andrus said:
Hixee said:
Now let's imagine these two coasters (with identical trains, heights and drop heights) have the same maximum drop angle, however one is like the SheiKra and has a very sharp radius and stays at the drop angle for a long time, the other is like Millennium Force and has a big sweeping radius and only stays at the drop angle for a moment. Using the same logic as above, the train on the sharper drop will enjoy less friction on the rails more quickly than the other train. The first train may be at 80° while the second train is only at 40°. It's hard to explain, but I hope I'm making sense?
Wouldn't it be the opposite? As long as a train is travelling on a flat piece of track (and the descent angel is less than 90°) there will always be a force on the track. However; if the coaster has a long sweeping radius à la Millenium Force the coaster train will be floating down the drop and experience 0G, ande hence the friction in the track will be minumum.
I don't really know! :lol: I sat and thought about it for a while, and eventually managed to convince myself of the 'answer' I gave before. However, I can see your point.
If we neglect the fact that the wheels maybe be spring-loaded (i.e. forced to stay in contact with the track, like on B&M wheel assemblies), the only force on the track from the train comes from the train's own weight. When falling vertically, the train (in theory) doesn't
need to be fastened to the track - in other workds the track is really just guiding it down, the force on the track is relatively minimal.
On a drop with a very sharp top (like the Dive Machines) the train is essentially falling vertically, with little friction between the track and the wheels, until the pull out. On a large sweeping drop (like on Millennium Force) the train is resting on the track while it accelerates, so there is always friction between the rails and the wheels. Then it pulls up, and the wheels are back in contact again. This does neglect the fact that the track will be experiencing a reduced force due to the 'zero-g' effect of the drop of course.
Now, I reasoned that (again, assuming the same trains, lift hill speed, etc) the vertical drop with the sharp entry would mean the train experienced less friction with the track, and therefore would accelerate faster. In reality, I don't suspect the difference would be all that much - and probably not noticeable in the small differences in drop angles of BigBad's original coasters. Overall, it's an extremely complex system to analyse. I suppose I could model two drops in NL2 and see what happens, that has a pretty good physics model! Truthfully, I don't know. I'm just hypothesising really, and like I said, I suspect the actual differences this would make are very minimal.
andrus said:
Another aspect of this, which I actually don't know and hence is more of a question, is whether a higher force exerted on the track also means higher friction? If so, a steeper pull-out from a drop (like that out of a dive machine) could potentially mean less speed at the bottom of the first drop than the speed of a coaster with a more shallow drop and hence also shallower pull-out angel.
Yes, higher force on the track equals higher friction.
Friction is a function of the normal (i.e. perpendicular) force between two objects. That's why, for example, an empty cardboard box is easier to push than a full one - there's more force on the ground, and so more friction (the force due to friction is a multiplier, called the coefficient of friction, of this force). It's also why pushing a box down a hill is easier. As the friction force is dependent on the normal (perpendicular) force between the two objects, the steeper the incline is the less force is acting between the slope and the box, and therefore the friction reduces (of course, if you're pushing a box up a hill, even though the friction force decreases, you're having to also push more of the box's own weight). I can do some diagrams if this doesn't make sense in text form. Anyway, I digress... As the friction force is a function of the [normal] force on the track, a more forceful pull out will mean more friction.
Now, where it get's complicated is when you think about how that friction force manifests itself. When pushing a box, it's simply the action of trying to move the two surfaces relative to each other that causes the friction. For a coaster train with rolling wheels and bearings, it's a bit more complicated. I think there are probably two or three main areas where the friction will manifest itself:
Rolling resistance - This is basically what I was describing above. Even though the wheels do roll, there is a friction force required for that to happen - if there wasn't, the wheels wouldn't roll, they'd just skid. That does therefore mean that there's a loss associated with the wheels rolling.
Deformation of the wheels - The rubber (polyurethane) on the wheels deforms under load and warms up. This is a manifestation of friction, both between the wheels and the track, and within the material itself. The higher force, the more deformation and therefore the more energy being absorbed in the rubber (and eventually re-emitted as heat). This is also true of the steel wheel 'hubs', but the effect is much less due to the stiffness of the steel. This is very apparent on coasters like I-305 which have to cool the wheels after each run, whereas on Great Pumpkin Coaster (picked as an example at the same park) they don't, even though their wheel make up is the same - the forces on the wheels are much less, and therefore don't cause as much energy loss in the wheels.
Bearing friction - This is actually a very complicated one to analyse, and I'm not going to go into it too much, but a bearing's resistance changes with the forces applied to it. Up to a point, the forces will help - they'll help lubricate the bearing and keep the grease warm and 'runny'. However, if you put too much force through a bearing you start to get strange things happening to the grease. The grease will begin to solidify, cavitate and generally start to do damage to the bearing. I studied some of this in my final year, so it's not straightforward to explain!
How that all impacts on the final speed at the bottom of the drop is complicated, but the simple answer is that I don't really know. I think for the most part, two drops with similar heights (within 5-10 ft) and similar drop angles (within 3-5°) will have final speeds so similar that they aren't worth really worrying about. As said previously, the effect of wind and 'time-of'day' will probably have a much more significant effect than most of the stuff I've touched on above. They will have an effect, but I suspect that on average the effect they do have is pretty similar for similar types of coaster.