# Area Analysis

The basic information and constraints for the search are:

- 80,000 square meters
- 2 hours or (60 sec / min * 60 min / hr * 2 hr) 7200 seconds
- 2 meter / second speed limit

Say you drop something on the floor and don't see it right away. You start to bend over or maybe get on your hands and knees to start searching carefully. You might start scanning back and forth in the search area. To get a feel for the sample search lets try that approach. To do that we'll assume we look at a 2 m^{2} area - because the 2 meter area makes the math really easy.

I'm also making a large number of other simplifications, which will become obvious as we go along. One to consider is how long does it take to turn the robot when it reaches a boundary? Ignoring them for now keeps us looking at the big, overall problem. It avoids getting buried in details even before we know if they apply.

With that assumption we have (80,000 m^{2} / 2 m^{2}) or 40,000 search units. (See the easy math!)The speed limit says we can search one 2 m^{2} section in a second so we can search the entire area in 40,000 seconds. That is a bit more than the 7200 secs allowed. It is clear we can't do a detailed search of the area.

Let's look at how big an area we have to examine each second to complete the search. We have 80,000 m^{2 }and 7,200 seconds so the area to examine is (80,000 m^{2} / 7,200 secs) 11.11 m^{2}/ sec. (If you aren't easily visualizing this in meters - which I am not - that is an area slightly larger than 33 ft x 3 ft, or, lets say, 100 ft^{2}, using the very rough approximation that 1 m is 3 ft and 99 is close enough to 100 to not worry about the difference. Again, we're just trying to get a feel for the size of the project.)

At this point you may be thinking, as I did, that this isn't bad. Surely a camera or two can see that area around the robot and see a sample. I went down that path for awhile before I realized the problem: **You have to search a new area every second!**

Aren't we? Nope! If the cameras are looking at the circle around the robot than the new area is toward the front. Any area toward the rear has already been examined. In fact, when looking at a circular area only the area immediately inside the front of the scan circle is new area. Visualizing that and calculating that area and its impact on the search gets into more detail than we need right now, so lets simplify.

The examination area we'll consider is the 2 m wide rectangle immediately to the left and right of the robot. Think of the area as wings stretching to the side of the robot. That area has to be 2 m wide and cover 11.1 m^{2} so the length is (11.1 m^{2} / 2 m) 5.6 m or roughly 3 m to each side of the robot.

It is useful now to touch base with reality by asking if a camera can see a sample if it is 3 m to the side of the robot. Surely a camera can see a sample across a typical bedroom so, fortunately, the answer is, it can. A better question to ask, at this point, is can the camera see a sample at twice that distance? The answer is, probably. Why twice that distance? Because with all our simplifications there are other issues that will consume time that are not being considered. By doubling the distance we reduce the search time required to (7200 sec / 2) 3600 secs thereby giving us a cushion for the issues we know we will encounter but just don't know what they are. (Hints: turning, time to pickup a sample, time to get to a sample, time to get back to the base platform.)

There is still a lot unknown about this aspect of the search but it at least seems feasible.

The calculations on this page, as for all other pages, are in a spreadsheet.