Talk:85: Paths - Revision history

42.book.addict at 23:00, 21 January 2025

2025-01-21T23:00:11Z

172.68.55.11 at 19:09, 24 November 2024

2024-11-24T19:09:57Z

172.70.91.78: Completion.

2024-10-08T17:38:15Z

Completion.

42.book.addict at 17:14, 8 October 2024

2024-10-08T17:14:38Z

CalibansCreations at 10:24, 2 October 2024

2024-10-02T10:24:40Z

DownGoer at 03:34, 26 June 2023

2023-06-26T03:34:21Z

Kazvorpal at 22:39, 14 August 2019

2019-08-14T22:39:59Z

172.68.62.4 at 08:39, 26 July 2018

2018-07-26T08:39:10Z

172.68.62.4 at 08:38, 26 July 2018

2018-07-26T08:38:21Z

Flewk at 17:10, 24 December 2015

2015-12-24T17:10:38Z

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 Along similar lines, [http://abcnews.go.com/Technology/story?id=97628&page=1 this mathematician's dog] uses Calculus (albeit at an intuitive, rather than mathematical level) to optimize the path that it takes to retrieve the ball from the water. -- mwburden [[Special:Contributions/70.91.188.49|70.91.188.49]] 21:27, 13 December 2012 (UTC)
-This particular situation is less interesting, since the walker's speed is the same for all three paths!  This is seen by the times being directly proportional to the distances.  Normally, the off-normal-path is at a lower speed, but some shorter path still gives the smallest time.DrMath 08:22, 14 October 2013 (UTC)
+This particular situation is less interesting, since the walker's speed is the same for all three paths!  This is seen by the times being directly proportional to the distances.  Normally, the off-normal-path is at a lower speed, but some shorter path still gives the smallest time.DrMath 08:22, 14 October 2013 (UTC) {{unsigned|DrMath}}
 Where do the equations come from to figure out #2 & #3 - can anybody derive it? {{unsigned ip|108.162.219.185}}

← Older revision		Revision as of 19:09, 24 November 2024
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	speedrun strats [[User:CalibansCreations\|CalibansCreations]] ([[User talk:CalibansCreations\|talk]]) 10:24, 2 October 2024 (UTC)		speedrun strats [[User:CalibansCreations\|CalibansCreations]] ([[User talk:CalibansCreations\|talk]]) 10:24, 2 October 2024 (UTC)
		+
		+	Randall Monroe attended Christopher Newport University, which has a large patch of grass ("The Great Lawn") surrounded by buildings much like this diagram shows. Would that be a worthwhile 'fun fact' to add to this entry? Granted, there have been no shortage of renovations and remodels in the the past 18+ years, so there might've been a more fitting grass feature at the time. [[Special:Contributions/172.68.55.11\|172.68.55.11]] 19:09, 24 November 2024 (UTC)

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 :Based on his times, it is two squares with the same side length. Base on that geometry, Path #2 will be the hypotenuse of a 45 degree right triangle. t = √(20^2 + 20^2) = 20 * √2 = 48.28. Path #3 would be t = √(20^2 + 40^2) = 20 * √5 = 44.72. Not sure where you got roughly 40 from. Were you thinking of the sin of 30 degree rule where the hypotenuse is double the opposite side? In this case, the adjacent is double the opposite which puts the hypotenuse at √5 times the opposite.[[User:Flewk|Flewk]] ([[User talk:Flewk|talk]]) 17:10, 24 December 2015 (UTC)
-Shouldn't this be added to Time management category? {{unsigned ip|172.68.62.4}}
+Shouldn't this be added to Time management category? {{unsigned ip|172.68.62.4|08:38, 26 July 2018}}
 What the hell, no unsigned templates here?

@@ Line 14: / Line 14: @@
 :Based on his times, it is two squares with the same side length. Base on that geometry, Path #2 will be the hypotenuse of a 45 degree right triangle. t = √(20^2 + 20^2) = 20 * √2 = 48.28. Path #3 would be t = √(20^2 + 40^2) = 20 * √5 = 44.72. Not sure where you got roughly 40 from. Were you thinking of the sin of 30 degree rule where the hypotenuse is double the opposite side? In this case, the adjacent is double the opposite which puts the hypotenuse at √5 times the opposite.[[User:Flewk|Flewk]] ([[User talk:Flewk|talk]]) 17:10, 24 December 2015 (UTC)
-Shouldn't this be added to Time management category? {{subst:Unsigned IP|172.68.62.4}}
+Shouldn't this be added to Time management category? {{unsigned ip|172.68.62.4}}
 What the hell, no unsigned templates here?

← Older revision		Revision as of 10:24, 2 October 2024
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	What the hell, no unsigned templates here?		What the hell, no unsigned templates here?
	Anyway, I'm wondering why there's any worry. Couldn't he just cut across the grass diagonally? Sidewalk-conformist. — [[User:Kazvorpal\|Kazvorpal]] ([[User talk:Kazvorpal\|talk]]) 22:39, 14 August 2019 (UTC)		Anyway, I'm wondering why there's any worry. Couldn't he just cut across the grass diagonally? Sidewalk-conformist. — [[User:Kazvorpal\|Kazvorpal]] ([[User talk:Kazvorpal\|talk]]) 22:39, 14 August 2019 (UTC)
		+
		+	speedrun strats [[User:CalibansCreations\|CalibansCreations]] ([[User talk:CalibansCreations\|talk]]) 10:24, 2 October 2024 (UTC)

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 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)
 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}}

← Older revision		Revision as of 22:39, 14 August 2019
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	:Based on his times, it is two squares with the same side length. Base on that geometry, Path #2 will be the hypotenuse of a 45 degree right triangle. t = √(20^2 + 20^2) = 20 * √2 = 48.28. Path #3 would be t = √(20^2 + 40^2) = 20 * √5 = 44.72. Not sure where you got roughly 40 from. Were you thinking of the sin of 30 degree rule where the hypotenuse is double the opposite side? In this case, the adjacent is double the opposite which puts the hypotenuse at √5 times the opposite.[[User:Flewk\|Flewk]] ([[User talk:Flewk\|talk]]) 17:10, 24 December 2015 (UTC)		:Based on his times, it is two squares with the same side length. Base on that geometry, Path #2 will be the hypotenuse of a 45 degree right triangle. t = √(20^2 + 20^2) = 20 * √2 = 48.28. Path #3 would be t = √(20^2 + 40^2) = 20 * √5 = 44.72. Not sure where you got roughly 40 from. Were you thinking of the sin of 30 degree rule where the hypotenuse is double the opposite side? In this case, the adjacent is double the opposite which puts the hypotenuse at √5 times the opposite.[[User:Flewk\|Flewk]] ([[User talk:Flewk\|talk]]) 17:10, 24 December 2015 (UTC)

−	Shouldn't this be added to Time management category?	+	Shouldn't this be added to Time management category? {{subst:Unsigned IP\|172.68.62.4}}
		+
		+	What the hell, no unsigned templates here?
		+	Anyway, I'm wondering why there's any worry. Couldn't he just cut across the grass diagonally? Sidewalk-conformist. — [[User:Kazvorpal\|Kazvorpal]] ([[User talk:Kazvorpal\|talk]]) 22:39, 14 August 2019 (UTC)

← Older revision		Revision as of 08:39, 26 July 2018
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	:Based on his times, it is two squares with the same side length. Base on that geometry, Path #2 will be the hypotenuse of a 45 degree right triangle. t = √(20^2 + 20^2) = 20 * √2 = 48.28. Path #3 would be t = √(20^2 + 40^2) = 20 * √5 = 44.72. Not sure where you got roughly 40 from. Were you thinking of the sin of 30 degree rule where the hypotenuse is double the opposite side? In this case, the adjacent is double the opposite which puts the hypotenuse at √5 times the opposite.[[User:Flewk\|Flewk]] ([[User talk:Flewk\|talk]]) 17:10, 24 December 2015 (UTC)		:Based on his times, it is two squares with the same side length. Base on that geometry, Path #2 will be the hypotenuse of a 45 degree right triangle. t = √(20^2 + 20^2) = 20 * √2 = 48.28. Path #3 would be t = √(20^2 + 40^2) = 20 * √5 = 44.72. Not sure where you got roughly 40 from. Were you thinking of the sin of 30 degree rule where the hypotenuse is double the opposite side? In this case, the adjacent is double the opposite which puts the hypotenuse at √5 times the opposite.[[User:Flewk\|Flewk]] ([[User talk:Flewk\|talk]]) 17:10, 24 December 2015 (UTC)

−	Shouldn't this ba added to Time management category?	+	Shouldn't this be added to Time management category?

← Older revision		Revision as of 08:38, 26 July 2018
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	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}}
	:Based on his times, it is two squares with the same side length. Base on that geometry, Path #2 will be the hypotenuse of a 45 degree right triangle. t = √(20^2 + 20^2) = 20 * √2 = 48.28. Path #3 would be t = √(20^2 + 40^2) = 20 * √5 = 44.72. Not sure where you got roughly 40 from. Were you thinking of the sin of 30 degree rule where the hypotenuse is double the opposite side? In this case, the adjacent is double the opposite which puts the hypotenuse at √5 times the opposite.[[User:Flewk\|Flewk]] ([[User talk:Flewk\|talk]]) 17:10, 24 December 2015 (UTC)		:Based on his times, it is two squares with the same side length. Base on that geometry, Path #2 will be the hypotenuse of a 45 degree right triangle. t = √(20^2 + 20^2) = 20 * √2 = 48.28. Path #3 would be t = √(20^2 + 40^2) = 20 * √5 = 44.72. Not sure where you got roughly 40 from. Were you thinking of the sin of 30 degree rule where the hypotenuse is double the opposite side? In this case, the adjacent is double the opposite which puts the hypotenuse at √5 times the opposite.[[User:Flewk\|Flewk]] ([[User talk:Flewk\|talk]]) 17:10, 24 December 2015 (UTC)
		+
		+	Shouldn't this ba added to Time management category?

← Older revision		Revision as of 17:10, 24 December 2015
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	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}}
		+	:Based on his times, it is two squares with the same side length. Base on that geometry, Path #2 will be the hypotenuse of a 45 degree right triangle. t = √(20^2 + 20^2) = 20 * √2 = 48.28. Path #3 would be t = √(20^2 + 40^2) = 20 * √5 = 44.72. Not sure where you got roughly 40 from. Were you thinking of the sin of 30 degree rule where the hypotenuse is double the opposite side? In this case, the adjacent is double the opposite which puts the hypotenuse at √5 times the opposite.[[User:Flewk\|Flewk]] ([[User talk:Flewk\|talk]]) 17:10, 24 December 2015 (UTC)