I. Start at the origin and search the cell number 1 and then number 2, 3. Now, the first revolution is completed. If the target not found, then go to step II.
II. Since, the target is not detected, then the searcher will do the second revolution which contain the cells number 4 to number 14 as in Fig.1(a). And so on until the hole will detect. Assuming that the target is randomly located in the space at the position which are independent and identically distributed random variables with joint probability density function where Since the space is divided into identical cubic cells with knowing side length then the target is equally probably in each cell and has the probability see Fig. (2). Here, the searching process is assumed to be continuous in both time and space. Also, let are complete revolutions of the spiral with line segment directory.
For the above Helix's path, the speed of the searcher is where the search path may be defined by its position vector . Thus, . To calculate the expected value of the time detection we need to find , that is; volume of the Helix's path along the directory of …show more content…
Thus, we need to know under what conditions should ?. In distribution theory we know that, except Cauchy distribution, the expected value of the independent and identically distributed random variables which have knowing trivariate distribution is finite (i.e., ). Then, the target will be met after finite number of revolutions that is make . On the other hand, the searcher take spiral with line segment directory in the space like the search plan which studied in the plane by El-Hadidy [2]. El-Hadidy showed that any spiral with line segment search plan to find a randomly located target in the plane must take at least arcs, where n is the number of