Thread: Getting confused with astronomy trig

1. Getting confused with astronomy trig

Hi, I'm not sure if this should be in the High School forum, but here goes:

I am currently writing some planetarium software for use on my website. I have calculated the Alt/Az of lots of objects, and used the conversion to spherical polars to plot them onto a cartesian xy plane (the screen!) with r=1. For those of you who are not sure, Alt is the altitude of the object above the horizon (0-90 deg) and Az is the azimuth (0-360 deg) measured East from North. The zenith (directly up) is at (0,0).

This all works fine. My problem is that I want to move my viewpoint around within the 'celestial sphere'. I can zoom in and out with no problem, and I can add small amounts to the the Azimuth of each object in order to rotate the 'sky'. What I cannot do is change the Alt of the point at which I am looking, the default is to look straight up at the zenith.

I assume to accomplish this, I need to add or subtract a certain value from the Alt and Az of each object. I cannot get my head around how to do this.

I have attached an image as an example. The circle shows a bottom-up view of a hemisphere. Firstly my viewpoint is centered on the zenith (0 deg, 0 deg). Here, A is at (20 deg, 0 deg) B is at (20 deg, 90 deg) C is at (20 deg, 180 deg) and D is at (20 deg, 270 deg). If I was to move my viewpoint to X (10 deg, 0 deg), then clearly A would become(10 deg, 0 deg) and C would be (30 deg, 180 deg). How would I find the apparent Alt/Az of B and D, and is there a generalised form of this that I could use using any offset to any initial position?

I might be making this much harder than it actually is, so if you can help or you have any questions, please give me a shout.

2. Originally Posted by djbog Hi, I'm not sure if this should be in the High School forum, but here goes:

I am currently writing some planetarium software for use on my website. I have calculated the Alt/Az of lots of objects, and used the conversion to spherical polars to plot them onto a cartesian xy plane (the screen!) with r=1. For those of you who are not sure, Alt is the altitude of the object above the horizon (0-90 deg) and Az is the azimuth (0-360 deg) measured East from North. The zenith (directly up) is at (0,0).

This all works fine. My problem is that I want to move my viewpoint around within the 'celestial sphere'. I can zoom in and out with no problem, and I can add small amounts to the the Azimuth of each object in order to rotate the 'sky'. What I cannot do is change the Alt of the point at which I am looking, the default is to look straight up at the zenith.

I assume to accomplish this, I need to add or subtract a certain value from the Alt and Az of each object. I cannot get my head around how to do this.

I have attached an image as an example. The circle shows a bottom-up view of a hemisphere. Firstly my viewpoint is centered on the zenith (0 deg, 0 deg). Here, A is at (20 deg, 0 deg) B is at (20 deg, 90 deg) C is at (20 deg, 180 deg) and D is at (20 deg, 270 deg). If I was to move my viewpoint to X (10 deg, 0 deg), then clearly A would become(10 deg, 0 deg) and C would be (30 deg, 180 deg). How would I find the apparent Alt/Az of B and D, and is there a generalised form of this that I could use using any offset to any initial position?

I might be making this much harder than it actually is, so if you can help or you have any questions, please give me a shout.
It's been eons ago since I've been to school where they taught us a little bit about spherical geometry so I cannot deal in spherical coordinates anymore. (If my life depends on it, I can always research the internet and study spherical geometry. )
So let us speak of plane geometry only.

I don't know if we can treat degrees like any ordinary units. A degree is not a "solid" unit, if you know what I mean.
For the purpose of your post here, let me consider degree as an ordinary "solid" unit. Then you might need to transform my explanations into spherical units of measures.

In your example in the diagram, when you moved to your new location X(10deg, 0deg), the original C(20deg, 180deg) became C((10deg +20deg), (0deg +180deg)) = C(30deg, 180deg). My understanding there is that you altered the original (alt, az) by adding alts and azs as if the degrees were "solid" units.

So, following that procedure, the new location of point B now would be:
Draw a straight line segment connecting X and B.

>>>For the new altitude, we will use the Pythagorean theorem,
(XB)^2 = (OX)^2 +(OB)^2
XB = sqrt[(10deg)^2 +(20deg)^2] = 22.36068 deg ------new alt of B.

>>>For the new azimuth of B, we find the angle that OB makes with say, OX.
tan(angle OXB) = (OB)/(OX) = (20deg)/(10deg) = 2
So, angle OXB = arctan(2) = 63.43495 deg
And so, new azimuth = 180deg -63.43495deg = 116.56505 deg

Therefore, new B would be B(22.36068deg, 116.56505deg) ----***

---------------------
Following same way, the new D would be
D(22.36068deg, 243.43495deg) ---------------***

======================
EDIT:

If my assumption above were correct, then the altitudes of some points would exceed 90 degrees because the hypotenuse is always longer than any leg of the right triangle.
If that is not allowed, then the altitude of the new B should not be governed by the hypotenuse XB. Maybe the altitude of the new B is still 20 degrees. Thus new B is B(20deg, 116.56505deg)?
And new D is (20deg, 243.43495deg)?

3. Hi, thanks very much for having a look at this. Unfortunately, I think it all needs to be done in Spherical polars, as moving from the zenith to an Alt of 10 degrees needs to move further in the y direction on the Cartesian plane than moving from 70-80 degrees in Altitude. In other words, the degrees get 'bunched up' near to the edge of the circle. Using standard Pythagorean trigonometry will not accomplish this, as it assumes adding x degrees to the altitude will move the point x units on the plane. This is not the case, the amount that the point will move on the flat plane will vary depending on how near to the edge of plane the point is.

I can only add 'Alt' to points A and C because they are exactly in the plane of movement in this example. For any point not on this plane, this would not work.

Do you know of any specialists in spherical coordinates (especially if they have some astronomy knowledge!) that could help?

Again, thank you for your prompt response.

4. Hi, just a note to say that I solved (well got around!) this problem, by converting everything into Cartesian (xyz) coordinates, then applying the transformations that can be seen in the attached image. I found these formula in an old games programming book.

Alpha, beta and gamma refer to rotations around the z, x and y axis respectively. Remember that the order that you do them in makes a huge difference to the outcome!

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