Duration of his civil dusk in march 21st?

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Homework Help Overview

The problem involves calculating the duration of civil dusk for an observer at latitude 45º on March 21st, when the declination of the sun is 0º. The discussion centers around the geometric relationship between the sun's position and the horizon during twilight.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to relate the hour angle and declination of the sun to determine the duration of civil dusk. Some participants question the relevance of certain relationships and explore the implications of the sun's trajectory on the celestial equator during the equinox.

Discussion Status

The discussion is ongoing, with participants providing insights into the geometric relationships involved. There is an exploration of the sun's path and its angular distance below the horizon, but no consensus has been reached on the best approach to calculate the duration of civil dusk.

Contextual Notes

Participants are working within the constraints of the problem, including the specific date and latitude, and are discussing the assumptions related to the sun's declination and its effect on twilight duration.

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Homework Statement



Evening civil twilight (civil dusk) begins at sunset and ends when the geometric center of the sun reaches 6° below the horizon. Calculate for an observer at latitude = 45º the duration of his civil dusk in march 21st (when \delta_{sun}=0º

Homework Equations



The equations of the transformation between horizontal coordinates and ecuatorial coordinates

"http://en.wikipedia.org/wiki/Celestial_coordinate_system#Equatorial_to_horizontal_coordinates"

The Attempt at a Solution



Let us do a = 0 (sun over the horizon). This implies cos h = -tan d tan f (h: hour angle, d: declination, f: latitude). If d = 0 (delination of the sun is zero on march 21st), then we have h = 90º. Thus A = 90º too (azimut). Is this correct?
doing the same for a =-6º I obtain that the difference between hour angles for the two positions (sun in a=0 and sun in a=-6) is 8.51º. How can I transform this into minutes?

Thanks
 
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360deg/24hr = 15deg/hr

24hr/360deg = 4min/deg = 240sec/deg
 
gneill said:
360deg/24hr = 15deg/hr

24hr/360deg = 4min/deg = 240sec/deg

I think that relation is useless here... am I wrong?
 
atomqwerty said:
I think that relation is useless here... am I wrong?

Not completely useless, just inconvenient. :smile:

On March 21st (equinox) the Sun follows a trajectory along the celestial equator, since the Sun is over the equator and the Earth's rotational inclination lies in a plane that is perpendicular to the Earth-Sun line.

So the Sun follows the projected arc of the equator on the sky. After sunset, when the Sun is so many vertical degrees below the local horizon, it actually followed the longer, slanting path of the equatorial arc, which it traversed at the rate of 15 degrees per hour.

If you have the angle between the equatorial plane and the local vertical, and you have the angular distance representing the twilight duration, then you can work out the angular length of the path of the Sun on the equatorial arc (approximated by a right angle triangle).

Unless, of course, I've mucked up in my thinking somewhere along the line...
 

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