The Accuracy of Astro / Celestial Navigation

plot 3 U.S. Navy and Royal Navy navigators are taught that the accuracy of astro navigation is ±1 minute of arc or 1 nautical mile and that where position lines are derived from astronomical observations, the resultant position is not known as a ‘fix’ but is known as an observed position and is marked on the chart as ‘Obs’.

A wise navigator will always be aware of the inherent errors in astro navigation and will never consider an observed position to be an absolute position.  An error of  ±1 nautical mile in an observed position in mid-ocean is of little consequence; however, when accurate navigation is necessary such as when approaching land, it can make the difference between life and death.

Errors in Astronomical Position Lines   Errors may be introduced into an astronomical position line by a number of contributory errors which are briefly described below.

Errors in the observed altitude.   Even when the sextant altitude has been corrected for index error, dip, refraction, semi-diameter and parallax, the resultant altitude reading may still be incorrect owing to a combination of other errors such as incorrect calculated values for dip and refraction.

A pronounced error in refraction is likely to occur when the altitude is below 15o.  The dip being affected by refraction is the most likely cause of error; when atmospheric conditions are abnormal, the actual value of dip may differ from the tabulated value by up to 10′.

An error in the observed altitude will lead to an error in the observed zenith distance.

Errors in the calculated altitude.   There are accumulative and unavoidable errors caused by the addition and rounding-off of quantities taken from the almanac.  There are also unavoidable errors in the method by which the zenith distance and hence the altitude are calculated.  These errors vary according to the method being used but generally, the inaccuracies arise because certain quantities are tabulated to the nearest minute and others are rounded-off.

Deck-watch error.  If the deck-watch error is incorrect, the GMT and the local hour angle will be incorrect.  An error in the LHA will lead to an error in the calculated altitude and this will cause the position line to be displaced.

Errors in the D.R. position.  Errors in the course and distance laid down on the chart may result from a combination of inaccurate plotting, compass error. the effects of wind and tidal stream and incorrect calculation of speed made good over the ground.

An error in the DR position and resultant assumed position will lead to errors in the estimated longitude and hence the local hour angle and this in turn will lead to an error in the calculated altitude.

Errors In Observed Positions Derived From More Than One Position Line.  Position lines obtained from three astronomical observations are not likely to pass through a common point.  The reasons for this are firstly, the observations are not likely to be taken simultaneously since it is not possible to take sextant readings of three different celestial bodies at the same instant.  The faster a vessel travels, the greater the movement of the observer between the three observations and the more significant this error becomes even when special methods of calculation such as ‘MOO’ are used.

Secondly, observed altitudes are very seldom correct and therefore, the resultant observed zenith distances will not be correct.  For these reasons, the resultant position lines will be displaced and a ‘cocked hat’ will be formed.

Thirdly, because the position within the triangle of the cocked hat is arrived at by guess-work, it is unlikely to be correct.

For these reasons, observed positions derived from more than one position line cannot be regarded as absolute positions.

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Finding Stars and Constellations – Part III

This post continues the series Finding Stars and Constellations.

Lyra, Cygnus and Aquila – The Summer Triangle.

The diagram below shows the constellations Lyra, Cygnus and Aquila.


If we follow the reference line ‘the Pointers’ from Ursa Major to Polaris, Vega, the brightest star in Lyra, can be found at about one hand-span from Polaris.  An alternative method is to join star 2 of Ursa Major to a point midway between stars 1 and 4 and this line, when extended, will point to Vega.

To the left of Lyra, at about a hand-span and at a slightly lower altitude, we will find the star Altair in the constellation Aquila.  The constellation Cygnus can be found slightly above a point midway between Lyra and Aquila.

The Summer Triangle.  The stars Deneb in the constellation Cygnus, Altair in Aquilla and Vega in Lyra form an astronomical asterism which is known as the ‘Summer Triangle’. The diagram  shows how the triangle is formed by imaginary lines drawn between these stars.   An alternative name given to this asterism by U.S. Air force pilots is the ‘Navigator’s Triangle’ because it’s bright stars provide a beacon in the sky.  The triangle is also popular with ‘seagoing’ navigators because it provides pointers to several constellations.

Sagittarius The Archer   Sagittarius is a large constellation lying over the southern hemisphere and is visible between latitudes +55 and -90.  For navigation purposes, it is best seen during evening nautical twilight in August.  It contains several bright stars, two of which, Nunki and Kaus Australis, are navigation stars.

In ancient Greek mythology, Sagittarius was said to represent the Archer, a beast called a Centaur which was half man and half horse.  In the representation, the Archer has a drawn bow with the arrow pointing to the star Antares, the heart of the scorpion which had been sent to kill Orion.

Finding Sagittarius.  The Summer Triangle provides a useful pointer to Sagittarius.  If we draw an imaginary line from the star Deneb through the star Altair in the Summer Triangle,  and extend that line by about 20o or one hand-span, it will point to the constellation Sagittarius as the diagram below shows.

triangle to sagittarius

Note. Nowadays, the Sun is over the constellation Sagittarius at the Winter Solstice on 21/22 December when the Sun’s declination reaches its southernmost latitude of 23.5o south. However, in ancient Greek times, the Sun passed through the constellation Capricornus at this time hence the reason for naming the latitude  23.5o south the Tropic of Capricorn.

 Scorpius (Scorpio)  The Scorpion   The constellation Scorpius lies above the southern hemisphere and is visible between latitudes +40 to -90.  For navigation purposes, it is best seen during evening nautical twilight in July.

Scorpius has several bright stars which, between them, form the shape of a scorpion.  The brightest star in Scorpius is Antares which is often mistaken for Mars because of its redish orange colour.  Antares is the 16th. brightest star in the sky and is a navigation star.  The second brightest star in Scorpius is Shaula which is said to represent the sting in the tail of the scorpion.  Shaula is also navigation star.

In Greek mythology, Scorpius represents the scorpion that the goddess Artemis sent to sting and kill Orion who had tried to ravish her.

Finding Scorpius.  The legend that the Archer’s arrow in Sagittarius points to the heart of Antares helps us to find and identify Scorpius. The line from Nunki to Kaus Media represents the arrow, the head of which points to Antares in Scorpio which is to the right of Sagittarius.  It also helps to remember that the orange star Kaus Media, points towards the red star Antares .  The bright red glow of Antares further helps us to identify Scorpius. The angular distance from Kaus Australis in Sagittarius to Shaula in Scorpio is approximately 10o  or roughly equivalent to the width of the palm of the hand when held at arms length.

Sagitarius to Scorpio

Notes.    Star maps can be very confusing because they are not drawn in the conventional way with east on the right and west on the left.  They are drawn as they would appear in the sky if we were lying down with our legs pointing to the south and looking upwards so that east would be on the left and west on our right.

Although this series of posts show the locations of certain constellations relative to other constellations in the sky, it does not necessarily indicate whether or not a certain constellation  will be visible above the horizon.  That will of course depend on its times of rising and setting at the  position of the observer.  ‘Risings and Settings’ will be the subject of another post.

Watch out for the next post in this series which will be issued shortly.


Astro Navigation Demystified

Finding Stars and Constellations Part I

Finding Stars and Constellations Part II

Rising and setting times of stars.

Latitude from Polaris

Astro Navigation Demystified on Amazon

The Application of Mathematics To Astro Navigation on Amazon

About Astro Navigation Demystified



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What’s The Point Of Calculating Our Latitude From The Altitude Of The Sun At Midday?

cover with Jester in useReferring to my earlier post ‘Calculating Latitude from the Midday Altitude of the Sun’, the question has been raised “what is the point of this if you already know your latitude”?  The answer to this question is quite simple, if we are using astro navigation at sea without reliance on GPS, we never know what our exact position is and the midday altitude gives us a handy check on our DR position.

As explained in the link above, to calculate latitude from the midday Sun, we need to know three things: the Sun’s declination, an accurate altitude reading and our approximate latitude.  Armed with this data, we can calculate a reasonably accurate latitude. (Note.  The only reasons that we need to know the approximate latitude is to enable us to know whether it is greater or less than the declination as explained in this link).

 Even in the best circumstances, the accuracy of astro navigation is only ±1 minute of arc or 1 nautical mile.  This level of accuracy may seem to be unacceptable in the light of modern electronic navigation systems such as GPS but before the advent of such systems, astro is all we had to rely upon.  Click here to learn more about the accuracy of astro navigation.

 It might be said “now that we have GPS, what’s the point of an inaccurate method such as astro navigation”?  The answer to that question is that we could lose GPS at any time through a variety of causes such as solar storms, cyber attacks, power failures, system failures, and so on (see my post ‘Could the Global Positioning System Fail’).  On the other hand, the Sun, Moon, stars and planets will always be there and so we will always be able to use astro navigation.

 Furthermore, in certain situations, it might not be advisable to use GPS anyway; just as a mobile phone can give our position away so can a GPS device.  GPS relies on two way communication with a satellite which means that your GPS device transmits radio signals and these can be easily detected with modern sophisticated electronic warfare systems.  So for special forces and those in ‘escape and evasion’ scenarios, it might be advisable to find another way of navigating (see my post ‘Astro Navigation in a Survival Situation’

It was not until John Harrison invented the chronometer in the 18th century that navigators had a method of calculating longitude). Before this time, navigators relied on dead reckoning to give them an approximate longitude.  However, they would have been able to calculate their latitude from the midday altitude of the Sun and they would able to use this to revise their DR position. To calculate latitude from the midday Sun, they would need to know two things, their approximate latitude and the declination of the Sun.  They would take their approximate latitude from the DR position and they would have had tables listing the Sun’s declination for each day.  So, the whole point of  knowing the approximate latitude was to enable them to calculate a more accurate one. 

 If we use astro navigation today, we ideally need to be able to take star and planet sightings during  nautical twilight.  If however, the sky is covered during these times, then we have to rely on the Sun and the Moon when they are visible.  In these circumstances, a midday sighting of the Sun is invaluable because not only does it give us our latitude, we can also use it to calculate our longitude as long as we have a chronometer.  When the Sun reaches its highest altitude from our position, we know that it is exactly over our meridian of longitude and so we know that, at that point, it is noon (local time) at our position.  A chronometer will give us GMT, so if we know when it is noon at our position, we can calculate our longitude.  In conclusion, the midday Sun enables us to check our DR position in terms of both latitude and longitude.

Midday Sun in Survival Situations.  In certain circumstances such as survival situations, we would probably not have access to declination tables but there is a ‘rule of thumb’ method which enables us to calculate the approximate declination for any day without having to rely on tables. 

Please note that this method is not very accurate and you are only be likely to use it in a survival situation.  We know what the Sun’s declination will be on four days of the year and with this simple knowledge, we can calculate an approximate value for declination which will help us to calculate our approximate latitude: 

At the Vernal Equinox (March 20/21) and at the Autumnal Equinox (September 22/23)  when the Sun is above the Equator, its declination will be 0o

At the Summer Solstice (June 20/21) when the Sun has reached the northerly limit of its path, its declination will be 23.5o north.

At the Winter Solstice (December 21/22) when the Sun has reached the southerly limit of its path, its declination will be 23.5o south.

Between these dates it moves north or south accordingly at an average rate of approximately 0.35o per day.

Armed with this information, we can calculate the Sun’s approximate declination for any day of the year.

For example, April 15 is 25 days after the Vernal Equinox so the declination on that day will be: 0o plus (25 x 0.35o) north = 8.75o north (approx.).

October 15 is 23 days after the Autumnal Equinox so the declination on that day will be:  0o plus (23 x 0.35o) south  =  8.05o south (approx).

If you check these answers with the Survival Declination Table, you will see that they are accurate to within one degree.

To learn how to calculate your lat and long in a survival situation, go to these links:  Lat from Sun    Lat from Polaris     Long.

To return to the original question, as long as we have an approximate latitude and an approximate declination to work on, we can  calculate our latitude from the midday Sun.  The whole point of doing this is to enable us to check our DR position and revise it if necessary.


Astro Navigation Demystified

What is Astro Navigation?

Accuracy of Astro Navigation – U.S. Navy

Astro Navigation in a Survival Situation

The Astronomical Position Line

Altitude and Azimuth

Altitude and Zenith Distance

Calculating Latitude from the Midday Sun

Calculating Longitude

Relationship Between Latitude, Longitude  & the Nautical Mile

Dead Reckoning

Lunar Distance

Planning Star and Planet Observations

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Finding Stars and Constellations Part II

Continuing the series ‘Finding Stars and Constellations’

combined with andromeda coloured mark 2

Cassiopeia.  This constellation is named after the vain queen Cassiopeia in Greek mythology.

The five stars in the constellation Cassiopeia appear in the celestial sphere in the shape of the letter W and can be observed in the northern hemisphere and down to 20oS.

Star 5 of Cassiopeia can be located along a line of reference from the Pole Star at an angle of 135o to the line of pointers as the diagram above shows.   As Ursa Major revolves around the Pole Star in an anti-clockwise direction, so do the five stars of Cassiopeia but star 5 keeps its position 135o from the line of pointers.  The angular distance of star 5 from the Pole Star is 30o or roughly one and a half hand-spans.

The brightest star in the constellation Cassiopeia  is Alpha Cassiopeiae otherwise known as Schedir which is a navigation star.

 Perseus. This constellation is named after the Greek mythological hero Perseus.

 If a line is drawn from star 3 to star 4 of Cassiopeia, it will point almost directly towards star 3 of the constellation Perseus at about one hand-span as shown in the diagram.

Perseus can be easily seen in the northern hemisphere during the winter months and in the northern areas of the southern hemisphere north of 35oS. during the summer.  Even though Perseus’ stars are bright relative to other constellations, Mirfak is its only navigation star.

Andromeda.  The diagram shows that, if the line from the Pole Star to star 5 of Cassiopeia is extended, it will point to star 3 of Andromeda at about one and a half hand-spans and that a line from the Pole Star through star 1 of Cassiopeia will point to star 1 of Andromeda. The diagram also shows that the 4 primary stars of the constellation Andromeda, when lined up, point to star 1 of Perseus.

The constellation Andromeda is named after Andromeda, the wife of Perseus in Greek Mythology and can be seen in the northern hemisphere and in the southern hemisphere as far south as 40oS.

The brightest star in Andromeda is Alpheratz which is a navigation star.

Watch out for the next post in this series which will be issued shortly.

Links: Finding Stars and Constellations

Rising and setting times of stars.

Latitude from Polaris

Astro Navigation Demystified on Amazon

The Application of Mathematics To Astro Navigation on Amazon

About Astro Navigation Demystified


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Finding Stars and Constellations

Astro Navigation Demystified Full Edition

The usual practice for distinguishing the more conspicuous constellations is to establish reference lines in a known constellation and from these to memorize the directions in which other constellations lie on the celestial sphere. For example, one of the most well known constellations isUrsa Major which is otherwise known by various names such as the Great Bear, the Big Dipper or the Plough.

ursa major

We know that Ursa Major contains a reference line known as the line of pointers.  In the next diagram, you will see that the line joining star 2 to star 1 forms the line of pointers which points to the Pole Star, otherwise known as Polaris.  Polaris is the brightest star in the constellation known as the Little Bear or Little Dipper.

ursa major and minor

Of course, it is not sufficient to know only the direction in which a constellation lies with respect to another constellation; it is also necessary to know the distance.  This presents us with a problem: whereas we measure distances on the surface of the Earth in miles or kilometres, in space these units of measurement have no significance. Instead, to measure distances between objects in space, we use degrees or angular distance as we say.

The following rules provide crude but surprisingly accurate methods of estimating the distance between two bodies:

  • The width of the first finger when held at arm’s length from the body subtends an angle of approximately 1 degree at the eye of the observer.
  • The width of the tips of the first three fingers subtends an angle of roughly 5 degrees.
  • The width of the palm of the hand will subtend an angle of approximately 10 degees.
  • The span of the hand will subtend an angle of roughly 20 degrees.

For example, the angular distance between Star 1 and Star 2  in Ursa Major is 6 degrees or roughly the width of three finger tips when held at arm’s length and the distance between Star 1 and the Pole Star is 28 degrees or roughly three palm widths at arms length.

When the line of pointers in Ursa Major is produced in the opposite direction to the Pole Star, they point to the constellation Leo or the Lion

ursa major to leo

Regulus, star1 and Denebola, star 2, which are the brightest two stars in the constellation Leo are about 25 degrees apart which is roughly a hand span at arm’s length.

The line joining star 4 to star 2 of Ursa Major points to the stars Castor and Pollux which form the eyes of the Heavenly Twins, otherwise known as the constellation Gemini.  Castor, star 1 of Gemini, can be found at 43 degrees (approximately 2 hand spans) from star 2 of Ursa Major.



The Southern Cross,  or Crux, is one of the best known constellations in the southern hemisphere, and is easily recognizable for the cross-shaped asterism formed by its four brightest stars. Crux (Latin for cross) is not visible north of +20° in the northern hemisphere but it is circumpolar south of 34° in the southern hemisphere which means that it never sets below the horizon there.


The cross  has four main stars marking the tips (alpha, beta, gamma and delta).

Alpha Crucis is also known as Acrux (a contraction of ‘alpha’ and ‘Crux’).  This is the brightest star in the constellation Crux.

Beta Crucis, also known as Mimosa or Becrux is the second brightest star in the constellation.

Gamma Crucis or Gacrux, is the third brightest star in Crux.

Delta Crucis or Palida is the fourth star and has variable levels of brightness .

The Pointers.  The constellation Centaurus contains two bright stars: alpha Centauri also known as Rigil Kentaurus and beta Centauri also known as Hadar, which we use as pointers to the Southern Cross.

How to find the direction of south by using the Southern Cross.  There are several methods of doing this but the simplest is as follows

  1. Make an imaginary line between Gacrux and Acrux.
  2. Extend this line from Acrux (the brightest star) for 4.5 times the length of the Southern Cross, as shown in the diagram. This will take you to the position of the South Celestial Pole in the sky.
  3. From the South Celestial Pole, drop a line down to the horizon. Where this line touches the horizon is the direction of south.


Links: Rising and setting times of stars.

Latitude from Polaris

Astro Navigation Demystified on Amazon

The Application of Mathematics To Astro Navigation on Amazon

About Astro Navigation Demystified

Methods of distinguishing other constellations will be the subject of future posts.

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Rising and Setting Times of Stars.

Astro Navigation Demystified Full EditionIt is very useful to be able to calculate the times at which the principal stars and constellations will rise in the east.  It is also helpful to be able to predict the approximate position of the star or constellation in question.  Professional astronomers and navigators have an array of tools such as almanacs, star-globes, computer software and the like at their disposal to help them make their predictions but such tools are not always available to all of us. Fortunately; however, there is a simple way of making these predictions without reliance upon published tables of data and other paraphernalia.

The Earth revolves from west to east about its axis of rotation which is a line joining the north and south poles and if this axis is produced far enough, it would cut the celestial sphere at a point marked by the North Star or Polaris, as shown in the diagram below.   Facing north on the Earth, the Pole Star appears stationary, and the other stars appear to rotate from east to west around the Pole Star although in fact the positions of the stars are fixed and it is the Earth which is revolving from west to east.

pole star


The time taken for a star to complete a circuit around the Pole Star is called a star’s day or sidereal day.  If the sidereal day were to be exactly 24 hours then the stars would rise and set at the same times every day.  However, the Earth completes each rotation about its axis in 23 hours and 56 minutes and so the stars will take the same amount of time to circuit the Pole Star and that is the length of the sidereal day. Therefore, if a star rises in the east at a certain time on a certain day, it will next do so  23 hours and 56 minutes later. In other words, the star in question will rise 4 minutes earlier the next day and this fact gives us the clue to the method of predicting the position and time of star rising and setting times.

Arcturus is the brightest star in the northern celestial hemisphere and so we shall use it for this demonstration.  Whereas the Pole Star is always above the horizon in the northern hemisphere, Acturus which is further south, will be below the horizon for part of its circuit of the Pole Star and so it will rise and set to observers in most parts of Europe and North America.  Whether we wish to observe Arcturus for navigation purposes or for ‘star gazing’ we need to know when it will be above the horizon and so we need to be able to calculate its rising and setting times.

As discussed above, professionals will have access to sets of published tables which tell them the times of rising and setting of all the major stars but we can manage without such tables.  If we note the time of rising of Arcturus on just one day, we can easily calculate the time that it will rise on all following days as the following example shows.

Say that Arcturus rises at 18.00 on a certain day; we know that it will rise again 23 hrs. and 56 mins. later so we can easily calculate that it will rise at 17.56 the next day (4 minutes earlier).  In 30 days time, we can calculate that it will rise 30 x 4 = 120 minutes or 2 hours earlier; that is at 1600.

It is easy to see that if we repeat this process for both rising and setting times for all the stars that we are interested in, we can easily construct our own table of data.

Part 2 of this series (Finding the Positions of Stars and Constellations) will be published in the next post.

Links:   Finding stars and constellations

How to Locate Polaris.

Calculating Latitude from Polaris.

A fuller explanation of this topic is given in ‘Astro Navigation Demystified’

Astro Navigation Demystified on Amazon


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Distance To The Horizon

Astro Navigation Demystified

Pythagoras provides us with a method calculating the distance to the horizon.

In the diagram below,

A represents a ship’s position.

B is the top of the lighthouse.

Bis the base of the lighthouse.

C is the Earth’s centre.

r is the radius of the Earth.

h is the height of the lighthouse.



By Pythagoras:

AB2  =  CB2 – AC2

→      AB2  =  (r + h)- r2

→      AB2  =  2rh + h2

→      AB   =  √(2rh + h2)

(Note. hcan be discounted since r is many, many times greater than h).

So the formula for calculating the distance from the top of an object to the horizon becomes:  √(2rh)

 Example.  Calculating the distance to an object on the horizon.

In the following example, we will imagine that the Pharos Lighthouse (ht. 120m.) is the object in question.

Note.  Since the heights of objects marked on modern navigational charts are usually given in metres, it would be easier to calculate the distance in kilometres instead of nautical miles.

Since we know that the mean radius of the Earth is 6367.45 Km. and the height of the Pharos light was 120m. we can put these values into the formula √(2rh) as follows:

Distance =√ (2×6367.45 x 120/1000)  (divide h by 1000 to convert to Km.)

= √(12734.9 x 0.12)

= √1528.18

= 39.09 Km.

To convert to nautical miles:  1 nautical mile = 18.5 Km. so, if we want to convert the answer to nautical miles, we must divide by 18.5.  (In fact, it is easier to multiply by 0.54 which is the inverse of 18.5).

So the answer to the above example in nautical miles becomes: 39.09 x 0.54 = 21.1 n.m.

 Therefore, assuming the observer is at sea level, the top of the lighthouse would become visible at a distance of 39.09 Km. or 21.1 n.m.

Summary of Formulas for finding the distance from the top of an object to the horizon, :

D =  √(2rh)

D = √(12734.9 x h/1000) Km.

D = √(12734.9 x h/1000) x 0.54 n.m.

What about the height of eye?   The above calculation was made with the assumption that the observer’s eye is at sea level.  Of course, the height of the observer’s eye will be several metres above sea level (say 3 – 4 m. in a yacht or small fishing vessel).  So the height of eye must obviously be taken into account and the problem now becomes as shown in the following diagram:

A represents the observer,

B represents the top of the lighthouse,

T represents the horizon,

h1 represents the observer’s height of eye (say 4m.)

h2 represents the height of the lighthouse (120m.).

So, we have  AB  =  AT + BT

To calculate the value of AB, we must first calculate the values of AT and BT.

Using the formula that we previously established for finding the distance from the top of an object to the horizon, we have:

AB  = √(2rh1) + √(2rh2)

Given that r = 6367.45 Km.   h= 4m.   and h= 120m.

we have:

D = (2 x 6367.45 x h1/1000) + (2 x 6367.45 x h2/1000)  Km.

(12734.9 x 4/1000) + (12734.9 x 120/1000)  Km.

(12734.9 x 0.004) + (12734.9 x 0.12)  Km.

= 7.1 + 39.09 Km.

=  46.19 Km.

To convert to nautical miles, we have:

D   = 46.19 x 0.54 = 24.9 n.m.

So we can see that taking the observer’s height of eye into account makes quite a difference to the result.

Summary of formula for calculating the distance that an object will appear on the horizon taking into account height of eye:

D = √(2rh1) + √(2rh2)

D = (12734.9 x h1/1000) + (12734.9 x h2/1000)  Km.

Or   D = (12734.9 x h1/1000) + (12734.9 x h2/1000)  x 0.54 n.m.

 What about the height of tide?

The height of objects shown on navigational charts is the height above the level for ‘mean high water springs’ (MHWS) which is used as the chart datum for heights.  When navigating in areas where there is a large tidal range, the height of tide in relation to MHWS must be taken into account if the charted height of the object is being used in navigational calculations.  For example, in the vicinity of the Channel Islands, the height of tide at low water springs can be in the region of 7.5 metres below the level for MHWS and in such circumstances, 7.5 metres would obviously need to be added to the charted heights of objects.

Revise Pythagoras

A fuller explanation of this topic is given in ‘Astro Navigation Demystified’

Astro Navigation Demystified on Amazon


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