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Question 190-1 : On a polar stereographic chart, the initial great circle course from a 70°n 060°w to b 70°n 060°e is approximately ? [ Level reports ]
030° t.
. 1386.conversion angle formula is not accurate for long distances..but we can try.conversion angle 1/2 g sin lm..g change of longitude 120°.lm = mean latitude 70°.conversion angle = 1/2 x 120 x sin70 = 56°..090° rhumb line 56° = 034°.
Question 190-2 : On a lambert conformal conic chart great circles that are not meridians are ?
Curves concave to the parallel of origin.
.the parallel of origin is approximately at half way between the two standard parallels. 1387. 1388.standard parallels are 20°n and 50°n, parallel of origin is 30°n..meridians, which are great circles, are straight lines..all other great circles are curved concave to the parallel of origin.. 2506
Question 190-3 : On a direct mercator projection, at latitude 45° north, a certain length represents 70 nm. at latitude 30° north, the same length represents approximately ?
86 nm.
.60 x cos 45 x x = 70..42.42 x x = 70.. x = 70 / 42.42 = 1.65..60 x cos 30 x 1.65 = 86 nm... ducksherminator.i'd just like to give the method i used to find the answer, i personnaly find it easier. 70/cos45° x cos30° = 85,73.
Question 190-4 : On a polar stereographic projection chart showing the south pole, a straight line joins position a 70°s 065°e to position b 70°s 025°w. the true course on departure from position a is approximately ?
225°.
.draw the situtation. 1389
Question 190-5 : 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 ?
1 3 500 000.
.scale = chart lenght/earth distance..earth distance = 5° x 60 nm x cos 60° = 150 nm..150 nm x 1.852 = 277,8 km..1 cm chart scale = 277.8 /8 = 34.725 km...1 cm 3 472 000 cm.
Question 190-6 : Two positions plotted on a polar stereographic chart, a 80°n 000° and b 70°n 102°w are joined by a straight line whose highest latitude is reached at 035°w..at point b, the true course is ?
203°.
.draw the situation. 1391.at the highest latitude 035°w , our true course is 270°..on a polar stereographic chart, convergency is equal to the change of longitude 102° 35° = 67°.270° 67° = 203°.
Question 190-7 : Assume a north polar stereographic chart whose grid is aligned with the greenwich meridian..an aircraft flies from the geographic north pole for a distance of 480 nm along the 110°e meridian, then follows a grid track of 154° for a distance of 300 nm. its position is now approximately. 2514 ?
80°00'n 080°e.
.the grid is aligned with the greenwich meridian...use meridian 090°e graduation to find distance 480 nm along the 110°e 1° = 60 nm. 480/60 = 8°...next step, your aircraft turn on a 154° heading grid track for a distance of 300 nm. 300/60 = 5°.. 1397.its position is now approximately 80°00'n 080°e.
Question 190-8 : The convergence factor of a lambert conformal conic chart is quoted as 0.78535. at what latitude on the chart is earth convergency correctly represented ?
51°45'.
.the convergency factor on a lambert chart is the sinus of the parallel of origin..n = sin l0..0.78535 = sin parallel of origin..parallel of origin = sin 1 0,78535 = 51,75 51°45'.
Question 190-9 : At 47° north the chart distance between meridians 10° apart is 12.7 cm. the scale of the chart at 47° north approximates ?
1 6 000 000
.scale = chart lenght/earth distance..earth distance = 10° x 60 nm x cos 47° = 409.2 nm..409.2 x 1.852 = 758 km..1 cm chart scale = 758 / 12.7 = 60 km..1 cm 6 000 000 cm... marcinkocybik.why we substract 758 / 12,7.. .we divide. because 12,7 cm on the chart corresponds to 758 km on earth, and we want to now the value for 1 cm.
Question 190-10 : On a direct mercator chart a great circle will be represented by a ?
Curve concave to the equator.
.on a direct mercator chart, meridians a parallel, equally spaced, vertical straight lines... /com en/com061 514.jpg.great circle is the shortest distance between two points on earth, but not on a direct mercator chart...for information. /com en/com061 394b.jpg..
Question 190-11 : 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 ?
23°18'.
Question 190-12 : On a lambert conformal chart the distance between meridians 5° apart along latitude 37° north is 9 cm. the scale of the chart at that parallel approximates ?
1 5 000 000
Scale = chart lenght/earth distance..earth distance = 5° x 60 nm x cos 37° = 239.6 nm..239.6 x 1.852 = 444 km..1 cm chart scale = 444 /9 = 49.33 km..1 cm 4 933 000 cm.
Question 190-13 : The great circle bearing from a 70°s 030°w to b 70°s 060°e is approximately ?
132° t.
. 1422..great circle direction at a = 090° rhumb line + conversion angle 1/2 g sin lm..conversion angle = 1/2 g sin lm.g change of longitude 90°.lm = mean latitude 70°..conversion angle = 1/2 x 90° x sin70°.conversion angle = 42°.great circle direction at a = 090° + 42° = 132°.
Question 190-14 : In a navigation chart a distance of 49 nm is equal to 7 cm. the scale of the chart is approximately ?
1 1 300 000
.49 nm x 1.852 = 91 km..91 / 7 = 13 km..1 cm 1 300 000 cm.
Question 190-15 : At 60°n the scale of a direct mercator chart is 1 3 000 000. what is the scale at the equator ?
1 6 000 000.
.scale at 60°n = scale at equator x 1/cos60° = 1/3 000 000....scale at equator = cos60°/3 000 000..scale at equator = 1/6 000 000.
Question 190-16 : What is the chart distance between longitudes 179°e and 175°w on a direct mercator chart with a scale of 1 5 000 000 at the equator ?
133 mm.
.scale = chart distance/earth distance..earth distance = 6º = 6° x 60 nm = 360 nm at the equator..360 nm = 667.3km = 667 300 000 mm..scale = chart distance / 667 300 000 = 1/5 000 000..chart distance = 667 300 000 / 5 000 000 = 133 mm... dalton.on a mercator chart, 179e and 175w are not distant by 6° but by 179+175=354°..... .earth is a globe, longitudes 179°e and 175°w are separated by 354° or 6°...
Question 190-17 : The total length of the 53°n parallel of latitude on a direct mercator chart is 133 cm. what is the approximate scale of the chart at latitude 30°s ?
1 26 000 000
.on a direct mercator chart, meridians are parallel, so all parallels of latitude will have the same lenght...scale = chart distance / earth distance..scale = 133 cm / 360 x 60 x cos30°..scale = 133 cm / 18706 nm = 133 cm / 34643.5 km.scale = 133 cm / 34 643 500 m.scale = 133 cm / 34 643 500 000 cm.1 cm / 26 047 744 cm.
Question 190-18 : A lambert conformal conic projection, with two standard parallels ?
The scale is only correct along the standard parallels.
.the lambert conformal is what most of today's aeronautical charts are based on.. 1776.on a lamberts chart, scale is only correct at the standard parallels where the cones slices through the surface of the globe and convergency is correct along the parallel of origin and the constant of the cone or convergence factor is the sine of the parallel of origin.
Question 190-19 : The constant of the cone, on a lambert chart where the convergence angle between longitudes 010°e and 030°w is 30°, is ?
0.75
Chart convergency = difference of longitude x constant of cone.difference of longitude = 10°e to 30°w = 40°..30° = 40° x constant of cone.constant of cone = 30/40 = 0.75.
Question 190-20 : A line drawn on a chart which joins all points where the value of magnetic variation is zero is called an ?
Agonic line.
. /com en/com061 87.jpg.agonic line a line which joins all points where the value of magnetic variation is zero.
Question 190-21 : The chart distance between meridians 10° apart at latitude 65° north is 9.5 cm. the chart scale at this latitude approximates ?
1 5 000 000
.scale = chart lenght/earth distance..earth distance = 10° x 60 nm x cos 65° = 254 nm..254 x 1.852 = 470 km..1 cm chart scale = 470 /9.5 = 49.47 km..1 cm 4 947 000 cm.
Question 190-22 : On a lambert conformal conic chart, with two standard parallels, the quoted scale is correct ?
Along the two standard parallels.
.the lambert conformal is what most of today's aeronautical charts are based on.. 1776.on a lamberts chart, scale is correct at the standard parallels where the cones slices through the surface of the globe and convergency is correct along the parallel of origin and the constant of the cone or convergence factor is the sine of the parallel of origin.
Question 190-23 : On a lambert conformal conic chart earth convergency is most accurately represented at the ?
Parallel of origin.
.the lambert conformal is what most of today's aeronautical charts are based on.. 1776.on a lamberts chart, scale is correct at the standard parallels where the cones slices through the surface of the globe and convergency is correct along the parallel of origin and the constant of the cone or convergence factor is the sine of the parallel of origin...convergency is the angle of inclination between two selected meridians measured at a given latitude.
Question 190-24 : A chart has the scale 1 1 000 000. from a to b on the chart measures 3.8 cm, the distance from a to b in nm is ?
20.5.
.1 cm = 1 000 000 = 10 km..3.8 x 10 km = 38 km..38 / 1.852 = 20.5 nm.
Question 190-25 : Contour lines on aeronautical maps and charts connect points ?
Having the same elevation above sea level
Question 190-26 : A rhumb line is ?
A line on the surface of the earth cutting all meridians at the same angle.
. /com en/com061 101.jpg..remember.rhumb lines or loxodromes are tracks of constant true course..great circle is the shortest distance between two points.
Question 190-27 : A straight line on a lambert conformal projection chart for normal flight planning purposes ?
Is approximately a great circle.
Meridians which are great circles are straight lines. all other great circles are almost straight lines but curved concave to the parallel of origin. rhumb lines loxodromes are curves concave to the pole.
Question 190-28 : An aircraft flies a great circle track from 56°n 070°w to 62°n 110°e. the total distance travelled is ?
3720 nm.
.070°w to 180°w/e = 110°.180°w/e to 110°e = 70°.difference of longitude = 180°..it seems that this great circle track will pass by the north pole..from position 56°n 070°w, we have 34° of latitude to reach the north pole. from north pole to position 62°n 110°e, we have 28° of latitude..34° + 28° = 62°.62° x 60 nm = 3720 nm.
Question 190-29 : Parallels of latitude on a direct mercator chart are ?
Parallel straight lines unequally spaced.
. 1753. direct mercator chart..parallels of latitude on a direct mercator chart are parallel straight lines unequally spaced. the distance between latitudes increases away from the centre..meridians are parallel, equally spaced, vertical straight lines.
Question 190-30 : The parallels on a lambert conformal conic chart are represented by ?
Arcs of concentric circles.
.the lambert conformal is what most of today's aeronautical charts are based on.. /com en/com061 699.jpg..
Question 190-31 : Approximately how many nautical miles correspond to 12 cm on a map with a scale of 1 2 000 000 ?
130.
12 cm x 2 000 000 cm = 24 000 000 cm..24 000 000 cm = 240 km..240 km / 1.852 = 130 nm.
Question 190-32 : 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 ?
034° w.
.convergency = 055° 043° = 12°..change in longitude = convergency / n..change in longitude = 12° / 0.75 = 16°..a is at 050°w and b 34°w 50° 16°.. minus 16° because we are heading east
Question 190-33 : A lambert conformal conic chart has a constant of the cone of 0.80..a straight line course drawn on this chart from a 53°n 004°w to b is 080° at a.course at b is 092° t..what is the longitude of b ?
Question 190-34 : What is the radial and dme distance from bel vor/dme n5439.7 w00613.8 to position n5410 w00710.. err a 061 241 ?
236° 44 nm
Question 190-35 : What is the radial and dme distance from bel vor/dme n5439.7 w00613.8 to position n5440 w00730.. err a 061 242 ?
278° 44 nm
.plot position n5440 w00730 and draw a line from belfast vor..center your protractor, you read 278°.. /com en/com061 242.jpg..use the latitude scale to find 44 nm.
Question 190-36 : What is the average track °m and distance between wtd ndb n5211.3 w00705.0 and ker ndb n5210.9 w00931.5 .. err a 061 244 ?
278° 90 nm.
Img /com en/com061 244.jpg..align your protractor with the average magnetic north between wtd and ker, magnetic track is 278°..use the latitude scale to find the distance, 90 nm.
Question 190-37 : What is the average track °m and distance between crn ndb n5318.1 w00856.5 and wtd ndb n5211.3 w00705.0 .. err a 061 246 ?
142° 95 nm.
.report the magnetic north tick from cml clonmel ndb, center your protractor, you read an average magnetic trak of 142°.. /com en/com061 246.jpg.use the scale to find the distance 95 nm.
Question 190-38 : What is the average track °m and distance between ker ndb n5210.9 w00931.5 and crn ndb n5318.1 w00856.5 .. err a 061 248 ?
025° 70 nm.
. /com en/com061 248.png..use the vertical scale to find the distance 70 nm. magnetic north is indicated over vors and ndbs. report a magnetic north tick, center your protractor, you read an average magnetic track of 025°.
Question 190-39 : 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 ?
122.3 nm.
.at latitude 15° 120 nm..at latitude 10° .. cos 10 / cos 15 x 120 = 122.3 nm...similar solution.lenght at the equator 120 nm / cos15 = 124.23nm...lenght at 10° of latitude 124.23 x cos10 = 122.34.
Question 190-40 : On a direct mercator chart at latitude 45°n, a certain chart length along 45°n represents a distance of 90 nm on the surface of the earth..the same length on a chart along latitude 30°n will represent a distance on the earth of ?
110 nm.
.60 nm x cos 45 x x = 90..42.42 x x = 90.. x = 90 / 42.42 = 2.12..60 x cos 30 x 2.12 = 110.15 nm.
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