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| picture courtesy of howstuffworks.com |
In the 1998-99 and 99-2000 school years, Justin Kriendel attended my AP physics classes. He was, even then, a roller coaster expert; by now, he says that he's ridden over 150 different roller coasters in the United States and abroad. Justin was particularly interested in the physics of roller coasters. In fact, for a reserach project he built a simple "linear synchrous motor," the same kind of motor that causes a roller coaster to shoot uphill out of its starting gate.
Since I just wrote two posts about roller coaster physics (look back a couple of days to see these), I sent Justin a link and asked for his input. Below is his response.
Interesting question which, you probably guessed, I've put a lot of thought into over the years. My brief thoughts at this late hour are these:
If you think of the center of mass as a point on the train (as opposed to where it really is, which is somewhere beneath the train when it is cresting an inverted U-shaped lift hill, I would think), the train always begins to accelerate when that center of mass is subject to a downward acceleration. That is, the moment it reaches a downward sloping portion of the lift hill that is steep enough to overcome friction. So that center of mass, I would think, accelerates to a speed of root 2gh, where h is the vertical distance of the drop, regardless of how the weight is distributed.
Another way I've though of it is in terms of work. I would think that the chain lift does the same amount of work to lift a train of mass m to a height h regardless of how the weight is distributed. Assuming a lift hill that starts on a flat section of track and then begins its ascent, I think you can look at it in terms of power necessary at various points of the train's ascent. Compare two trains, one with weight in front, and one with the weight in the back, and think about what happens when each train first hits the lift hill. More power is required initially on the front-loaded train, since that weight hits the ascent first. The once the front begins to crest the lift hill, less power is needed to the push the rest of the train over the lift. For the back-loaded train, I feel like the opposite happens. Less power at first as most of the weight is in the back of the train, which takes some time to actually reach the ascending portion of the lift (remember, it starts flat). But for whatever time in the beginning where less power was needed because the weight of the train was still on a flat portion of the track, you make up for it when the chain lift needs to keep chugging with more power until the very end of the train crests the summit. I hope I'm making sense here. Either way, the total amount power necessary (and in turn, the total amount of work done) seems like it would be the same regardless of the weight distribution.
As for Splash Mountain, it's not the best example to test your theory. Disney greatly controls how wet you get. A lot of the splash is from jets on the side of the bottoming out portion of the drop. When it's cold, they just turn these jets off. They can also control the depth of the water at the bottom of the drop. Shallower water means less splash as well. But, assuming all other things equal, I think it's difficult to test this either way. Sure, when everyone's sitting up front, you're probably getting wetter. But I feel like this is because the splash is produced by the front of the boat hitting the water, regardless of where the weight is distributed. So with the splash being created by the front of the boat, it would make sense that those sitting at the front of the boat would get wetter, as that is where the splash "is". With a full boat, this is very evident. The people sitting up front often end up much wetter than those sitting in back, as they "block" the water from reaching those in the back. Which is why it's fun on a log flume to try and duck down low when you hit the water, much to the dismay of those behind you, especially when you're know they've strategically chosen to sit toward the back to avoid getting as wet. Or maybe just I enjoy doing that.
Regarding the post about "the back going faster," your analysis is exactly right. When you're going over a drop, for instance, the radius of that drop stays fixed, but the speed of the train is changing as it traverses any given point of track. So if a point on the track is the beginning of a drop, then the front of the train is going much more slowly over that point than the back of the train. Hence more negative G's (or "airtime") in the back of the train going over a drop. The reverse happens of course when you crest a hill: the front of the train experiences more airtime at the crest than the back of the train.
-- Justin Kreindel
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