So, I was writing an intensely long physics lesson for y'all about what some of these numbers mean, but I don't know that it'll be useful since I doubt anyone would get through it. That said, let me share its highlights:
-I've studied mechanically engineering and am responsible for the design and manufacture of competition robots.
-I teach high school students about physics and engineering while I do this.
-I don't know anything, but am pretty capable of inferring things; at least, moreso than many.
-Everest is listed as a 120' lift.
-Given that, a 112' drop represents only an 8' net descent throughout the first part of the ride, including both switchbacks and an upward helix.
-All roller coasters have a 'friction slope,' or the average rate of descent required over a given length to ensure the train is not slowed by friction to such a point where it stops moving forward. Failure to meet this friction slope means your ride will not work -- unless it's a shuttle coaster, anyway. If you look at Everest in part, it can be a shuttle coaster due to its switchbacks, but as a whole, it is a continuous circuit and must answer to the friction slope.
-No point of the coaster can be higher than its lift. Newton said so.
-The second switchback is just before the drop and is, therefore, at or above a height of 112'.
-A 1/2% friction slope would mean that there's no less than 1600' of track before the drop -- 6" net descent per 100' of track.
-Without seeing the blueprints, the Common Point files, or knowing both Vekoma's typical friction slope and the length of track between the top of the lift and the top of this drop, there's no good way of inferring its true height. Even with that information, it'd require some pretty big assumptions.
Also, for the record, turns will not slow a coaster down appreciably more than it would slow in a straight line. Friction in the wheel assemblies and friction between those wheels and the track slow the train, as well as friction against the air. In reality, a train passing through a curve is typically exerting something greater than 1G of force through its wheels and the resultant force of friction increases as a result. Thus, appropriately banked curves can slow the ride down a bit faster.
The S-turns you referred to when speaking about aircraft are not, to my knowledge, to reduce airspeed but to increase separation between aircraft. If a plane ahead hasn't cleared the runway quickly enough, or if the following aircraft is traveling much faster than that in front of it, the separation must be increased or maintained, respectively, to fall within FAA guidelines and keep things safe. The plane travels at the same speed, but does so over a longer distance, thus taking longer to reach its destination.
I wouldn't say unequivocally that the turns do not do something to slow the plane down when examined in isolation, however. I just don't know enough about airplane behavior to go that far.
