| ||||||
OCR Rendition - approximate298 Art of Travel. 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 r ; 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 P PH PD L-cos HPL- Cos J RPS; the consequence of which is that P L exceeds P D by onesixth, one-half as much again, or twice as much again, according as R P s = 60°, 90°, 120°, or 140.° 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.
* The words "least distance," mean the least distance that the traveller can specify with absolute confidence, as that within which the path, &c., he wishes to regain, is situated. ).F_ www.fastio.com |