Layout and specs for Toverland's GCI now online

Where did that "180 foot" thing start, anyway? Maybe they have different rulers in Europe (and I'm not talking metric)?

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Rob Ascough said:
Wish it was closer to home, though.

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Of the eleven GCI coasters all bar two of them are in the USA. Don't be greedy :)

It's only a ninety minute flight from here plus a quick drive in a rental car. Toverland is a great little park; the Booster Bike there is pretty sweet.

Just thinking aloud (meh, I have worse habits, LOL). Could 180' be the trackage from the top of the lift to the bottom of the first drop?

At 105' tall (GCI's tallest, BTW), it seems ABOUT reasonable for the drop LENGTH to be about 180'. Just a thought...

Yes, that is very reasonable, Bill.

^ Don't you ever call me that again! ;) :)

Yeah, how dare you insult him like that? That's not the Bill I know ;)

That would seem logical...except if you check the construction pics the ride is being build on a flat plot of land and the footer foundation is already poured.

No Moosh. He's talking about the length of track along the slope of the drop. Not the height of the drop.

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Acoustic Viscosity said:
All of GCI's coasters have at least one straight hill.

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WHERE then is this STRAIGHT hill on OzCat? 'Cause I don't remember feeling or seeing one.

Oh! Well why didn't he SAY that? ;)

Ok maybe not perfectly straight. The last three bunny hills on the ride are almost "straight" hills, but it looks like each of them has a slightly banked entrance and exit. So three hills that are about 75% straight. :) Oh wait. There is that micro hop on the way to the lifthill. lol :) *** Edited 1/30/2007 6:04:30 AM UTC by Acoustic Viscosity***

With a lift of 105' and assuming Chuck's comment was about the length of the drop's track - 180' - that would mean the bottom of the first drop is a distance of 208.38' as measured along the ground [not linearly of course as the drop curves] from the apex of the lift.

[105 x 105] + [180 x 180] = ~[208.38 x 208.38]

That's all I remember from the Pythagorean Theorem. Anyone want to calculate the angle of the drop given those numbers? ;)

Moosh, your figures are a bit off. the 105 is the y component ('a'), but if the 180 is the length of the track in the drop, that would be the hypotenuse in in the Pythagorean Theorem (the 'c'). The hypotenuse (or "the slant" in a right triangle) is always the longest side.

a²+b²=c²
105²+b²=180²
11025+b²=32400
b²=21375
b=146.20

Keeping 3 significant figures, that gives us approximately 146' laterally, as measured along the ground (or "in the x direction").

Now that we have that calculation, the angle of the drop uses the famous "SohCahToa" rule: Sin = Opposite/Hypotenuse; Cos=Adjacent/Hypotenuse; Tan=Opposite/Adjacent

Sin(x)=105/180; Six(x)=.583; Sin(x)=35.7°
Cos(x)=146/180; Cos(x)=.811; Cos(x)=35.8°
Tan(x)=105/146; Tan(x)=.719; Tan(x)=35.7°

Given rounding errors, these all arrive at approximately the same figure, which for a 105' tall coaster, would be one not-thrilling drop (although Silver Bullet does pack a bunch with a low-angled drop!)

[/emgee]

Edit: If, on the other hand, the 180 was the x-component, then your figure is correct of 208 (would would be the actual length of track traversed from the top of the drop to the bottom as if it were laid out straight), and that would make the angle of descent (using tangent) of 105/180, or 30.3°.. still in the same ballpark, but just as boring-sounding. :) *** Edited 1/30/2007 12:12:51 PM UTC by dannerman***

That's some pretty fancy math there, LOL, but the drop is neither *straight*, nor is in consistent in the angle. That would be a drop in one "plane", and would look more like the Beast's second drop than anything else. Instead, the drop curves on the way down, so the angle would INCREASE to accommodate the extra trackage length (beginning to think I need to learn another language to explain myself, LOL)...

Since I brought it up, I had to check the angle of Beast's second drop....rcdb says it's a whopping 18* - and I doubt anyone finmds THAT drop "un-thrilling".... ;)

Thanks Tim! Guess that's what happens when the last math class one had was over 20 yrs ago! lol ;)

Like Gator said, the Pythagorean Theorem assumes a straight line from the apex of the drop to the bottom, but we know it curves and doesn't come to a point at either end, so the drop will be longer than the straight line calculation.

By the way, I teach that in my math class. ;)

^ I teach it here...but somehow I'd bet YOUR students understand it better in the end...LOL!

It appears that Troy's unfinished (and it's important to note, not-yet-laterally-supported) lift hill has collapsed due to wind.

Before:
http://www.toverland.info/content/woodynews/images/cam1.jpg

After:
http://www.toverland.info/content/woodynews/images/cam2.jpg

Here's some information copied from Justin Garvanovic's post on the ecc mailing list. It's a reply from GCI regarding the incident:

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At around 12:30 today, a gale-force wind swept through the Sevenum area. The
wind blew over 13 bents of the lift structure, but thankfully all employees
were off-site eating lunch at the time of damage.

Initial inspections of the damage reveal that the foundation and anchors are
intact, and most of the lumber contained in the affected bents is
salvageable. We will have to replace several large legs and batters with new
material.

It is important to note that a coaster is not as strong during the
construction process as it is when completed, since all supporting batters,
bracing and track are not yet in place.

We hope to have all affected areas rebuilt within a week or so.

Again, we are blessed that no employees were injured during this unforseen
event.
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Crazy stuff.

Weird indeed.

Didn't something similar happen during SOB's construction?