TTD launch length

pyrocoasterkid

Saturday, April 8, 2006 9:52 PM

Sorry if this question has been asked before, kind of searched, but don't really know what to search for... Anyway, my cousin and I were comparing a launch of on aircraft carrier in comparison to the launch on TTD. Everywhere always states the time, instead of the length of the launch track. I mean, you could guesstimate, but I'm curious if anyone knows the exact length of the launch track?

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CP03

Saturday, April 8, 2006 11:10 PM

Sorry pyrocoasterkid I don't know the exact length of the launch track, but I'm Interested to here the length of it also.

As a side note the total length of TTD track is 2,800 feet.


RIDE ON, and ON and ON and ON and ON and ON...

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Gomez

Sunday, April 9, 2006 12:26 AM
Gomez's avatar

I'm not sure on the exact length, but it is around 600 feet.


-Craig-
2008:Magnum XL-200 | Top Thrill Dragster
2007:Corkscrew | Magnum XL-200 | Maverick

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Paradroid[DK]

Sunday, April 9, 2006 12:45 AM

My figures (roughly):

Total length: 2,800 feet

Top hat: 2 x 400 (drop length) + maybe 50: 850 feet

- that leaves 1950 feet of track. Finish track + turnaround + loading station: around 1200 feet, I would say. That leaves 750 feet for the launch track. But maybe 600 is closer, I really don´t know. ;)

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Purdue University Engineer

Sunday, April 9, 2006 3:13 AM

The 600 foot figure is reasonable for the distance from the launch position to hydraulic room, but the actual portion of the launch track that is used for accelerating the train is just under 350 feet. That is where the catch car brakes start (half way through the 9th piece of launch track). The rest is used for bringing the catch car to a stop.


Maverick '07 Crew (1, 2, 3, 4, 5...oh no...)
Los Alamos National Lab '04-'07 (LoA to finish Masters Degree)
TTD '03 Crew (76 Launches - 71 Complete Circuits)

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pyrocoasterkid

Sunday, April 9, 2006 7:23 AM

Wow, so after 350ish feet the catch car disengages? So the best guess so far is 350 feet? Thats pretty impressive

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Twisted Wicker

Wednesday, July 12, 2006 5:46 AM

The launch track is 497 feet.

The park sweeps carry a book with all sorts of weird trivia in it, and this is one of the trivia.


Not to be confused with Twisted Wicker 08 from 2002

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DBCP

Wednesday, July 12, 2006 6:02 AM

Yes, but you're talking about the same book that, in 2005, said that Top Thrill Dragster seated 10 rows of passengers, 4 across, with three 36 passenger trains... I don't know what's worse, the fact that it has 6 trains and doesn't seat 4 across, or that 10 times 4 is not 36... :)


2007: Millennium Force, 2008: Millennium Force ATL, 2009: Top Thrill Dragster
www.pointpixels.com | www.parkpixels.com

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TTD 120mph

Wednesday, July 12, 2006 6:30 AM
TTD 120mph's avatar

How about both!:)


-Adam G- The OG Dragster nut

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DBCP

Wednesday, July 12, 2006 6:35 AM

Well, you know they like to be very precise.... ;)

Like how it lists all the coasters' speeds and then for Wildcat it says "varies throughout ride." :)


2007: Millennium Force, 2008: Millennium Force ATL, 2009: Top Thrill Dragster
www.pointpixels.com | www.parkpixels.com

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Maverick07

Wednesday, July 12, 2006 3:03 PM

It also says that for CCMR

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Top Thrill 182

Wednesday, July 12, 2006 4:01 PM

I've always found that quite amusing. I would have never suspected it! How fast does Wildcat go?


Thrills Around the Corner!

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JuggaLotus

Wednesday, July 12, 2006 4:07 PM
JuggaLotus's avatar

According to RCDB - 40mph.


Goodbye MrScott

John

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Vince982

Thursday, July 13, 2006 4:38 AM
Vince982's avatar

According to the ride op's speil on Monday, 47mph.


We'll miss you MrScott and Pete

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DBCP

Thursday, July 13, 2006 6:04 AM

Hmm... It says "speed varies" on the website for both of them too.


2007: Millennium Force, 2008: Millennium Force ATL, 2009: Top Thrill Dragster
www.pointpixels.com | www.parkpixels.com

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Jessai

Thursday, July 13, 2006 11:33 AM

Assuming the launch goes as advertised (120MPh in 4 seconds):

120MPh = 53.645 Meters Per Second

Acceleration = Velocity / Time

Acceleration = (53.645m/s)/4s

Acceleration = 13.41125m/s²

x = (initial velocity)(time) + (1/2)(acceleration)(t²)

x = (0m/s)(4s) + (1/2)(13.41125m/s²)((4s)²)

x = 107.29 meters

107.29 meters = 352 feet

This of course assumes the acceleration is constant the whole time, which it isn't. But it's close. :)

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Bodes

Thursday, July 13, 2006 2:26 PM

Twisted Wicker said:


The park sweeps carry a book with all sorts of weird trivia in it, and this is one of the trivia.

I want that book! haha. I love weird little trivia like that.


2009--Dragster Photo

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