up" (e.g. at a Declination that is the same at Waterloo's latitude to object is directly overhead at Elevation of 90 degrees, but at all or any Azimuth... Azimuth is poorly defined for such objects at that time). Use this plot to answer the following questions: (a) On June 21 the Sun is at a Declination of 23.5 degrees. Estimate the time of sunrise in Waterloo (e.g. how many hours before noon the Sun at Elevation = 0. (b) How long is the day in Waterloo on June 21? (c) What is the time of sunrise on September 21 when the Declination of the sun is 0? (d) On Dec 21 the Sun is at a Declination of -23.5 degrees. Estimate the time of sunrise on that date and the length of the day. 2) The figure below shows the positions of objects in the sky as seen from Waterloo in "natural" coordinates of Azimuth (0 degrees is to the North, 90 degrees is East, 180 degrees is South) and Elevation (0 degrees is the horizon and 90 degrees is directly overhead). This is just half of the sky but the other half (from Azimuth of 180 to 360 is a perfect mirror reflection of this). Note that Elevation of 0 is the position where an object rises above the horizon (the Elevation of 0 on the reflected figure shows the position at setting). Azimuth and Elevation from UWaterloo 50 deg 80 60 deg 60 60 Elevation 8 70 deg 20 1 -1 hr -2 hr 1 -3 hr 1 -4 hr -5 hr 1 1 -6 hr -7 hr -8 hr -9 hr 50 100 Azimuth 40 deg 30 deg 20 deg 、 . 10 deg IC 0 deg 10 deg -20 deg 150 The blue lines show the Declinations of the objects and the red lines mark the time in hours, where 0 hours is "noon" for object, or the time (in Sidereal Time) where the object is at its highest Elevation - note that 0 hours is not marked as the plot "blows

up" (e.g. at a Declination that is the same at Waterloo's latitude to object is directly overhead at Elevation of 90 degrees, but at all or any Azimuth... Azimuth is poorly defined for such objects at that time). Use this plot to answer the following questions: (a) On June 21 the Sun is at a Declination of 23.5 degrees. Estimate the time of sunrise in Waterloo (e.g. how many hours before noon the Sun at Elevation = 0. (b) How long is the day in Waterloo on June 21? (c) What is the time of sunrise on September 21 when the Declination of the sun is 0? (d) On Dec 21 the Sun is at a Declination of -23.5 degrees. Estimate the time of sunrise on that date and the length of the day. 2) The figure below shows the positions of objects in the sky as seen from Waterloo in "natural" coordinates of Azimuth (0 degrees is to the North, 90 degrees is East, 180 degrees is South) and Elevation (0 degrees is the horizon and 90 degrees is directly overhead). This is just half of the sky but the other half (from Azimuth of 180 to 360 is a perfect mirror reflection of this). Note that Elevation of 0 is the position where an object rises above the horizon (the Elevation of 0 on the reflected figure shows the position at setting). Azimuth and Elevation from UWaterloo 50 deg 80 60 deg 60 60 Elevation 8 70 deg 20 1 -1 hr -2 hr 1 -3 hr 1 -4 hr -5 hr 1 1 -6 hr -7 hr -8 hr -9 hr 50 100 Azimuth 40 deg 30 deg 20 deg 、 . 10 deg IC 0 deg 10 deg -20 deg 150 The blue lines show the Declinations of the objects and the red lines mark the time in hours, where 0 hours is "noon" for object, or the time (in Sidereal Time) where the object is at its highest Elevation - note that 0 hours is not marked as the plot "blows

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