Top Thrill Dragster question

I was wondering if anyone knows why Cedar Point didn't post the maximum g-forces of Top Thrill Dragster. I know that they design and plan coasters years before they begin to build them, so I don't know why they wouldn't know. Thanks in advance.

Well that I know of Millie and Twisted Wicker didn't have the g-forces posted. I guess CP just really doesn't want anyone to know about g-forces on their rides. I doubt it will be too strong though maybe around 5.

Most people in their demographic don't know what G's are. Secondly, If they listen to propaganda from markey and other sources the g amount might scare them off.

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"I remember the first time I had sex - I kept the receipt."- Groucho Marx

You can use a physics formula to figure it out, which i dont know of hand. The only thing i know about coasters involving physics G forces is that if you go through a perfect circular vertical loop, you will be 6 Gs at the bottom.

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So you believe that you are studying us, then kindly explain why you are the ones trapped in your seats.

That's totally untrue.

Force = mv^2/r

where r is the radius of curvature and m is the mass and v is the velocity.

Of course, if you're trying to calculate G's then the m will drop out and in a vertical loop at the bottom you'll need to add 1 and at the top you'll need to add cos(angle).

If you're looking at the pullout for TTD, you're going to have to estimate the radius of curvature. Then use the above formula: mv^2/r. Note that v is not constant because of the upward slope of the track.

Also note: 120MPH = ~54m/s

By conservation of energy, we have:

mgh=0.5mv^2

m's drop out.

So: v(h) = 54-((2gh)^0.5).

Plugging that into the original equation, we get:

F = m*(54-((2gh)^0.5))^2 / r

This is an a gravity free environment.

To include gravity, add cos(angle)*mg:

F = m*(54-((2gh)^0.5))^2 / r + cos(angle)*mg

g is the gravitational constant, h is the height of the train in m., r is the radius of curvature of the track in m., angle is the angle of the track.

If you want to calculate the G's, divide by m*g (the force of gravity at 1G).

That concludes today's physics lesson.

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Be polite and ignore the idiots. - rollergator
You must be this dumb to ride Viper. -SFGAdv.

*** This post was edited by ApolloAndy on 1/29/2003. ***

You could use spreadsheets to figure it out, but only if you knew the exact height of the track at each point. You'd use centripetal acceleration equations and some other nice physics-y ideas. Needless to say, we don't have enough hard information to make a perfect number of g-forces at any given time on that curve.

I'm going to say, based on how big that curve looks, that the forces involved in guiding the train upwards can't be more than 4 Gs at any instant, but certainly not for the whole trip up the curve.

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"Well, I'm sure I'd feel much worse if I weren't under such heavy sedation." - David St. Hubbins, Spinal Tap
http://www.loopscrew.com

I'm sure someone can make an educated guess at the radius of curvature at particular heights (which, as shown is all that's needed, unless you really want to add friction into the mess).

I was mightily impressed by many of the analyses of the height of TTD while it was being built. Can someone do something similar for this?

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Be polite and ignore the idiots. - rollergator
You must be this dumb to ride Viper. -SFGAdv.

A rep from Cedar Point said at a recent Coaster Con (For ACE members) that G-forces will top out at 4 on the vertical turns, and -2 on the top hat. Who knows if those are right, but that is what the representitve said.

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Jes
Jes's Roller Coasters DJ Jes
Six Flags Worlds Of Adventure Ride-Ops Crew 2002-2003(Have Fun Trying To Find Me!)