Appendix V Polaris Altitude

AV.1 Schematic of the geometry used to prove that Polaris (or the south celestial pole) lies above the horizon at an angle, or has an altitude, roughly equal to the observer’s latitude. An observer at a latitude of θ° N (green), has an angle of φ° (red) between starlight from Polaris and a line connecting the observer to the center of the earth. Also, because the line connecting the observer to the center of the earth is oriented at a right angle, or 90°, to the observer’s local horizon, then the angle between starlight from Polaris and the observer’s local horizon must also be θ° (green). Finally, since the altitude of Polaris must be equal to the angle between starlight from Polaris and the observer’s local horizon, then we have our proof that the altitude of Polaris is equal to the observer’s latitude!

Here, we prove that Polaris (or the south celestial pole) lies above the horizon at an angle, or has an altitude, roughly equal to the observer’s latitude. First, note in Figure A.V.1 that since Polaris is so far away, all its light arrives at Earth in the form of parallel rays. For an observer at a latitude of θ° N (green), then, the angle between starlight from Polaris and a line connecting the observer to the center of the earth is φ° (red).

Next, since the sum of the interior angles of a right triangle must add up to 180°, we see that:

or,

Also, because the line connecting the observer to the center of the earth is oriented at a right angle, or 90° to the observer’s local horizon, then the angle between starlight from Polaris and the observer’s local horizon must also be θ° (green).

Finally, since the altitude of Polaris must be equal to the angle between starlight from Polaris and the observer’s local horizon, then we have our proof that the altitude of Polaris is equal to the observer’s latitude!