© My star trail image on 23rd April from The Oaks, NSW, Australia. Field of view approx. 98º x 75º.
Exposed with a Tamron 10mm lens on my Canon 60D camera, at f/4.5, ISO 640. The total duration of the trails was 50.5 minutes, consisting of 98 x 25 second exposures taken 5 seconds apart. 3 polluted images were discarded – hence some gaps in the trails.
The brightest trail is the planet Jupiter and the bright trail above it is Spica.
North is bottom right, South is top left.
Whilst a wide-field lens like this will naturally distort any image, the curved sidereal trails clearly indicate the location of the celestial equator (where the trail appears as a straight line between the opposing curves). The celestial equator is the mid point between the two celestial poles. If you stood on the Earth's equator, the celestial equator would pass through the zenith.
The meridian can also be implied.
Relative to the observer, the meridian is an imaginary line connecting True South and True North, passing directly overhead i.e. through the zenith. The stars and planets pass across it daily, from east to west, so the very bright object (Jupiter) has moved towards the meridian (from right to left) in this image.
Imagine yourself facing East, with your arms outstretched to your sides. One arm points North, the other points South - and directly above your head is the zenith. Connect north and south through the zenith and that line is the meridian. It is static and the stars move across it. Sidereal time (a measure of celestial time, comparable to longitude) occurs at the meridian as the stars move across it.
The star, Spica is the fairly bright star above Jupiter. It's location in the sky includes a Right Ascension of 13 hr 26m 06s, which also indicates the sidereal time at the moment it crosses the meridian.
Unfortunately, my camera lens was not aiming directly north-south, as I tried to guess it and I was a few degrees out. Hence the meridian would be a diagonal line on my image.
One day I will take another similar shot but aim it more accurately at the intersection of the celestial equator and the meridian - to get a truly symmetrical image. To do that I will need to orient the camera exactly towards true north and aim it at 34º down from the zenith (or 56º above the horizon).