Re: Mir seen at -0.8 degree?

Leigh Palmer (
Sun, 29 Dec 1996 17:47:21 -0800

>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

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.