ere the other point touches on the latitude scale,
will be the latitude desired. For the longitude, do exactly the same
thing, but use a meridian of longitude instead of a parallel of latitude
and read from the longitude scale at the top or bottom of the chart
instead of from the side.

2. Bearing and distance of a known object, the height of which is known.

Take a bearing of, say, a lighthouse the height of which is known. The
height of all lighthouses on the Atlantic Coast can be found in a book
published by the U.S. Dept. of Commerce. Correct the bearing, as
mentioned in case No. 1. Now read the angle of the height of that light
by using your sextant. Do this by putting the vernier 0 on the arc 0,
sliding the limb slowly forward until the top of the lighthouse in the
reflected horizon just touches the bottom of the lighthouse in the true
horizon. With this angle and the known height of the light, enter Table
33 in Bowditch. At the left of the Table will be found the distance off
in knots. This method can be used with any fairly perpendicular object,
the height of which is known and which is not more than 5 knots away, as
Table 33 is not made out for greater distances.

3. Two bearings of the same object, separated by an interval of time and
with a run during that interval.

Take a compass bearing of some prominent object when it is either 2, 3
or 4 points off the bow. Take another bearing of the same object when it
is either 4, 6 or 8 points off the bow. The distance run by the ship
between the two bearings will be her distance from the observed object
at the second bearing. "The distance run is the distance off."

A diagram will show clearly just why this is so:

[Illustration]

The ship at A finds the light bearing NNW 2 points off her bow. At B,
when the light bears NW and 4 points off, the log registers the distance
from A to B 9 miles. 9 miles, then, will be the distance from the light
itself when the ship is at B. The mathematical reason for this is that
the distance run is one side of an isosceles triangle. Such triangles
have their two sides of equal length. For that reason, the distance run
is the distance off. Now the same fact holds true in running from B,
which is 4 points off the bow, to C, which is 8 points off the bow, or
directly abeam. The log shows the distance run between B and C is 6.3
miles. Hence, the ship is 6.3 miles from the light when directly abeam
of it. This last 4 and 8 point