Celestial Navigation in 60 Min

8/3/2019 Celestial Navigation in 60 Min

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Let's show this by the example of anot

imagine you observe a lighthouse from

the sextant, you measure the angle alphheight of the lighthouse seen from you

If you know the height h, you can find

lighthouse. On a chart, you can draw alighthouse with a radius d. You are som

This is yourcircle of position.

A second observation gives you a second circle of position. You are at the intersection of the circles of positIn fact, there is most often 2 intersections but yourestimated position or a third observation will help you to

you don't observe a lighthouse but the angle between your horizon and a star, you are doing celestial naviga

observation gives you a second circle of position. You are at the intersection of the circles of position.In fact, there is most often 2 intersections but yourestimated position or a third observation will help you to

you don't observe a lighthouse but the angle between your horizon and a star, you are doing celestial naviga


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Of course, at this stage we need to look a little bit closer at the celestial mechanics to

understand how we can calculate the exact position of a heavenly body (star, planet,

moon, sun) in your local sky at any given time and which mathematical relation islinking the altitude of a body to a circle of position. Unfortunately, it is not as simple

as tan(alpha) = h / d.

Celestial mechanics

Imagine the Earth in space surrounded by a celestial sphere on which all the heavenly bodies are moving. Trepresentation of the universe, but this is enough for our purposes. We are just poor seamen (correction, I amcentred on Earth with the celestial equator passing through the Earth equator and the axis Earth centre C to N

axis of reference of the celestial sphere. A plan of reference defined on Earth is also used on the celestial sp

meridian. On the celestial sphere, we show a star S and its hour circle.

The star in the sky is like the lighthouse of your previous example.

Let's put an observeron the Earth. The point Z on the celestial sphere, which is directly above the head of th

zenith. The distance on the celestial sphere between the zenith Z and the star S is the zenith distance z. The


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distance d from the lighthouse of the previous example. We can draw a circle centred on the star S with a rad

on the projection of this circle on the Earth. This is our new circle of position. Let's put an observeron th

the celestial sphere, which is directly above the head of the observer, is called his zenith. The distance on ththe zenith Z and the star S is the zenith distance z. The zenith distance is like the distance d from the lightho

example. We can draw a circle centred on the star S with a radius z. We are somewhere on the projection oThis is our new circle of position.

The position of the projection of the star on the Earth (latitude, longitude) and the position of the star S itselare identical (same angles, same reference plans: the celestial equator and the Greenwich meridian).

The position of the projection of the star on the Earth (latitude, longitude) and the position of the star S itselare identical (same angles, same reference plans: the celestial equator and the Greenwich meridian).

The position of the star on the celestial sphere is

given by its declination delta (90N - 90S) and itsGreenwich Hour Angle (GHA, 0 - 360).

We can find both values for any given time in the Nautical Almanac (or with the kind help of ASNAv). Kno

measuring the zenith distance z, we can find our circle of position and finally our position. We can find bothin the Nautical Almanac (or with the kind help of ASNAv). Knowing delta & GHA and measuring the zenit

our circle of position and finally our position.

Well, very nice... :-) Let's enter into the details now. :-(

Well, very nice... :-) Let's enter into the details now. :-(

Take the previous figure and wipe-off the surplus.Note the latitude and co-latitude (90 - lat), the declination delta andpolar distance Delta (90 - delta). Take

wipe-off the surplus.

Note the latitude and co-latitude (90 - lat), the declination delta andpolar distance Delta (90 - delta).

The figure is using the equatorial coordinate system (the reference is the celestial equator). But what we n

to our local coordinate system: the horizontal coordinate system (the reference is the local apparent horizo

The figure is using the equatorial coordinate system (the reference is the celestial equator). But what we n

to our local coordinate system: the horizontal coordinate system (the reference is the local apparent horizo


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At our position (Observer) on the Earth, we canimagine theplan of our horizon with the 4 cardinal

points. If we move this plan to the centre of the

Earth and redraw the figure using this plan asreference, we get the next drawing.

The reference is now the horizontal coor

When we observe a star from our local horizon, we can define its position in the sky by its Azimuth (0

(0 - 90).The reference is now the horizontal coor

When we observe a star from our local horizon, we can define its position in the sky by its Azimuth (0(0 - 90).

The zenith distance z is 90 - h.

In practice, we are only measuring the altitude h of the star, its azimuth is calculated.

The zenith distance z is 90 - h.

In practice, we are only measuring the altitude h of the star, its azimuth is calculated.


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Aha! Interesting. We already seen that our circle of position is centred on (delta, GHA) with a radius z.Thanks to the Nautical Almanac, we can define the position of the star S on the celestial sphere (declinationAngleGHA). Thanks to our local observation of the star, we can measure its altitude and deduce its zenith d

Interesting. We already seen that our circle of position is centred on (delta, GHA) with a radius z.

Thanks to the Nautical Almanac, we can define the position of the star S on the celestial sphere (declination

AngleGHA). Thanks to our local observation of the star, we can measure its altitude and deduce its zenith dInteresting. We already seen that our circle of position is centred on (delta, GHA) with a radius z.

Thanks to the Nautical Almanac, we can define the position of the star S on the celestial sphere (declination

AngleGHA). Thanks to our local observation of the star, we can measure its altitude and deduce its zenith d


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We just need to find a mathematical relation between what we know (delta, GHA, h) and what we are looki

Gw ) and we can solve the problem of the celestial navigation! We just need to find a mathematical relation(delta, GHA, h) and what we are looking for (latitude, longitude Gw ) and we can solve the problem of the c

just need to find a mathematical relation between what we know (delta, GHA, h) and what we are looking foand we can solve the problem of the celestial navigation!

The hatched triangle on the top of the celestial sphere is the one we will use to solve the celestial navigation

The 3 sides of the triangle are: col, the co-latitude (90 - latitude); z, the zenith distance (90 - altitude h)(90 - declination delta).

The angle of the triangle opposite to the side z is called the polar angle P (180W - 180E). This angle is al

D at the celestialHere: PE = g w - GHA*

It's a spherical triangle, not a plane triangle. We all remember (aren't we?) the formula to solve a triangle in

is the one in case of spherical geometry?


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The triangle in plane geometry, for old times' sake:

The cosinus rule for the spherical geometry in the general case.


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The application of the general case to our problem:

We found a mathematical relation between what we know (delta, GHA, h) and what we are looking for(latitude, longitude). With 2 observations,

we get a system of 2 equations with 2 unknowns that we are able to solve.

The celestial navigation problem is thus resolved.What? You don't like my equation?

Well, I agree that its resolution is not so simple... For a computer, the process is quite straightforward:

solving the system by an iterative method

using the estimated latitude and longitude as starting values. With more than 2 observations, it's even

possible to improve the traditional method

and to perform a statistical analysis, in other words:

to give a certain weight to each observation according to its reliability in the normal law

model;

to compute and eliminate the possible systematic error of the observer;

to correct the assumed course and speed if enough observations are provided (exactly the

same way the GPS is able to give the

course and speed of the vessel if enough satellites are visible).

To do this, ASNAv is using the least-square method with iterative weighting adjustment by the

Biweight function on a system of equations given

by the differential correction method. Each equation i looks like:


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Traditional method : lines of position (LOPs) and intercepts

This method was invented in 1875 by the admiral Marcq de Saint-Hilaire (some other sources say Y.

Villarcau and A. de Magnac).

The true line of position A, tangent to the circle of position, can be merged into the line of position B

because the estimated position e

is close to the true position O.

On the line of position B, the intercept is the difference between the true (observed) altitude and the

calculated altitude:


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Practically, the procedure is as follow:

1. find the estimated position with an accuracy of 50 nautical miles (in order to get a fix with

1 nautical mile maximum

error due to the method itself);2. observe with the sextant a star altitude Hs at the time C (GMT);

3. correct the sextant altitude Hs with the instrumental error, the dip of the apparent horizon,the terrestrial refraction,

the astronomic refraction, the parallax, the semi-diameter of the star (if needed) to get Ho

(observed altitude);

4. compute the azimuth of the star using the estimated position and the data's of the Nautical

Almanac at the time C;

5. compute the calculated altitude Hc;

6. compute the intercept ITC = Ho - Hc;7. plot the line of bearing (azimuth of the star) from the estimated position;

8. plot the line of position perpendicular to the line of bearing, at a distance ITC from the

estimated position,

away if Hc > Ho;

9. start over again the steps 2-8, at least once, to get the drawing below:


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O is the observer true (astronomic) position.

Note: ITC is used in this text as foreshortening forInTerCept and is not an abbreviation. Don't confuse

the intercept

ITC with ITP - the Intercept Terminal Point.

ITP is the point through which the circle of position passes. The LOP is tangent to the circle of position

at this point.

LOPs, cocked hat and common sense

Common sense is judgment without reflection, shared by an entire class, an entire nation, or the entire

human race(Giambattista Vico (1688-1744), Italian philosopher)

The observer astronomic position O is at the intersection of the 2 LOPs.

With 3 sextant observations, you get 3 LOPs and if you are very good and very lucky at the same time,

you could end up

with LOPs intersecting like this:


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Most often, however, you will get LOPs intersecting like this:

This triangle is known as a cocked hat after its resemblance to the common three-cornered

hat of the times

when these navigation techniques were developed.

Where is exactly the observer astronomic position O?

Well, the common sense is telling us that O is exactly in the middle of the triangle:


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Unfortunately, the common sense is seriously misleading here...

If (and only if) the observations azimuths are spread over more than 180, then the most probable position

(MPP) is inside the cocked hat, but with a probability of only 25%!

Errors... and how to live with them

The reason why the 3 LOPs don't intersect as a point but as a triangle is the observations errors.

The observations errors are:

systematic error

random errors

The systematic error is the algebraic sum of the uncorrected index error of the sextant and the

observer personal error.

If not equal to zero, a personal error shows the observer inclination to always overestimate or

underestimate the stars

altitudes of a definite value.

The random errors depend on the observer experience and the observation conditions (bad horizon,

rolling ship,

abnormal atmospheric refraction, ...).

If you are really experienced (and lucky) and there are no random errors, then the systematic error can b

eliminated by taking the cocked hat centre as True Position ONLY IF the observations azimuths are

spread over more than 180.


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If the observations azimuths are not spread over 180, the True

Position is NOT the Cocked Hat Centre.


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In this second case, to say that the True Position is the cocked hat centre, you need to correct 2 LOPs

by moving them

backwards and 1 LOP by moving it forward.

This is impossible because the systematic error is a constant of the same sign.

We have here an 'outside' fix: the True Position is outside the cocked hat.

Knowing his own personal error (inclination to always overestimate or underestimate the stars altitudes

of a definite value)

is the only way to find the True Position.

If there are random errors (and there will be, no matter how good observer you are), then the situation

is even worse...

Random errors and common senseRandom errors are inevitable... We can find in the Bowditch chapter 16 (Nathaniel Bowditch,

The American Practical Navigator, an Epitome of Navigation, pub. n9 NIMA, USA, 1995)

19 possible errors when observing the celestial body height and 30 possible errors until the LOP can bedrawn on the chart.

In case of random errors, without further statistical analysis, the True Position can be on the left or on

the right of each LOP,


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with an equal probability.

Each LOP divides the world in 2 areas and the True Position has exactly 50% of chance to be in one of

them.

Let's call (arbitrarily) the zone inside the triangle +++. The names of the other zones follow directly:

There is no - - - zone. The True Position just cannot be in the - area of each LOP and the cocked hat

still be shaped

as drawn!So we know that the True Position is in the + area of at least 1 position line.

The True Position can be found by flipping a coin twice (head side is +, reverse side is - and eachoccurrence

has a 50% probability).1) If the True Position is in the +++ zone, it needs to be in 3 + areas:

we know that we are in 1 + area

flip the coin, there is 1 chance among 2 to be in a second + area

flip the coin again, there is also 1 chance among 2 to be in a third + area

so there is 1/2 x 1/2 = 1/4 chance to be in the +++ zone = 25% probability

2) If the True Position is in a -++ zone, it needs to be in 2 + areas and 1 - area:


flip the coin, we are in 1 + area or 1 - area flip the coin again, there is now 1 chance among 2 to be in the remaining + or- area

so there is 1/2 chance (50%) to be in a -++ zone

as there are 3 -++ zones, 50% divided by 3 = 16.67% probability

3) If the True Position is in the --+ zone, it needs to be in 1 + area and 2 - areas:


flip the coin, there is 1 chance among 2 to be in a - area


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The confidence ellipse will normally overlap the cocked hat partly.

Another advantage of the confidence ellipse is that this ellipse can be drawn for any number of LOPs

and

therefore give a visual representation where the cocked hat fails to do so.

See the confidence ellipse

ASNAv is able to draw the confidence ellipse around the MPP. It gives also the radius of the circle

of equivalent probability (as this is easier to plot on the chart).

Celestial Navigation in 60 Min

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