<p>Let's assume you want to observe stars at Kitt Peak (32 degrees North) at an airmass above 2, for 2 hours (30 degrees of Earth rotation). </p>

<p>My attempt:</p>

<p>The angle between the zenith and celestial equator is 32 so the altitude to the celestial equator is 58 degrees and a star with a declination of 28 degrees south will cross the meridian at 30 degrees altitude at Kitt Peak (assuming my calculations are correct)... but will not be above the required airmass for 2 hours.</p>

<p>Here is where I'm stuck:</p>

<p>We know that one hour of time corresponds with 15 degrees. In this case, would I just subtract 15 degrees from the 28-degree star? i.e. 15 degrees to go up and 15 back = 2 hours. If this is correct the final southernmost declination star would be 13 degrees south. I'm not sure if my thinking is correct and I don't know how to check. </p>

CC BY-SA 4.0
2020-01-01T02:26:48.130Z
Are you asking *How to calculate declination limits for a given airmass maximum, observation location and duration?* If so, then the first step is to calculate what elevation (altitude) above the horizon corresponds to an airmass of 2 and the second step is to calculate what declination will have that elevation at an hour angle of 15 degrees. This is a (hopefully) helpful hint, not an answer.
uhoh
CC BY-SA 4.0
2020-01-01T17:22:08.920Z
Step 1: An elevation or altitude of 30 degrees corresponds with an airmass of 2 (90 - sin^(-1)(1/2)).
Step 2: A declination of -28 (28 S) will be at 30 degrees for the provided altitude (90-32-30). I'm not sure where to go from there. ie. That the star will be visible for at least 2 hours or 30 degrees of rotation.
Astroturf
<observational-astronomy><amateur-observing><coordinate>
How can one determine the declination range of stars visible for a certain amount of time above a specific airmass at a certain location?
Astroturf
# How can one determine the declination range of stars visible for a certain amount of time above a specific airmass at a certain location?

(Asked by: username_0 on 2020-01-01)

Let's assume you want to observe stars at Kitt Peak (32 degrees North) at an airmass above 2, for 2 hours (30 degrees of Earth rotation).

My attempt:

The angle between the zenith and celestial equator is 32 so the altitude to the celestial equator is 58 degrees and a star with a declination of 28 degrees south will cross the meridian at 30 degrees altitude at Kitt Peak (assuming my calculations are correct)... but will not be above the required airmass for 2 hours.

Here is where I'm stuck:

We know that one hour of time corresponds with 15 degrees. In this case, would I just subtract 15 degrees from the 28-degree star? i.e. 15 degrees to go up and 15 back = 2 hours. If this is correct the final southernmost declination star would be 13 degrees south. I'm not sure if my thinking is correct and I don't know how to check.

**Tags:** observational-astronomy, amateur-observing, coordinate
> **Comments:**
> **username_1**: Are you asking *How to calculate declination limits for a given airmass maximum, observation location and duration?* If so, then the first step is to calculate what elevation (altitude) above the horizon corresponds to an airmass of 2 and the second step is to calculate what declination will have that elevation at an hour angle of 15 degrees. This is a (hopefully) helpful hint, not an answer.  
> **username_0**: Step 1: An elevation or altitude of 30 degrees corresponds with an airmass of 2 (90 - sin^(-1)(1/2)). Step 2: A declination of -28 (28 S) will be at 30 degrees for the provided altitude (90-32-30). I'm not sure where to go from there. ie. That the star will be visible for at least 2 hours or 30 degrees of rotation.  

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