Editing 85: Paths

{{comic
| number    = 85
| date      = April 5, 2006
| title     = Paths
| image     = paths.jpg
| titletext = It's true, I think about this all the time.
}}

== Explanation ==
This comic centers around the consideration of what is the shortest path available to a person traveling by foot. [[Cueball]] has to travel across a rectangular distance, which has an established path around the periphery. When Cueball follows these paths, he has to walk for 60 seconds. He realizes that by ignoring the paths and taking the {{w|Desire path|desire lines}} from corner to corner, his route will be shorter, and he calculates that he could cut up to 26% of his time. As a result, every time he has to travel this rectangle, he worries about the extra time taken as a result of following the path.

Each path has labels for the time it takes (e.g. Path 2 takes 48.2 seconds) and the time compared to the longest path (e.g. Path 3 takes 74% as long as Path 1). Each path also has a corresponding equation for in the upper-right corner representing the time each path would take if Path 1 takes ''t'' seconds (instead of 60).

===Paths===
Each path represents a different way of traveling on/through the two squares that make up the rectangle.

Path 1 takes the long way around both squares. It takes 60 seconds in total, meaning it takes Cueball 20 seconds to walk across each of the three sides. By definition, it takes ''t'' seconds to walk the whole path and ''t''/3 seconds to walk each side.

Path 2 takes the long way around one square but cuts diagonally across the other. Since each side takes 20 seconds, the total time is 20 (one side) plus 20√2 (the diagonal) seconds, which adds up to about 48.28 seconds. This is about 80.5% of the full, 60-second path. More generally, it takes t/3 + (t/3)√2, or t(1+√2)/3, seconds to walk the second path (though the percentage never changes).

Path 3 cuts diagonally across the rectangle. The total time is the length of the diagonal, which is 20√5 (44.72...) seconds, per the Pythagorean theorem. This is about 74.5% of the the full path. Generally, it takes t√5/3 seconds to walk path 3. (As with path 2, and any other path that scales linearly with the full path length, the percentage doesn't change.)

==Transcript==
:[Blueprint of a campus. Two buildings in the upper and lower left corners, respectively, and a rectangular lawn. A road encloses the lawn, another road traverses horizontally through the center of the lawn. The character is in the lower left and the upper right corner, where it says "my apartment".]
:[Dashed line 1, from the lower-left along the road to the top-left corner, then to the top-right corner.] 60 seconds
:[Dashed line 2, from the lower-left along the road up to the center crossroads, then diagonally over the lawn to the top-right corner.] 48 seconds (80%)
:[Dashed line 3, diagonally from the lower-left to the top-right corner.] 44.7 seconds (74%)
:My apartment
:#1=t
:#2=(t*(1+√2))/3
:#3=(t*√5)/3
:When I'm walking, I worry a lot about the efficiency of my path.

{{comic discussion}}
[[Category:Geometry]]
[[Category:Comics featuring Cueball]]
[[Category:Maps]]
[[Category:Time management]]
[[Category:Comics with lowercase text]]