Air Mass

Introduction

The magnitudes of stars are listed as they would be seen outside the Earth's atmosphere. These are known as extraterrestrial magnitudes and are the values found in books and tables that list a star's extraterrestrial magnitude.

When starlight passes through the Earth's atmosphere it is attenuated. This attenuation, known as atmospheric extinction, is caused by the absorption of some of the starlight by the atmosphere. The more atmosphere starlight goes through, the greater the extinction. Observations made at observatories located at high altitude have less extinction than those at a lower altitude. This is one reason why major observatories are located on mountain tops. Observations made directly overhead at the zenith have the least extinction for a given location. The extinction increases as a star gets further from the zenith and is maximum at the horizon. The section of atmosphere that the starlight travels through is known as air mass.

One of the first steps in determining a star's magnitude for a measurement at the Earth's surface is to know the extinction. The key factor in knowing the extinction is in knowing the star's air mass at the time of the observation.

The following is an explanation on how to determine air mass for a given location, time and star. While seemingly simple, there are many pitfalls in determining the air mass.

Getting Started

A star's air mass, X, is the effective path length of air through which starlight must pass to reach the observer.
By definition, X=1 at the zenith and increases as one looks at stars closer to the horizon. Letting Z be the angular distance of a star from the zenith (0 degrees <= Z <= 90 degrees), then the simplest relationship is

X = secZ (Eqn. 1).

Eqn. 1 would be correct if the Earth and its atmosphere were flat. However, the Earth's curvature causes the relationship to overestimate the airmass for large zenith distances. The correct configuration is illustrated in Figure 1.

Figure 1
Illustration of a Star's Airmass

Two equations are in common use that take into account not only the curvature, but also the refraction, of the atmosphere:

X = secZ * (1 - 0.0012 * (sec^2Z - 1 )) (Eqn. 2)

X = secZ - 0.0018167 * (secZ - 1 ) - 0.002875 * (secZ - 1 )^2 - 0.0008083 * (secZ - 1 )^3 (Eqn. 3)

where the value of secZ depends only on the location of the observer and the position of the star in the sky and can be determined by:

secZ = (sin(LAT) * sin(Declination) + cos (LAT) * COs(Declination) * COs(HA))^-1 (Eqn. 4)

In this equation LAT is the observer's latitude while declination and HA are the star's declination and hour angle, respectively. All values are in decimal degrees.

NOTE: The Z in Eqn. 1 refers to the apparent zenith distance of a star (i.e., taking into account atmospheric refraction). In physical terms, this distance is equivalent to the angle between the optical axis of a telescope and a plumb bob hanging from the mount. However, direct determination of this angle is both cumbersome and inherently inaccurate. In contrast, the Z in Eqns. 2 through 4 refer to the true zenith distance, which assumes that no atmospheric refraction is present. Use of the latter value is preferred as it can be readily calculated from available parameters.

Star's Declination

Figure 2
Illustration of a Star's Declination

The declination, of a star is the angular distance above or below the celestial equator (which is the projection of the Earth's equator on the celestial sphere). The declination of a star on the celestial equator is 0 degrees while that of a star at the north celestial pole is +90 degrees. Stars between the celestial equator and the north celestial pole (NCP) have 0 degrees< declination < +90 degrees. Similarly, a star at the south celestial pole (SCP) has declination = -90 degrees while those between that pole and the celestial equator have -90 degrees < declination < 0 degrees.

Determining a Star's Hour Angle (HA)

Figure 3
Illustration of a Star's Hour Angle for Northern Hemisphere
Observers Facing Their Southern Horizon

The observer's celestial meridian is the north-south line passing directly overhead (i.e., through the zenith). The hour angle, HA, of a star is the amount of time since the star crossed the celestial meridian or until the crossing will occur. Stars east of the meridian are designated either as a negative value or with the symbol E while stars to the west have positive values or a symbol W. The HA of a star increases with time as the celestial sphere rotates.

Mathematically, the hour angle is defined as

HA = (LST - RA) hours (Eqn. 5)

Note: HA in Eqn. 5 is in hours, but must be converted to degrees to be used in Eqn. 4. To produce this multiply HA in hours by 15 (15 degrees per hour).

HA = (LST - RA) * 15 degrees (Eqn. 5a)

where LST is the local sidereal time and RA is the star's right ascension. The LST is simply the right ascension of a star on the observer's celestial meridian at the time of the observation.

Note: The star on the meridian may or may not be the star being observed and is referred to only to define LST.

Determining Local Sidereal Time (LST)

LSTcan be calculated, however high accuracy is needed to determine an accurate LST. Most computers and computer programs have a difficulty in proving sufficient accuracy. Another approach is to look up the Greenwich Mean Sidereal Time (GMST) from an Astronomical Almanac for UT= 00:00:00 (GMS0hUT), subtract (if west of Greenwich or add if east of Greenwich) the observation longitude (in hours) and then add the UT of the observation times 1.00274.

LST = (GMS0hUT - Long + 1.00274 * UT) (Eqn. 6)

where

GMS0hUT is Greenwich Mean Sidereal Time at UT= 00:00:00
Long is the Longitude of the observer's site converted from degrees to time from GMT.
UT is the Universal Time of the observation the 1.00274 is to convert UT solar time to UT sidereal time.

Note: Convert all times to hours decimal. (Hours + minutes/60 + seconds/3600)

For example:

Kitt Peak has a longitude of 111 degrees 35' 52" or 111.59777778 degrees.
This is equivalent to 7h 26m 23s or 7.423555556 hours from Greenwich.

Long (degrees) = 111.597777778 (degrees) * 24/360 (hours/degree)

Long (time)= 7.423555556 hours

If the observation was made 4h 24m 16s (4.406154 hours) on 9 September 1990,
GMS0hUT = 23h 11m 52s or 23.18555556 hours UT= 4.406154 hours

Thus from Eqn.6:

LST = 23.18555556 - 7.423555556 + (1.00274 * 4.406154)

LST = 20.180227 hours

If one is to use a computer program to calculate this, it may be desirable to create a look up table of all values of LST at UT= 00:00:00 for a ranges of dates. While individual observation UTs can be entered and LST determined, it is not practical to create a table of such times. Better to have just one LST per date and location at UT= 00:00:00 and then add or subtract the (UT * 1.00274) to get the LST. This table can be created using the U.S. Naval Observatory Multiyear Interactive Computer Almana (MICA) 1800 - 2050 program available from Willmann Bell, Inc. (www.willbell.com). I will be happy to supply a text file with tab separated data to anyone who would like it. Just tell me the date range an the observatory longitude.

Creating an LST Table Using MICA

The date and observation location (a different table is needed for different locations) are specified. The location's LST for that date at UT= 00:00:00 is then looked up. The observation's UT times 1.00274 is then added for each observation time.

Figure 4
Sample MICA Created Table for LST at HPO for the Month of October 2005

Example:

For 21 October 2005 at UT= 07h 10m 00s or 7.166667 hours.

Using the MICA Table in Figure 4, the lookup Local Mean Sidereal Time for 21 October 2005 for HPO at longitude = 112 degrees 13' 22.0' west and latitude 33 degrees 30' 06.0" north is 18h 29m 16.1843s or 18.487829 hours.

LST = 18.4897829 + 1.00274 * 7.1666667

LST = 25.674132 hours

Note: To test and adjust LST
If (LST Lookup + UT * 1.00274) =>24
Use (LST Lookup + UT * 1.00274) - 24
Otherwise use LST Lookup + UT * 1.00274.

Since the above value is greater than 24, 24 hours must be subtracted from it to give

LST = 1.674132 hours

Determining the Hour Angle

From Eqn. 5:

HA = LST - RA

To determine the Observation's hour angle (HA ) the observed star's right ascension is subtracted from the Local Sidereal Time (LST).

Assume star alpha Aur

RA = 5h 16m 41.3s or 5.278139 hours
Dec = 45 degrees 59' 53.0" or 45.998056 degrees

For LST= 1.674132 hours
HA= 1.674132- 5.278139
HA = -3.604007 hours

To convert HA from hours to degrees:
HA = -3.604007 * 15 degrees
HA = - 54.060105 degrees

Alternatively, the hour angle of a star can be determined directly from the telescope's right ascension setting circle (if so equipped) by noting the difference in right ascension between the star and the meridian.

Because the celestial equator intersects the observer's horizon at the due east and west points, a star with declination = 0 degrees (on the celestial equator) is above the horizon for exactly 12 hours; the star rises with HA=-6 hour and sets with HA =+6 hour. For observers in the Earth's northern hemisphere, a star with declination < 0 degrees spends less than 12 hours above the horizon each day while a star with declination > 0 degrees is visible more than 12 hours per day. As a final comment, note that circumpolar stars never rise nor set, so that their hour angle can range from -12 hours (12 hours East) to +12 hours (12 hours West). The hour angle is zero for stars at upper culmination.

Determining the Air Mass

Now that we know the HA for the observation the value of secZ and thus the air mass X can be calculated from Eqn 3 and 4.

From Eqn. 4:

secZ = (sin(Lat) * sin(Declination) + COs Lat * COs(Declination) * COs(HA))^-1

Latitude for HPO
Lat= 33 degrees 30' 06.0" north or + 33.50166667 degrees
alpha Aur Declination = 45.998056 degrees
HA = - 54.060105 degrees

thus

secZ= 1.4061942273781

To determine Air Mass X:

X = secZ - 0.0018167 * (secZ - 1 ) - 0.002875 * (secZ - 1 )^2 - 0.0008083 * (secZ - 1 )^3

X = 1.404928

Figure 5 HPO
Program for Calculating Air Mass

Present Page Version as of 8 December 2005

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