# Radial Search Analysis

The previous analysis on searching used two rectangular wings off each side of the robot. In that discussion I alluded to searching by looking toward the front of the robot. For lack of a better term, lets call it *radial search*. There is some math involved in this so the spreadsheet page Radial Search presents that calculations.

Assume that the robot can see the semi-circle toward its front. As the robot moves forward 2 meters a new area, darker gray in the diagram to the right, becomes visible.

Recall that in the Area Analysis we determined that the 'bot needed to scan a new 2 meter area that was 11.1 m^{2} each second to complete the search in the 2 hours allowed. (I also suggested the search needed to be twice that to allow for all the other activities of the 'bot but will stick with the basics for now.)

What is the radius of the semi-circle such that the new search area is 11.1 m^{2} and can it be covered by a camera?

A quick (30,000 foot) answer is reached by calculating the difference in area between two circles with a difference in 2 meters in their radii. The area of a 3 meter radius circle is (3*3*π) is 28 m^{2}. A 5 meter circle has an area of (5*5*π) 79 m^{2}. The area difference is (79 - 28) 50 m^{2}. A semi-circle is half that area so we have 25 m^{2}. Since we've already assumed that a camera can see 3 meters (across a bedroom) this is well within range with the addition of being twice as large as the 11.1 m^{2} required. The exact number is going to be somewhat smaller since the front-most circle will only be 3 meters in radius but there appears to be enough leeway to not be concerned.

At least so I thought. <sigh> After staring at the diagram for a bit I realized there is much more area being contributed by the outer circle as it wraps around toward the 'vertical' diameter of the inner circle. A quick analysis shows that the upper and lower limbs of the outer semi-circle sweep two 4 m^{2} areas. That is 8 m^{2} that is subtracted.

I pondered the diagram and thought more about the problem. I am pretty sure, subject to a comment correcting me, that the curvature is irrelevant. With the 'bot moving forward we are still seeing a new area that is the length forward and the diameter of the circle, or (2 x 6) 12 m^{2}. Lets go with that as the conclusion: the new area searched is determined by the diameter of the semi-circle times the speed.

Again pondering the diagram, notice that there is a considerable difference in the amount of time a sample is visible directly in front of us versus to the side.