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Kings Island | Banshee | B&M Invert

Re: Kings Island | "Unknown" | "Unknown"

^ Normally, never.

However, Cedar Point saw a web leak similar to this one week before announcing Gatekeeper last year. I would imagine marketing and IT to double down on their efforts within the chain to ensure it wouldn't happen again. The coincidence of it is simply too good.
 
Re: Kings Island | "Unknown" | "Unknown"

1. Someone who decides to build a ride 199ft tall is a dick
2. It's still not as TALL as Wild Eagle (blah blah drop blah blah terrain)

It would be genius from King's Island if this is all a wind up; first they put out a very promising inverter layout, but most are a little disappointed it's not a giga, then they release the wingrider thing and we all go back to hoping it's an inverter, but it turns out to be a giga all along. Unfortunately I have a feeling this is real. Still, the comb makes no sense...
 
Re: Kings Island | "Unknown" | "Unknown"

I would honestly prefer to have a wing coaster just an hour away from my house (just one look at my top 10 shows that I'm a total sucker for wing coasters :p), especially one as massive as GateKeeper.

However, I'm taking this with a grain of salt. Those blueprints looked professional, and this Bat logo looks very professional, but since they contradict each other, at least one has to be fake, and I think it would be easier to fake the Bat logo than those blueprints.

And I do agree that some of the theming could hint towards this Bat coaster, but the one thing that almost completely kills this is the comb. Only way this could be true with the comb is that Kings Island is deliberately screwing with us enthusiasts, and it seems like a long way to go to throw off a couple of guys on a forum.

I hope it's real, though! I seem to be the only person here that liked GateKeeper. While Wild Eagle had better forces, GateKeeper was the better ride, and if they could combine GateKeeper's near misses and massive size with Wild Eagle's forces this could be amazing!
 
Re: Kings Island | "Unknown" | "Unknown"

Not sure if its been mentioned, but if you look at the layout picture it's gatekeeper...
 
Re: Kings Island | "Unknown" | "Unknown"

^FPO = For position only. It's just an image reference
 
Re: Kings Island | "Unknown" | "Unknown"

I don't think the 14-minute video of planning documents was a red herring, so if this is a Wing, it'll be the first one to have a vertical loop...

Yes/No?
 
Re: Kings Island | "Unknown" | "Unknown"

^^ Nope, wild eagle had a vertical loop straight after the initial drop

But to be honest, Kings Island is probably yanking our chains because for one a 55 degree angled drop on it? I don't know wild eagles or fly over the rainforest's but that seems rather low. And the coaster would more than likely travel faster than 70 mph cosidering the height of the drop according to the "source"
 
Re: Kings Island | "Unknown" | "Unknown"

I'll be disappointed with an invert and disappointed with a wing rider as well, just because wing riders don't really provide the intensity or thrill that I look for in a coaster, and inverts are just out of style IMO. I'm too picky. I'll second Casio's comment about the 199 ft tall thing though. That would bug the living crap out of me if they decided not to make it 200 ft.
 
Re: Kings Island | "Unknown" | "Unknown"

Hyde244 said:
Snoo said:
Bat name can EASILY be drawn to all that.. halloween and such are synonymous with Bats.
Click to expand...
Well, not really.

A Banshee is an irish mythological creature, a woman who shrieks when someone is about to die. Comb, celtic, usage of the verb shrieking.

I cannot see how Bat can be linked to the combs with bits of hair that were sent to the local press, or the new celtic-named tombstones that have appeared.

For the record though, this Bat logo is very cool.
Click to expand...

Ever heard of the Ahool? It is a cryptid that is said to be very bat like. Just watch this and see how it works :3

[youtube]http://www.youtube.com/watch?v=mV4iaMC3yJI[/youtube]
 
Re: Kings Island | "Unknown" | "Unknown"

I've gone on record on here that I'm not a huge invert fan, but I'll tell you now I like the inverts I've ridden more than the wing riders I've ridden.

tl;dr

Inverts>Wing Riders
 
Re: Kings Island | "Unknown" | "Unknown"

My thoughts on the 55 degree drop angle would be that it could perhaps be a twisting drop. It could be the vertical angle while it is twisting and highly banked. Sort of like an old school flared B&M drop but not as steep and larger in turn radius.

I think I mentioned this a while ago, but a 199 foot drop would need a massive pullout since the trains, in addition to being heavier, put more force on the track since the load has greater distance from the track.

WARNING: Enginerd alert! Enginerd alert!

I hate to go all engineer on here, but consider the formula w=F*d, where w is work, or how much the train deflects the track.

Since the d (distance) is larger than a traditional coaster, this number has to be made smaller by reducing F, or force. The force is represented by the formula F=mg, or Force equals mass times gravity. The mass might as well be a constant, so g would have to be smaller. In other words, the drop can't pull as many vertical Gs, and this can be done by elongating the pullout from the drop.

So I think that if it is a wing coaster called The Bat and it has this drop angle, it'll be a bit of flared drop with a large, elongated pullout.
 
Re: Kings Island | "Unknown" | "Unknown"

Actually...thinking back to the blue prints leak, the vertical loops support structure seemed oddly wide for a standard Inverted Coaster.

While I'm not joining in with the Wing crowd, maybe it could have a newer train than classic 4 across?
 
Re: Kings Island | "Unknown" | "Unknown"

King's Island just posted a tweet about Thursday asking whether or not to "Wing" the speech... Hmmm.. I see disappointment coming
 
Re: Kings Island | "Unknown" | "Unknown"

2012Jarrett said:
My thoughts on the 55 degree drop angle would be that it could perhaps be a twisting drop. It could be the vertical angle while it is twisting and highly banked. Sort of like an old school flared B&M drop but not as steep and larger in turn radius.

I think I mentioned this a while ago, but a 199 foot drop would need a massive pullout since the trains, in addition to being heavier, put more force on the track since the load has greater distance from the track.

WARNING: Enginerd alert! Enginerd alert!

I hate to go all engineer on here, but consider the formula w=F*d, where w is work, or how much the train deflects the track.

Since the d (distance) is larger than a traditional coaster, this number has to be made smaller by reducing F, or force. The force is represented by the formula F=mg, or Force equals mass times gravity. The mass might as well be a constant, so g would have to be smaller. In other words, the drop can't pull as many vertical Gs, and this can be done by elongating the pullout from the drop.

So I think that if it is a wing coaster called The Bat and it has this drop angle, it'll be a bit of flared drop with a large, elongated pullout.
Click to expand...


Careful, Spongebob! You're getting your physics in your mechanics of materials.

Now you're right about the train causing track deflection. Wherever the train is, the track will encounter both bending and shear stresses, but even those aren't the culprit to the size of the pullout. The track causes the train to undergo centripetal acceleration (kind of. A pullout isn't perfectly circular so the instantaneous center will change as the train moves) which causes a force to act back on the track. We can easily calculate the g-force simply by taking a ratio of the force exerted by the train in the pullout over the free standing weight of the train. The max g-force is usually around 5g. Now that a design constraint has been established, the pullout can be designed around that. Have you ever noticed how B&M coasters' spines get thicker in pullouts than at the tops of elements? That's because stress is simply force over cross sectional area (in its simplest form. Bending and shear stresses are a tad bit different). Make the spine thicker and your stresses decrease. That way, the roller coaster can have a tighter pullout and higher g-forces.
 
Re: Kings Island | "Unknown" | "Unknown"

Antinos said:
2012Jarrett said:
My thoughts on the 55 degree drop angle would be that it could perhaps be a twisting drop. It could be the vertical angle while it is twisting and highly banked. Sort of like an old school flared B&M drop but not as steep and larger in turn radius.

I think I mentioned this a while ago, but a 199 foot drop would need a massive pullout since the trains, in addition to being heavier, put more force on the track since the load has greater distance from the track.

WARNING: Enginerd alert! Enginerd alert!

I hate to go all engineer on here, but consider the formula w=F*d, where w is work, or how much the train deflects the track.

Since the d (distance) is larger than a traditional coaster, this number has to be made smaller by reducing F, or force. The force is represented by the formula F=mg, or Force equals mass times gravity. The mass might as well be a constant, so g would have to be smaller. In other words, the drop can't pull as many vertical Gs, and this can be done by elongating the pullout from the drop.

So I think that if it is a wing coaster called The Bat and it has this drop angle, it'll be a bit of flared drop with a large, elongated pullout.
Click to expand...


Careful, Spongebob! You're getting your physics in your mechanics of materials.

Now you're right about the train causing track deflection. Wherever the train is, the track will encounter both bending and shear stresses, but even those aren't the culprit to the size of the pullout. The track causes the train to undergo centripetal acceleration (kind of. A pullout isn't perfectly circular so the instantaneous center will change as the train moves) which causes a force to act back on the track. We can easily calculate the g-force simply by taking a ratio of the force exerted by the train in the pullout over the free standing weight of the train. The max g-force is usually around 5g. Now that a design constraint has been established, the pullout can be designed around that. Have you ever noticed how B&M coasters' spines get thicker in pullouts than at the tops of elements? That's because stress is simply force over cross sectional area (in its simplest form. Bending and shear stresses are a tad bit different). Make the spine thicker and your stresses decrease. That way, the roller coaster can have a tighter pullout and higher g-forces.
Click to expand...

Yuck! I actually took a great picture of Kings Island that captures that on Diamondback's bunny hop into the final helix. How could I not think of that? I just thought that the force caused by the train in the pullout would be greater with the weight further out.

I sort of get what you're saying but I sort of don't. Are these formulas right? (I struggle with reading my problems in school and do much better when I can actually see it)

Let Ftrain stand for the centripetal force put out by the train when it goes through the pullout

Let Mrest stand for the mass of the train at rest.

Ftrain/Mrest=5 Gs

Correct? Sorry, I'm just trying to understand what you're saying. I'm working on this degree to get a job in the amusement industry doing this and I want to make sure I know as much as I can about coaster physics ASAP.
 
Re: Kings Island | "Unknown" | "Unknown"

There is a chance that there could be a twisting drop. Outside of what will be the station area, there are quite a few footers in a small area. These COULD be for a twisting drop or an inversion. Unfortunately, i do not have a picture of the area.
 
Re: Kings Island | "Unknown" | "Unknown"

^ Yes. That would offer a completely different experience? :L
 
Re: Kings Island | "Unknown" | "Unknown"

^^not exactly, for one it would already provide the sensation of now track below or above you on a regular wing rider and supports would be a nightmare
 
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