The distance between two meridians, spaced 2° longitude, is according to the chart 70 NM. What is the scale of the chart in the middle of the square described?
A) 1 : 756 000
B) 1 : 1 056 000
C) 1 : 1 233 000
D) 1 : 1 984 000
Scale = Chart distance (CD) / Earth distance (ED)
CD = 112mm
ED = 2° = 120 NM (2° x 60) = 222240000mm
222240000mm : 112mm = 1 984 285
► On a Lambert conformal chart the distance between two parallels of latitude having a difference of latitude = 2° , is measured to be 112 millimetres. The distance between two meridians, spaced 2° longitude, is according to the chart 70 NM. What is the latitude in the centre of the described square?
A) 49°
B) 38°
C) 54°
D) 42°
Departure = Change in longitude (minutes) x Cos(latitude)
70 NM = (2° x 60) x Cos(latitude)
Latitude = Cos-1(70NM / 120NM)
Latitude = 54.3°
► On a Lambert conformal chart the distance between two parallels of latitude (difference of latitude = 2°), is measured to be 112mm. The distance between two meridians, spaced 2° longitude, according to the chart is 70 NM. The parallel of origin (selected parallel) runs through the middle of the described square, what is the convergence for a d-long of 15° on this map?
A) 9,23°
B) 14,56°
C) 12,18°
D) 7,50°
First find the parallel of origin in the same way we did in the question above
(Departure = Ch Long x Cos(Lat)) = 54.3°
Constant of cone = Convergence / Ch long
Constant of cone = Sin(latitude of parallel of origin) = Sin(54.3°) = 0.812
Ch long = 15° (given in question)
Convergence = 0.812 x 15° = 12,18°
► On a polar stereographic chart the scale at the pole is 1 : 5 mill. Calculate the scale of the chart at 65N.
A) 1 : 4.213 mill
B) 1 : 5.250 mill
C) 1 : 5 mill
D) 1 : 4.766 mill
For a Polar chart the formula is a bit different:
Scale at latitude = Scale at the pole x (Cos(90° - Latitude / 2) )
Scale at (65°N) = 5 000 000 x (Cos(90° - 65° / 2))
Scale at (65°N) = 4 881 480 (closest answer being 4 766 000)
► A straight line drawn on a chart measures 4.63 cm and represents 150 NM. The chart scale is:
A) 1 : 3 000 000
B) 1 : 6 000 000
C) 1 : 5 000 000
D) 1 : 1 000 000
Scale = CD / ED
CD = 4.63 cm
ED = 150NM = 27780000 cm
27780000 cm : 4.63 cm = 6 000 000
► On a direct Mercator projection, the distance measured between two meridians spaced 5° apart at latitude 60°N is 8 cm. The scale of this chart at latitude 60°N is approximately:
A) 1 : 4 750 000
B) 1 : 3 500 000
C) 1 : 7 000 000
D) 1 : 6 000 000
Departure = Ch long (minutes) x Cos(lat)
Departure = 300 (5° x 60) x Cos(60°)
Departure = 150 NM = 27780000 cm (= ED)
CD = 8 cm
27780000 cm : 8 cm = 3 472 500 (closest answer being 3 500 000)
► At 60 degrees the scale of a Direct Mercator chart is 1 : 3 000 000 . What is the scale at the equator?
A) 1 : 1 500 000
B) 1 : 3 000 000
C) 1 : 3 500 000
D) 1 : 6 000 000
Scale at B = Scale at A x (Cos(latB) : Cos(latA) )
Scale at (0°N/S) = 3 000 000 x (Cos(0°N/S) : Cos(60°) )
Scale at 0°N/S = 6 000 000
► The total length of the 53 degrees N parallel of latitude on a direct Mercator chart is 133 cm. What is the approximate scale of the chart at latitude 30°S?
A) 1 : 26 000 000
B) 1 : 30 000 000
C) 1 : 18 000 000
D)1 : 21 000 000
Scale at B = Scale at A x (Cos(latB) : Cos(latA) )
First find Scale at A (53°):
CD = 133cm
ED = Departure = Ch long in minutes (360° x 60) x Cos(53°)
Departure (ED) = 12 999 NM (2407452673cm)
They told us "Total length of the 53° parallel" which is all around the earth or 360° change in longitude
Scale at A (53°) = 1 : 18 101 147
Scale at B = 18 101 147 x (Cos(30) : Cos(53) )
Scale at B = 26 000 000
► The chart distance between meridians 10° apart at latitudes 65° North is 3.75 inches. The chart scale at this latitude approximates:
A) 1 : 6 000 000
B) 1 : 5 000 000
C) 1 : 2 500 000
D) 1 : 3 000 000
Same "Scale = CD/ED" formula, but be careful that everything is in the same units!
CD = 3.75 inches
ED = departure = (10° x 60) x Cos(65°)
ED = 253 NM (or 18488680 inches)
18488680 inches : 3.75 inches = 4 930 314 (closest answer being 5 000 000)
► A chart has the scale 1 : 1 000 000. From A to B on the chart measures 1.5 inches (one inch equals 2.54 centimetres). The distance from A to B in NM is:
A) 44.5
B) 38.1
C) 20.5
D) 54.2
1 / 1 000 000 = 3.81cm (CD) / ED
or
1 x ED = 1 000 000 x 3.81cm
ED = 3 810 000 cm = 20.5 NM
► On a chart, the distance along a meridian between latitudes 45°N and 46°N is 6 cm. The scale of the chart is approximately:
A) 1 : 1 000 000
B) 1 : 1 850 000
C) 1 : 185 000
D) 1 : 18 500 000
Scale = CD/ED
CD = 6 cm
ED = Ch Latitude = 1° (60 NM) = 11 112 000 cm
Scale = 11 112 000 cm : 6 cm = 1 852 000
► Assume a Mercator chart. The distance between positions A and B, located on the same parallel and 10° longitude apart, is 6 cm. The scale at the parallel is 1 : 9 260 000. What is the latitude of A and B?
A) 45° N or S
B) 30° N or S
C) 0°
D) 60° N or S
1 / 9 260 000 = 6cm / ED
ED = 6 cm x 9 260 000
ED = 300 NM
Departure (ED) = Ch long (min) x Cos(lat)
300 NM = 600 (10° x 60) x Cos(lat)
Latitude = Cos-1(300 / 600)
Latitude = 60°
► A Lambert conformal conic chart has a constant of the cone of 0.75 . The initial course of a straight line track drawn on this chart from A (40°N 050°W) to B is 043°(T) at A; course at B is 055°(T). What is the longitude of B?
A) 41°W
B) 36°W
C) 38°W
D) 34°W
Constant of cone (0.75) = Convergence (055° - 043°) / Ch long
Ch long = 12° / 0.75
Ch long = 16°
A is at 50°W and your initial course is 043° which means you are flying in an easterly direction thus 50°W - 16° = 34°W
► The convergence factor of a Lambert conformal conic chart is quoted as 0.78353. At what latitude on the chart is Earth convergence correctly represented?
A) 38° 15min
B) 51° 45min
C) 52° 05min
D) 80° 39min
Constant of cone (0.78353) = Sin(latitude of parallel of origin)
Latitude of parallel of origin = Sin-1(0.78353)
Latitude of parallel of origin = 51° 45min
► What is the constant of the cone for a Lambert conic projection whose standard parallels are at 50° N and 70° N?
A) 0.500
B) 0.941
C) 0.866
D) 0.766
(50° + 70°) / 2 = 60°
Constant of cone = Sin(60°)
Constant of cone = 0.866
► Calculate the constant of cone on a Lambert chart given chart convergency between 010°E and 030°W as being 30°.
A) 0.40
B) 0.75
C) 0.50
D) 0.64
Constant of cone = Convergence / Ch long
Constant of cone = 30° / 40°
Constant of cone = 0.75
► A course of 120°(T) is drawn between X (61°30min N) and Y (58° 30min N) on a Lambert conformal conic chart with a scale of 1 : 1 000 000 at 60°N. The chart distance between X and Y is:
A) 33.4 cm
B) 66.7 cm
C) 38.5 cm
D) 36.0 cm
Both points are on different latitudes so we are going to use trigonometry for this one;
Construct the triangle and plot the values that we know;
- The angle at X = 180° - 120° = 60°
- The distance vertically between 61°30'N and 58°30'N = 3° x 60 = 180 NM
As we know the angle at X , the adjacent side (X-Z) we can find the hypotenuse (X-Y) with CAH:
COS(60°) = Adjacent (180) x Hypotenuse
or
180NM / COS(60°) = X-Y
Distance X-Y (earth distance) = 360 NM (666.7 km)
Now plot ED into the scale formula to find CD:
1 / 1 000 000 = CD / 66672000 cm CD = 66672000cm / 1 000 000 CD = 66.7cm
A) 0.60
B) 0.39
C) 0.92
D) 0.42
Constant of cone = Sin(latitude of parallel of origin)
Latitude of parallel of origin = (07°40'N + 38°20'N) : 2 = 23°N
Constant of cone = Sin(23) = 0.39
► The constant of cone of a Lambert conformal conic chart is quoted as 0.3955. At what latitude on the chart is earth convergency correctly represented?
A) 23° 18 minutes
B) 66° 42 minutes
C) 68° 25 minutes
D) 21° 35 minutes
On a Lambert conformal conic chart, the earth convergency is correct at the parallel of origin:
Constant of cone = Sin(latitude of parallel of origin)
0.3955 = Sin(latitude of parallel of origin)
Sin-1(0.3955) = 23° 18 minutes
► A Mercator chart has a scale at the equator = 1 : 3 704 000. What is the scale at latitude 60°S?
A) 1 : 1 852 000
B) 1 : 7 408 000
C) 1 : 3 208 000
D) 1 : 185 200
Scale at B = Scale at A x (Cos(latB) : Cos(latA) )
Scale at (60°S) = 3 704 000 x (Cos(60°S) : Cos(0°N/S) )
Scale at 60°S = 1 : 1 852 000
► On a Direct Mercator chart at latitude 15°S, a certain length represents a distance of 120 NM on the earth. The same length on the chart will represent on the earth, at latitude 10°N, a distance of:
A) 122.3 NM
B) 117.7 NM
C) 124.2 NM
D) 118.2 NM
Distance at B = Distance at A x (Cos(latB) : Cos(latA) )
Distance at (10°N) = 120 x (Cos(10°N) : Cos(15°S) )
Distance at 10°N = 122.3 NM
► On a Lambert conformal chart the distance between two parallels of latitude having a difference of latitude = 2° , is measured to be 112 millimetres. The distance between two meridians, spaced 2° longitude, is according to the chart 70 NM. What is the scale of the chart in the middle of the square described?
A) 1 : 756 000
B) 1 : 1 056 000
C) 1 : 1 233 000
D) 1 : 1 984 000
Scale = Chart distance (CD) / Earth distance (ED)
CD = 112mm
ED = 2° = 120 NM (2° x 60) = 222240000mm
222240000mm : 112mm = 1 984 285
► On a Lambert conformal chart the distance between two parallels of latitude having a difference of latitude = 2° , is measured to be 112 millimetres. The distance between two meridians, spaced 2° longitude, is according to the chart 70 NM. What is the latitude in the centre of the described square?
A) 49°
B) 38°
C) 54°
D) 42°
Departure = Change in longitude (minutes) x Cos(latitude)
70 NM = (2° x 60) x Cos(latitude)
Latitude = Cos-1(70NM / 120NM)
Latitude = 54.3°
► On a Lambert conformal chart the distance between two parallels of latitude (difference of latitude = 2°), is measured to be 112mm. The distance between two meridians, spaced 2° longitude, according to the chart is 70 NM. The parallel of origin (selected parallel) runs through the middle of the described square, what is the convergence for a d-long of 15° on this map?
A) 9,23°
B) 14,56°
C) 12,18°
D) 7,50°
First find the parallel of origin in the same way we did in the question above
(Departure = Ch Long x Cos(Lat)) = 54.3°
Constant of cone = Convergence / Ch long
Constant of cone = Sin(latitude of parallel of origin) = Sin(54.3°) = 0.812
Ch long = 15° (given in question)
Convergence = 0.812 x 15° = 12,18°
► On a polar stereographic chart the scale at the pole is 1 : 5 mill. Calculate the scale of the chart at 65N.
A) 1 : 4.213 mill
B) 1 : 5.250 mill
C) 1 : 5 mill
D) 1 : 4.766 mill
For a Polar chart the formula is a bit different:
Scale at latitude = Scale at the pole x (Cos(90° - Latitude / 2) )
Scale at (65°N) = 5 000 000 x (Cos(90° - 65° / 2))
Scale at (65°N) = 4 881 480 (closest answer being 4 766 000)
► A straight line drawn on a chart measures 4.63 cm and represents 150 NM. The chart scale is:
A) 1 : 3 000 000
B) 1 : 6 000 000
C) 1 : 5 000 000
D) 1 : 1 000 000
Scale = CD / ED
CD = 4.63 cm
ED = 150NM = 27780000 cm
27780000 cm : 4.63 cm = 6 000 000
► On a direct Mercator projection, the distance measured between two meridians spaced 5° apart at latitude 60°N is 8 cm. The scale of this chart at latitude 60°N is approximately:
A) 1 : 4 750 000
B) 1 : 3 500 000
C) 1 : 7 000 000
D) 1 : 6 000 000
Departure = Ch long (minutes) x Cos(lat)
Departure = 300 (5° x 60) x Cos(60°)
Departure = 150 NM = 27780000 cm (= ED)
CD = 8 cm
27780000 cm : 8 cm = 3 472 500 (closest answer being 3 500 000)
► At 60 degrees the scale of a Direct Mercator chart is 1 : 3 000 000 . What is the scale at the equator?
A) 1 : 1 500 000
B) 1 : 3 000 000
C) 1 : 3 500 000
D) 1 : 6 000 000
Scale at B = Scale at A x (Cos(latB) : Cos(latA) )
Scale at (0°N/S) = 3 000 000 x (Cos(0°N/S) : Cos(60°) )
Scale at 0°N/S = 6 000 000
► The total length of the 53 degrees N parallel of latitude on a direct Mercator chart is 133 cm. What is the approximate scale of the chart at latitude 30°S?
A) 1 : 26 000 000
B) 1 : 30 000 000
C) 1 : 18 000 000
D)1 : 21 000 000
Scale at B = Scale at A x (Cos(latB) : Cos(latA) )
First find Scale at A (53°):
CD = 133cm
ED = Departure = Ch long in minutes (360° x 60) x Cos(53°)
Departure (ED) = 12 999 NM (2407452673cm)
They told us "Total length of the 53° parallel" which is all around the earth or 360° change in longitude
Scale at A (53°) = 1 : 18 101 147
Scale at B = 18 101 147 x (Cos(30) : Cos(53) )
Scale at B = 26 000 000
► The chart distance between meridians 10° apart at latitudes 65° North is 3.75 inches. The chart scale at this latitude approximates:
A) 1 : 6 000 000
B) 1 : 5 000 000
C) 1 : 2 500 000
D) 1 : 3 000 000
Same "Scale = CD/ED" formula, but be careful that everything is in the same units!
CD = 3.75 inches
ED = departure = (10° x 60) x Cos(65°)
ED = 253 NM (or 18488680 inches)
18488680 inches : 3.75 inches = 4 930 314 (closest answer being 5 000 000)
► A chart has the scale 1 : 1 000 000. From A to B on the chart measures 1.5 inches (one inch equals 2.54 centimetres). The distance from A to B in NM is:
A) 44.5
B) 38.1
C) 20.5
D) 54.2
1 / 1 000 000 = 3.81cm (CD) / ED
or
1 x ED = 1 000 000 x 3.81cm
ED = 3 810 000 cm = 20.5 NM
► On a chart, the distance along a meridian between latitudes 45°N and 46°N is 6 cm. The scale of the chart is approximately:
A) 1 : 1 000 000
B) 1 : 1 850 000
C) 1 : 185 000
D) 1 : 18 500 000
Scale = CD/ED
CD = 6 cm
ED = Ch Latitude = 1° (60 NM) = 11 112 000 cm
Scale = 11 112 000 cm : 6 cm = 1 852 000
► Assume a Mercator chart. The distance between positions A and B, located on the same parallel and 10° longitude apart, is 6 cm. The scale at the parallel is 1 : 9 260 000. What is the latitude of A and B?
A) 45° N or S
B) 30° N or S
C) 0°
D) 60° N or S
1 / 9 260 000 = 6cm / ED
ED = 6 cm x 9 260 000
ED = 300 NM
Departure (ED) = Ch long (min) x Cos(lat)
300 NM = 600 (10° x 60) x Cos(lat)
Latitude = Cos-1(300 / 600)
Latitude = 60°
► A Lambert conformal conic chart has a constant of the cone of 0.75 . The initial course of a straight line track drawn on this chart from A (40°N 050°W) to B is 043°(T) at A; course at B is 055°(T). What is the longitude of B?
A) 41°W
B) 36°W
C) 38°W
D) 34°W
Constant of cone (0.75) = Convergence (055° - 043°) / Ch long
Ch long = 12° / 0.75
Ch long = 16°
A is at 50°W and your initial course is 043° which means you are flying in an easterly direction thus 50°W - 16° = 34°W
► The convergence factor of a Lambert conformal conic chart is quoted as 0.78353. At what latitude on the chart is Earth convergence correctly represented?
A) 38° 15min
B) 51° 45min
C) 52° 05min
D) 80° 39min
Constant of cone (0.78353) = Sin(latitude of parallel of origin)
Latitude of parallel of origin = Sin-1(0.78353)
Latitude of parallel of origin = 51° 45min
► What is the constant of the cone for a Lambert conic projection whose standard parallels are at 50° N and 70° N?
A) 0.500
B) 0.941
C) 0.866
D) 0.766
(50° + 70°) / 2 = 60°
Constant of cone = Sin(60°)
Constant of cone = 0.866
► Calculate the constant of cone on a Lambert chart given chart convergency between 010°E and 030°W as being 30°.
A) 0.40
B) 0.75
C) 0.50
D) 0.64
Constant of cone = Convergence / Ch long
Constant of cone = 30° / 40°
Constant of cone = 0.75
► A course of 120°(T) is drawn between X (61°30min N) and Y (58° 30min N) on a Lambert conformal conic chart with a scale of 1 : 1 000 000 at 60°N. The chart distance between X and Y is:
A) 33.4 cm
B) 66.7 cm
C) 38.5 cm
D) 36.0 cm
Both points are on different latitudes so we are going to use trigonometry for this one;
Construct the triangle and plot the values that we know;
- The angle at X = 180° - 120° = 60°
- The distance vertically between 61°30'N and 58°30'N = 3° x 60 = 180 NM
As we know the angle at X , the adjacent side (X-Z) we can find the hypotenuse (X-Y) with CAH:
COS(60°) = Adjacent (180) x Hypotenuse
or
180NM / COS(60°) = X-Y
Distance X-Y (earth distance) = 360 NM (666.7 km)
Now plot ED into the scale formula to find CD:
1 / 1 000 000 = CD / 66672000 cm CD = 66672000cm / 1 000 000 CD = 66.7cm
