Now, if a man can answer all three questions, A, B, to within eight
points of the compass, and C, he is four and a half times as well off as
if he could only answer A; as will be seen by the following
considerations. A knowledge of B in addition to A, is of only one-third
the use that it would be if C also were known.
1. Let P (fig. 1) be the point where the traveller finds himself at
fault, and let P D to be a distance within which the path certainly lies;
then the circle, E D F, somewhere cuts the path, and the traveller
starting from P must first go to D, and then make the entire circuit, D E
H F D, before he has exhausted his search. This distance of P D + D E H F
D = P D + 6 P D nearly, = 7 P D altogether, which gives the length of
road that the man must be prepared to travel over who can answer no other
than the question A. Of course, P D may cut the path, but I am speaking
of the extreme distance which the lost man may have to travel.
[Sketch as described above].
Supposing that question B can be answered as well as question A, an that
the direction of the line of road lies certainly within the points of the
compass, P S and P R. Draw the circumscribing parallelogram, G L H E M,
whose sides are respectively parallel to P S and P R. Join L M. By the
conditions of this problem, the path must somewhere cut the circle E D F;
and since L M cuts L H, which is a tangent to it, it is clear it must cut
every path--such as a a, parallel to L H, or to P R--that cuts the
circle. Similarly, the same line, L M, must cut every path parallel to P
S, such as b b. Now if L M cuts every path that is parallel to either of
the extreme directions, P R or P S, it is obvious that it must also cut
every path that is parallel to an intermediate direction, such as c c,
but
PL = PH/cos HPL = PD/cos 1/2 RPS;
The consequence of which is that P L exceeds P D by one-sixth, one-half
as much again, or twice as much again, according as R P S = 60 degrees,
90 degrees degrees, or 140 degrees.
The traveller who can only answer the questions A and B, but not C, must
be prepared to travel from P to L, and back again through P to M, a
distance equal to 3 P L. If, however, he can answer the question C, he
knows at once whether to travel towards L or towards M, and he has no
return journey to fear. At the worst, he has simply to travel the
distance P L.
The probable distance, as distinguished from the utmost possible distance
that a man may have to travel in the three cases, can be calculated
mathematically. It would be out of place here to give the working of the
little problem, but I append the rough numerical results in a table.
[Table as referred to above].
The epitome of the whole is this:--1. If you can only answer the question
A, you must seek for the lost path by the tedious circle plan; or, what
is the same, and a more manageable way of setting to work, by travelling
in an octagon, each side of which must be equal to four-fifths of P D.
(See fig. 2.)
[Fig. 2].
That is to say, look at your compass and start in any direction you
please; we will say to the south, as represented in the drawing. Travel
for a distance, P D; then supposing you have not crossed the path, turn
at right angles, and start afresh--we will suppose your present
direction to be west--travel for a distance 4/10 of P D, which will take
you to 1; then turn to the N.W. and travel for a distance 8/10 of P D,
which will take you to 2; then to the N. for a similar distance, which
will take you to 3; and so on, till the octagon has been completed. If
you know B to eight points, and not C, adopt the L M system; also, if you
know A and C, and B to within thirteen points (out of the sixteen that
form the semicircle), you may still adopt the L M system; but not
otherwise. A rough diagram scratched on the ground with a stick would
suffice to recall the above remarks to a traveller's recollection.