Once upon a time people thought that the earth was the centre of the universe and that the sun and stars all revolved around us (sounds a bit like one of my ex-boyfriends).

These ancient people started mapping the sun and the stars on to a sphere – the Celestial Sphere – basically it is a concept.

So imagine the Earth is a small ball, and then imagine we are hovering inside a big balloon. Now imagine that all the stars are painted on the inside of the balloon – that balloon is the Celestial Sphere.

It is much easier to imagine the stars mapped like this – than a billion stars in an infinite universe.

Here on earth we have got a grid pattern to identify whereabouts we are on the globe – this is called Latitude and Longitude. The Celestial Sphere – or the inside of that balloon also has a grid which maps the location of all the Celestial Bodies. Instead of being called Latitude and Longitude the grid is called Declination and Hour Angle. The Sphere also has a celestial equator and celestial poles.

So let’s regroup for a minute:

- Declination – is equivalent to parallel lines of Latitude
- Greenwich Hour Angle (GHA) – is equivalent to meridians of Longitude
- Celestial Equator – is equivalent to the earth’s equator
- Celestial Poles – are equivalent to the earth’s poles

Are you with me so far?

There has to be exceptions (of course…) With Longitude on earth we measure that in degrees East or West of Greenwich – up to 180º in each direction. Whereas with Greenwich Hour Angle (GHA), we measure that westward to 360º – from the Prime Meridian.

You can work out the difference between Longitude and Greenwich Hour Angle as follows:

GHA = Longitude W or 360º – Longitude E

*(Don’t worry if you don’t understand this just yet – I am hoping it will all become clear soon!)*

Stop for a minute, go outside and look straight up in to the sky – directly above your head, the area on the celestial sphere directly above you is called the **Zenith – z** – and the area where you are standing is called the **Ground Position.**

Likewise the point directly beneath a star on the Earth’s surface is the Ground Position of that star.

When we use our Sextant to measure the angle between the horizon and a star, we are measuring it’s **Altitude.**

When we note the exact UTC time that we take that measurement, we can work out the position of that star – or it’s ground position in that moment of time.

Our aim is to measure the shortest distance between the ground position of the star and the ground position of the observer.

Now if you paid more attention than I did in high school maths, then you might want to do some more research on all the geometry behind this concept.

But very basically if you look at the diagram above you can see that the angle between the Zenith of the observer and the Horizon is 90º – ok got that.

We then measure with our sextant that the angle between the horizon and the star is say 40º. We then deduct that from 90º to obtain the missing angle between the star and our Zenith – which in this case must be 50º.

By using tables we can figure out the ground position of that star at that particular time. If we set our dividers to a distance of 50º *(more info on how to do that later)* and draw a circle around the ground position of the star – we have got a position circle – we are somewhere on that line.

The trouble is that this circle could be a huge diameter and if our position could be anywhere on there it isn’t a particularly accurate fix! With one sight we can obtain a position line but to get a fix we need an intersection of two or more lines as per the diagram above – this can come from two or more different celestial bodies at the same time, or the same body at different times.

**Quick Quiz**

- What is a Zenith?
- What is Declination?
- What is an Altitude?
- How do you work out the difference between the Altitude and the observer’s zenith?
- What does GHA stand for?
- Describe the celestial sphere
- What is the equivalent of Latitude on the Celestial sphere?
- What is the difference between Longitude and GHA?

Tune in next time when we go back to basics with plotting positions, dead reckoning, position circles and all that kind of thing.

If you want to revise the Sextant post – click here

If you want to swot up about Time and why that is so important – click here

Thanks for following! Please comment, make suggestions and ask questions in the section below. 🙂

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If the ground position of the star is known (from the celestial sphere / tables), then why not just take an equally accurate compass bearing to the star (which will be the same as to its ground position) as this singularly intersects the circle (obtained from the angle between it and the observer’s zenith) at the observer’s ground point ?

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Good question…! Does that work? I guess it would?

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I don’t know, as I’m here on your website to learn. ( Thank you 😉 ) Tbh., I’ve only handled a sextant once, many, many years ago, and I don’t remember seeing a compass built into it. It just seemed logical when I looked at your diagram, as it would be very much more accurate to use a radial line (intersecting your circle at 90 degrees) than two other arcs whose angle of crossing is so small.

I haven’t yet read all your pages on the subject, so I don’t yet know how your position is worked out when a sighting is just taken on the sun ..as no other celestial bodies can be seen.? It’s time for me to do some more study within your pages ..to try and find the answer to that one :))

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Lol! I’ll do some research too. I hope you enjoy my notes. I’m definitely not an expert on celestial nav. Just a very keen student who is always learning! I’m heading away racing this weekend, but will dig out my books when I get back and ponder your question!

Cheers Viki 🙂

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There is no possibility of “an equally accurate compass bearing” when compared to the radius of the circle (the observer’s zenith distance) Compasses are accurate to ±½° at best while a sextant is accurate to ±1/2 minute of arc or better. an example is that you could know the radius of the circle to an accuracy of ±1/2 nm, but the position along the circumference of the circle to ±30nm. A 60nm range in position is just not good enough. That is why we shoot multiple stars so that the 1 nm thick circles can intersect.

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Thank you

I’m in the middle of an Astro navigation course and was struggling with the EP CP calculations.

Your clear instructions was very helpful 👍

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Thanks Keith! 🙂

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what is observer’s zenith?

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I think it’s the point in the sky directly overhead? Could be wrong? It’s been a while since I did my study. What do you think?

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