An introduction to astronomy: designed as a text-book for the use of students in college. By Denison Olmsted ...

34 THE EARTH.
41~ 4' from the pole.* It should have passed the meridian at
the same distance below the pole, but the distance was found to
be only 40~ 57' 35". Hence, 4:1i 4' 40~ 57' 35"=6' 25" is the
refraction due to that altitude, that is, at the altitude of
7~ 46'=(48~ 50'-41~ 4'). By taking similar observations in
various places situated in high latitudes, the amount of refraction might be ascertained for a number of different altitudes,
and thus the law of increase in refraction, as we proceed from
the zenith towards the horizon, might be ascertained.
91. Another method of finding the refraction at different
altitudes, is as follows. Take the altitude of the sun or a star,
whose right ascension and declination are known, and note the
time by the clock. Observe also when it crosses the meridian,
and the difference of time between the two observations will
give the hour angle ZPx (Fig. 9). In this triangle ZPx we
Fig. 9.
z
P
E
Z'
also know PZ the co-latitude and Px the co-declination. Hence
we can find the co-altitude Zx, and of course the true altitude.
Compare the altitude thus found with that before determined
by observation, and the difference will be the refraction due
to the apparent altitude.
Ex. On iMay 1, 1738, at 5h. 20m. in the morning, Cassini
* For the polar distance of the placc=90-48~ 50'=41~ 10'; and 410 10-6'
=410 4'.