Wow, I thought our cars were fast off the line...

[mdickie]

I am up in Ohio, visiting my cousin's family for the weekend, and today I got a chance to take a ride on Top Thrill Dragster at Cedar Point. This, for a time, was the tallest roller coaster in the world at 420'.

From a standing start, you are launched from 0-120 in 3.8 seconds, then sent up a 90 degree hill for 420', then you go over the top and back down at 90 degrees the same 420'.

What a rush...

Mark

[Zippy]

Cedar Point is the absolute best for roller coaster freaks.
Did you ride any of the other coasters?

Zipster

[jub jub]

I saw that coaster on "Travel" the other day. First thing I thought was "How do I get a ride on that?"

That one is going on the bucket list for sure!

[ZO6 AL-X]

We have the Superman ride here at a local park. Does a crazy 0-60 mph like that one.

[RocketSled]

V = 120MPH = 176 F/S
t = 3.8S

What is the acceleration necessary to achieve this speed in this time?

V = 1/2at^2, so "a" = 2V/t^2 = 2*176/(3.8*3.8) = 24 F/S/S.

1 "g" is 32 F/S/S so (assuming the initial acceleration occurs over a flat section of track before the coaster starts up the first hill) you're experiencing roughly 2/3rds of a "g" of acceleration to get to 120 MPH in 3.8 seconds. By comparison, if you jumped off a tall building you'd hit 157MPH in that same 3.8 second fall (ignoring drag).

According to WikiPedia:

Indy Car driver Kenny Bräck crashed on lap 188 of the 2003 race at Texas Motor Speedway. Bräck and Tomas Scheckter touched wheels, sending Bräck into the air at 200+ mph, hitting a steel support beam for the catch fencing. According to Bräck's site his car recorded 214 g.

That's 321x the force experienced on the roller coaster. Ouch.

[blkZ06]

Originally Posted by RocketSled

V = 120MPH = 176 F/S
t = 3.8S

What is the acceleration necessary to achieve this speed in this time?

V = 1/2at^2, so "a" = 2V/t^2 = 2*176/(3.8*3.8) = 24 F/S/S.

1 "g" is 32 F/S/S so (assuming the initial acceleration occurs over a flat section of track before the coaster starts up the first hill) you're experiencing roughly 2/3rds of a "g" of acceleration to get to 120 MPH in 3.8 seconds. By comparison, if you jumped off a tall building you'd hit 157MPH in that same 3.8 second fall (ignoring drag).

According to WikiPedia:

Indy Car driver Kenny Bräck crashed on lap 188 of the 2003 race at Texas Motor Speedway. Bräck and Tomas Scheckter touched wheels, sending Bräck into the air at 200+ mph, hitting a steel support beam for the catch fencing. According to Bräck's site his car recorded 214 g.

That's 321x the force experienced on the roller coaster. Ouch.

ok dude what do you do for a living?

[wvphoto]

Originally Posted by RocketSled

V = 120MPH = 176 F/S
t = 3.8S

What is the acceleration necessary to achieve this speed in this time?

V = 1/2at^2, so "a" = 2V/t^2 = 2*176/(3.8*3.8) = 24 F/S/S.

1 "g" is 32 F/S/S so (assuming the initial acceleration occurs over a flat section of track before the coaster starts up the first hill) you're experiencing roughly 2/3rds of a "g" of acceleration to get to 120 MPH in 3.8 seconds. By comparison, if you jumped off a tall building you'd hit 157MPH in that same 3.8 second fall (ignoring drag).

According to WikiPedia:

Indy Car driver Kenny Bräck crashed on lap 188 of the 2003 race at Texas Motor Speedway. Bräck and Tomas Scheckter touched wheels, sending Bräck into the air at 200+ mph, hitting a steel support beam for the catch fencing. According to Bräck's site his car recorded 214 g.

That's 321x the force experienced on the roller coaster. Ouch.

damn
he beat me to it.. that was my thought exactly !!!!

except.. 79%# v 2>2

[ZO6 AL-X]

[itchynackers]

I was just there in July. When you're in line next to the ride you can't believe how fast it goes. Also while standing in line, the girl in the front car came flying by with her halter top down. True story. Good looking too!

[novetteyet]

V = 1/2at^2, so "a" = 2V/t^2 = 2*176/(3.8*3.8) = 24 F/S/S.

I like the idea, but one small trouble: your formula isn't right . You're thinking of distance traveled accelerating from zero speed (a is the 2nd derivative of distance with respect to t, but the first derivative of velocity WRT t). One way to see the mismatch is to work out the units. In SI units, V is m/s, and A is m/s^2. Thus the left side unit of "a = 2V/t^2" is in m/s^2 and the right side unit is in m/s^3, so there's a conformance error. It helps to not drop the units in the intermediate steps; that'll also catch non-conformance unit mismatches.

What you want is A = (V1-V0) / T, which gives average acceleration over a period of time. Again in SI units, deltaV=120 MPH, which is about 53.6 m/s, so our accel is 53.6 (m/s) / 3.8s, or 14.1 m/s^2. g=9.8m/s^2, so that's around 1.4g on average for the coaster. The instantaneous acceleration might or might not be even larger at any point in time.

Interestingly, although the acceleration experienced by the car vs post driver was probably a bit less than 214 g's due to a crumpling car, duration also matters. If you experienced the acceleration that driver did for the duration of the coaster ride instead of a fraction of a second, you'd surely die!

*EDIT* and Cedar Point rocks.

[RocketSled]

Originally Posted by blkZ06

ok dude what do you do for a living?

I make little beanie hats with propellers that stick out from their tops...

:pp: