Bearing is the direction of something in relation to a fixed point. A bearing can be measured in degrees in any direction and in any plane.
For example, in marine navigation, relative bearing is measured in the horizontal plane in relation to the ships heading from 0^{o} to 180^{o} to either port (red) or starboard (green). For example, the relative bearing of an object on the port beam would be Red 90^{o} and an object on the starboard bow would be Green 45^{o}.
Note. The definition given above is according to the Admiralty Manual of Navigation. However, it is acknowledged that there are other definitions which state that relative bearings are measured from ‘right ahead’ in a clockwise direction from 0^{o }to 360^{o}. It is little wonder that there is so much confusion surrounding this topic.
Azimuth is a specific type of bearing which measures the direction of an object in relation to true north, in the horizontal plane, clockwise from 0^{o} to 360^{o}. For example, in terms of azimuth, due east is 090^{o} and due west is 270^{o}.
Azimuth is measured by use of either a magnetic compass or a gyro compass. A gyro compass is a form of electrically driven gyroscope which measures azimuth in relation to true north. Such an azimuth measurement is known as true azimuth. A magnetic compass employs a magnetised needle which is used to measure the azimuth of an object in relation to magnetic north. Magnetic compass readings must be corrected for variation and deviation in order to convert them to true azimuth.
Azimuth Angle (see diagram above). Another area of confusion is the difference between azimuth and azimuth angle. In astro navigation, when we calculate the azimuth of a celestial body, the result is expressed as an azimuth angle. Azimuth angle is measured from 0^{o} to 180^{o} either westwards or eastwards from either north or south. If the observer is in the northern hemisphere, the azimuth angle is measured from north and if in the southern hemisphere, it is measured from south. For example, if the true azimuth of an object is 225^{o}, the azimuth angle for an observer in the northern hemisphere will be N135^{o}W but for an observer in the southern hemisphere, it will be S045^{o}W.
To Convert Azimuth Angle to True Azimuth. The rules for converting azimuth angle to true azimuth are summarised in the following table:
Rules for converting Azimuth Angle (Z) to True Azimuth (Zn) 

Lat. North  Lat. South  
LHA>180^{o}  Zn = Z  Zn = 180^{o} – Z 
LHA<180^{o }  Zn = 360^{o}Z  Zn = 180^{o} + Z 
Examples:
If latitude is 55^{o}N, LHA is 145^{o} and azimuth angle is 120^{o} then true azimuth is 360^{ o }– Z i.e. 360^{o} – 120^{o} = 240^{o}.
If latitude is 25^{o}S, LHA is 245^{o}, and azimuth angle is 075^{o} then true azimuth is 180^{o }– Z i.e. 180^{o} – 075^{o} = 105^{o}.
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It is the year 1400 and Alaka’i, an experienced and highly respected Hawaiian navigator is sailing Manu, a Polynesian voyaging canoe, home to Hawaii after a visit to Tahiti, the ancient homeland of his people. The distance from Tahiti to Hawaii is over 2000 nautical miles so, along the way, Alaka’i has used several small islands and atolls as ‘waypoints’ to rest the crew and to replenish their supplies of food and water. The atoll that we now know as Palmyra Atoll is about two thirds of the way from Tahiti to Hawaii and is the last of these waypoints.
Alaka’i uses the time at the atoll to prepare for the last leg of the journey to Hawaii. He has no compass, no chart and none of the hydrographical paraphernalia that we have today. However, he does have his own homemade ‘star compass’ and in his mind, he has a plan of mathematical perfection involving hundreds of complex calculations, intuitively made without formal mathematical training or any form of mathematical notation. He unconsciously uses these calculations to constantly visualize the position of his craft and in this way, he practices ‘dead reckoning’ in his head instead of on a chart. His calculations involve many factors including speeds; distances; wind and leeway; current set and drift; points of sail; courses and headings; wave patterns and shapes; swells and swell deflections and the directions of the Sun and stars.
He is able to recognise many stars along with their parent constellations and he knows the ‘on top’ stars for the major islands. The star compass shows the rising and setting points of important stars and this enables him to choose appropriate ‘steering stars’ to suit his course. The star compass also shows the rising and setting points of the Sun at the equinoxes and the solstices. It is shortly before the Summer Solstice and Alaka’i knows that the Sun will be to the north of Hawaii at this time of the year.
Alaka’i selects a prominent position on the atoll and from this position, at midday when the Sun is at its zenith, he scratches a line on a rock pointing towards it to indicate the direction of North. When the Sun is very high in the sky, as it is at noon, it is very difficult to judge its direction but Alaka’i has a method to overcome this problem. He takes a piece of semitransparent cloth which is stretched over a bamboo frame; he uses this as a filter to enable him to look at the Sun and point a finger towards it. He then lowers his hand to indicate the Sun’s direction on the horizon.
At sunrise and sunset, he makes further scratches on the rock to indicate the rising and setting points of the Sun. When this is done, he places the star compass on the rock and aligns the mark for Polaris to the scratch mark showing the direction of the midday Sun. He notices that the scratch marks for the rising and setting points of the Sun coincide roughly with the rising and setting points of Altair on the star compass and he will use this knowledge to help to orientate the canoe to its course during the voyage.
Arcturus is the ‘Star on Top’ for Hawaii and Alaka’i knows, that when he sees the star is immediately above him at its zenith, he will be on the same latitude as the island but he won’t know whether he is to the east or the west of it. If he were to be in that situation, he could easily sail off in the wrong direction and then have to spend days or possibly weeks sailing back and forth until he found land or else become hopelessly lost. His solution to this problem is to deliberately steer to one side of the island so that he will know in which direction to turn.
He knows that, in the Summer, the direction of the winds in the region of Hawaii fluctuate between NE and ENE for 90% of the time and that the current sets in the same direction. If he deliberately steers to the west of the island, he will face the laborious task of beating against the wind to reach land. He decides therefore, to aim for a point to the east of Hawaii which will enable him to sail downwind and so make a ‘windward landfall’, a technique that was described in the post ‘Pillars of the Sky’.
Hawaii lies roughly 870 nautical miles away in a direction between Polaris and the rising point of the star Kochab (approximately 008^{o}). However, because he aims to be upwind of Hawaii, he plans to make for a point 100 miles further to the east. He calculates the course to this imaginary point to be in the direction midway between Kochab and Dubhe (roughly 015^{o}) and he makes a further scratch on the rock to mark this.
The final task in Alaka’i’s preparations is to set up transit marks to help him to orientate the canoe to the planned course when he sails from Palmyra. He instructs crewmen to cut two tall poles and directs them to stand them in the ground in line with the scratch mark that indicates the course.
Alaka’i sails Manu from Palmyra in an ENE breeze using the transit poles to set the canoe on course. He watches the transit poles closely to help him to gauge the set and drift of the current which he judges to be setting East at about half a knot. He observes that the waves are steeper and taller than could be accounted for by the wind speed and this confirms for him that the current is setting in the opposite direction to the wind. By observing the wake and bowwave he estimates that the speed through the water is approximately 4 knots. He also observes the angle between the wake and the fore and aft line of the boat to estimate that the leeway caused by the wind is approximately 5 degrees.
Without the aid of navigation charts and relative velocity diagrams, he takes all of these factors into account to intuitively calculate a course to steer and he puts the canoe on a close reach, sailing on the starboard tack with a heading of NbE (approx. 010^{o})
Alaka’i is aware that swells, created by distant winds, may travel in a completely different direction to waves that are driven by local winds and that they do not change direction frequently in the way that local winds and waves do. He notes that the swell is coming from just abaft the port beam so he instructs the helmsman to maintain this orientation in order to keep the canoe on the correct heading. He does not orientate the canoe by the local wind and waves because he knows that these can change direction quickly. At midday, he is pleased to see that, when the Sun is at its zenith, its direction is about 10^{o} on the Port bow which tells him that the helmsman is keeping the canoe on the correct heading of 010^{o}.
Alaka’i consults the star compass to devise a plan to keep Manu orientated to the correct course by day and by night. He calculates that the Sun should be about 20^{o} ahead of the starboard beam at sunrise; 10^{o} on the port bow at midday and on the port beam at sunset. During darkness, Polaris should be about 10^{o} on the port bow; the rising point of Dubhe should be about 15^{o} on the starboard bow and its setting point 35^{o} on the port bow. Altair should rise about 10^{o} ahead of the starboard beam and set abeam to port. Acrux will rise on the starboard quarter and will slowly lead the Southern Cross westward over the southern sky and then set again on the port quarter.
Kingman Reef is a hazardous, partly submerged reef which lies 36 nautical miles northwest from Palmyra Atoll. Alaka’i is confident that his course will take him well clear of the reef but even so, he is constantly on the alert for signs to warn him of the danger.
Kingman reef does not host land based birds and it is too low to produce the cloud effects referred to in Pillars of the Sky but Alaka’i has another trick up his sleeve for detecting land that may be over the horizon. Sometimes a swell may not be visibly detected but the Polynesian voyagers developed techniques for sensing weak swells, often by lying in the bottom of the boat and feeling for faint variations in its movement. Because they are no longer driven by the winds that created them, swells are easily deflected by land and if these deflected swells can be detected, they may indicate, to a skilled Polynesian navigator, that land is nearby. Alaka’i uses this ability to sense deflected swells to ensure that he does not sail too close to Kingman Reef.
By nightfall, Manu is well clear of Kingman Reef and Alaka’i is able to relax. Before lying down to get some well earned sleep, he instructs the helmsman to hold the course by keeping the canoe on a close reach on the starboard tack with Polaris about 10^{o} on the port bow. He does not sleep for long because, as well as checking the bearings of Kochab, Dubhe, Altair and Acrux when they rise and set, he needs to begin to observe the star Arcturus at its zenith so that he will know when he has reached the latitude of Hawaii. (Arcturus is the ‘star on top’ for Hawaii).
Alaka’i knows that, at about 200 miles north of Palmyra atoll, the set of the current will change from easterly to westerly as Manu leaves the ocean current that we know as the Equatorial Counter Current and enters the North Equatorial Current. As expected, during the morning of the third day out from Palmyra Atoll, Alaka’i notices turbulence in the water and confused wave shapes and patterns from which he deduces that he has entered the region between the two currents. Towards the end of the morning, the current is setting westwards with a drift that Alaka’i knows from experience, will be about half a knot. The wind is ENE (about 070^{o}) but its speed has increased to about 15 knots causing a leeway angle of roughly 10^{o}. Alaka’i takes these new factors into account and in an effort to keep to the planned course, he calculates that the course to steer should be towards the rising point of Dubhe (025^{o}). In an effort to keep to the new course to steer, he sails the canoe as close to the wind as possible so that it is now closehauled on the starboard tack. At midday, Alaka’i is relieved to see that the Sun is about 25^{o} on the port bow and this tells him that Manu is sailing in the direction of Duhbe’s rising point and so he is confident that the helmsman is keeping the canoe on the correct heading.
During the night of the ninth day of the voyage, Alaka’i observes that Arcturus is immediately above him at its zenith; this tells him that he has reached the latitude of Hawaii and so he alters course to the west in order to make a ‘windward landfall’. However, he knows that Hawaii Big Island is the southernmost of the Hawaiian islands and he doesn’t want to risk missing it altogether by steering too far south. With this in mind, he adjusts the heading to WNW so that, with Maui and the other Hawaiian islands strung out in a line to the northwest, he will have a block of islands to aim for.
He orientates the canoe by keeping the North Star just before the starboard beam in order to maintain his WNW course during the night. He knows that Arcturus will set right ahead on this course which is convenient because he can use it as the ‘steering star’ as well as the ‘on top star’. With the wind from the NE, he puts the canoe on a broad reach, sailing on the starboard tack. With the wind and current helping them, he calculates that Manu is making about 8 knots over the ground and he hopes to sight land before the end of the next day.
At dawn on the tenth day, Alaka’i observes that the Sun is on the starboard quarter when it rises at EbN and this tells him that Manu is on course. At noon, one of the crewmen excitedly points to a number of boobies and frigate birds hovering over a shoal of fish. This is encouraging news for Alaka’i because he knows that the maximum range of these birds from land is around 50 to 60 miles. At about this time, long white clouds can be seen just above the horizon. Alaka’i believes that these are the cloudtrails which appear during the day as the heat of the land forces moist sea air up over Hawaii and Maui’s high volcanoes. From experience, he knows that, in good visibility, the cloudtrails can be seen over 50 miles away and he is now certain that land is over the horizon.
Towards late afternoon, large numbers of birds can be seen including noddies and white terns which are normally found within 20 miles of land. He can now make out the silhouette of the volcanoes against the Sun which is beginning to set. The distance to the horizon from a voyaging canoe is around 10 miles and because he cannot yet see the land beneath the volcanoes, he judges that the shoreline is about 20 miles distant.
He is wary of approaching a lee shore in darkness so at dusk, he decides to heaveto for the night. With the current and the wind as they are, he estimates that the canoe will drift towards the islands at about one knot at the most and that it will not reach land before dawn. When Arcturus reaches its zenith during the night, Alaka’i observes that it is still immediately overhead and this tells him that the canoe is keeping to the latitude of Hawaii.
As day breaks on the eleventh day, there is great excitement amongst the crew as the shoreline of Hawaii can now be seen about 6 miles ahead. Alaka’i steers Manu towards Waiakea Bay on Hawaii Big Island, navigates her through a small channel in the reef and heads for his home village of Waiakea.
Data used for the construction of the star compass are listed below.
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The last post in this series discussed the Polynesian’s Star Compass and their use of ‘steering stars’ for direction finding. Although the Polynesians made extensive use of a form of dead reckoning to estimate position, as far as we know, they did not have a method of fixing a vessel’s position at sea. If they did, we will never know because their methods were closely guarded secrets which were known only to elite groups of navigators and were never recorded. However, David Lewis, in his book ‘We The Navigators’ discusses how early Polynesian navigators pinpointed the position of certain islands by what they called the ‘Star on Top’.
If the latitude of a certain island coincides with the declination of a star, it stands to reason that when the star crosses the meridian of that island, it will be immediately above it. So, to an observer on the island, the star will be overhead when it reaches its zenith; in other words, it will be the ‘Star on Top’. It was believed that the ‘on top’ stars for all the islands were held up in the sky by pillars and that the sky was supported by these pillars which were known as ‘Pillars of the Sky’.
So, how would the Polynesian navigators have used an island’s ‘star on top’ to help them to navigate towards it? Let’s try an example:
Palmyra is a tropical Atoll located roughly half way between Hawaii and Samoa. Although it has no indigenous population now, it may well have had in the past and its position would have made it a suitable waypoint for voyages between Hawaii and the South Pacific islands, particularly since it has an abundant supply of fresh water, coconut palms, many species of nesting birds and lagoons teeming with fish. It is highly likely therefore, that ancient Polynesian wayfarers would have made voyages between Palmyra and Hawaii in their large outrigger canoes. We will use such a voyage for this example.
When Polynesian exploration was at its height around one thousand years ago, the declination of Arcturus was just to the north of Hawaii but due to precession, it slowly moved south and now sits above the southern tip of Hawaii (19.2^{o}N). So, for many centuries, Arcturus would have been the ‘star on top’ for Hawaii and could still serve that purpose today.
Hawaii lies about 1,000 miles to the north of Palmyra and so the obvious course to steer from Palmyra to Hawaii would seem to be north. However, for two reasons this would not have been the chosen course.
Firstly, for the first part of the voyage, the Equatorial Counter Current would set the canoe eastwards, but from about half way, the North Equatorial Current would set it westwards again. Assuming that the voyage was conducted during Hawaii’s summer months, the prevailing Trade Winds would also set the vessel westwards. So, taking the winds and currents into account, the navigator would have to lay off a course to the east of North.
The second reason is this. If the navigator sailed north until Arcturus was directly overhead at its zenith, all he would be able to tell from this would be that his canoe was on the same latitude as Hawaii but he would not know if he was to the east or the west of the island. If his chosen course caused the boat to finish downwind of the island, he would then have the difficult task of beating against the wind and tide to reach his goal. However, if he if he deliberately steered a course that would take him upwind of the islands, he would then be able to sail downwind while maintaining latitude by keeping Arcturus ‘on top’. This technique of deliberately steering so as to finish upwind of the target island was called ‘Windward Landfall’.
A star compass which shows the rising and setting points of the stars which would likely to be of help for a voyage from Palmyra to Hawaii has been constructed below.
The navigation plan for the first part of the voyage would probably be to sail from Polymyra on a course of roughly NNE keeping Polaris on the port bow with the Southern Cross on the starboard quarter during the hours of darkness. (Note that the Southern Cross is not circumpolar north of 34^{o} South and that Acrux, its brightest star, would rise at roughly SSE). Rigel Kentaurus and Hadar in the constellation Centaurus rise on approximately the same bearing as Acrux but shortly after it. These two stars are known as the ‘Pointers’ because an imaginary line from Rigel Kentaurus to Hadar will point towards the Southern Cross.
The direction from which the North East Trade winds blow fluctuates between NE and and ENE so with any luck, the canoe would be able to complete the whole of this leg of the journey on the starboard tack without the need to beat upwind. The heading would be checked by aligning the canoe with the star Dubhe when it rose at approximately NNE. (Dubhe is in the constellation Ursa Major (Great Bear) and is not circumpolar south of 38^{o} North). At the time of aligning the canoe with Dubhe, the angle between the direction of the advancing waves and the fore and aft line of the canoe would be noted and the navigator would use this information for guidance during daylight hours.
When a point was reached where Arcturus (declination 19.2^{o}N) was immediately overhead at its zenith, the course would be changed to westerly for the second part of the voyage to sail downwind to Hawaii.
You will see from the star compass that the stars Alnilam and Altair set approximately due west and so they would make suitable ‘steering stars’ for this part of the voyage. Alnilam is a winter star and Altair is a summer star so one of them will always be visible at night. The plan for the second part of the voyage would probably be to use Alnilam or Altair as the ‘steering star’ while keeping Polaris on the starboard beam and the Southern Cross to port. The latitude of Hawaii would be maintained by keeping Arcturus ‘on top’.
Quite how one can tell what point is ‘immediately overhead’ from a canoe which is rocking and rolling in a choppy sea is not clear but according to Lewis, the Polynesian navigators had several secret methods such as lying in the bottom of the canoe facing upwards. Another method was the ‘floating cane’ which Lewis vaguely describes. Apparently, a cane would be cut below two consecutive growth rings so that a short length of cane which was sealed at one end and open at the other was obtained. A small weight would be attached to the sealed end and the cane would then be filled with water. In theory, when the cane was placed in a container of water, it would remain vertical in spite of the movement of the canoe.
Probably, for Polynesian navigators, the first indication that the canoe was approaching land would be the sighting of certain birds such as terns, noddies, boobies and frigate birds which are landbased and therefore fly out from the land in the mornings and return to it in the evenings thereby giving the navigator indications not only of the nearness of land but also its direction. Pelagic species such as the albatross which roam freely over the open ocean would obviously be of no navigational use and so it would have been important to have the ability to recognise the different species.
Clouds would have been another indication of approaching land. David Lewis gives an indepth discussion of ‘cloud lore’ which was developed by Polynesian sailors over many centuries. There is no space here to discuss this topic fully but there are a few useful tips that modern day navigators could take from the Polynesians. Firstly, although an island may be below the horizon, clouds above it may be visible. Drifting clouds tend to slow down and ‘stick’ over an island for a while and then pick up speed again. Some islands, particularly those with mountains or volcanoes will often appear to have a permanent cloud above them as moist air rising above them condenses and then evaporates again as it descends.
*Ref. Lewis, David, 1972. ‘We the Navigators’, Honolulu: The University Press of Hawaii.
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Nainoa Thompson tells us how that, for centuries before European sailors reached the Pacific Ocean, the South Sea Islanders accurately found their way from island to island without the aid of magnetic compasses, sextants or any other navigational equipment. They navigated the Pacific by using their knowledge of natural phenomena such as the directions of the winds and waves, the flight paths of birds, cloud formations and the colour of the sea. Their most important technique however, was the ‘Star Compass’ which was not a physical tool but a mental construct based on their knowledge of the directions that certain stars would rise and set.
Such knowledge was acquired over many hundreds of years and passed down by word of mouth and example with each generation committing it to memory. It is a tragedy that because the navigational knowledge and techniques of the early Polynesians and Melanesians were not recorded, they have largely been forgotten or would have been if it were not for the enthusiastic work of Nainoa Thompson, Mau Piailug and others.
In modern times, the majority of navigators rely on GPS to find their way although many still keep the traditional art of Astro (Celestial) Navigation alive. How many could navigate the oceans without either of these methods though?
It is well known that we could be deprived of the GPS at any moment for many reasons including the ever increasing threat of cyber attack, coronal mass ejections, equipment failure or shipwreck. How would we cope in such situations? Would it not be a good idea to construct and then memorise our own star compasses for the latitudes at which we sail? This would not be as difficult as it may seem at first. Let us try an example:
Sailor Jerry plans to conduct an experiment by sailing from Fogo Island, Newfoundland (49.8^{o}N) to Mullion Cove (50.02^{o}N) in the British Isles without the aid of GPS or any other navigational equipment, not even a magnetic compass (no jokes about rum line sailing please). He plans to to sail due East along parallel 50^{o}N using only a star compass until he makes landfall, hopefully at one of the following: the Scilly Isles (49.9^{o}N), Lands End (50.1^{o}N) or Lizard Point (49.95^{o}N). Once he has made landfall, he will be able to set a course for Mullion Cove.
Jerry is aware that strong ocean currents in the North Atlantic will make it difficult for him to stay on track; at the beginning of his voyage, he will be pushed south by the Labrador and Irminger currents, later he will be pushed east and then northeast by the North Atlantic Current and then he may get caught by the Canary Current which will push him south east. To add to his difficulties, he knows that although the star compass will provide him with directional information, it will not help him to find his position. To overcome these problems so that he can try to stay on track, he plans to frequently calculate his latitude from the North Star by the method explained here. It is likely that he will not have a sextant in a survival situation so he plans to measure the altitude of the North Star using a home made clinometer for that purpose.
The method he uses to construct the star compass is simple. Firstly, he needs to select several bright stars which he could use to guide him in the right direction (preferably, stars with a magnitude of 1 or less). The lower the star is to the horizon, the better it is to indicate direction so his next step is to calculate the azimuth of the chosen stars when they rise and set from his latitude of 50^{o}N.
He doesn’t have time to observe and memorise the directions of the stars as the South Sea Islanders did so he allows himself the luxury of a nautical almanac to help him with this task.The following table shows the data that Jerry collected to help him to construct his star compass. Notes: 1. To make his calculations of azimuth, he uses the method explained here).
2.To calculate the azimuth of a star when it is on or just above the horizon, Jerry uses an LHA of approximately 271^{o} for dawn and 089^{o} for dusk. (The method of calculating a star’s LHA is shown at step 1 here).
3. Bearings of the Sun at sunrise and sunset are included in the table but only for the equinoxes and the solstices since these are the only four dates on which we can be sure of the Sun’s declination without the aid of an almanac. This information is not included in Jerry’s star compass on this occasion because he will be making the voyage in August.
Below is a diagram of the star compass that Jerry constructed. Because he wants to steer an easterly course, he has selected stars that rise between northeast and southeast and set between northwest and southwest. The North Star (Polaris) and Kochab in Ursa Minor (Little Dipper) are also shown to help him find the direction of north.
Additional Links: Star Compass 1, Survival Astro Navigation Predicting rising and setting times of stars, Locating the North Star, What is Astro navigation?
*Reference for the “know the stars” quote: Michael Punk. 2002. The Revenant
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This post brings together all of the information from parts 1, 2 and 3 of this series to demonstrate the full procedure for establishing an astronomical position line.
.Links: Astro Navigation In A Nutshell Part One
Astro Navigation In A Nutshell Part Two
Astro Navigation In A Nutshell Part Three
Please note. There is not sufficient scope in this post to fully explain this topic; however, there are indepth expositions in my books ‘Astro Navigation Demystified’ and ‘Celestial Navigation – The Ultimate Course’.
Demonstration of the procedure:
Scenario:
Date: 18 July
D.R. Position at Zone Time: 16^{h} 44^{m}: 52^{o }N 21^{o }43.1’W.
Time Zone +1
Deck Watch Time (DWT): 17^{h} 50^{m} 28^{s}.
Deck Watch Error (DWE) 40^{s} fast (40^{s})
Body observed: Sun lower limb.
Sextant Altitude at true position: 32^{o} 10.’4 = 32^{o}.173
Compass Bearing at true position: 261^{o } (for rough check on azimuth)
Index error: +0’.54. Ht. of eye: 8m.
Temperature: 28^{o}C. Pressure: 991mb.
Step 1. Note Lat and Long of DR Position.  
Lat: 52^{o} N  
Long: 21^{o }43.1’W  
Step 2. Calculate PZ. (90 – Lat).
PZ = 90^{o} – 52^{o} = 38^{o} 

Step 3. Calculate Greenwich Date at time of observation.  
Date: 18 July  
Zone time: 16^{h} 44^{m}  
Zone correction: +1^{h}  
Universal Time (GMT): 17^{h} 44^{m}  
Deck watch time: 17^{h} 50^{m} 28^{s}  
Deck watch error: 40^{s}  
Greenwich date: 18^{d} 17^{h} 49^{m} 48^{s }July  
Step 4. Calculate Greenwich Hour Angle and Declination.  
Date: 18 July  
GHA Dec  
UT 17^{h} 73^{o} 26’.1 N20^{o} 54’.7 (d:0’.5 decreasing)  
Inc. 49^{m} 48^{s}: 12^{o} 27’.0 0’.4  
85^{o} 53’.1 N20^{o} 54’.3  
= 85^{o}.885 = N20^{o}.9  
Step 5. Determine if Lat and Dec are ‘Same’ or ‘Contrary’.  
Lat = N
Dec = N Therefore Lat and Dec are same. 

Step 6. Calculate PX
(Lat and Dec Same therefore PX = 90 – Dec). PX = 90^{o} – Dec. = 90^{o} – 20^{o}.9 = 69^{o}.1 

Step 7. Calculate the Local Hour Angle (LHA). (Longitude combined with GHA should equal LHA as a whole number of degrees).  
DR Long: 21^{o }43.1’W  
GHA: 85^{o}.885  
DR Long: 21^{o}.718 West ()  
LHA: 64^{o} .167  
Step 8. Determine Angle ZPX.
ZPX = LHA = 64^{o} .167 

Step 9. Calculate True Altitude at True Position (Observed altitude corrected for IE, Dip, Parallax and Refraction).  
Sextant Altitude = 32^{o} 10′.40
Index error (IE) = +0′.54 Observed Altitude = 32^{o} 10′.94 Dip (ht. 8m.) = 5′.00 (table 6a) Apparent Altitude = 32^{o} 05′.94 Altitude correction = +14′.50 (table 6d) Added refraction (28^{o}/991mb) = +0′.10 (table 6c) True Altitude = 32^{o} 20′.54 = 32^{o}.34 

Step 10. Calculate Zenith Distance at True Pos. (90^{o} – Altitude).  
Zenith Dist = 90^{o} – 32^{o}.34 = 57^{o}.66 = 3459.6′  
Step 11. Calculate Zenith Distance at DR Position. (ZX).  
Lat. = 52^{o}N
Declination = N20^{o}.9 (From Step 4) Lat and Dec Same (From Step 5) ZPX = 64^{o} .167 ^{ }(From Step 7) PZ = 38^{o} (From Step 2) PX = 69^{o}.1 (From Step 5) 

Cos (ZX) = [Cos(PZ) x Cos(PX)] + [Sin(PZ) x Sin(PX) x Cos(ZPX)]
= [Cos(38) x Cos(69.1)] + [Sin(38) x Sin(69.1) x Cos(64.167)] 

= 0.53  
ZX = Cos^{1} (0.53) = 57^{o}.99
Zenith Distance at DR position = 57^{o}.99 = 3479′.4 

Step 12. Calculate Azimuth at DR Position (PZX)  
Cos PZX = Cos(PX) – [Cos(ZX) x Cos(PZ]
[Sin(ZX) x Sin(PZ)] 

Cos PZX = Cos(69.1) – [Cos(57.99) x Cos(38]
[Sin(57.99) x Sin(38)] 

= 0.119  
PZX = Cos^{1}(0.119) = 96^{o}.8 ≈ 97^{o}
Azimuth at DR position = 97^{o} 

Step 13. Convert azimuth angle to true bearing (ZN):  


DR Lat. = 52^{o}.0N  
Azimuth (Z) = 97^{o} (from step 12)  
LHA = 64^{o} (from step 7)  
Therefore ZN = 360^{o} – 97^{o} = 263^{o}
Therefore true bearing of body at DR position = 263^{o} Compass Bearing at true position: 261^{o }(for rough check on azimuth) 

Step. 14. Note observed compass bearing at true position and compare with true bearing at DR position for rough check.  
Observed compass Bearing at true position: 261^{o}
True Bearing of body at DR position = 263^{o} 



Step 16. Plot the position line. (Reminder: Plot intercept from DR position along azimuth line).
Latitude: 52^{o }00’N Longitude: 21^{o} 53’W Intercept = 19.8 n.m. towards 263^{o }(From step 15) Note. Drawing not drawn to scale. 
INSERT PLOT
This topic is explained in far greater depth in my books ‘Astro Navigation Demystified’ and ‘Celestial Navigation – The Ultimate Course’.
Many thanks to Jeremy Parker for his help with this post.
Celestial Navigation at Amazon.com
Celestial Navigation at Amazon.uk
Astro Navigation Demystified at Amazon.com
Astro Navigation Demystified at Amazon.uk
Applying Mathematics to Astro Navigation at Amazon .com
Applying Mathematics to Astro Navigation at Amazon .uk
Astronomy for Astro Navigation at Amazon.com
Astronomy for Astro Navigation at Amazon.uk
web: http://www.astronavigationdemystified.com
e: astrodemystified@outlook.com
The Azimuth is the angle PZX in the diagram (it is, the angle between the observer’s celestial meridian and the vertical circle through the celestial body). The Azimuth is similar to the bearing in that it is the angle between the observer’s meridian and the direction of the celestial body. However, whereas bearings are measured clockwise from north from 0^{o} to 360^{o}, azimuth is measured from 0^{o} to 180^{o} from either north or south. If the observer is in the northern hemisphere, the azimuth is measured from north and if in the southern hemisphere, it is measured from south. For example, if the true bearing is 045^{o}, in terms of azimuth it is either N45^{o}E for an observer is in the northern hemisphere or S135^{o}E for an observer in the southern hemisphere.
Local Hour Angle (LHA). LHA is the angle ZPX; that is the angle between the observer’s celestial meridian and the meridian of the celestial body.
Relationship between LHA and Azimuth. When the LHA is less than 180^{o}, the celestial body lies to the west of the observer’s meridian and when it is greater than 180^{o} it lies to the east. (Remember LHA is measured westwards from the observer’s meridian). It follows that if the celestial body is to the west of the observer’s meridian, the azimuth must be west and when to the east, the azimuth must be east. So we have the rule: LHA = 0^{o} – 180^{o }: Azimuth West. LHA = 180^{o }– 360^{o} : Azimuth East
The Altitude is the angle AOX in the diagram; (it is the angle from the celestial horizon to the celestial body measured along the vertical circle through the celestial body).
The Zenith Distance is the angular distance ZX measured along the same vertical circle from the zenith to the celestial body; that is the angle XOZ.
Relationship between Altitude and Zenith Distance Since the celestial meridian is a vertical circle and is therefore, perpendicular to the celestial horizon, it follows that angle AOZ is a right angle and angles AOX and XOZ are complementary angles. From this we can deduce that: Zenith Distance = 90^{o} – Altitude and Altitude = 90^{o} – Zenith Distance
Example. Suppose that from our true position, the calculated altitude of a celestial body is 32^{o}.34 and the compass bearing is 261^{ o}. As explained above, Zenith Distance = 90^{o} – Altitude Therefore zenith distance = 90^{o} – 32^{o}.34 = 57^{o}.66 = 3,459.6′
Since 1 minute of arc subtends a distance of 1 nautical mile on the Earth’s surface, we know that the distance of the true position from the geographical position of the celestial body will be 3,459.6 nautical miles in the approximate direction of the compass bearing 261^{ o} which when converted to azimuth is N99^{o}W for an observer in the northern hemisphere. (More on this in the next post).
However, because of the impracticability of plotting a position line of such an immense distance on a large scale navigational chart, this does not tell us our exact position.
Where to buy books of the Astro Navigation Demystified series:
Celestial Navigation at Amazon.com
Celestial Navigation at Amazon.uk
Astro Navigation Demystified at Amazon.com
Astro Navigation Demystified at Amazon.uk
Applying Mathematics to Astro Navigation at Amazon .com
Applying Mathematics to Astro Navigation at Amazon .uk
Astronomy for Astro Navigation at Amazon.com
Astronomy for Astro Navigation at Amazon.uk
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Link: Astro Navigation In A Nutshell Part One
Suppose we are in a yacht and we measure the altitude of the Sun and find it to be 35^{o}; what does this tell us? All that we know is that the yacht lies somewhere on the circumference of a circle centred at the geographical position of the Sun. Such a circle is known as a ‘position circle’ since our position is known to lie somewhere on its circumference. The diagram shows that, at any point on the circumference of the circle, the Sun’s altitude will be 35^{o} and our distance from the GP will be equal to the radius of the circle. The problem is to establish at which precise point on the position circle the yacht lays.
At first, it might seem that all we need to do is to observe the bearing of the Sun at the same time that we measure its altitude and then draw the line of bearing on the chart along with the position circle. In this way, it would seem that our true position would correspond to the intersection of these lines on the chart. However, building on the work covered in Part One, we will find that there is a problem with this idea which makes it impracticable. Because of the great distance of the Sun from the Earth, the radius of the position circle will be very large (approximately 3000 nautical miles or so). A chart on which such a large circle could be drawn would require such a small scale that accurate positionfixing would be impracticable.
However there is another way of solving the problem. We cannot physically measure the distance from the yacht to the GP but we can measure the altitude of the Sun at the true position and from that we can calculate the zenith distance as can be explained with the diagram below.
The true position of the yacht is represented by A in the diagram
Z represents the zenith of the true position
X represents the position of the Sun
U represents the geographical position of the Sun
ZX is the zenith distance and AU is equal to the angular distance ZX in nautical miles.
We can see that the zenith distance is equal to 90^{o} – Altitude
So, measuring the altitude gives us a method of calculating the zenith distance and the zenith distance gives us the distance AU in nautical miles.
The Intercept. Our aim is to calculate the azimuth and altitude of a celestial body from our DR position at the time that we accurately measure the altitude at the true position. The azimuth will give us a line of direction between the DR position of the yacht and the geographical position of the celestial body. (We use the azimuth rather than the compass bearing because a magnetic compass is not very accurate whereas the azimuth can be accurately calculated. By finding the difference between the two altitudes and hence the difference between their zenith distances, we can calculate the intercept which is the distance from the DR position to the circumference of the true position’s positioncircle.
Astronomical Position Line. We draw the position line at a point where the intercept intersects with the position circle. Since the circumference of a circle at any point is at rightangles to the radius at that point, no accuracy will be lost by drawing the position line as a straight line at rightangles to the intercept. The navigator will then be able to draw a short, straight line on the chart along which he knows his position lies; such a line is known as an astronomical position line.
In celestial navigation, just as in coastal navigation, we need to calculate the intersection of three or more position lines in order to obtain a position fix (or observed position as it is called in celestial navigation).
Please Note. This topic is explained in far greater depth in my books ‘Astro Navigation Demystified’ and ‘Celestial Navigation – The Ultimate Course’.
Many thanks to Jeremy Parker for his help in updating this page.
Celestial Navigation at Amazon.com
Celestial Navigation at Amazon.uk
Astro Navigation Demystified at Amazon.com
Astro Navigation Demystified at Amazon.uk
Applying Mathematics to Astro Navigation at Amazon .com
Applying Mathematics to Astro Navigation at Amazon .uk
Astronomy for Astro Navigation at Amazon.com
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web: http://www.astronavigationdemystified.com
e: astrodemystified@outlook.com
Links: Astro Navigation In A Nutshell Part One
Astro Navigation In A Nutshell Part Two
There are several ways of calculating the azimuth and altitude at the assumed position; these include the use of sight reduction methods and software solutions. However, the traditional method is by the use of spherical trigonometry which is demonstrated below.
In the diagram above,
PZ is the angular distance from the Celestial North Pole to the zenith of the observer and is equal to 90^{o} – Lat.
PX is the angular distance from the Celestial North Pole to the celestial body and is equal to 90^{o} – Dec.
ZX is the Zenith Distance and is equal to 90^{o} – altitude.
Therefore, altitude is equal to 90^{o} – ZX
The angle ZPX is equal to the Local Hour Angle of the Celestial Body with respect to the observer’s meridian.
The angle PZX is the azimuth of the body with respect to the observer’s meridian.
Summary.
PX = 90^{o} – Dec.
PZ = 90^{o} – Lat.
ZX = 90^{o} – Alt.
Alt = 90^{o} – ZX
<PZX = Azimuth.
<ZPX = Hour angle.
In order to calculate the azimuth and altitude of a celestial body we must solve the triangle PZX in the diagram above. Specifically, we must calculate the angular distance of side ZX so that we can find the altitude and we must calculate the angle PZX so that we can find the azimuth.
However, because the triangle PZX is on the surface of an imaginary sphere, we cannot solve this triangle by the use of ‘straight line trigonometry’; instead we must resort to the use of ‘spherical trigonometry’ which is explained here.
Example of the use of spherical trigonometry to calculate the azimuth and altitude of celestial bodies.
Note. Traditionally, the ‘halfhaversine’ formula was used for this task but this formula does not lend itself well to solution by electronic calulator; therefore, the following solutions involve the cosine formula.
Example: Star Sight.
Scenario: Greenwich date: 30 June 18hrs 05 mins 33 secs
Assumed Position: Lat. 30^{o}N Long. 45^{o}W
Selected body: Alioth
SHA: 166
Declination: 56^{o}N
GHA Aries: 250
SHA Alioth: 166
Add GHA Aries: 166 + 250 = 416
Subtract Long(W) = 416 – 45 = 371
Subtract 360 = 11
Therefore, LHA = 11W
(all results in degrees)
PZ = 90^{o} – 30^{o} = 60^{o} ∴PZ = 60^{o}
PX= 90^{o} – 56^{o} = 34^{o} ∴PX = 34^{o}
ZPX = LHA = 11^{o }west
As explained here, the formula for calculating side ZX is:
Cos (ZX) = [Cos(PZ) . Cos(PX)] + [Sin(PZ) . Sin(PX) . Cos(ZPX)]
∴To calculate zenith distance of Alioth:
Cos (ZX) = [Cos(PZ) . Cos(PX)] + [Sin(PZ) . Sin(PX) . Cos(ZPX)]
= [Cos(60^{o}) . Cos(34^{o})] + [Sin(60^{o}) . Sin(34^{o}) . Cos(11^{o})]
= [0.5 x 0.829} + [0.866 x 0.559 x 0.982]
= 0.415 + 0.475
Cos (ZX) = 0.89
∴ ZX = Cos^{1} (0.89) = 27^{o}
Altitude = 90^{o} – ZX = 90^{o} – 27^{o} = 63^{o}
As explained here the formula for calculating angle PZX is:
Cos PZX = Cos(PX) – [Cos(ZX) . Cos(PZ)] / [Sin(ZX) . Sin(PZ)]
∴To calculate azimuth of Alioth:
Cos PZX = Cos(34) – [Cos(27) . Cos(60)] / [Sin(27) . Sin(60)]
= 0.829 – [0.89 x 0.5)] / [0.454 x 0.866)]
= (0.829 – 0.445) / 0.393
= 0.384 / 0.393 = 0.977
Cos(PZX) = 0.977
∴ PZX = Cos^{1}(0.977) = 12.31
∴ Azimuth = N12^{o}W (since LHA is west)
In terms of bearing, the azimuth is 348^{o}.
LHA = 11^{o }west
Declination = 56^{o}N
Azimuth at assumed position = N12^{o}W
Altitude at assumed position = 63^{o}
Please Note. This topic is explained in far greater depth in my books ‘Astro Navigation Demystified’ and ‘Celestial Navigation – The Ultimate Course’.
Where to buy books of the Astro Navigation Demystified series:
Celestial Navigation at Amazon.com
Celestial Navigation at Amazon.uk
Astro Navigation Demystified at Amazon.com
Astro Navigation Demystified at Amazon.uk
Applying Mathematics to Astro Navigation at Amazon .com
Applying Mathematics to Astro Navigation at Amazon .uk
Astronomy for Astro Navigation at Amazon.com
Astronomy for Astro Navigation at Amazon.uk
web: http://www.astronavigationdemystified.com
e: astrodemystified@outlook.com
Although the imaginary Mean Time gives us an accurate measurement of time, it presents the navigator with a problem. When fixing his position by an observation of the Sun, he measures the altitude of the True Sun which keeps apparent solar time. However, he notes the time of the observation from a deck watch that keeps mean solar time. To enable us to connect mean solar time with apparent solar time, we have the Equation of Time which is defined as follows:
Equation of Time = Mean solar time – Apparent Solar Time
In other words, the equation of time is the difference between apparent solar time and mean solar time taken at the same instant at one place.
The equation of time can be either positive or negative depending on the time of the year.
Nautical Almanac.
The Equation of Time for 00^{h} (lower meridian) and 12^{h} (upper meridian) for each day is printed at the foot of the Nautical Almanac daily page as shown in this extract. The Local Mean Time of the Sun’s Meridian Passage is shown in the column to the right of the EOT. (This is the apparent time of Mer Pas adjusted for EOT to give the LMT and rounded up to the nearest minute).
If the mean time of Mer. Pas is shown to be greater than 1200 then the EOT must be negative, indicating that apparent time is slow compared to mean time. Conversely, if the mean time of Mer. pas. is less than 1200 then EOT is positive, indicating that apparent time is ahead of mean time.
To calculate longitude we simply find the difference between LMT of Mer. Pas. and the GMT of our observation of Mer. Pas. then, by converting the time difference to arc we are able to find the difference in degrees of longitude.
Let’s try an example:
Date: 22 June. Zone Time: 1140 (+4). DR Pos: 32^{0} 30’N. 61^{0} 55’W. At meridian passage, the deck watch time was 16^{h} 08^{m}^{ }25.1^{s} and the Deck Watch Error was 05.0^{s}. The daily page for that date shows that the Eqn. of Time is 02^{m} 02^{s} and that Mer. Pas. is 1202 indicating that EOT is negative.
Procedure:
Deck Watch Time: 16^{h} 08^{m}^{ }25.1^{s}
DWE: 05.0^{s}
GMT: 16^{h} 08^{m}^{ }20.1^{s}
LMT Mer.Pas: 12^{h}^{ } 02^{m} 00^{s}^{ }
Time Diff: 04^{h} 06^{m}^{ }20.1^{s}
(Longitude West, GMT Best)
4^{h}^{ }^{ }= 4 x 15 = 60^{o }00’ 00”
06^{m}^{ }= 6 ÷ 4 = 1^{o} 30’ 00”
20.1^{s} = 20.1 ÷ 4 = 0^{o} 05’ 01″.5
Therefore, Long = 61^{o }^{ }35’ 01″.5 W
The navigator seldom requires the time of meridian passage to accuracies greater than one minute. Therefore, use the time listed under the “Mer. Pass.” column unless extreme accuracy is required.
Alternative Method. Notwithstanding the slight inaccuracy caused by the rounding up of the listed time of Mer. pas., practicing navigators may prefer the above method but for students and tutors of astro navigation, it provides very little understanding of the equation of time. To overcome this problem, it may be more beneficial for students to calculate the time of Mer. Pas themselves by applying the equation of time. There is also the benefit of a greater degree of accuracy.
So if, in the above example, we wished to recalculate the time difference by applying the equation of time to the local apparent noon (LAN) instead of simply using the published time of Mer. Pas., we would proceed as follows:
On 22 June, the EOT is 01^{m} 55^{s} at 00^{h} and it is 02^{m} 02^{s }at 12^{h} so the hourly rate of change is (02^{m} 02^{s} – 01^{m} 55^{s}) ÷ 12 = 0.58^{s}. Therefore, at 16^{h} 00^{m}, the EOT will be 02^{m }02^{s} + (4 x 0.58^{s}) = 02^{m} 04.32^{s} which we can approximate to 02^{m} 04^{s}. The table shows that Mer. Pas is greater than 1200 indicating that apparent time is slow compared to mean time. This means that EOT is negative and must be added to the apparent time to give mean time.
Deck Watch Time: 16^{h} 08^{m}^{ }25.1^{s}
DWE: 05.0^{s}
GMT Mer. Pas: 16^{h} 08^{m}^{ }20.1^{s}
LAN: 12^{h}^{ } 00^{m} 00^{s}^{ }
EOT: 02^{m} 04^{s}^{ }
LMT Mer. Pas.: 12^{h}^{ } 02^{m} 04^{s}^{ }
Time Diff (GMT – LMT) = 04^{h} 06^{m} 16.1^{s}
(Longitude West, GMT Best)
4^{h}^{ }^{ }= 4 x 15 = 60^{o }00’ 00”
06^{m}^{ }= 06 ÷ 4 = 1^{o} 30’ 00”
16.1^{s} = 16.1 ÷ 4 = 0^{o} 04’ 01″.5
Therefore, Long = 61^{o }^{ }34’ 01″.5 W
Links: Understanding Meridian Passage, Meridian Passage Short Method, Meridian Passage Long Method What is the point of meridian passage? Zone Time, Local Hour Angle and Greenwich Hour Angle, Converting GMT to GHA , Altitude Corrections
Where to buy books of the Astro Navigation Demystified series:
Celestial Navigation at Amazon.com
Celestial Navigation at Amazon.uk
Astro Navigation Demystified at Amazon.com
Astro Navigation Demystified at Amazon.uk
Applying Mathematics to Astro Navigation at Amazon .com
Applying Mathematics to Astro Navigation at Amazon .uk
Astronomy for Astro Navigation at Amazon.com
Astronomy for Astro Navigation at Amazon.uk
web: http://www.astronavigationdemystified.com
e: astrodemystified@outlook.com
As I have done in all of my books, I will attempt to avoid overcomplicated and stilted academic language and as far as possible, give my explanations in straightforward, plain English.
Links: Understanding Meridian Passage, Meridian Passage Short Method, Meridian Passage Long Method What is the point of meridian passage? Zone Time, Local Hour Angle and Greenwich Hour Angle, Converting GMT to GHA , Altitude Corrections
Measuring the Distance Between Meridians of Longitude Along a Parallel of Latitude. In the diagram below, PBC and PAD lie on separate meridians of longitude.
The arc BA is the distance between these meridians measured along a certain line of latitude. The arc CD is the distance between the same meridians measured along the Equator. Clearly, the distance CD is much greater than the distance BA
To Calculate The Distance Between Two Meridians Along A Parallel Of Latitude. The following formulas are used for calculating the difference in distance along a parallel of latitude (Ddist) corresponding to a difference in longitude (Dlong) and vice versa.
Ddist = Dlong x Cos Lat. and Dlong = Ddist ÷ Cos Lat.
Since the secant is the inverse of the cosine, the formula for Dlong can be simplified to: Dlong = Ddist x Sec Lat.
The Rhumb Line. If a ship were to steer a steady course, that is one on which her heading remains constant, her track would cut all meridians at the same angle, as the next diagram shows. Such a line on the Earth’s surface is called a rhumb line.
When the rhumb line cuts all meridians at 90^{o}, it will coincide with either a parallel of latitude or with the Equator. When the angle is 0^{o}, the rhumb line will be along a meridian of longitude.
A vessel’s course will always be a rhumb line; thus the course to be steered to travel from one place to another will refer to the angle between the rhumb line joining the places and any meridian.
Calculating the distance between two points along a rhumb line. In the next diagram, A, B, C, D and Z are meridians of longitude; the lines aB, bC, and cD are different parallels of latitude; and the line ABCDZ is a rhumb line. A series of rightangled triangles have been constructed along the rhumb line AZ and in each triangle, one short side lies along a meridian of longitude, one lies along a parallel of latitude and the hypotenuse lies along the rhumb line.
It can be seen from the diagram that the eastwest distance between two points along the rhumb line is the sum of the distances along the parallels of latitude corresponding to the difference in longitude in each of the rightangled triangles. This eastwest distance is known as the departure
Middle Latitude. If we were to calculate the departure along each of the parallels of latitude aB, bC, cD, we would find that they would not be equal and so the task of calculating the total departure would be complicated. In practice, the total departure is taken to be the eastwest distance along the intermediate of these parallels which is known as the ‘middle latitude’.
By the formula established for Ddist above, we can derive a formula to calculate departure as follows: Departure = d.long cos(middle latitude).
Mean Latitude. In most cases, the arithmetic mean of the two latitudes can be used as the middle latitude without appreciable error, so the approximate formula dep.= d.long cos(mean lat) may be used.
When the difference of latitude is large (over 600 n.m.) or the latitudes are close to either of the poles, the middle latitude must be used instead of the mean latitude and in these cases, we have the more accurate formula: Dep. = d.long cos(mid lat).
The difficulty lies in the task of calculating the middle latitude which involves finding the mean of the secants all the intermediate latitudes by integration. Such methods are obviously impracticable in situations where courses and distances have to be calculated rapidly at sea. For this reason, tables of corrections to be applied to the mean latitude are contained in various collections of nautical tables. Since, celestial navigation involves short distance sailing calculations, it is not intended to copy middle latitude correction tables here; however, the following example demonstrates their use:
Suppose a ship sails from position 50^{o}N, 32^{o}E., to 70^{o}N., 15^{o}E.
The d.long is 17^{o} and the mean latitude is 60^{o}.
The formula for calculating departure using the mean lat. is: dep.= d.long cos(mean lat)
Using this formula we have:
Dep. = 17^{o} cos(60)
= 1020’ cos(60)
= 510’ or 510 n.m.
In the tables for converting mean latitude to middle latitude, the correction for a mean latitude of 60^{o} and a difference of latitude of 20^{o} is +1^{o} 09’. So the middle latitude = 61^{o}.15.
The formula for calculating departure using the middle latitude is: Dep. = d.long cos(mid lat) = 1020 cos (61.15) = 492.17 n.m.
By comparing these results, we can see that there is a significant difference between calculations involving the mean latitude on one hand and the middle latitude on the other.
Summary of Formulas. The formulas so far derived in this appendix are summarized below:
Ddist = Dlong x Cos Lat.
Dlong = Ddist ÷ Cos Lat = Ddist x Sec Lat. .
dep.= d.long cos(mean lat) (for distances 600 n.m. or less).
Dep. = d.long cos(mid lat). (for distances over 600 n.m.).
The Rhumb Line Formulas. With the next diagram, we expand on the work above:
Consider triangle ABa in the diagram above: AB is the distance made good, aB is the departure along a parallel of latitude, angle aAB is the course angle.
Therefore, Sin(course angle) = departure ÷ dist. This formula applies to all of the small triangles since they are equal.
By transposition, the above formula becomes: Dep = Dist x sin(course)
The departure between A and Z therefore, is the sum of the departures of all of the small triangles. Therefore, by addition:
aB + bC + cD + …. = (AB + BC + CD + …. x sin(course)
i.e. Dep = Dist sin(course)
If we again consider triangle ABa, Aa = AB cos (course)
But Aa is the difference in Latitude (D.Lat) between A and B
So D.Lat = AB cos(course)
Again, this formula applies to all of the small triangles since they are equal. Therefore, by addition, the total D.Lat corresponding to the total distance between A and Z becomes:
D.Lat = Dist cos(course)
We have established formulas to calculate Dep and D.Lat; we now need a formula to find the course.
If we return to triangle ABa, we can see that the course angle can be found by the formula: tan(course) = Dep ÷ D.Lat.
As before, this formula applies to all of the small equal triangles. So, by addition, the rhumb line course between A and Z can be found by the formula:
Tan(course) = Dep ÷ D.Lat
Short Distance Sailing. Short distance sailing is a term which is applied to sailing along a rhumbline for distances less than 600 nautical miles. From the formulas derived above, the following are used extensively in short distance sailing:
To Calculate Departure when the course is not known: dep.= d.long cos(mean lat)
To Calculate Departure when the course is Known: Dep = Dist x Sin(course)
To Calculate Distance when departure and course are known:
Dist = Dep ÷ Sin (course)
To Calculate Dlat when the distance and course are known:
DLat = Dist x Cos(course)
To Calculate Course to Steer (the rhumb line course between two points)
Tan(course) = Dep ÷ D.Lat
To calculate Dlong (difference in longitude corresponding to the departure):
DLong. = Dep. x Sec(Mean.Lat) or Dlong = Dep ÷ Cos(Mean.Lat)
Worked Example. What is the rhumb line course to steer and the distance to travel from position 40^{o}.5N, 43^{o}.0W to position 42^{o}.25N 41^{o}.8W?
Solution:
Dlat = 42^{o}.25N – 40^{o}.5N = 1^{o}.75N = 105’N
Mean Lat = 40^{o} 30’N + 52’.5 = 41^{o} 22’.5N Dlong = 43^{o}.0W – 41^{o}.8W = 1^{o}.2E = 72’E Dep = d.long x cos(mean lat) = 72 cos(41.38) = 54’.02 Tan(course) = Dep ÷ D.Lat = 54.02 ÷ 105 = 0.51 Therefore course = N27^{o}E = 027^{o} Dist = Dep ÷ Sin (course) = 54.02 ÷ Sin(27) = 120’ Course to steer = 027^{o } Distance to new position = 120 n.m. 
Where to buy books of the Astro Navigation Demystified series:
Celestial Navigation at Amazon.com
Celestial Navigation at Amazon.uk
Astro Navigation Demystified at Amazon.com
Astro Navigation Demystified at Amazon.uk
Applying Mathematics to Astro Navigation at Amazon .com
Applying Mathematics to Astro Navigation at Amazon .uk
Astronomy for Astro Navigation at Amazon.com
Astronomy for Astro Navigation at Amazon.uk
web: http://www.astronavigationdemystified.com
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