As regards B, if the man knows the course of the path to within eight points of the compass (or one-fourth of the whole horizon), it is a great gain; or even if he knows B to within twelve points, say 120 degrees, or one-third of the whole horizon, his knowledge is available. For instance, let us suppose a man's general idea of the run of the path to be, that it goes in a northerly and southerly direction: then if he is also positive that the path does not deviate more than to the N.E. on the one side of that direction, or to the N.W. on the other, he knows the direction to within eight points. Similarly he is sure to twelve points, if his limits, on either hand, are E.N.E. and W.N.W. respectively.
C requires no further explanation.
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.
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.