It's not only the Spring Equinox but also the Autumnal Equinox where that happens, and it's because of the motion of the Sun across the equator.
What your missing is that different points in the day, the angle subtended between the ecliptic and the celestial equator changes - because the Earth doesn't rotate inline with the ecliptic but an angle of 23.5 degrees - and this is most noticeable towards the solstices.
Since none of us are managing to convince you in terms you can understand, download a copy of Stellarium http://stellarium.org and see for yourself by advancing minute by minute or hour by hour until you are satisfied that is true...
Sorry I haven't had a chance to sit down and write a reply earlier. Here's how I reason it. I hope someone will jump in if I'm mistaken.
Start with an observer on the equator on a perfectly flat plane. If the pole star had a declination of exactly +90* (* = degrees) it would sit on the horizon due north of the observer, neither rising nor setting (I'm taking it to be a mathematical point with no size and am, of course, ignoring refraction) . Likewise a hypothetical object at -90* would sit on the horizon due south of the observer. Halfway between, objects with a declination of 0* would rise due east, past directly overhead and set due west. An object at +20* would rise 20* north of east, be 20* from the zenith at it highest point (ie when it culminates) and set 20* north of west. Similarly for a star of any declination. The star at 0* would be in the sky for 12 hours and all other star for less time. I think that is fairly intuitive.
Now lets consider an observer at 30* south. For this observer, an object with declination -90* is stationary 30* above the horizon. In fact all stars south of -60* never touch the horizon, they are circumpolar (eg the Southern Cross). A star touching the horizon due south of the observer has a declination of -60*. Similarly stars north of +60* never rise and a star on the horizon due north has a declination of +60*. So the arc of 180* on the ground from due north to due south only covers 120* of declination! Now symmetry dictates that an object at 0* dec. must rise half way between +60* and -60*, which is half way between due north and due south and so is due east of the observer. No matter where you are stars with a dec. of 0* rise due east (an set due west).
Next, consider the line going from due south, through the zenith to due north. A star on that line on the northern horizon has a dec. of +60*. A star on that line 10* above the horizon has a dec. of +50* and a star 60* above the horizon has a dec. of 0*. So now the star with a dec. of 0* rises due east but appears to move to the left as it rises and culminates 30* from the zenith (and 60* from the northern horizon). A star with dec. of -30* rises south of east (see Note 1), moves to the left as it rises and passes directly overhead. A star with dec. of -50* rises in the far south and culminates 70* above the southern horizon.
Note 1: I can't work out the exact geometry but it isn't 30* south of east. If you look at Stellarium with both equatorial and azimuth grids visible you will see that the relationship between dec. and azimuth isn't linear. If I find out how to compute the azimuth exactly I'll update this post.
I hope this helps rather than hinders your understanding!
The star at 0* would be in the sky for 12 hours and all other star for less time. I think that is fairly intuitive.
All stars would be in the sky for 12 hours, their tangental velocity is just slower and so they travel less distance. Or maybe I'm misinterpreting what you said?
Thanks Joshua, you are right. At the equator stars rising anywhere along the horizon have the same RA and so will be in the sky for the same length of time. It's away from the equator where stars rising in different directions have different RA and so will be in the sky for different periods of time.
Your explanation way to complicated for me. You lost me before I even finishing reading the first line.
I'm going over and over everybody's explanations hoping it will click in my head.
Just a quick comment. I notice that earlier you said:
Quote:
Originally Posted by j.skett
Now I need to work on understanding why the Sun comes up in the SE and sets in the NW during Perth's summer.
Is this part of your confusion, or is it a typo? The sun rise in the SE and sets in the SW (not NW) in Perth's summer...
Have you tried using Stellarium yet? If you open it, set your location and hit the "e" button. It will then show the projected latitude and longitude lines. Just imagine we are inside a big ball made out of these lines: the axis of the ball goes through the N and S poles. If you follow the line the sun is sitting on at any give date you will see how it crosses the sky, and where it rises and sets. (Remember, we are the ones in the "ball": not the sun- so as the "ball" rotates around its axis, the sun and stars stay on the same latitude line each day- and during the year the sun "moves" between the latitude lines of (approx) 23N and 23 S because our axis is tipped over as we go around the sun each year. The axis always points towards the same point of the sky as the earth moves around the sun (PS: nobody say anything about precession please: that just confuses the issue!), and the southern side of the earth is now "tipping" towards the sun for our summer, so the sun appears to be moving south to the southern latitude lines...)
Your explanation way to complicated for me. You lost me before I even finishing reading the first line.
I'm going over and over everybody's explanations hoping it will click in my head.
Jim
So where, specifically, do you have trouble? If I know that we can go from there.
I put an hour into that post, I'd like to think you put a commensurate effort into understanding it.
David
Appreciate the effort you put into your post, and I have read through it several times, however it goes over my head like a F18 Hornet, especially the continued mention of Declination.
I think the only way I will understand is to sit down with somebody and have them go through it with me using maybe a globe and a light source representing the Sun or old school with white board and marker.
Once again thanks for taking the time to try and explain it to me.
I have downloaded Stellarium and now trying to work out how to use it to help me visualise things.
Dean
Yes it was a slip of the finger, meant to SW not NW.
As mentioned to David I have downloaded Stellarium and as you said hit the E button which brings up the LAT/LONG lines. I'm now trying to get my head around the rest of your instructions. Will let you know how I go.
... however it goes over my head like a F18 Hornet, especially the continued mention of Declination.
Hi Jim,
Substitute "latitude" for "declination", and "longitude" for "RA (right ascension)". Not exactly the same thing, but gives you the idea. Alternatively "up and down (North/South)" for "declination", and "sideways (East/West)" for RA...
Next time you are in Adelaide you can sit in my planetarium dome and I will show you!
I'd start with the notion that the sun is a long long way away and it is very very big. So the rays of light hitting earth are essentially parallel. This diagram might help. http://solar.steinbergs.us/images/jp...tate-R-sun.jpg
Now draw in a spot for your location in summer (on the left) and winter (on the right) at the midday position. (The diagram is labelled for Norther Hemisphere)
Then draw in a line that traces the path of the sun through the day. It should be horizontal.
Now imagine a map overlaid (or maybe use Google earth) and see where the path of the sun intersects the day/night boundary.
You should see that the intersection point is at SE/SW in summer relative to your location.
It also shows why there are 24 hour days and nights at the poles
Disclaimer: I'm making this up as I go and I get confused about it all too
David
Appreciate the effort you put into your post, and I have read through it several times, however it goes over my head like a F18 Hornet, especially the continued mention of Declination.
Ah ha. Now we're getting somewhere. You need to start by understanding the basics of positional astronomy. That will stand you in good stead, not just for this question but for many other uses (finding objects, working out when they are visible etc). Years ago I wrote the 'Sky this Month' for my local club. It was pre-computer and so everything was done manually - tables of siderial time from Nortons Star Atlas, semi-diurnal arcs from the ASNSW Yearbook (itself produced manually by the great Ed Lumley) and poring over the Skalnate-Pleso Star Atlas. That really banged the concepts into my boney skull.
Quote:
Originally Posted by SkyWatch
Hi Jim,
Substitute "latitude" for "declination", and "longitude" for "RA (right ascension)". Not exactly the same thing, but gives you the idea. Alternatively "up and down (North/South)" for "declination", and "sideways (East/West)" for RA...
Yes, RA and Dec are analogous to longitude and latitude (respectively) but they are not the same. One applies to the surface of the Earth and the other to the celestial sphere. If you don't separate the two you will soon confuse yourself. If you want to know whether an object of a given declination is visible from your latitude it will confuse the hell out of you if you start thinking 'what latitude is visible from my latitude?'
I suppose you understand this but, for the purposes of positional astronomy, we are on a sphere inside a sphere. A lot of this discussion revolves around what part of the outer (celestial) sphere is visible from our position on the inner sphere (the Earth). And, if the lines of RA and Dec were painted across he sky how would they appear and how does that change as we move to a different latitude.
Don't give up, if I could eventually understand it, so will you!
Dean
Thanks for the encouragement. Yes I am confident if I keep at it, it will eventually click until then I remain frustrated that I can't grasp it.
Jim
The reason for the apparent rise of the sun in the SE and the SW in summer is due to the axial tilt of the Earth. See the first image. This is set up for the Northern hemisphere but if you imagine that the S and N are switched that will explain things for you. We only see part of the ecliptic and that is the path the Sun appears to follow, remember we are orbiting the Sun, but the apparent view is that the sun is orbiting us. In the top part of the image the dotted line is the ecliptic as traced onto the celestial sphere. We are observing that path from essentially a flat plane. Even though the earth is curved it gives the apparent view of being flat because of its size and our size.
In the bottom part of the image the ecliptic is traced again by the dotted line. As you can see it again appears to move from South to North (remember I said to swap the S and N around).
The second image demonstrates the tilt of the earths axis and how the light rays are essentially parallel.
Lines of declination are the like the lines of latitude as drawn on the earth but are extended outward into space. Lines of latitude are drawn parallel from the equator (0 degrees) down to the poles (90 degrees).
As to why we see different constellations over the entire year. Imagine you are on a merry go round which is slowly turning. Looking out as it travels around. Now imagine that the circular path you are taking on the merry go round is the Earth's orbit around the sun. You will note that the view on the merry go round shows something different all the time until you are back to the starting point. Hence why we see different constellations. We are on a giant merry go round.
