Tag: Polar Express
I love the emails I get from readers of this blog. Yesterday morning, I got the following from Shane Schieffer of FitTrip and I just saw it as I was grinding through my morning backlog from the weekend.
“Hi Brad, Happy holidays. My friend and I were watching Polar Express with our kids last night, and he commented on the impossibility of skiing down the top of a train towards the engine, even on a crazy steep slope. I know you like to geek out on software and things such as figuring out the algorithms behind slip numbers, how about physics? The question was bothering me so I took a shot at it this morning. Thought I’d share the result. I bet users of your forum could top my work, or find errors, or extend it to further interesting observations. Kind of a fun holiday geeky thing to contemplate, so I thought I’d send it your way.”
This reminded me of a great story from Amy. On her very first trip to Dallas from Boston to meet my parents, she was sitting on a plane next to a guy who was doing a bunch of math sheet of paper. She asked him what he was doing. He said he was calculating how much fuel the airplane needed to get from Boston to Dallas. It turned out to be a guy named Christopher Couch, who was an undergrad at MIT that had crossed paths with me for some reason I can no longer remember. She and Chris had a great time on the plane together talking about all kinds of nerdy things. The entire memory made smile.
Oh – here’s the answer.
“Assuming just the train engine (not cars, and cargo) that the polar express was modeled after, which weighs 361,136kg and has an approximate cross sectional surface area of 10m^2, at freezing (0-deg Celsius) and 1atm of pressure (sea level) on a -128.5 gradient (what the sign said in the movie which equates to a -52 degree slope) assuming frictionless tracks, and a hill tall enough to induce free fall, where no braking nor engine acceleration is applied, and the coefficient of drag is based on a rectangular shape, the train would be traveling 1,434 mph. This is clearly much faster than the terminal velocity of a 6-foot tall man with a boy on his shoulders (say a combined 3m), standing up, weighing a combined 240lbs, with a cross sectional surface area of 2.25m^2, whose terminal velocity would only be 52.5 mph on that same slope. Of course the train itself would create a wind draft that would lessen the difference, but either way the man and boy are going off of the back end of that train. Unless the man is a spirit…”