I often get searches for blank sky maps, so I thought I'd do just that. Problem is, how blank, in what coordinate system, and in which projection. See, there are quite a lot of knobs to twiddle. So let me just give a few examples here. Suggestions/requests are welcome.

First: Equatorial coordinates, in Winkel-Tripel projection with coordinate grid on white background. Projection-wise, probably the best compromise between length- area- and shape-distortions that come inevitably with every map projection.

Galactic coordinates in Mollweide projection. Nothing better than Mollweide (or Hammer-Aitoff, see below) for galactic maps, with the origin, i.e. the center of the milky way, right in the middle.

Ecliptic coordinates, this time on a black background, with the simplest of all projection, equirectangular, thats, really just a linear mapping of coordinates to area.

And something completely different, supergalactic coordinates in Hammer-Aitoff projection. The supergalactic plane is the one in which the local filament of galaxy clusters seems to be oriented, visualized here with all galaxies brighter than 12th magnitude in red.

Stellar data from the Hipparcos catalogue, with stars down to magnitude 7 shown, galaxies from the handy Saguaro Astronomy Club Database, and map projections according to USGS's fabulous Map Projections: A Working Manual by John P. Snyder. And milky way data from PP3

Awesome site, thanks!

ReplyDeleteAwesome site, thanks!

ReplyDeleteSay it was midnight on the vernal equinox for an observer standing at 45°N: if you were to overlay his horizon(s) on the Equatorial coordinates, in Winkel-Tripel projection with coordinate grid (image/fig 1); would the resulting image be 45° sine wave (blocking everything towards the bottom of the image)?

@Lynn, thanks.

ReplyDeleteNot exactly a sine wave, as it is distorted by the projection, but essentially, yes, with the southernmost point at latitude minus 90º, in this case 45º south. I made an image here as illustration, assuming the zenith to be at 12h/45ºN.

Thanks. Now what would, say 15° increments of elevation (both above and below) the horizon then look like? Using your (awesome) horizon diagram image as a base. Use the color green and show -15, +15, +30, etc. to +90 (zenith) which should be a dot. I'll explain where this is going...

ReplyDeleteIs this what you wanted? I made the elevation lines different shades of green, from darkest at -15ª to brightest at 90ª (the green dot). The 45ª line appears to be cut by the upper border, thats because the projection stretches the poles from points to the lines top and bottom.

ReplyDeleteThat's fantastic, Olaf; thank you! I've been studying your newest diagram (thanks again!) for some time and it's exactly what I needed to explain the problem I've been tasked with: Given one's latitude, what time of year does the sun transition from sunset to "astronomical twighlight" (-18° degrees) the quickest?

ReplyDeleteTo me the answer comes down to: What time of year (the sun on the ecliptic) does the line of declination that the sun is on, make the highest angle between the horizon and the -18° line of elevation?

I have a running argument with an astronomy blogger who claims that *all* lines of declination make the same angle with the horizon.

In looking at your latest diagram, do you agree or disagree with this bloggers statement? I don't agree and I believe that the line of 0°Dec makes the highest angle therefore at the equinoxes is when this transition occurs the quickest. What are your thoughts? ...And thanks again!

ReplyDeleteHi Lynn, glad to be of help. Your comment got me thinking how best to illustrate the problem, and I came up with this. It shows the same altitude lines as before, only in azimuthal projection, so this represents what you would see with your zenith at 12h/45º, plus the part below the horizon to -15º elevation. The gray graticule represents celestial coordinates. This makes the declination lines nicely comparable, and you can see that they indeed cross the horizon (2nd outermost green line) at different angles.

ReplyDeleteI've also included the ecliptic as the blue line. It's northernmost point is where it crosses 6h right ascension, it's southernmost point at 18h, slightly outside the graphic. If you follow the declination lines at those points, you can see that their path's are indeed longer than the equatorial line (marked by the RA numbers), the northernmost line very much so, and therefore astronomical twilight lasts longer.

So you win! With the caveat that this is only valid form mid latitudes.At low latitudes between the tropics the length of twilight is pretty much the same (short) all year, because the sun goes straight down, while close to the poles twilight can last for months.

It sounds like you got it, but unfortunately I cannot see the gray lines or chars on my LG Pad 7.0. Ill try on my laptop, but do you think you could recreate these lines and figures in black? Meanwhile based on what you're saying you must have nailed it! I've always told my twin brother (where the question originated) that everything changes between the Tropics and between the Artic and Antarctic circles and their respective poles. But I could never prove it, lacking the graphic tools and talent you obviously posses! This has been at least a six year journey with searching the internet for this diagram it sounds like you've created. I can't thank you enough for staying with me on this Olaf! Next I'm going to bring my brother into this thread. Thanks again!

ReplyDeleteOh well, here you go. And just for the fun of it, I'll throw in the situation at the equator and at the north pole! Where the elevation and declination lines run concentric, which might be a bit confusing, but with the different colors clear enough, I think..

ReplyDeleteNope, that was the wrong link for the first map, this is the right one.

ReplyDeleteThat's it! That's the image/chart I've always wanted to create. I got to soak this in for a bit. Thanks!

DeleteHere's my thoughts: tangents!

ReplyDeleteDue to my in ability to communicate in math, the only way I can visualize how to render the angle formed by two intersecting circles is thru tangents drawn at the point of intersection off of both circles. Your thoughts?

Hi Lynn,

ReplyDeleteyou just hit upon the very definition of the intersection angle between two circles. Check the second image here:

http://www.had2know.com/academics/intersection-angle-two-circles.html.

So yeah, you're right.