The earth can be regarded as a spherical object, and you will most likely be found on the surface of this sphere while using another system of coordinates, that covers our planet with imaginary lines called meridians and parallels. All these lines together provide the grid which would enable you to describe any position in Latitudes and Longitudes. It takes the earth 24 hours for a full rotation of 360°, every hour we rotate 15° of longitude.
When it is 12:00 Greenwich Mean Time (GMT) anywhere in the world, it is 12:00 Local Time in Greenwich and 24:00 Local Time at the other side of the planet: 180° E or 180° W of the date line. Crossing this meridian changes not only the hour but also the date. The North Pole has a latitude of 90° N and the South Pole 90° S. The meridians cover twice this angle up to 180° W or E. Meridians converge at the poles, where as parallels run parallel to each other and never meet. All meridians and the equator the biggest parallel form great circles, and the remaining parallels form so called small circles. A great circle divides the earth in two exact halves. On small scaled charts you want to be accurate within one minute or one nautical mile. On larger scaled charts the accuracy is more likely to be within a tenth of a mile (a cable).
If the earth were a perfect sphere with a circumference of roughly 40000 kilometres all great circles, meridians plus the equator, would have the same length and could be used as a distance unit when divided into 360 degrees, or 360° x 60' = 21600' minutes. In 1929, the international community agreed on the definition of 1 international nautical mile as 1852 metres, which is roughly the average length of one minute of latitude one minute of arc along a line of longitude (a meridian). You should now able to describe any position in latitudes and longitudes. Now you can state the distance between two of those positions using nautical miles or minutes. All you need now is a way to define speed. For that, you use the term knots, the number of nautical miles an hour.
Parallels: Circles parallel to the equator, ranging from 0° to 90° N or S. Only the equator is a great circle.
Meridians: Half-circles converging at the poles, ranging from 0° to 180° E or W. Each pair of opposing meridians forms a great circle.
Prime meridian: 0° or the Greenwich meridian which, together with the date line meridian, divides the Western and Eastern hemispheres.
Great circle: The intersection of a sphere and a plane that passes through the sphere's center.
Small circle: The intersection of a sphere and a plane that doesn't pass though the sphere's center.
Time zones: By convention 24 zones, each 15° longitude wide. Noon at Greenwich gives you midnight at 180° E.
GMT: Greenwich Mean Time, UTC or Zulu, which is the local time at Greenwich. For example, local time in Seattle, Wa. = GMT + 8.
Date line: The 180° meridian which extends from or is opposite to the prime meridian. Here, not only the hour changes when crossing the meridian, but also the date.
Latitude: Position defined by the number of degrees north or south of the equator, varies from 0° to 90°.
Longitude: Position defined by the number of degrees east or west of the prime meridian, varies from 0° to 180°.
Position: Latitude first and longitude second. For example: 42° 21'.5 N , 71° 03'.6 E.
Nautical mile: One nm is one minute (') on the vertical scale on the chart. 1' equals 1852 metres. Nautical miles are divided into 10 cables. A cable is equal to one tenth of a nautical mile.
Knots: Nautical miles per hour.
Wednesday, February 11, 2009
Tuesday, February 10, 2009
The Nautical Almanac, Pub. 229 and Longitude
The Nautical Almanac is published annually by the Naval Observatory in the United States and Her Majesty's Stationery Office in the United Kingdom. It provides the navigator with the declination and Greenwich hour angle of all the celestial bodies normally used in navigation for an entire year. In order to find these coordinates, the exact Greenwich mean time (GMT) of the observation must be known. The Almanac contains the information needed for solution of navigation problems in addition to the Greenwich hour angles and declinations of celestial bodies.
The 1981 edition is used by the Coast Guard in all license exam questions requiring information from the Almanac. Each pair of pages covers a three day period.
The left hand daily pages provide:
1. The GHA of the hour circle of the first point of Aries for each hour of GMT.
2. The GHA's and declinations of the planets Venus, Mars, Jupiter, and Saturn for each hour of GMT.
3. The sidereal hour angles (SHA) and declinations of the 57 navigational stars. The SHA's and declinations change only slightly from one three day period to the next, they are only listed once for each three-day period rather than for each hour of GMT.
4. A tabulation of the local mean time of meridian passage and sidereal hour angle (SHA) of the planets is in the lower right corner of the page. This information is used in star identification.
The right-hand daily pages provide:
1. The GHA and declination of the sun for each hour of GMT.
2. The GHA and declination of the moon for each hour of GMT.
3. Local mean times of sunrise, sunset, twilight, moonrise, and moonset.
4. The mean times of the sun's meridian passage, moon's meridian passage, and the equation of time are in the lower right corner.
In order to obtain accuracy the exact value of the GMT of the observation to the nearest second must be used. The daily pages provide coordinates for whole hours, and the increments and corrections pages provide corrections for minutes and seconds of GMT, which must be added to the GHA and declination values given for a body in the "daily pages".
In the increments and corrections section, the table is entered with minutes and seconds of GMT. Included in the Almanac is a table for each minute from zero to 60, and each minute's table has a line for zero through 60 seconds. Reading each line from left to right are the GHA corrections for that number of minutes and seconds for the sun and planets, Aries, and the moon. The portion of the table to the right labeled "v" or "d" Corr" (correction) provides corrections for declination and for "excess of motion". Standard sight forms cannot be taken in the testing room for use in license exams. Using standard forms in learning how to solve celestial problems is recommended.
Finding your Location
Step 1: You can express your position by Latitude and Longitude.
Step 2: Chartmakers or geographers use an imaginary grid to locate a place on a map. This grid is made up of intersecting lines of latitude and longitude.
Step 3: Lines of latitude run east and west around the globe. The latitude is measured in degrees north and south from the Equator. Lines of longitude that run north and south from North pole to the South pole are measured in degrees east and west from the Prime Meridian.
Step 4: Prime Meridian runs through Greenwich, England. It is the planet’s Home Base.
Step 5: You can find the longitude of your location with an accurate sense of time. Before you can do any longitude calculations, you must convert your local zone time, as shown on your watch, to GMT (Greenwich Mean Time, the clock time at Greenwich).
Step 6: The local zone time is the local clock time which is the mean solar time of central meridian of your local time zone.
In several places in the world, hundreds of different times were adopted, each one corresponding to its own meridian. To simplify this situation, the Earth surface was divided into 24 time zones, each one delimited by two meridian forming a hour angle of 1 hour at the poles. The mean solar time of the central meridian of each time zone was assigned by convention to all places belonging to the time zone.
Three Basic Ideas to find Longitude:
1. The first of these ideas is the relationship between time and the rotation of the Earth. It takes an average time of 24 hours for the Earth to rotate 360 degrees. If you divide the number degrees in a circle by the number of hours in a day, you will find that the Earth turns 15 degrees every hour. 360° / 24 hours = 15° per hour. You can take this a step further and say that the Earth turns one degree in four minutes. 1 hour = 60 minutes / 15° = 4 minutes per degree.
2. The second idea is that you have to be careful about the difference between the events and time. Events like sunrise in the east always happen before the same event in the west. But time as shown on eastern clocks is later than on western clocks at the same instant. Local time earlier, position is westward. Local time later, position is eastward.
3. The third and last idea needed for longitude is the applying of Equation of Time.
Clock time and Sun time are different by as much as 16.5 minutes. The important thing is that if you're going to compare Sun time to the chronometer's clock time, you have to change the chronometer's clock time to Sun time so that you're comparing like terms. And that's what the Equation of Time does. You can find the Equation of Time from the Nautical Almanac. By applying the Equation of Time to the chronometer's clock time, you convert Greenwich Mean Time (GMT:Clock time.) to Greenwich Apparent Time (GAT:Sun time). GAT is simply the Sun time back at Greenwich, England.
Now you can observe Local Apparent Noon and do your simple subtraction of GAT to find your longitude. It is noon at the very instant that the sun were right over your head. Local Apparent Noon is simply noon for your exact location, and sets your watch to 12:00 based on Sun Time. The time of Local Apparent Noon, recorded as 12:00 local time, is compared to the time back in Greenwich.
To solve the navigational triangle to find computed altitude (Hc) and azimuth (Zn) for a celestial sight. By comparing Hc with observed altitude (Ho), the navigator finds the altitude intercept (a). With the assumed position coordinates, azimuth, and intercept, a line of position can be established. There are two basic methods for solving a navigational triangle or other spherical triangle by sight reduction table or by mathematical solution.
A sight reduction table presents solutions of the navigational triangle for computation of the altitude and azimuth of a celestial body. The variable components are co-latitude, polar distance, and meridian angle "t". Their values depend on time of the observation and the observer's position. The parts of the triangle are computed co-altitude (in order to find Hc) and azimuth angle (in order to determine Zn). Mathematical solutions use the same formulas used to pre-compute the values found in the sight reduction tables. Final results, found through the correct use of either method, seldom differ by more than one-tenth minute of arc for intercept values, or one tenth degree for azimuth values.
Sight Reduction Table - Pub. 229
The Sight Reductton Table for Marine Navtgatton (Pub 229) is a form of sight reduction table which is produced by the Defense Mapping Agency. They are designed to help you in solution of navigational triangles to find computed altitude (Hc) and azimuth (Zn). There are six volumes of Publication 229, each volume covering 15 degrees of latitude. All of the problems used in USCG license exams, which may be solved with Publication 229, are in latitudes 15 degrees to 30 degrees covered by Volume II of the tables.
The entering arguments for 229 are assumed latitude, the whole degree LHA which results when assumed longitude is applied to GHA, and declination. The main part of the tables is divided in half. The first half of Volume II covers latitudes 15 degrees through 22 degrees, and the second half covers latitudes 23 through 30 degrees. The latitude which is used to enter the table may be either north or south. Each half covers LHA's from 0 degrees through 360 degrees.
There are two pages for each LHA value from 0 through 90 degrees, and 270 through 360 degrees. Left-hand pages are for situations where declination and latitude are both north or both south, they are said to have the "same name." Each "same name" page covers declinations from 0 through 90 degrees.
The adjacent right-hand pages cover "contrary name" situations (meaning the declination is north and latitude south or vice versa) for the LHA values at the top of the page, and "same name" situations when the LHA values are between 90 degrees and 270 degrees located at the bottom of the page. The dividing line between "contrary" and "same name" situations is the step-like line across the right-hand page. The step-like line across the right-hand page represents the horizon. Entry values of LHA, latitude, and declination (same or contrary) should never result in crossing this line, as this would mean the body is below the horizon.
The inside front and back covers of each 229 are an interpolation table to be used for correcting the the altitude and azimuth angle obtained in the main part of the table for the declination increment. Declination increment is the minutes and tenths of minutes part of the declination of the body. For example, if the body's declination is N 43°23.7', the declination increment is 23.7'. In other words, declination is tabulated on each page in whole degrees, and you must interpolate between those whole degrees for your actual declination value. Declination increments 00.0 through 31.9 minutes are located inside the front cover, and declination increments 28.0 through 59.9 minutes are inside the back cover.
The 1981 edition is used by the Coast Guard in all license exam questions requiring information from the Almanac. Each pair of pages covers a three day period.
The left hand daily pages provide:
1. The GHA of the hour circle of the first point of Aries for each hour of GMT.
2. The GHA's and declinations of the planets Venus, Mars, Jupiter, and Saturn for each hour of GMT.
3. The sidereal hour angles (SHA) and declinations of the 57 navigational stars. The SHA's and declinations change only slightly from one three day period to the next, they are only listed once for each three-day period rather than for each hour of GMT.
4. A tabulation of the local mean time of meridian passage and sidereal hour angle (SHA) of the planets is in the lower right corner of the page. This information is used in star identification.
The right-hand daily pages provide:
1. The GHA and declination of the sun for each hour of GMT.
2. The GHA and declination of the moon for each hour of GMT.
3. Local mean times of sunrise, sunset, twilight, moonrise, and moonset.
4. The mean times of the sun's meridian passage, moon's meridian passage, and the equation of time are in the lower right corner.
In order to obtain accuracy the exact value of the GMT of the observation to the nearest second must be used. The daily pages provide coordinates for whole hours, and the increments and corrections pages provide corrections for minutes and seconds of GMT, which must be added to the GHA and declination values given for a body in the "daily pages".
In the increments and corrections section, the table is entered with minutes and seconds of GMT. Included in the Almanac is a table for each minute from zero to 60, and each minute's table has a line for zero through 60 seconds. Reading each line from left to right are the GHA corrections for that number of minutes and seconds for the sun and planets, Aries, and the moon. The portion of the table to the right labeled "v" or "d" Corr" (correction) provides corrections for declination and for "excess of motion". Standard sight forms cannot be taken in the testing room for use in license exams. Using standard forms in learning how to solve celestial problems is recommended.
Finding your Location
Step 1: You can express your position by Latitude and Longitude.
Step 2: Chartmakers or geographers use an imaginary grid to locate a place on a map. This grid is made up of intersecting lines of latitude and longitude.
Step 3: Lines of latitude run east and west around the globe. The latitude is measured in degrees north and south from the Equator. Lines of longitude that run north and south from North pole to the South pole are measured in degrees east and west from the Prime Meridian.
Step 4: Prime Meridian runs through Greenwich, England. It is the planet’s Home Base.
Step 5: You can find the longitude of your location with an accurate sense of time. Before you can do any longitude calculations, you must convert your local zone time, as shown on your watch, to GMT (Greenwich Mean Time, the clock time at Greenwich).
Step 6: The local zone time is the local clock time which is the mean solar time of central meridian of your local time zone.
In several places in the world, hundreds of different times were adopted, each one corresponding to its own meridian. To simplify this situation, the Earth surface was divided into 24 time zones, each one delimited by two meridian forming a hour angle of 1 hour at the poles. The mean solar time of the central meridian of each time zone was assigned by convention to all places belonging to the time zone.
Three Basic Ideas to find Longitude:
1. The first of these ideas is the relationship between time and the rotation of the Earth. It takes an average time of 24 hours for the Earth to rotate 360 degrees. If you divide the number degrees in a circle by the number of hours in a day, you will find that the Earth turns 15 degrees every hour. 360° / 24 hours = 15° per hour. You can take this a step further and say that the Earth turns one degree in four minutes. 1 hour = 60 minutes / 15° = 4 minutes per degree.
2. The second idea is that you have to be careful about the difference between the events and time. Events like sunrise in the east always happen before the same event in the west. But time as shown on eastern clocks is later than on western clocks at the same instant. Local time earlier, position is westward. Local time later, position is eastward.
3. The third and last idea needed for longitude is the applying of Equation of Time.
Clock time and Sun time are different by as much as 16.5 minutes. The important thing is that if you're going to compare Sun time to the chronometer's clock time, you have to change the chronometer's clock time to Sun time so that you're comparing like terms. And that's what the Equation of Time does. You can find the Equation of Time from the Nautical Almanac. By applying the Equation of Time to the chronometer's clock time, you convert Greenwich Mean Time (GMT:Clock time.) to Greenwich Apparent Time (GAT:Sun time). GAT is simply the Sun time back at Greenwich, England.
Now you can observe Local Apparent Noon and do your simple subtraction of GAT to find your longitude. It is noon at the very instant that the sun were right over your head. Local Apparent Noon is simply noon for your exact location, and sets your watch to 12:00 based on Sun Time. The time of Local Apparent Noon, recorded as 12:00 local time, is compared to the time back in Greenwich.
To solve the navigational triangle to find computed altitude (Hc) and azimuth (Zn) for a celestial sight. By comparing Hc with observed altitude (Ho), the navigator finds the altitude intercept (a). With the assumed position coordinates, azimuth, and intercept, a line of position can be established. There are two basic methods for solving a navigational triangle or other spherical triangle by sight reduction table or by mathematical solution.
A sight reduction table presents solutions of the navigational triangle for computation of the altitude and azimuth of a celestial body. The variable components are co-latitude, polar distance, and meridian angle "t". Their values depend on time of the observation and the observer's position. The parts of the triangle are computed co-altitude (in order to find Hc) and azimuth angle (in order to determine Zn). Mathematical solutions use the same formulas used to pre-compute the values found in the sight reduction tables. Final results, found through the correct use of either method, seldom differ by more than one-tenth minute of arc for intercept values, or one tenth degree for azimuth values.
Sight Reduction Table - Pub. 229
The Sight Reductton Table for Marine Navtgatton (Pub 229) is a form of sight reduction table which is produced by the Defense Mapping Agency. They are designed to help you in solution of navigational triangles to find computed altitude (Hc) and azimuth (Zn). There are six volumes of Publication 229, each volume covering 15 degrees of latitude. All of the problems used in USCG license exams, which may be solved with Publication 229, are in latitudes 15 degrees to 30 degrees covered by Volume II of the tables.
The entering arguments for 229 are assumed latitude, the whole degree LHA which results when assumed longitude is applied to GHA, and declination. The main part of the tables is divided in half. The first half of Volume II covers latitudes 15 degrees through 22 degrees, and the second half covers latitudes 23 through 30 degrees. The latitude which is used to enter the table may be either north or south. Each half covers LHA's from 0 degrees through 360 degrees.
There are two pages for each LHA value from 0 through 90 degrees, and 270 through 360 degrees. Left-hand pages are for situations where declination and latitude are both north or both south, they are said to have the "same name." Each "same name" page covers declinations from 0 through 90 degrees.
The adjacent right-hand pages cover "contrary name" situations (meaning the declination is north and latitude south or vice versa) for the LHA values at the top of the page, and "same name" situations when the LHA values are between 90 degrees and 270 degrees located at the bottom of the page. The dividing line between "contrary" and "same name" situations is the step-like line across the right-hand page. The step-like line across the right-hand page represents the horizon. Entry values of LHA, latitude, and declination (same or contrary) should never result in crossing this line, as this would mean the body is below the horizon.
The inside front and back covers of each 229 are an interpolation table to be used for correcting the the altitude and azimuth angle obtained in the main part of the table for the declination increment. Declination increment is the minutes and tenths of minutes part of the declination of the body. For example, if the body's declination is N 43°23.7', the declination increment is 23.7'. In other words, declination is tabulated on each page in whole degrees, and you must interpolate between those whole degrees for your actual declination value. Declination increments 00.0 through 31.9 minutes are located inside the front cover, and declination increments 28.0 through 59.9 minutes are inside the back cover.
Monday, February 9, 2009
Using Gnomonic Charts for Voyage Planning
There are three types of gnomonic charts depending on the location of the point of tangency. In equatorial gnomomic charts, the point where the plane is tangent is located at some point on the equator. A polar gnomonic chart is tangent at one of the poles. Oblique gnomonic charts have their point of tangency at some latitude and longitude between the equator and the poles.
The land features of gnomonic charts become more distorted as the distance from the point of tangency increases. The most often used gnomonic charts are oblique charts with a point of tangency located in the center of the ocean basin. Gnomonic tracking charts WOXZC 5270 and WOXZC 5274, have their points of tangency in the North Pacific and North Atlantic Ocean basins.
The advantage of gnomonic charts is that a straight line between two points represents the great circle between the points. If it is desired to proceed by great circle, all you need do is connect the points of departure and arrival with a straight line. The latitudes and longitudes of points at intervals along this line can be measured on the gnomonic chart and then transferred to a Mercator chart. When these points are connected with straight lines (rhumb lines), the navigator has the legs of the great circle track. Once transferred to the Mercator chart, the course on each leg may be determined, and the sum of the distances along each leg gives the total great circle distance. The main disadvantages of the gnomomic projection are that courses, direction and distance can not be readily measured. For these reasons, gnomonic projections are not used for day to day navigation and plotting, and should be regarded as an aid in voyage planning.
Using Gnomonic Charts For Track Planning
The use of gnomonic charts for voyage planning provides for quick checking of planned great circle tracks for intervening land, iceberg hazards, and other factors. For example, if you wanted to determine the advisability of proceeding by great circle from Cape Flattery, at the entrance to Puget Sound, to the entrance to Tokyo Bay, Japan, you would draw a straight line between the two places on chart WOXZC 5270. The chart would show that such a track would require your vessel to pass through the Aleutian Chain twice, and part of the voyage would be in the Bering Sea. Knowledge of the average poor weather, violent storms, uncharted rocks, and fog in this far northern waters makes the saving in distance of a great circle over a rhumb line less attractive. The great circle chart can be a big help to plan composite sailing tracks. A composite sailing consists of proceeding from the point of departure until the great circle reaches a limiting latitude. You then proceed by rhumb line (parallel sailing) along that latitude to the point where a great circle from the point of arrival is tangent to the limiting latitude line.
Great Circle Tracking Chart Questions On The License Exam
The Coast Guard uses questions which refer to great circle tracking charts. An applicant testing for Mates, or Master unlimited could easily have one of these questions on the exam. I have six example problems in another blog dated on 7 February 2009, titled "Great Circle Tracking Charts" with a full explanation. These examples should give a idea of how to reach the answers to problems of this type. If you can get copies of charts WOXZC 5270 and WOXZC 5274 so that you can follow the explanations with each example problem.
Saturday, February 7, 2009
Great Circle Tracking Charts
The Coast Guard uses questions which refer to great circle tracking charts. If you are testing for a mates license you could easily have one of these questions on your exam. Below are six example problems with a explanation. These examples should give you a sense of how to reach the answers to problems of this type. If you can get charts WOXZC 5270 and WOXZC 5274 you can follow the explanations with each of the 6 example problems below.
Example Problem 1:
A great circle track would be most advantageous when compared to the rhumb line track on which of the following routes? (Use gnomonic tracking chart WOXZC 5274 - North Atlantic)
A. Cayenne (Lat. 4°40.0' N, Long. 52°15.0' W) to Sao Tome (Lat. 0°00.0', Long. 3°15.0' W)
B. Palm Beach, FL to the English Channel
C. Natal, Brazil to Reykjavick, Iceland
D. Recife, Brazil to Monrovia, Republic of Liberia
Step 1: Using a pencil and straightedge lightly draw in a line between the points listed in each of the above choices.
Step 2: Eliminate choice A. Cayenne and Sao Tome are both so near the equator (which is a great circle) that a rhumb line track between them would not be appreciably longer than a great circle.
Step 3: Choice B is the correct choice. It represents the situation where a great circle is advantageous, a voyage between points in the middle latitudes with a significant difference of longitude between the points.
Step 4: Eliminate choice C. Natal, Brazil is almost due south of Iceland, the rhumb line between the places does not diverge significantly from being a meridian which is a great circle.
Step 5: Choice D may be eliminated because the two points are in the vicinity of the equator rather than the middle latitudes. In addition, the distance between the points is relatively short which decreases the advantage of great circle over rhumb line tracks.
Example Problem 2:
You are planning a voyage from Cape May (Lat. 38°45.0' N, Long. 74°45.0' W) to Lisbon (Lat. 38°37.0' N, Long. 9°45.0' W). Which of the following is true? (Use gnomonic chart WOXZC 5274)
A. Because the latitudes are almost the same, a great circle track approximates parallel sailing.
B.The northern hemisphere vertex is approximately at longitude 42°10.0 W
C.The distance is measured by using the length of one degree of the meridian at the position of the vertex.
D. A composite sailing must be plotted to remain south of a limiting latitude of 44° N
Step 1: Use a straight edge to draw a line between the two points on the chart.
Step 2: Eliminate choices A, C, and D for the following reasons.
Choice A - The great circle track diverges significantly, from parallel sailing. This can be seen if a line is drawn along latitude 38°40 to compare with the great circle. A rhumb line track like a parallel sailing would be much longer.
Choice C - The distance is not measured by the method stated. Measurement of distance on a Gnomonic chart can not be done by direct methods.
Choice D - The latitude of the vertex of this great circle is 43.8° N. This is less than the limiting latitude of 44° N. So composite sailing is not required.
Step 3: Choice B is correct. The great circle track reaches its maximum north latitude 43.8 degrees north (vertex) at approximately 42.1° W longitude.
Example Problem 3:
On a voyage via the southern tip of Nova Scotia (Lat. 43°20.0' N, Long. 65°35.0' W) you wish to sail the shortest route to La Corunna, Spain (Lat. 43°20.0' N, Long. 8°24.0' W). Which of the following will require you to plot a composite sailing? (Use gnomomic sailing chart WOXZC 5274)
A. Shoals extending 15 miles from Sable Island
B. Sea ice reported 68 miles ESE of St. Johns, Newfoundland
C. Icebergs reported extending to west to west-northwest from Lat. 47°00.0' N, Long. 35°00.0' W
D. Naval exercises using live ammunition being conducted within a 150 mile radius of Lat. 49°00.0' N, Long. 20°00.0' W.
Step 1: Use a straightedge to draw a great circle track between the two points.
Step 2: Choice A may be eliminated because your great circle track will pass more than one degree of latitude (60 miles) north of Sable Island avoiding any shoals extending 15 miles from the island.
Step 3: Choice B is eliminated because this great circle track passes well away from the sea ice reported 68 miles ESE (112.5° T) from St. John.
Step 4: The icebergs reported extending west to west-northwest (270° to 292.5° T) from Lat. 47°00.0' N, Long. 35°00.0' W lie astride the track. A composite sailing can be used to avoid this hazard. This is the correct answer.
Step 5: Choice D can be eliminated as your track takes you south of the naval exercises extending a 150 mile radius around Lat. 49°00.0' N, Long. 20°00.0'W.
Example Problem 4:
You are planning a voyage from departure Seattle (Lat. 48°30.0' N, Long. 125°00.0' W) to a position at Lat. 44°00.0' N, Long. 161°00.'0 E. Which of the following statements is true? (Use gnomonic chart WOXZC 5270)
A. You must plot a composite sailing to remain south of the Aleutian Islands.
B. The northern hemisphere vertex lies to the west of your arrival position.
C. Military exercises north of 53° N between longitudes 150° W to 165° W will not effect your voyage.
D. At your highest latitude, the sun will be visible at upper and lower / transit if the voyage occurs near 21 June.
Step 1: Use a straightedge to draw a line bewteen the points of departure and arrival which were given.
Step 2: Eliminate choice A. The great circle track passes south of the southern most of the Aleutian Islands.
Step 3: Eliminate choice B. The northern hemishere vertex is at about Lat. 52°45.0' N, Long. 155°00.0' W between the points of departure and arrival.
Step 4: Choice C is the correct choice. The statement that the voyage will not be effected by military exercises is true. The great circle track passes just south of the exercice area above 53° N between 150° W tp 165° W.
Step 5: Eliminate choice D. In order to see the sun at upper and lower transit about June 21, the track would have to go above latitude 66°30.0' N (the Arctic Circle).
Example Problem 5:
You are planning a voyage by great circle from Lat. 59°00.0' N, Long. 07°00.0' W via Lat. 38°00.0' N, Long. 61°30.0' W. Which of the following statements is true? (Use gnomonic tracking chart WOXZC 5274)
A. You are east of the northern hemisphere vertex.
B. When plotted on a Mercator chart the track line will be concave to Cape Farwell.
C. All courses are in the southwest quadrant of the compass.
D. Distance is measured by using the length of a degree of latitude at the midpoint of the trackline.
Step 1: Use a straight edge to draw a line between the coordinates of the points given.
Step 2: Eliminate choice A. The northern hemisphere vertex is east of the point of departure. It is located approximately 59°15.0' N, 01°00.0' E.
Step 3: Eliminate choice B. Great circles between points which lie in the northern hemisphere are concave to the equator. Therefore, this trackline is convex, not concave, to Cape Farwell (south tip of Greenland) which lies to the north of the track.
Step 4: Choice C is the true statement. Inspection of the angle at which the trackline intersects each meridian between departure and arrival indicates all the courses to be steered will fall in the southwest quadrant (between 180° T and 270° T).
Step 5: Eliminate choice D. Distance may not be measured directly on gnomonic charts. The method given for measurements is true for Mercator chart projections not gnomonic projections.
Example Problem 6:
In planning a North Pacific voyage, you wish to steam the minimum distance from Lat. 48°30.0' N, Long. 124°45.0' W, to Lat. 44°00.0' N, Long. 150°00.0' E, while remaining south of 51° N latitude. Which of the following tracks meets these requirements? (Use gnomonic tracking chart WOXZC 5270)
A. A Mercator sailing from departure to the mid-longitude at 51° N, which is great circle to arrival.
B. A great circle between departure and arrival with parallel sailing between the longitudes where the great circle intersects 51°N.
C. A great circle, tangent to 51°N, from departure to the mid-longitude, then a great circle to arrival.
D. A great circle from departure to Lat. 51° N, Long. 148°, parallel sailing to Lat. 51° N Long. 171° W, then a great circle to arrival.
The shortest distance between two positions when there is a limiting latitude is called a "composite sailing". Choice D exactly describes the method of plotting a composite sailing track and it is the correct choice.
Choice A has the ship traveling by rhumb line for 42.6 degrees of longitude between departure and the mid-longitude. This is half the trip by a less economical route.
Choice B is even less efficient than A. On the direct great circle route the ship would reach the limiting latitude of 51° N at 133° W. It would then travel by rhumb line until the direct great circle returns to 51° N at 168° E. This would mean traveling for 59 degrees of the total difference of longitude, which exceeds 85 degrees, by rhumb line.
Choice C is a fairly efficient route but its combination of two great circles lies south of the composite sailing route for its entire length. This indicates it is longer than the composite sailing. Which means the composite route lies closest to the direct great circle route and will save the greatest distance while conforming to the limiting latitude restriction.
Example Problem 1:
A great circle track would be most advantageous when compared to the rhumb line track on which of the following routes? (Use gnomonic tracking chart WOXZC 5274 - North Atlantic)
A. Cayenne (Lat. 4°40.0' N, Long. 52°15.0' W) to Sao Tome (Lat. 0°00.0', Long. 3°15.0' W)
B. Palm Beach, FL to the English Channel
C. Natal, Brazil to Reykjavick, Iceland
D. Recife, Brazil to Monrovia, Republic of Liberia
Step 1: Using a pencil and straightedge lightly draw in a line between the points listed in each of the above choices.
Step 2: Eliminate choice A. Cayenne and Sao Tome are both so near the equator (which is a great circle) that a rhumb line track between them would not be appreciably longer than a great circle.
Step 3: Choice B is the correct choice. It represents the situation where a great circle is advantageous, a voyage between points in the middle latitudes with a significant difference of longitude between the points.
Step 4: Eliminate choice C. Natal, Brazil is almost due south of Iceland, the rhumb line between the places does not diverge significantly from being a meridian which is a great circle.
Step 5: Choice D may be eliminated because the two points are in the vicinity of the equator rather than the middle latitudes. In addition, the distance between the points is relatively short which decreases the advantage of great circle over rhumb line tracks.
Example Problem 2:
You are planning a voyage from Cape May (Lat. 38°45.0' N, Long. 74°45.0' W) to Lisbon (Lat. 38°37.0' N, Long. 9°45.0' W). Which of the following is true? (Use gnomonic chart WOXZC 5274)
A. Because the latitudes are almost the same, a great circle track approximates parallel sailing.
B.The northern hemisphere vertex is approximately at longitude 42°10.0 W
C.The distance is measured by using the length of one degree of the meridian at the position of the vertex.
D. A composite sailing must be plotted to remain south of a limiting latitude of 44° N
Step 1: Use a straight edge to draw a line between the two points on the chart.
Step 2: Eliminate choices A, C, and D for the following reasons.
Choice A - The great circle track diverges significantly, from parallel sailing. This can be seen if a line is drawn along latitude 38°40 to compare with the great circle. A rhumb line track like a parallel sailing would be much longer.
Choice C - The distance is not measured by the method stated. Measurement of distance on a Gnomonic chart can not be done by direct methods.
Choice D - The latitude of the vertex of this great circle is 43.8° N. This is less than the limiting latitude of 44° N. So composite sailing is not required.
Step 3: Choice B is correct. The great circle track reaches its maximum north latitude 43.8 degrees north (vertex) at approximately 42.1° W longitude.
Example Problem 3:
On a voyage via the southern tip of Nova Scotia (Lat. 43°20.0' N, Long. 65°35.0' W) you wish to sail the shortest route to La Corunna, Spain (Lat. 43°20.0' N, Long. 8°24.0' W). Which of the following will require you to plot a composite sailing? (Use gnomomic sailing chart WOXZC 5274)
A. Shoals extending 15 miles from Sable Island
B. Sea ice reported 68 miles ESE of St. Johns, Newfoundland
C. Icebergs reported extending to west to west-northwest from Lat. 47°00.0' N, Long. 35°00.0' W
D. Naval exercises using live ammunition being conducted within a 150 mile radius of Lat. 49°00.0' N, Long. 20°00.0' W.
Step 1: Use a straightedge to draw a great circle track between the two points.
Step 2: Choice A may be eliminated because your great circle track will pass more than one degree of latitude (60 miles) north of Sable Island avoiding any shoals extending 15 miles from the island.
Step 3: Choice B is eliminated because this great circle track passes well away from the sea ice reported 68 miles ESE (112.5° T) from St. John.
Step 4: The icebergs reported extending west to west-northwest (270° to 292.5° T) from Lat. 47°00.0' N, Long. 35°00.0' W lie astride the track. A composite sailing can be used to avoid this hazard. This is the correct answer.
Step 5: Choice D can be eliminated as your track takes you south of the naval exercises extending a 150 mile radius around Lat. 49°00.0' N, Long. 20°00.0'W.
Example Problem 4:
You are planning a voyage from departure Seattle (Lat. 48°30.0' N, Long. 125°00.0' W) to a position at Lat. 44°00.0' N, Long. 161°00.'0 E. Which of the following statements is true? (Use gnomonic chart WOXZC 5270)
A. You must plot a composite sailing to remain south of the Aleutian Islands.
B. The northern hemisphere vertex lies to the west of your arrival position.
C. Military exercises north of 53° N between longitudes 150° W to 165° W will not effect your voyage.
D. At your highest latitude, the sun will be visible at upper and lower / transit if the voyage occurs near 21 June.
Step 1: Use a straightedge to draw a line bewteen the points of departure and arrival which were given.
Step 2: Eliminate choice A. The great circle track passes south of the southern most of the Aleutian Islands.
Step 3: Eliminate choice B. The northern hemishere vertex is at about Lat. 52°45.0' N, Long. 155°00.0' W between the points of departure and arrival.
Step 4: Choice C is the correct choice. The statement that the voyage will not be effected by military exercises is true. The great circle track passes just south of the exercice area above 53° N between 150° W tp 165° W.
Step 5: Eliminate choice D. In order to see the sun at upper and lower transit about June 21, the track would have to go above latitude 66°30.0' N (the Arctic Circle).
Example Problem 5:
You are planning a voyage by great circle from Lat. 59°00.0' N, Long. 07°00.0' W via Lat. 38°00.0' N, Long. 61°30.0' W. Which of the following statements is true? (Use gnomonic tracking chart WOXZC 5274)
A. You are east of the northern hemisphere vertex.
B. When plotted on a Mercator chart the track line will be concave to Cape Farwell.
C. All courses are in the southwest quadrant of the compass.
D. Distance is measured by using the length of a degree of latitude at the midpoint of the trackline.
Step 1: Use a straight edge to draw a line between the coordinates of the points given.
Step 2: Eliminate choice A. The northern hemisphere vertex is east of the point of departure. It is located approximately 59°15.0' N, 01°00.0' E.
Step 3: Eliminate choice B. Great circles between points which lie in the northern hemisphere are concave to the equator. Therefore, this trackline is convex, not concave, to Cape Farwell (south tip of Greenland) which lies to the north of the track.
Step 4: Choice C is the true statement. Inspection of the angle at which the trackline intersects each meridian between departure and arrival indicates all the courses to be steered will fall in the southwest quadrant (between 180° T and 270° T).
Step 5: Eliminate choice D. Distance may not be measured directly on gnomonic charts. The method given for measurements is true for Mercator chart projections not gnomonic projections.
Example Problem 6:
In planning a North Pacific voyage, you wish to steam the minimum distance from Lat. 48°30.0' N, Long. 124°45.0' W, to Lat. 44°00.0' N, Long. 150°00.0' E, while remaining south of 51° N latitude. Which of the following tracks meets these requirements? (Use gnomonic tracking chart WOXZC 5270)
A. A Mercator sailing from departure to the mid-longitude at 51° N, which is great circle to arrival.
B. A great circle between departure and arrival with parallel sailing between the longitudes where the great circle intersects 51°N.
C. A great circle, tangent to 51°N, from departure to the mid-longitude, then a great circle to arrival.
D. A great circle from departure to Lat. 51° N, Long. 148°, parallel sailing to Lat. 51° N Long. 171° W, then a great circle to arrival.
The shortest distance between two positions when there is a limiting latitude is called a "composite sailing". Choice D exactly describes the method of plotting a composite sailing track and it is the correct choice.
Choice A has the ship traveling by rhumb line for 42.6 degrees of longitude between departure and the mid-longitude. This is half the trip by a less economical route.
Choice B is even less efficient than A. On the direct great circle route the ship would reach the limiting latitude of 51° N at 133° W. It would then travel by rhumb line until the direct great circle returns to 51° N at 168° E. This would mean traveling for 59 degrees of the total difference of longitude, which exceeds 85 degrees, by rhumb line.
Choice C is a fairly efficient route but its combination of two great circles lies south of the composite sailing route for its entire length. This indicates it is longer than the composite sailing. Which means the composite route lies closest to the direct great circle route and will save the greatest distance while conforming to the limiting latitude restriction.
Thursday, February 5, 2009
Tuesday, February 3, 2009
Subscribe to:
Posts (Atom)