This is my first post so please be kind. Cedar Point, an amusement park in Ohio, recently built a new roller coaster that is the fastest in the world. It is said to go 193 km/h in under 4 seconds. Assuming that statement is true and that it reaches 193 km/h in exactly 4 seconds, how many Gforces do people on the ride experience? Would someone please include the formula for calculating Gforces? There is probably already a thread on this subject, but I cannot figure out how to search for topics. I am also at work so I do not have access to my Physics text from college nor have I found a website with the equation I need. Thanks.
The acceleration experienced on a roller coaster is mostly due to the curves and loops, not to the straght line motions of the little cart. That being said, if V(0) = 0, the acceleration is just v/t = 13.4 m/s/s, or about 1.4 gravities.
If you do an advanced Google search using “G force, formula” and say it must include the words “physics, math” you will get plenty of formulas. To me the problem is that your information will only tell you how many G’s the riders will experience as they accelerate. They will probably pull more G’s when they reach the bottom of a slope, during the ride. Good luck! 
Firstly convert the speed to ms^-1 (simply because I don’t know off the top of my head what g is in km/h and it’s better work in the units your used to to avoid mistakes):
(193 * 1000)/3600 = 53. 6 ms^-1 (correct to three sig figs) divide it by 4 (secs) to find the (average) accelration :
53.6/4 = 13.4 ms^-2
Now divide this by g (9.81 ms^-2) to find the no. of Gs :
13.4/9.81 = 1.37
Thanks for the quick response! I thought, incorrectly, that just because the coaster was 420 feet tall and went 120 mph that it would have more Gforce. A coaster that was shorter and a lot slower could create more Gforces than this one, right? If it had loops and turns, there would be more force?
Here’s a site with lots of great Roller Coaster physics: http://www.vast.org/vip/BOOK/HOME.HTM
Others have correctly calculated the value of the acceleration in g’s for you, but this may not actually be the answer to your question. The acceleration from 0 to maximum speed is typically experienced during the initial drop of a roller coaster, when you would feel it working against gravity. So the people in your coaster may actually be feeling 0.4 g’s up.
To calculate the force experienced coming out of the drop, you would need to know the radius of curvature of the track there.
Wow - it looks like signatures default to “on” now. Sorry about that!
The coaster starts straight then goes up the lift hill at an angle of 90 degrees. It goes straight up to 420 feet, turns back at a descent angle of 90 degrees. It looks like an intense ride :
http://www.cedarpoint.com/public/inside_park/rides/thrill/ttd/specs/index.cfm
Am I missing something? An object dropped from 0 initial velcity straight down for a distance of 128 meters (420 ft.) will reach a velocity of 127 km/hr ignoring air resistance. It does this in 3.61 seconds. In order to reach 193 km/hr it needs to start with an initial velocity of 66 km/hr.
If this is an ordinarly coaster that is pulled up by a chain engaging a pawl on the cars and gains its velocity by a drop the max acceleration is 9.8 m/sec[sup]2[/sup] or 1g. If the downslope is at 70[sup]o[/sup] the max acceleration is 1g*cos(70) or about 0.94 g.
I don’t see how the thing can reach 193 km/hr. unless it goes over the crest of the first drop at 66 km/hr (about 40 mph).
As has been pointed out, the accelerations experienced do not come from gaining the initial velocity. In fact, on the drop you really don’t experience much acceleration since 1g down is the max you can get. And that leaves you with no acceleration toward the seat so you need a seat belt.
If the velocity at the entry to the first curve is 127 km/hour the curve needs a radius of about 42 m(150 ft.) to keep the acceleration to a max of 3 g’s. I would think that would be a pretty good maximum unless you give everyone a physical and have them sign a release form at the start of the ride.
Or am I overlooking something obvious?
While I was going back and forth to Mathcad snapthe9 showed us a picture. So the answer is fairly easy. On the vertical drop you get 1 g and are weightless.
The G Force does not have the 0-193 km/h in 4 secs along the freefall section. It acheives this along the initial straight (horizontal) section of track before going up (and down) the hill.
A hydraulic system (big cylinder) is responsible for this force.
The acceleration (1.37 G’s as calculated above) is higher than could achieved in freefall, so external force is definately required.
-Jeff
Must be one hell of a cylinder. For a 4 sec. constant acceleration of 1.4 g’s the distance traveled is:
s = (a*t[sup]2[/sup])/2 which works out to 110 meters.
It actually accelerates the train with a cable wound up around a spool turned by a hydraulic motor. I was at Cedar Point last year but didn’t get to ride Top Thrill Dragster, since it was intermittently out of order with mechanical problems for a good part of the Summer.