Celestial Navigation in 60 mins Introduction
The principles of celestial navigation are simple. The harder mathematics come in details of calculations, and a com like ASNAv will take care of this for you, so don't be afraid to read this page. It's pleasant to understand how it is po position on Earth just be looking at a few stars… The principles of celestial navigation are simple. The harder mathematics come in details of calculations, and a com like ASNAv will take care of this for you, so don't be afraid to read this page. It's pleasant to understand how it is po position on Earth just be looking at a few stars…
The celestial mechanics is precision mechanics. It is possible to calculate the exact position of a heavenly body (star sun) in the sky at any given time. Knowing the position of the star in the sky, the measure of the angle between the h observer and the star, using a sextant, is enough to determine the observer position in latitude and longitude (in fact, at least two measures are needed).
The celestial mechanics is precision mechanics. It is possible to calculate the exact position of a heavenly body (star sun) in the sky at any given time. Knowing the position of the star in the sky, the measure of the angle between the h observer and the star, using a sextant, is enough to determine the observer position in latitude and longitude (in fact, at least two measures are needed).
Let's show this by the example of another situa imagine you observe a lighthouse from a certai the sextant, you measure the angle alpha corres height of the lighthouse seen from your positio If you know the height h, you can find your dis lighthouse. On a chart, you can draw a circle ce lighthouse with a radius d. You are somewhere This is your circle of position.
A second 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 your estimated position or a third observation will help you to choose you don't observe a lighthouse but the angle between your horizon and a star, you are doing celestial navigation. Tha 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 your estimated position or a third observation will help you to choose you don't observe a lighthouse but the angle between your horizon and a star, you are doing celestial navigation. Tha
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 is linking 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. This is a representation of the universe, but this is enough for our purposes. We are just poor seamen (correction, I am). The c centred on Earth with the celestial equator passing through the Earth equator and the axis Earth centre C to North Po axis of reference of the celestial sphere. A plan of reference defined on Earth is also used on the celestial sphere: the 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 observer on the Earth. The point Z on the celestial sphere, which is directly above the head of the observ zenith. The distance on the celestial sphere between the zenith Z and the star S is the zenith distance z. The zenith di
distance d from the lighthouse of the previous example. We can draw a circle centred on the star S with a radius z. W on the projection of this circle on the Earth. This is our new circle of position. Let's put an observer on the Earth. the celestial sphere, which is directly above the head of the observer, is called his zenith. The distance on the celestia the zenith Z and the star S is the zenith distance z. The zenith distance is like the distance d from the lighthouse of th example. We can draw a circle centred on the star S with a radius z. We are somewhere on the projection of this cir This 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 itself on the are 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 itself on the are 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 (90°N - 90°S) and its Greenwich 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). Knowing de measuring the zenith distance z, we can find our circle of position and finally our position. We can find both values in the Nautical Almanac (or with the kind help of ASNAv). Knowing delta & GHA and measuring the zenith distan 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 and polar distance Delta (90° - delta). Take the prev wipe-off the surplus. Note the latitude and co-latitude (90° - lat), the declination delta and polar distance Delta (90° - delta).
The figure is using the equatorial coordinate system (the reference is the celestial equator). But what we need in fa to our local coordinate system: the horizontal coordinate system (the reference is the local apparent horizon).
The figure is using the equatorial coordinate system (the reference is the celestial equator). But what we need in fa to our local coordinate system: the horizontal coordinate system (the reference is the local apparent horizon).
At our position (Observer) on the Earth, we can imagine the plan 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 as reference, we get the next drawing.
The
reference
is
now
the
horizontal
coordinate
When we observe a star from our local horizon, we can define its position in the sky by its Azimuth (0° - 360°) (0° - 90°). The reference is now the horizontal coordinate
When we observe a star from our local horizon, we can define its position in the sky by its Azimuth (0° - 360°) (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.
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 (declination delta, G Angle GHA). Thanks to our local observation of the star, we can measure its altitude and deduce its zenith distance 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 delta, G Angle GHA). Thanks to our local observation of the star, we can measure its altitude and deduce its zenith distance 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 delta, G Angle GHA). Thanks to our local observation of the star, we can measure its altitude and deduce its zenith distance
We just need to find a mathematical relation between what we know (delta, GHA, h) and what we are looking for (l Gw ) and we can solve the problem of the celestial navigation! We just need to find a mathematical relation between (delta, GHA, h) and what we are looking for (latitude, longitude Gw ) and we can solve the problem of the celestial n just need to find a mathematical relation between what we know (delta, GHA, h) and what we are looking for (latitu and 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 problem The 3 sides of the triangle are: col, the co-latitude (90° - latitude); z, the zenith distance (90° - altitude h); Delta, (90° - declination delta).
The angle of the triangle opposite to the side z is called the polar angle P (180°W - 180°E). This angle is also the a D at the celestial Here: 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 plane g is the one in case of spherical geometry?
The triangle in plane geometry, for old times' sake:
The cosinus rule for the spherical geometry in the general case.
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:
For a no human b solving celestial navigati equation will tak
To check manually the results of ASNAv, we can use the traditional method of the Lines of Position (LOPs). However, this traditional method is not able to correct errors in the estimated course and speed. See an example of th skills to understand the differences between the ASNAv method and the traditional method. To check manually the ASNAv, we can use the traditional method of the Lines of Position (LOPs). However, this traditional method is not able to correct errors in the estimated course and speed. See an example of th skills to understand the differences between the ASNAv method and the traditional method.
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:
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:
O is the observer true (astronomic) position. Note: ITC is used in this text as foreshortening for InTerCept 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:
Most often, however, you will get LOPs intersecting like this:
This triangle is known as a cocked hat hat of the times
after its resemblance to the common three-cornered
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:
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 be eliminated by taking the cocked hat centre as True Position ONLY IF the observations azimuths are spread over more than 180°.
If the observations azimuths are not spread over 180°, the True Position is NOT the Cocked Hat Centre.
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 sense Random errors are inevitable... We can find in the Bowditch chapter 16 (Nathaniel Bowditch, The American Practical Navigator, an Epitome of Navigation, pub. n°9 NIMA, USA, 1995) 19 possible errors when observing the celestial body height and 30 possible errors until the LOP can be drawn 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,
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 each occurrence 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: •
we know that we are in 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: •
we know that we are in 1 + area
•
flip the coin, there is 1 chance among 2 to be in a - area
•
flip the coin again, there is also 1 chance among 2 to be in a second - area
•
so there is 1/2 x 1/2 = 1/4 chance (25%) to be in a --+ zone
•
as there are 3 --+ zones, 25% divided by 3 = 8.33% probability
So by 1, 2 and 3, we can draw:
Conclusion The True Position is the cocked hat centre if there are no random errors and the observations azimuths are spread over more than 180°. If there are random errors, then the most probable position (MPP) is inside the cocked hat but with a probability of only 25%. In plain English, this means that there is a 75% probability for True Position to be outside the cocked hat...
Random errors - a solution Random errors are inevitable. With random errors, there is a 75% probability for True Position to be outside the cocked hat. How to trust the ship position drawn on the chart, then? More helpful than the cocked hat or the MPP (Most Probable Position) by itself is the confidence ellipse. The confidence ellipse defines the area within which the True Position lies with a given probability (95% or 99% for instance). A statistical analysis is needed to be able to draw this ellipse. Confidence ellipse characteristics: •
its centre is the MPP
•
its size depends on the size of the random errors and on the chosen probability
•
its shape depends on the number of observations and distribution of the azimuths.
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).
