¡dn ʇǝƃ ʇ,uɐɔ ı puɐ uǝllɐɟ ǝʌ,ı
Canada Chestermere Alberta
There are 10 kinds of people who understand binary: Those who do, and those who don't.

Tangentially from The Riddler, however. One of the links below their main article got me to this interesting puzzle:
You are in an airport, and you are walking from security to your gate. The distance includes stretches of travelator, or moving walkway, and stretches of floor. You have a constant walking speed, u, and the travelators also have constant speed, v. So, when you are on the travelator you are traveling at u + v.
The Question: You have enough energy to walk at faster speed, w, for a fixed time. In order to get to the gate quicker, do you increase your speed when you are on the travelator, off the travelator, or does it make no difference?
Spoiler (click to reveal) You get to the gate quicker if you temporarily walk faster when on the floor.
u: human speed
w: higher human speed
v: travelator speed
Df: Floor distance
Dtr: travelator distance
Let T = time spent at higher speed w.
Total time if speeding on floor: T + (Df – Tw)/u + Dtr/(u + v)
Total time if speeding on travelator: (Df/u) + T + (Dtr – T(w + v))/(u + v)
Since w > u then w/u > (w + v)/(u + v) and this* shows that total time is less when you walk fast on the floor.
*somehow

Andy Andersen
United States Michigan

It's just easier to take the train. All this math gives me a headache

David Jones
United States Wilsonville Oregon

MABBY wrote: *somehow
This somehow reminded me of one of the many controversies that ended up in Marilyn Vos Savant's columns many years ago. The question (simplified) was: Suppose a race car goes around a track in one minute at 60MPH. How fast does he have to go the second time around the track in order to average 120MPH for both trips, to which the answer is, its impossible. To travel the track at 120MPH, you would have to do two laps in one minute, but your one minute is already over. You can't average speeds the way you average other numbers. But the point I want to move towards here is that most people really don't understand how numbers behave when you start increasing denominators in fractions. Since speeds are derived from fractions (time = distance / speed), both the racetrack problem and your problem boil down to issues of how you manipulate denominators. If you happen to remember graphing the equation y=1/x, you will recall that moving twice the distance across the horizontal axis does not push the curve down twice as far. This isn't a behavior that the average person sees in ordinary day to day life, so answers to these kinds of questions are not intuitive without some additional math literacy. That said, this might be an easier way to think about your problem:
Spoiler (click to reveal) Use real numbers instead of variables. Suppose you are going 10 mph and can add another 10 mph to your speed. You've doubled your speed, so now you go twice as fast. Now suppose you are going 50 mph and can add the same 10 mph  you've not doubled your speed. Going from 50 to 60 only increases it by 20%. If you only get to use that boost once, its better to use it when it doubles your speed.

Josh Jennings
United States San Diego CA

Yeah, the math checks out. I decided to calculate an example and these are the results I got:
u (walking speed) = 5 feet/second w (running speed) = 10 feet/second v (travelator speed) = 10 feet/second
Df (Floor Distance) = 2000 feet Dt (Travelator Distance) = 1500 feet
T (Time you can spend running) = 30 sec
No running: Time taken on floor + Time taken on travelator Df/u + Dt/(v + u) 2000/5 + 1500/(10 + 5) 400 + 100 = 500 seconds
Running on travelator: Time walking on floor + time running on travelator + time walking on travelator Df/u + T + (Dt  T*(v+w))/(v + u) 2000/5 + 30 + (1500  30(10 + 10))/(10 + 5) 400 + 30 + (1500  600)/15 400 + 30 + 60 = 490 seconds
Running on floor: Time running on floor + time walking on floor + time walking on travelator T + (Df  (T*w))/u + Dt/(v + u) 30 + (2000  (30*10))/5 + 1500/(10 + 5) 30 + (2000  300)/5 + 100 30 + 340 + 100 = 470 seconds

Jack van Riel
Netherlands Tilburg It's wurples all the way down.
My heart is a fish / Hiding in the watergrass / In the green, in the green.

It depends on the direction of the travelator.
I think it's easier to look at it not as where does the faster speed have a bigger effect, but how much speed do you need for the same effect. In one case you need some desired percentage times u, and in the other that percentage time u+v.

¡dn ʇǝƃ ʇ,uɐɔ ı puɐ uǝllɐɟ ǝʌ,ı
Canada Chestermere Alberta
There are 10 kinds of people who understand binary: Those who do, and those who don't.

It would be an interesting twist if you calculated things if you stopped walking, once you got onto the travelator, but then you'd need to know the actual physical length of the travelator and the distances to it from the start point and the distance to the airport gate after you got off of it and continued walking.

Andy Andersen
United States Michigan

Actually the problem really doesn't exist. There is no need to get there in a hurry
When was the last time a plane took off on time?

Matt
United States Central Coast California
0110100110010110

Orangemoose wrote: Actually the problem really doesn't exist. There is no need to get there in a hurry
When was the last time a plane took off on time?
The only time that happens is when I have a 40minute layover in Phoenix (which has dwindled to 25 minutes after struggling to get off the first plane) and I have to run across the entire airport.
Then my plane leaving on time is the MOST important g**dammed thing on American Airline's list of things to do.

Geeky McGeekface
United States Manassas Virginia
Calling all Spielfrieks! It's time to vote for the Meeples Choice Awards at the Yahoo User Group!

I think the key insight is that by walking faster when on the floor, you minimize the amount of time that (slower) part of the trip takes. The less time you spend on the part of the journey where you aren't being assisted by technology, the better.

wayne mathias
United States Niceville Florida

Let's rewrite the problem.
A 30 mile river. 15 miles with a 5mph current = walking. 15 miles with a 15mph current = travelator. You can row at 10mph for 1 hour.
If you row during the 5mph section that 15 miles will take 1 hour and the other 15 miles also 1 hour for 2 hours total.
If you row during the 15mph section it takes 36 minutes for that 15 miles, then for 24 minutes you are still rowing in the 5mph section (going 6 miles) then the remaining 9 miles take an hour and 48 minutes. 2 hours and 48 minutes total.
So you row (or run) where the speed is otherwise slowest.

"L'état, c'est moi."
Canada Vancouver BC
Roger's Reviews: check out my reviews page, right here on BGG!
Caution: May contain wargame like substance

Hey Mabby, you know about this show, right?
This episode is about a saving someone in a pool when you run along the pool at x and swim at y, what's the optimal point to hop into the pool...


