# Calculation of right ascension and declination

If I see an object from Mount Teide (longitude is 16"30'E and latitude is 28"18'N) that passes the meridian (azimuth=0) at 5h (am) UTC, and I also know that the elevation of the star is 43"40', the stellar time at Greenwich at 0h UTC is 22h20min.

How can I calculate the right ascension and declination?

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Have you read this question and its answers yet? If so, is there something still unclear to you which you could state more explicitly? – called2voyage Nov 27 '13 at 17:32
When you say elevation, do you mean altitude? – astromax Nov 27 '13 at 18:10
@astromax I think that the elevation is synonymous with altitude. – Jackson Hart Nov 27 '13 at 20:27
@called2voyage I am very confused with the basic equations in which to use to solve this problem. I do not have a background in this field. I guess the task just seems daunting right off the bat, but I understand that it is just a regular coordinate transformation – Jackson Hart Nov 27 '13 at 20:28
Also, are you using arcminutes/arcseconds to describe latitude and longitude? – astromax Nov 30 '13 at 1:50

The answers to this question are very clear and much better than I could have written.

I'm not sure what you mean by "the stellar time at Greenwich at 0h UTC is 22h20min", but I'm assuming that you mean that that is the amount of time since the culmination of Aries over the Prime Meridian (Greenwich). That being the case, the Right Ascension (RA) of your body can be easily calculated. All we need to do is figure out what RA is passing your meridian at the time in question. This is probably a good time to point out that the longitude of Mt Teide is 16° 30'W not E.

We already know that RA 22H 20m is passing over Greenwich at the time of interest (or, at least that is what I am assuming as your explanation is not at all clear.), as we are to the West of Greenwich, then we know that any RA on our meridian is earlier and we can calculate how much earlier simply by dividing our longitude by 15 (as the earth spins at a nice steady 15°/hour):-

diff in RA = Long/15
= 16° 30'/15
= 1H 06m


Then:-

RA = RA at Greenwich - diff in RA
= 22H 20m - 1H 06m
= 21H 14m


Now for Declination:-

The first stage is to take a piece of ruled or graph paper and near the middle mark a short, horizontal line. This is your horizon, and is marked 'H'. Now draw upwards 9 lines and mark the top ,'Z' and downwards 9 lines and mark the bottom 'N'. You should end up with something that looks like this:-

Which is, basically a picture of your meridian as you sit and gaze south at it. 'Z' is your zenith, the point above you in the sky and 'N' is the nadir, it's opposite out on the celestial sphere through the earth beneath your feet.

Now we need to add a little more detail. Imagine that the celestial sphere is actually a glass sphere surrounding the earth with the stars and stuff stuck to it that we can project anything we want on to (we'll ignore porn and coke adverts for now), where would the celestial equator show and where would it go on our picture of the meridian? Well, it would be a distance above the horizon equal to our latitude (lets round it off to 30°), so we now have:-

Now we put in the body we are looking at. We know it's elevation, just over 43°, we can just call it x as you didn't specify a body. We can also label some of the distances we know, to end up with:-

We already knew that Lat was H -> Q, that's how we place Q, we know that elevation, or altitude is between the horizon and the body and that declination is measured from the celestial equator. We want to find dec and the diagram shows us that:-

dec = Alt - Lat
= 43° 40' - 28° 18'
= 15° 18'N


As 'x' is to the north of 'Q', then Declination is north.

So now we know that we are looking at a body with RA 21H 14m and Dec N 15° 18'.

This is much more complex to explain here than it is to do, or to teach in person. With a bit of practice you'll be doing it in your head. I hope this has helped a bit.

I should point out that I have ignored observation errors, such as height of eye, which may be significant from the place you specify and parallax but I wanted to keep it as simple as possible.

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Your answer is very diligently written, but it only answers half of the question. – Alexey Bobrick Dec 4 '13 at 21:33
Yes, I give a reason for that in the second paragraph. – vascowhite Dec 4 '13 at 22:41
Yes, absolutely, it is fair and I have read it. However, the other half of the question still needs to be answered by someone. – Alexey Bobrick Dec 4 '13 at 23:40
Oh no! Sorry for misunderstanding. I put a downvote on your post, and then I wrote why. I think thew answer is good, but another answer is needed to complement yours. – Alexey Bobrick Dec 5 '13 at 12:57
Oh, I see. Sorry. Obviously your votes are yours to dish out as you see fit, but I feel it is a little unfair to down vote for something that is clearly explained in the answer. Will you also down vote the person who provides the other half of the solution? – vascowhite Dec 5 '13 at 14:20