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Verifying Stats

BigBad

Mega Poster
Who verifies stats? I'm curious if, for instance, Cedar Point could have gotten away with MF's drop being 299 feet instead of their claimed 300. (I've never heard this as a rumor, but it seems like something that could happen.)
 

GuyWithAStick

Captain Basic
I think it's in the original simulation, the given stats.

Or you can go up top of MF's lift, and drop a 300 foot tape measure off and see if it touches the ground. Granted, you'd probably get a park ban for life, but at least you can rest easy. :p
 

Hixee

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No one, I suspect.

Rides that are breaking a record will be verified by the body that gives the award. How they do this, I'm not sure, but I highly doubt they get a tape measure out. In construction (well, engineering in general) the design drawings would be enough to prove this. The design drawings have to be followed to make sure the ride is built properly, so simply reviewing the final plans (likely more complex plans that we'd ever get our hands on - I'm talking proper construction detail 'blueprints') would probably be enough to award the coaster the record. That being said, I doubt Joe Bloggs at The Local Newspaper reporting on the "fastest new coaster on the West Coast" is going to have access to these. He'll just take someone's word for it.

For rides that aren't breaking a record, why does it matter? I mean, it's not like I'd be able to tell the difference between a 299ft drop and a 300ft drop... :lol:
 

BigBad

Mega Poster
Height wasn't a good example because, yes, that's locked in by the blueprints. I started thinking about this when I saw some stats about speed, though.

According to Wikipedia
Bizarro: 221ft at 72 degrees = 77mph
Nitro: 215ft at 68 degrees = 80mph
Behemoth: 230ft at 75 degrees = 77mph

Nitro, which I really like, is listed as having the shortest and shallowest drop, yet it's also listed as the fastest. There's more to it than just the drop height and maximum angle, and while I've studied a lot of science in my life, I've managed never to have a serious discussion about how air resistance works, but that struck me as strange, particularly comparing the two B&Ms.
 

Hixee

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2-3mph difference on a ride breaking 75mph is really not as significant as it might seem at first. An average head wind could easily slow a train down by that much, or a tail wind speed it up. The drop height and angle do matter, but the physical difference between a drop of 68° and one of 72° is probably quite insignificant.

Air resistance isn't a particularly straightforward phenomenon, and when interacting with wheel friction, train weight and drop angle becomes a complex thing to analyse. However, the general rule of thumb is that the resistive force exerted on the train is a function of the velocity squared. A few miles an hour difference can have surprisingly big effects on the forces felt by the trains (for example, increasing the speed of Behemoth from 77mph to 80mph increases the force due to air resistance by ~8%, that's quite significant). It's worth noting that these differences can easily come from the wind!

A thought experiment. Imagine two coasters with identical trains (aerodynamic profile, weight, bearing friction etc) and exactly the same height and drop height. One has a vertical drop (90°), the other a very shallow drop (lets say 20°). In theory, a train dropping on a vertical section of track will only see the air resistance slowing it down. There won't be any friction as there's no force on the track, and therefore no force to cause friction. When the drop levels out is the only the time the train experiences friction on the track. Now think about what happens to the train on the shallow drop. All the way down the drop, it's experiencing the friction of the rails AND the air resistance. It's bound to be going slower by the bottom, as it's had more forces acting on it all the way down the drop, even though the starting heights are the same and at any given speed the force due to air resistance will be the same.

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?

So, what if Nitro happens to get to it's 68° quicker (ie. has a sharper radius at the top of the drop) than Behemoth? Can you see how the example above comes into play here?

Of course, there are tons of other factors that come into play. To list a few (that come to mind straight away):

Chain speed - This could easily add a couple of mph to one coaster vs another.
Wind - Again, a 2-3mph headwind can have drastic effects on the speed.
Bearing friction - I couldn't speak for the numbers exactly, but there are probably noticeable differences between the rolling resistances of the bearings on 'cold-first-run-of-the-day' compared to 'hot-run-mid-afternoon'. I mean, we know it's true rides 'warm up' during the day, right? We've all experienced it.
Train weight - This one is interesting, as the actual impact of train weight is complex. Lighter train means less friction, but less momentum to hammer through the wind resistance. Heavier train will experience more friction, but will be able to push through more air resistance. (Imagine a toy car vs a real car rolling down a hill. The toy car will accelerate faster, but as soon as the ground levels out it'll probably stop more quickly too).
Train length - This is another interesting one. A long train (like those on Bizarro and Behemoth, compared to Nitro) will experience 'hang-time'. A short train doesn't. This probably means that the train is able to accelerate faster into the steepest section of the drop. Whereas a long train might still being doing the lift hill speed by the time it gets to the steepest section of drop. Of course, the rest of the train will eventually follow, but maybe that slight delay in accelerating is enough to slow it down a bit more? I'd need to have a good think about that one, as it's difficult to think about quickly.

In summary, this is a very complex question and there isn't a straight answer. I've gone on a tangent, but I'm sure you can appreciate the point I'm getting at? Trains speeds vary from train to train, day to day and coaster to coaster. It's probably just a one of those things that Behemoth is slower than Nitro despite being taller and steeper. 3mph isn't enough to notice at 80mph (although I realise the irony of this lengthy post relating to this subject).

Finally, a quick thought on how they measure it? The design software will probably give a pretty good estimate - it'll have basic friction and air resistance models built into it (in other words it has a series of factors that it applies, rather than running a CFD and FEA analysis on every new coaster). This sort of thing is available in NL2, so they sure as hell have it on their proper software. The trims and sensors are able to give speed readings pretty easily, so they could use those, or they just get a radar gun (like the police use to catch speeding motorists) and actually measure it. I don't suspect anyone really 'verifies' that they've got it exactly right, as they're unlikely to get the same reading twice. I think it's more of a guideline "this coaster is in the 70+ category". And finally, does it really matter if they're a few mph out?

I presume that didn't really answer your question, but maybe... :lol:
 

BigBad

Mega Poster
The physics behind all of this is definitely cool. I've had an interest in curvature and mechanics for a long time, though I never got into physics deep enough to have serious discussions about how air resistance works or what happens when we realize that objects in life AREN'T points.

I kind of had a feeling that no one verified most stats.
 

Hixee

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BigBad said:
or what happens when we realize that objects in life AREN'T points.
In my experience, we then do everything we can to try and model them as points again. At the very least regular cuboids... :lol:
 

andrus

Giga Poster
Very interesting read Hixee! =D>

I never got this far into physics, but I still have some thought on the topic:

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.

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.

Hixee said:
Bearing friction - I couldn't speak for the numbers exactly, but there are probably noticeable differences between the rolling resistances of the bearings on 'cold-first-run-of-the-day' compared to 'hot-run-mid-afternoon'. I mean, we know it's true rides 'warm up' during the day, right? We've all experienced it.
This is probably it in the case of the three coasters BigBad posted. I know that B&M uses wheel bearings with different resistance due to control the speed througout the course. I bet they used wheels with lower friction on Nitro to make the coaster run faster (or at correct speed) in the final sections of the ride. While on Behemoth (which is a later coaster) they probably played it safe with a slightly taller lift hill and higher friction wheels instead. Or it could be that the final helix of Behemoth pulled too many G's, and B&M put on higher friction bearings to reduce the speed at the end of the ride to lower the G's here. Two plausible scenarios. Bizarro is from another manufacturer, but i assume Intamin employs the same strategy.
 

Hixee

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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.
 

BigBad

Mega Poster
I think the issue of the sharp dive vs. the sweeping MF-style dive is that it takes longer (even if just slightly) to get down.

***MATH***
It's more complicated than what I'm about to say, because the vector field changes as time progresses (higher speed equals more wind resistance, plus whatever happens with the wheel friction), but this is a line integral through the wind resistance field, so the longer path results in more opportunity for forces to slow the train.
***MATH***

The sweeping drop, however, can produce airtime, and the zero-G moment provided by the curvature could reduce friction, since the 100,000 N from the loaded train is no more. Even if it's spring-loaded, now it's just the force from the springs.
 
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