Voyage Planning, Position Fixing, Gyro and Magnetic Compasses Flashcards
If the observer is at 5˚ South and the sun’s declination is 10˚ North, what will be the bearing of the body during meridian passage? …………..Sun 10° N ^ The Sun is North ……………. ^ of the Observer, ……………Obs 5° S ^ therefore, the ……………………………….. bearing of the …………………………….. …body is NORTH. NOTE: In answering this kind of question, you have to plot the Position of the “OBSERVER” and the Position of the “SUN”. …………If the “SUN” is North or above the “OBSERVER”, the bearing is NORTH or 000°T. ………..If the “SUN” is South or below the “OBSERVER”, the bearing is SOUTH or 180°T.
North
If the observer is at 5˚ South and the sun’s declination is 10˚ North, what will be the bearing of the body during meridian passage?
000 deg.T
If the observer is at 5˚ North and the sun’s declination is 10˚ South, what will be the bearing of the body during meridian passage?
South
If the observer is at 5˚ North and the sun’s declination is 10˚ South, what will be the bearing of the body during meridian passage?
180 deg.T
If the observer is at 5˚ North and the sun’s declination is 10˚ North, what will be the bearing of the body during meridian passage?
North
If the observer is at 10˚ North and the sun’s declination is 5˚ North, what will be the bearing of the body during meridian passage?
South
If the observer is at 10˚ North and the sun’s declination is 5˚ South, what will be the bearing of the body during meridian passage?
180 deg.T
If the observer is at 5˚ North and the sun’s declination is 10˚ North, what will be the bearing of the body during meridian passage?
000 deg.T
If the observer is at 10˚ North and the sun’s declination is 15˚ North, what will be the bearing of the body during meridian passage?
North
If the observer is at 10˚ North, and the sun is at summer solstice, what will be the bearing of the body during meridian passage?
North
If the observer is at 40˚North, and the sun is at summer solstice, what will be the bearing of the body during meridian passage?
South
If the observer is at 10˚ North, and the sun is at summer solstice, what will be the bearing of the body during meridian passage?
000 deg.T
If the observer is at 40˚ North, and the sun is at summer solstice, what will be the bearing of the body during meridian passage?
180 deg.T
If the observer is at 10˚ South, and the sun is at summer solstice, what will be the bearing of the body during meridian passage?
North
If the observer is at 10˚ South, and the sun is at summer solstice, what will be the bearing of the body during meridian passage?
000 deg.T
If the observer is at 10˚ South, and the sun is at winter solstice, what will be the bearing of the body during meridian passage?
South
If the observer is at 10˚ South, and the sun is at winter solstice, what will be the bearing of the body during meridian passage?
180 deg.T
If the observer is at 40˚ South, and the sun is at winter solstice, what will be the bearing of the body during meridian passage?
North
If the observer is at 40˚ South, and the sun is at winter solstice, what will be the bearing of the body during meridian passage?
000 deg.T
Find the latitude of the observer if the sun’s altitude at upper transit is 75˚ bearing North and the declination is 10 deg North.
5 deg. S
Find the latitude of the observer if the sun’s altitude at upper transit is 50˚ bearing South and the declination is 5˚ South.
35 deg. N
Find the latitude of the observer if the sun’s altitude at upper transit is 50˚ bearing South and the declination is 10˚ South.
30 deg. N
Find the latitude of the observer if the sun’s altitude at lower transit is 10˚ bearing South and the declination is 70˚ South.
30 deg. S
At meridian passage, upper transit, Ho of the sun is 39˚19.3’, bearing North. Declination of the sun is 23˚38.2’ South. Find latitude at transit(U.T.) …….. .. Ho = 39° 19.3’ N…………………….. …..(-) 90° ( always - 90 ) MZD = 50° 40.7’ S same sign +… . Dec = 23° 38.2’ S (+) diff sign - .Lat = 74° 18.9’ S ……………… ………………………………………………………………..NOTE: If the sign of Ho is “North”, then the sign of MZD will be “South” . If Ho is “South”, MZD will be “North”.(Always reverse the sign)
74˚18.9’ S
At meridian passage, upper transit, Ho of the sun is 48˚11.6’, bearing South. Declination of the sun is 1˚46.3’ North. Find latitude at transit.
43˚34.7’ N
At meridian passage, upper transit, Ho of the sun is 76˚46.5’, bearing North. Declination of the sun is 20˚10.5’ North. Find latitude at transit.
9˚57.0’ N
The Observed altitude (Ho) of Star Kochab at meridian passage, lower transit is 16˚11.5’. The star’s declination is 74˚16.2’ North. Find latitude (L.T.) …………………………………………………………… Dec = 74° 16.2’ N .. ( - ) 90° ………………….. ……… .PX = 15° 43.8’ ……….. . . … Ho = 16° 11.5’ ( always + )…. …. … Lat = 31° 55.3’ N (same name as Dec)
31˚55.3’ N
The Observed altitude (Ho) of Star Deneb at meridian passage, lower transit is 15˚23.5’. The star’s declination is 45˚16.6’ North. Find latitude.
60˚06.9’ N
On April 22, in longitude 098˚East, the observed altitude (Ho) of the Sun’s U.L. at meridian passage was 54˚42.2’ bearing North, declination is 11˚42.1’ N. Calculate the observer’s latitude.
23˚35.7’ S
On Jan. 16, in longitude 132˚54’W, the observed altitude (Ho) of the Sun’s L.L. at meridian passage was 64˚04.2’ bearing North, declination is 21˚58.5’S. Calculate the observer’s latitude.
47˚54.3’ S
The observed meridian altitude (Ho) fo the star Sirius (lower transit) was 14˚03.9’ bearing South. Find the observer’s latitude if the declination is 16˚43’ S.
87˚20.9’ S
At meridian passage, upper transit, the observer’s latitude was found to be 43˚34.7’ North, Declination is 1˚46.3’ North. Find the Observed Altitude (Ho).
48˚11.6’ S
At meridian passage, upper transit, the observer’s latitude was found to be 9˚57’ North, Declination is 20˚10.5’ North. Find the Observed Altitude (Ho).
79˚4.5’ N
The Meridian Altitudes of a celestial body are 12˚ and 68˚, North and South respectively from the navigator. Calculate the latitude.
62˚ N
The Meridian Altitudes of a celestial body are 12˚ and 68˚, North and South respectively from the navigator. Calculate declination of the celestial body.
40˚ N
The Meridian Altitudes of a celestial body are 12˚ and 68˚, North and South respectively from the navigator. Calculate the true altitude of the celestial body when it crosses the prime vertical.
46˚43’ N
What is the latitude of a place where the sun is exactly at the zenith of the observer at Local Apparent Noon (LAN) of June 21?
23˚27’ N
What is the latitude of a place where the sun is exactly at the zenith of the observer at Local Apparent Noon (LAN) of December 22?
23˚27’ S
What is the latitude of a place where the sun is exactly at the zenith of the observer at Local Apparent Noon (LAN) of March 21?
at the equator
What is the latitude of a place where the sun is exactly at the zenith of the observer at Local Apparent Noon (LAN) of September 22 or 23?
0 deg. Latitude
On June 10 in Longitude 058˚42’E, the sextant meridian altitude (Hs) of the sun’s U.L. was 87˚32.0’ bearing South, I.E. 1.3’ on the arc. DIP correction is 4.8’, Main corr. Is 15.9’. Calculate the latitude of the observer if the sun’s declination from the almanac indicates 23˚17.4’ North.
26˚07.4’ N
The observed meridian altitude (Ho) fo the star Sirius (lower transit) was 14˚03.9’ bearing South. Find the observer’s latitude if the declination is 16˚43’ South.
87˚20.9’ S
On Jan. 20, at GMT 22h 53m 40s, in longitude 041˚26’W, the observed altitude (Ho) of the star Polaris was 35˚54.9’ . The correction values determined by Polaris tables are Ao = 0˚ 53.2’, A1 = 0.4’ A2 = 0.7’. Calculate the observer’s latitude.
35˚49.2’ N
On Sept. 12, at GMT 10h35m 00s LZT in longitude 057˚ 58’ W, the Ho of star Polaris was 35˚50’. The correction values from the Polaris Tables: Ao = 1˚22.1’, A1 = 0.4’, A2 =0.9’ Find the latitude of the observer.
36˚13.4’ N
Find the latitude of the observer if the true altitude of Polaris is 34˚17.6’. The correction values from the Polaris Tables: Ao = 0˚58.5’, A1 = 0.6’, A2 = 0.9’
34˚17.6’ N
The mean radius of the Earth is 3440 nautical miles, find the radius of the parallel of Latitude of Manila (Lat. 14˚30’ N approximately)
3330.4 nm
The Earth is not a true sphere so that the equatorial radius is larger than the polar radius. The difference is _____.
13.5 miles
Given equatorial radius of the Earth as 3444 nm, find radius of a parallel of latitude at 70˚N or S
1177.9 nm
With an equatorial radius of the Earth of 3444nm find circumference of a parallel of latitude 30˚N
18740.2 nm
At what rate per hour is the Royal observatory at Greenwich (Latitude 51˚28.5’ N) being carried around the Earth’s axis?
560.6 knots
At what rate per hour is the Latitude 23˚20’ N being carried around the Earth’s axis?
826.4 knots
At what rate per hour is latitude 60˚00’ S being carried around the Earth’s axis?
450 nm/hr
At what rate per hour is latitude 45 deg. North being carried around the Earth’s axis?
636.4 nm/hr
At what rate per hour is latitude 36˚38’ South being carried around the Earth’s axis?
722.22 nm/hr
With a given mean radius of the Earth as 3440 nm, calculate the radius of a parallel of latitude at 46˚24’ North.
2372.3 nm
Assuming the equatorial radius of the Earth as 3444 nm, find te circumference of the equator?
21639.3 nm
The distance between two meridians is 427 nautical miles in Latitude 50˚20’ N. What is the angle at the pole?
11˚08.9’
The distance between two meridians is 600 nautical miles in Latitude 45˚20’ N. What is the angle at the pole?
22˚16.9’
The distance between two meridians is 248 nautical miles in Latitude 63˚20’ N. What is the angle at the pole?
9˚12.6’
The distance between two meridians is 127 nautical miles in Latitude 39˚30’ N. What is the angle at the pole?
2˚44.5’
The distance between two meridians is 724 nautical miles in Latitude 20˚50’ N. What is the angle at the pole?
12˚54.6’
The distance between two meridians is 600 nautical miles in Latitude 45˚20’ N. Find the Difference of Longitude (Dlo) between the 2 meridians.
22˚16.9’
In what latitude will a departure of 300 nm corresponds to a Dlo of 6˚40’?
41˚24.6’ N or S
In what latitude will a departure of 250 nm corresponds to a Dlo of 6˚40’?
51˚19.1’ Nor S
In what latitude will a departure of 200 nm corresponds to a Dlo of 4˚16’ ?
38˚37.5’ N or S
On a certain parallel, the distance between two meridians is 250 nm while the Dlo between the meridians is 12˚30’. What is the latitude?
70˚31.9’ N or S
On a certain parallel, the distance between two meridians is 340 nm while the Dlo between the meridians is 18˚30’. What is the latitude?
72˚09.8’ N or S
In Latitude 50˚10’ N, the departure between two meridians is 360 nautical miles. What is the Difference of Longitude?
9˚22’
In Latitude 45˚20’ N, the departure between two meridians is 430 nautical miles. What is the Difference of Longitude?
10˚11.7’
In Latitude 20˚40’ N, the departure between two meridians is 130 nautical miles. What is the Difference of Longitude?
2˚19’
In Latitude 38˚20’ N, the departure between two meridians is 530 nautical miles. What is the Difference of Longitude?
11˚15.6’
A ship steams on a course of 090˚T from port A in latitude 23˚30’ N; Long. 059˚10’E to Port B in Latitude 23˚30’ N; Longitude 065˚30’E. How far did she steam?
348.5 nm
A ship steams on a course of 270˚T from port A in latitude 23˚30’ N; Long. 059˚10’E to Port B in Latitude 23˚30’ N; Longitude 065˚30’E. How far did she steam?
348.5 nm
In Latitude 50˚20’N, a vesse steams from Long. 015˚46’W to Long. 031˚18’W. What distance was made good?
594.9 nm
On a certain parallel, the distance betweeen two meridians is 150 nm. On the equator the distance between the same two meridians is 235 nautical miles. What is the latitude of the parallel?
50˚20.1’ N or S
On a certain parallel, the distance betweeen two meridians is 400 nm. On the equator the distance between the same two meridians is 470 nautical miles. What is the latitude of the parallel?
31˚40.3. N or S
On a certain parallel, the distance betweeen two meridians is 250 nm. On the equator the distance between the same two meridians is 330 nautical miles. What is the latitude of the parallel?
40˚44.9’ N or S
On a certain parallel, the distance betweeen two meridians is 25 nm. On the equator the distance between the same two meridians is 30 nautical miles. What is the latitude of the parallel?
33˚33’ N or S
On a certain parallel, the distance betweeen two meridians is 50 nm. On the equator the distance between the same two meridians is 150 nautical miles. What is the latitude of the parallel?
70˚31.7’ N or S
On a certain parallel, the distance betweeen two meridians is 130 nm. On the equator the distance between the same two meridians is 330 nautical miles. What is the latitude of the parallel?
66˚48’ N or S
On a certain parallel, the distance betweeen two meridians is 15 nm. On the equator the distance between the same two meridians is 30 nautical miles. What is the latitude of the parallel?
60˚00’ N or S
On a certain parallel, the distance betweeen two meridians is 40.8 nm. On the equator the distance between the same two meridians is 81.6 nautical miles. What is the latitude of the parallel?
60˚00’ N or S
On a certain parallel, the distance betweeen two meridians is 20 nm. On the equator the distance between the same two meridians is 80 nautical miles. What is the latitude of the parallel?
75˚31.3’ N or S
On a certain parallel, the distance betweeen two meridians is 120 nm. On the equator the distance between the same two meridians is 280 nautical miles. What is the latitude of the parallel?
64˚37.4’
On a certain parallel, the distance betweeen two meridians is 2350 nm. On the equator the distance between the same two meridians is 4860 nautical miles. What is the latitude of the parallel?
61˚05’ N or S
On a certain parallel, the distance betweeen two meridians is 2123 nm. On the equator the distance between the same two meridians is 8130 nautical miles. What is the latitude of the parallel?
74˚51.8’ N or S
On a certain parallel, the distance betweeen two meridians is 235 nm. On the equator the distance between the same two meridians is 350 nautical miles. What is the latitude of the parallel?
47˚49.3’ N or s
On a certain parallel, the distance betweeen two meridians is 320 nm. On the equator the distance between the same two meridians is 680 nautical miles. What is the latitude of the parallel?
61˚55.6’ N or S
On a certain parallel, the distance betweeen two meridians is 240 nm. On the equator the distance between the same two meridians is 670 nautical miles. What is the latitude of the parallel?
69˚00’ N or S
A vessel steers a course of 146˚T from Lat. 35˚10’N to Lat. 8˚46’ N. How far did she steam?
1910.7 nm
A vessel sails from Lat.21˚45’N; Long.023˚56’ W on course 146˚ T for a distance of 300 nm. Find the Latitude and Longitude of Arrival by mercator Sailing.
Lat. 17˚36.3’ N; Long. 026˚53.1’W
Find the DLAT and Depature made good if a vessel steams for 1936 nautical miles on course 248˚ T
Dlat 725.2’ S; Dep. 1795.1’
Find the DLAT and Depature made good if a vessel steams for 435 nautical miles on course 026˚ T
Dlat 391.0’ N; Dep. 190.7’
Find the DLAT and Depature made good if a vessel steams for 341 nautical miles on course 215˚ T
Dlat 279.3’ S; Dep 195.6’
A vessel makes a Dlat of 289.4’ N and a departure of 203.2 nautical miles. Find the course and distance.
Course 324˚55.5’; Dist. 353.6’
In latitude 50˚20’ N, a vessel steams 270˚T from Long. 015˚ 46’ W to Long. 031˚18’ W. Find the distance made good.
594.9 nm
A vessel steams 470 nautical miles along the parallel of “x” North from Long.015˚35’ W to Long. 027˚20’ W. What is the latitude of “x” ?
Lat. 48˚ 11.3’ North
From Latitude 39˚00’N; 033˚10’W, a ship steams on course 270˚T at 10 knots for 3 days and 8 hours. Find arrival position.
Lat. 39˚00’ N; Long. 050˚ 19.4’ W
Two ports A and B are in the Northern Hemisphere. On the parallel of Port A, the distance between their meridians is 250 nm. On the parallel of Port B it is 350 nm and on the equator it is 400 nm. What are the Latitudes of the ports?
Lat 51˚19.1’ N for Port A Lat 28˚57.3’ N for Port B
A ship steams 090˚T for 200 nautical miles in Lat. 49˚10’ N. By how much will her clocks have to be advanced?
20m 23s
Your vessel receivesa distress call from a vessel reporting her position as LAT 35˚01’ S; LONG. 018˚51’ W. Your position is LAT 35˚01’ S; LONG.021˚42’ W. Determine what will be your True Course and distance to the vessel in distress by parallel sailing method.
090˚T, 140 nm
A vessel in Latitude 55˚12’ N sails on course 270˚T and made a Dlo of 21˚36.6’. If the time taken was 3 days 2 hours, find the vessel’s speed.
10 knots
On a certain parallel, a vessel must steam one nautical mile to alter her longitude by 2 minutes. What is the latitude of the parallel?
Lat. 60˚ N or S
A ship “A” is on the equator steering 090˚T at 16 knots; while a ship “B” is on a parallel of North latitude, steering 270˚T at 12 knots. When “A” makes a Dlo of 1’, “B” makes a Dlo of 48’. Calculate the latitude of “B”.
Lat 20˚ 22’
A ship in Latitude 59˚40’ sailed on a certain course until the D.M.P. was twice the DLat. Calculate the Latitude reached.
Lat. 60˚ 20’ N or S
By sailing due East for a distance of 245 nautical miles, a ship alters her longitude by 7˚25’. Find Latitude of the ship.
Lat. 56˚ 36’ N or S
Find the distance between initial position at Lat1. 50˚0’ N; 178˚0’ W and Lat2. 50˚00’ N; Long. 179˚ 00’E.
116 nm
Your ship departs Yokohama, Japan from position Lat. 35˚27’ N; Long. 139˚39’ E bound for San Francisco, California,USA. At position Lat. 37˚48.5’ N, Long. 122˚24’ W. Determine the course and distance by Mercator sailing only.
088.3˚T; dist. 4738.4 miles
Your ship departs Yokohama, Japan from position Lat. 35˚27’ N; Long. 139˚39’ E bound for San Francisco, California,USA. At position Lat. 37˚48.5’ N, Long. 122˚24’ W. Determine the distance by Great Circle sailing.
4473 miles
Your ship departs Yokohama, Japan from position Lat. 35˚27’ N; Long. 139˚39’ E bound for San Francisco, California,USA. At position Lat. 37˚48.5’ N, Long. 122˚24’ W. Determine the initial course by Great Circle sailing.
054.3˚T
Your ship departs Yokohama, Japan from position Lat. 35˚27’ N; Long. 139˚39’ E bound for San Francisco, California,USA. At position Lat. 37˚48.5’ N, Long. 122˚24’ W. Determine the Latitude of the Vertex (Lv) by Great Circle sailing.
Lv = 48˚35.9’ N
Your ship departs Yokohama, Japan from position Lat. 35˚27’ N; Long. 139˚39’ E bound for San Francisco, California,USA. At position Lat. 37˚48.5’ N, Long. 122˚24’ W. Determine the Longitude of the Vertex (Lv) by Great Circle sailing.
Longv = 169˚14’ W
Determine the great circle distance and intial course from Lat. 27˚51’ N; Long. 071˚41’ W to Lat. 49˚ 45’ N; Long. 006˚14’ W ?
3214 nm, 046.9˚T
Determine the great circle distance and intial course from Lat. 36˚00’ S; Long. 056˚00’ W to Lat. 34˚ 00’ N; Long. 018˚15’ E ?
3557 nm; 112˚T
Determine the great circle distance and intial course from Lat. 24˚52’ N; Long. 078˚27’ W to Lat. 47˚ 19’ N; Long. 006˚ 42’ W ?
3593 nm; 048.1˚T
The great circle distance from Lat. 35˚57.2’ N, Long. 005˚45.7’ W to Lat. 24˚25.3’ N; Long 083˚02.6’ W is 3966.5 nautical miles and the initial course is 283.7˚T. The latitude of the vertex is 38˚09.4’ N. What is the longitude of the vertex?
028˚ 18.5’ W
The great circle distance from Lat. 08˚50’ N, Long. 080˚21’ W to Lat. 12˚36’ N; Long 128˚16’ E is 8664 nautical miles and the initial course is 306.6˚T. The latitude of the vertex is 37˚39.6’ N. What is the longitude of the vertex?
157˚44’ W
You are on a great circle track departing from LAT. 25˚50’ N; LONG. 077˚00’ W and your initial course is 061.7˚T. The position of the vertex is LAT 37˚35.6’ N; LONG. 025˚57.8’ W. What is the distance along the great circle track between the point of dep
2664.9 nm
The great circle distance from Lat. 35˚08’ S, Long. 019˚26’ E to Lat. 33˚16’ S; Long 115˚36’ E is 4559 nautical miles and the initial course is 121˚T. Determine the latitude of the vertex?
45˚30’ S
The latitude and longitude of the vertex along your great circle course is 38˚15’S, 168˚19’W. Your vessel is on course 102˚T. Which of the following is your course when crossing the equator?
051.8˚T
The latitude and longitude of the vertex along your great circle course is 38˚15’S, 168˚19’E. Your vessel is on course 060˚T. Which of the following is your longitude when crossing the equator?
101˚21’ W
The latitude and longitude of the vertex along your great circle course is 43˚32’S, 039˚18’ E. Your vessel is on course 246˚T. Which of the following is your course when crossing the equator?
313.5˚ T
A great circle crosses the equator at 134˚ E. It will also cross the equator at what other longitude?
046˚ W
A great circle crosses the equator at 114˚ E. It will also cross the equator at what other longitude?
066˚ W
A great circle crosses the equator at 114˚ W. It will also cross the equator at what other longitude?
066˚ E
A great circle crosses the equator at 086˚ E. It will also cross the equator at what other longitude?
094˚ W
A great circle crosses the equator at 157˚ W. It will also cross the equator at what other longitude?
023˚ E
A great circle crosses the equator at 049˚ W. It will also cross the equator at what other longitude?
131˚ E
A great circle crosses the equator at 102˚ E. It will also cross the equator at what other longitude?
078˚ W
A great circle crosses the equator at 012˚ 35’ E. It will also cross the equator at what other longitude?
167˚ 25’ W
A great circle crosses the equator at 032˚ 35’ W. It will also cross the equator at what other longitude?
147˚25’ E
A great circle crosses the equator at 078˚ 45’ E. It will also cross the equator at what other longitude?
101˚15’ W
A great circle crosses the equator at 120˚ E. It will also cross the equator at what other longitude?
060˚ W
A great circle crosses the equator at 120˚ W. It will also cross the equator at what other longitude?
060˚ E
The longitude of the upper vertex of a great circle track is 169˚ E. What is the longitude of the lower vertex?
011˚ W
The longitude of the upper vertex of a great circle track is 011˚ W. What is the longitude of the lower vertex?
169˚ E
The longitude of the upper vertex of a great circle track is 134˚ E. What is the longitude of the lower vertex?
046˚ W
The latitude of the upper vertex of a great circle is 36˚ N. What is the latitude of the lower vertex?
36˚ S
The latitude of the upper vertex of a great circle is 63˚ N. What is the latitude of the lower vertex?
63˚S
The latitude of the upper vertex of a great circle is 27˚ N. What is the latitude of the lower vertex?
27˚S
The latitude of the upper vertex of a great circle is 54˚ S. What is the latitude of the lower vertex?
54˚N
The latitude of the upper vertex of a great circle is 46˚ S. What is the latitude of the lower vertex?
46˚N
The latitude of the upper vertex of a great circle is 30˚ S. What is the latitude of the lower vertex?
30˚N
What is the difference of longitude between the intersection of the great circle and the equator to the lower vertex?
90 deg.
What is the difference of longitude between the intersection of the great circle and the equator to the upper vertex?
90 deg.
The vertex of a great circle track in Long. 109˚ E. An eastbound vessel will cross the equator in what longitude?
161˚ W
The vertex of a great circle track is Long. 121˚ E. An eastbound vessel will cross the equator in what longitude?
149˚ W
The vertex of a great circle track is Long. 134˚ E. An eastbound vessel will cross the equator in what longitude?
136˚W
The vertex of a great circle track is Long. 029˚ E. An eastbound vessel will cross the equator in what longitude?
119˚ E
The vertex of a great circle track is Long. 079˚ E. An eastbound vessel will cross the equator in what longitude?
169˚ E
The vertex of a great circle track is Long. 029˚ W. An eastbound vessel will cross the equator in what longitude?
061˚ E
The vertex of a great circle track is Long. 029˚ W. A westbound vessel will cross the equator in what longitude?
119˚ W
The vertex of a great circle track is Long. 134˚ E. A westbound vessel will cross the equator in what longitude?
044˚E
The vertex of a great circle track is Long. 109˚ E. A westbound vessel will cross the equator in what longitude?
019˚ E
The vertex of a great circle track is Long. 121˚ E. A westbound vessel will cross the equator in what longitude?
031˚ E
The vertex of great circle track is Long. 121˚ W. An eastbound vessel will cross the equator in what longitude?
031˚ W
The upper vertex of great circle track is at Long. 158˚ W. An eastbound vessel will cross the equator in what longitude?
068˚ W
The upper vertex of great circle track is Long. 144˚ W. A westbound vessel will cross the equator in what longitude?
126˚ E
The vertex of great circle track is Long. 168˚ W. An eastbound vessel will cross the equator in what longitude?
078˚ W
The difference of longitude between the upper vertex and the lower vertex of a great circle is _____.
180 deg.
A ship following a great circle track crosses the equator at long. 135˚ 00’ E on course 051.8˚T. If she continues along this great circle track, find the latitude and longitude of the upper vertex.
Lat. 38˚12’ N; Long. 135˚ 00’ W
Determine the latitude and longitude of the vertex along your great circle track when going from Lat. 35˚ 17.6’ N; Long.144˚ 23’ E to Lat.47˚ 36’ N; Long. 124˚ 22’ W.
Lat. 52˚47.8’ N; Long. 158˚ 07.3’ W
Determine the great circle distance and initial course from Lat 08˚ 36’ N, Long. 126˚ 17’ E to Lat. 02˚12’ S, Long. 081˚ 53’ W.
9076 miles, 079˚T
The great circle distance from Lat. 38˚17’ N; Long. 123˚ 16’ W to Lat. 35˚ 01’ N; Long. 142˚ 21’ E is 4330 nautical miles and the initial course is 300.9˚ T Determine the latitude and longitude of the vertex.
Lat. 47˚39.5’ N; Long. 167˚ 18.5’ W
You receive a distress call from a vessel reporting her position as Lat. 30˚21’ N; Long. 088˚34’ W. Your position is at Lat. 24˚30’ N; Long. 083˚00’ W. Determine the true course and distance to the distress scene by Mercator Sailing.
320˚ T; dist. 460 nm