# 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.

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}. Let's 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.

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.)