there are digital levels available.  i see them here in the u.s.a 
at hardware and building supply stores.  you might even find a 
protractor of some sort, where you could set a predetermined angle
then look at the bubble level that is on it.
dan cade


--- In [email protected], "music_ki" <[EMAIL PROTECTED]> 
wrote:
>
> Hi Dan,
> 
> you're close.  Basically, my problem is that even if I know the 
> degrees in CDC, when I go outside, I find that the object is much 
> farther above the horizon than I expected.  
> 
> I actually want to fram objects while they are closer to the 
horizon 
> (primarily the moon, or maybe even exclusively) so that I don't 
only 
> get the detail of the moon, but the "cut out" of the foreground in 
> front, or in the case of daytime/twilight moons, some detail in the 
> foreground, rather than just the moon itself.
> 
> While it would be nice to be able to get out every night, and have 
> perfect viewing conditions, that unfortunately isn't possible, 
> especially in Central Europe where I see a lot of clouds.  So 
> usually I walk around, pick my spots and try to keep them as a 
> reference for the future when it is clear.  My problem is even 
> though I know the rise/set positions, I can't accurately predict 
the 
> vertical position of the moon when I'm just walking around looking 
> for some good vantage points.  I guess what I'm almost asking, is 
if 
> there is something like a compass, but for vertical degrees, that I 
> could use for this kind of stuff, in order to better plan in 
advance.
> 
> thanks for the help so far!
> 
> --- In [email protected], "dan" <hitthepin59@> 
> wrote:
> >
> > 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" 
<music_ki@> 
> > 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!
> > >
> >
>


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