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02 January 2011

More roller coasters -- do heavier riders go faster?

When I went to amusement parks as a middle- and high- schooler, my friends and I strategized carefully for the log flume.  We always put the heaviest people in the front.  Our reasoning:  we figured that way we'd go faster and get wetter.  Funnily enough, I don't remember ever performing a controlled experiment, or recording our results.  Perhaps I should return my Nerds of America membership card.

In retrospect, I think we had half of our conclusion right. 

I *think* that putting heavier people in the front does cause a bigger splash when the log flume hits the bottom, because the front end penetrates the water sufrace more easily -- fluid resistance generally depends on mass and surface area.  I have a wee bit of experimental evidence from last week's Disney trip. 

"Splash Mountain," pictured to the right, is essentially a sophisticated log flume.  As the temperature in Orlando had risen only to about 50 degrees farenheit, we wanted to minimize how wet we got.  And, as the temperature had risen only to about 50 degrees, we could ride pretty much as many times in a row as we pleased -- hardly anyone else was dumb enough to be in line.  We always sat in the last of the four rows.  It seemed as if there was a bigger splash when the front rows were occupied, though I cannot quantify the effect.  The only statement I can make for sure is that when people sat in the front row, they always got substantially wetter than we did.  So teenaged Greg was possibly correct about heavier people in front leading to wetter riders.

On the other hand, we were likely dead wrong on the issue of speed at the bottom of the ride.  The analysis will be different, I think, for a log flume and a roller coaster.  For a log flume, where the "log" is not particularly long compared to the length of the hill, one can probably consider the log as a point mass released from rest at the top.  The frictionless case is well studied in introductory physics -- the speed at the bottom will be root 2gh, where h is the vertical height of the hill, and so does NOT depend on the mass of the riders or the mass distribution.  (With air and water resistance the answer may change, but I'm not exactly sure how or how much.)

The roller coaster problem is more interesting.  Consider Big Thunder Mountain Railroad (covered in yesterday's post).  We do have to consider the mass distribution.  

On BTM, the train is hoisted up to the top of a hill, then released approximately from rest -- but not until most of the train has already gone well over the crest of the hill.  If the train is empty, or if the train has an approximately uniform distribution of riders from front to back, then its center of mass is somewhere near the center; the speed at the bottom will be root 2gh, with h measured from the center of mass's position at the top of the hill to the bottom of the hill.

Consider what happens, though, if we load up the heaviest riders in the front.  That moves the center of mass closer to the front.  When the train is released at the top, the center of mass will be lower than for a uniform rider distribution.  Thus, the h term is smaller, and the speed at the bottom is smaller.

If my analysis is correct, then the fastest speed should be obtained by putting heavy riders toward the back, so that they are right on top of the hill when the hoisting mechanism lets go of the train.  I don't have any experimental evidence, not even anecdotally.  (That's primarily because 37-year-old Greg can't ride even mild roller coasters more than once or twice without barfing.)

I know that amusement park physics is a popular endeavor.  A decade ago, I even took a few classes to the parks myself.  Now that we have wireless, portable, standalone probes like Labquest or the equpiment with Pasco's iphone app, it sure seems like hard data should be easy to acquire.  Has anyone done this?  I'd love to find out...



  1. Yes ur thesis is sound; more mass up front moves the center of mass up front, making for a faster/better transfer of energy to the water. Result: BIG SPLASH!

  2. I thought even with friction acceleration down a hill only relied on the angle of the hill?

    mgcos(theta) - mu mgsin (theta) = ma