hi if i understand your question, you're wanting to know how many degrees and object is above the horizon so you can photograph it while it's up high as opposed to near the horizon in the "dirty air"? i can offer you a couple ideas to help.
1, make sure your chart is showing azimuthal grids. these relate to the horizon (altitude) and direction(azimuth...north,east,south,west) the altitude grids start at 0.00 for the horizon and go up to 90 degrees for the zenith or straight up. the azimuth grids begin at 0 degrees for due north and work their way around to 90 for east, 180 for south, 270 for west and up to 360 for north (which is also zero). 2. if you use the equatorial grids, zero in declination is the celestial equator. negative degrees are below the equator closer to the southern horizon. it goes up from zero to 90 at the celestial north pole. the north star is at 89 degrees. your right ascension grids are based on the 24 hour clock and are continually changing with the seasons. what you can do is select an object, right click it and select identification of nearest object. when you read down the information, it will give rise and set times and culmination, which is when it reachs it's highest point. this is the best time to image if possible. also, if you need field of view information, check out ron wodaski's ccd calculator. there is a free version for download. you can enter info from your camera and lenses or telescope and it will give the field of view. you can then enter this into cdc on the eyepiece menu and show your field of view right on the chart to help choose focal length. hope this helps, dan cade --- In [email protected], "music_ki" <[EMAIL PROTECTED]> wrote: > > This may be a ridiculous questions, but I'm wondering about the > vertical positions of objects. In terms of position relative to the > horizon it's easy enough to use a compass, but I find that I'm always > off on the position vertically when I look at the CDC charts. I > always think that something is going to be lower to the horizon than > it actually is. I'm asking this specifically for photographic > purposes, otherwise it wouldn't really matter I guess. > > Thanks for any help, I'm in love with this program! >
