Editing Talk:85: Paths
Please sign your posts with ~~~~ |
Warning: You are not logged in. Your IP address will be publicly visible if you make any edits. If you log in or create an account, your edits will be attributed to your username, along with other benefits.
The edit can be undone.
Please check the comparison below to verify that this is what you want to do, and then save the changes below to finish undoing the edit.
Latest revision | Your text | ||
Line 9: | Line 9: | ||
Where do the equations come from to figure out #2 & #3 - can anybody derive it? {{unsigned ip|108.162.219.185}} | Where do the equations come from to figure out #2 & #3 - can anybody derive it? {{unsigned ip|108.162.219.185}} | ||
:The equation #2 comes from the second route. t(1+√2)/3 is how far the second path takes the guy. If each block is a unit square, the diagonal to the corner is √2 while the next part is 1. The t/3 part is making it comparable to the first one (the first one is t despite it being 3 unit squares). Equation #3 is t√(5)/3. Plugging 1, 2, and √5 into wolfram|alpha for triangle side lengths makes it a right triange, so the √5 comes from the side length (assuming unit squares) while the t/3 makes it comparable to the first one.[[User:Mulan15262|Mulan15262]] ([[User talk:Mulan15262|talk]]) 02:36, 1 December 2014 (UTC) | :The equation #2 comes from the second route. t(1+√2)/3 is how far the second path takes the guy. If each block is a unit square, the diagonal to the corner is √2 while the next part is 1. The t/3 part is making it comparable to the first one (the first one is t despite it being 3 unit squares). Equation #3 is t√(5)/3. Plugging 1, 2, and √5 into wolfram|alpha for triangle side lengths makes it a right triange, so the √5 comes from the side length (assuming unit squares) while the t/3 makes it comparable to the first one.[[User:Mulan15262|Mulan15262]] ([[User talk:Mulan15262|talk]]) 02:36, 1 December 2014 (UTC) | ||
− | |||
Interestingly enough, if the three sides are equal in time taken (20 seconds each), the time it would take for path #2 would be 20rt2 + 20, and path three would be roughly 40, which comes out to 60, 48.28, and 40 seconds by using very simple geometry. {{unsigned ip|108.162.237.179}} | Interestingly enough, if the three sides are equal in time taken (20 seconds each), the time it would take for path #2 would be 20rt2 + 20, and path three would be roughly 40, which comes out to 60, 48.28, and 40 seconds by using very simple geometry. {{unsigned ip|108.162.237.179}} | ||
− | |||
− | |||
− | |||
− | |||
− | |||
− |