>I just went out and did my best to determine the actual >horizon for the setting point for the Mir observation >yesterday. I measure it to be approximately -0.4 deg. >This was done using a level and then coming home and >recreating the angle observed, then measuring and >calculating by simple trig. You can't measure the altitude of the horizon directly because of refraction. The apparent altitude of the horizon will be above the geometrical altitude of the horizon. You must calculate it if it is at a different elevation, and the calculation is not trivial if the elevation difference is large because of refraction, but I can do it, at least approximately. Refraction is not an unvarying quantity. An angle that small won't require any trig. What are the relevant landforms? Your elevation and that of the horizon are all that will be necessary. My son in law says Sterling CO is about 3200 feet elevation, but I don't have a better reference. Using 2073 m for your elevation and 970 m for the elevation of your horizon (based on my son in law's guess) your geometric horizon lies just over a degree below true horizontal. Refraction will raise that, but not by as much as 0.6 degree as your measurement would suggest. Thus it is possible that, if your measurement is really good to 0.1 degree, the plain has a higher elevation than I have assumed. If you're on the Front range at 2073 meters elevation you can't see as far as Sterling anyway. If there is an identifiable landmark on your horizon that will help me calculate what it is, by the way. >Estimated angle of the cloud bank above the horizon >(disappearance of Mir) is 0.2 degree. Thus (ignoring >the errors) the observed elevation of Mir was about >-0.2 degree. This means that is was below my true >horizon (elevation = 0 deg) That should read "altitude = 0 degree". Elevation refers to height above a reference datum (mean sea level here in BC) and is measured in meters or feet. Altitude refers to the apparent position coordinate of an object above the true horizon and is measured in angular measure. You can measure the angular height of the cloud bank easily. Use a ruler held at arm's length and measure the height in inches or millimeters. Measure the distance (or have someone else measure it) from the ruler to your eye in the same units. The angle representing the apparent height of the clouds in degrees is given by the quotient of those two numbers multiplied by 57.3. The only way you can be sure that the cloud bank is above the horizon, however, rather than being nearer to you than the horizon, is that you can see the horizon itself clearly. >Assuming a refraction angle of 0.5 degree yields an >actual elevation of -0.2 minus 0.5 equals -0.7 deg. >This is consistent with the values calculated by >STSPLUS and SKYMAP using post-burn elements. Rather than relying on your elements to give you the geometric altitude of Mir at a given time, you should measure an appulse with a prominent star at a higher altitude. This will tell you whether Mir is early or late, and by how much. The time difference between the appulse and the disappearance will be much more accurate than the clock time. >I am using the term "elevation" to represent vertical >angle from the horizon, where the local true horizon >is 0 degrees and the zenith (overhead) is 90 deg. As I mentioned, that is an unusual meaning. The conventional coordinates are altitude and azimuth. >To make this obs, I travelled to a point about 1 mile >SE of my normal location to improve the horizon. > >It may just be a matter of semantics, but I am assuming >that the elevation that would be appropriate for use >in deriving element sets would be the -0.7 deg value. As I've mentioned before, I'm a relatively inexperienced user of these elements and I don't know how the number you mention would be used in deriving them. Leigh