Great+Bear

= Shiloh, Aaron, Ian, and Katie =

Introduction:
 The Great Bear is an inverted steel roller coaster named after the large constellation, Ursa Major. It is one of the many coasters at Hershey Park, rating at the top of the list scoring a "5" qualifying it as an aggressive thrill ride. Created by Bolliger and Mabilliard, Hershey added this $13 million roller coaster in May 1998 to be a family thrill ride, but as it turns out, the ride is far more thrilling than originally thought. It is a host to several unique elements that invert the dangling train four times while the riders experience uncommon forces due to the centripetal force of several of the turns. The noise created by the coaster gives no doubt to the ride being named properly, as it roars through the park being one of the most famous attractions at this magnificent park.

The Great Bear: A Summary of the Ride:
The Great Bear is known as a hanging coaster because the rider is strapped into the car from the top while the feet dangle below. The rider is initially lifted 90 feet up a slope. Immediately afterward they are dropped into a helix turn, which drops them while spinning them in a circle like a drain. They proceed into a descent followed by a vertical loop, where the feet fly in the air above the rider’s heads. The rider returns to the ground then into an immelman, which loops the rider up then flips them over. From the immelman they enter a flat roll. The flat roll twists the rider horizontally around the track. This is followed by an S-turn, corkscrew, then S-turn combination until forty seconds after the ride began the rider is deposited back in the loading station.

Number of Cars: 8 cars Car Length: 1.6m Train Length: 12.8m Number of Trains: 2 Riders Per Train: 32 Track Length: 853.44m Capacity: 1025 riders/hour Ride Length: 2min, 15 sec Maximum Speed: 58mph; 25.93m/s Maximum Height: 37.8m Maximum Drop: 27.4m Approximate Mass of Train: 22400lbs; 11.2ton; 10160.6kg Approximate Total Mass(With Passengers at 150lbs/68kg): 27200lbs; 12336.6kg Designer: Bolliger & Mabilliard Year Installed: 1998

[|Helix Turn] - Max G Forces measured at **2.8x**
- The helix turn creates a centripetal force upon the riders that is constant throughout the turn and releases the train as the track straightens out.

[|First Loop] - Max G Forces measured at **2.82x**


- Measured at 30.48m, this first loop brings the train up 100' and back down in 6 seconds, causing high Forces upon the body.

[|Immelman(Second) Loop] - Max G Forces measured at **3.35x**

-

Immelman(Second) Loop Exit/Drop - Max G Forces measured at **.11x**

-

[|Flat Roll] - Max G Forces **measured** at **2.78x** -

[|0 Gravity Roll] - Max G Forces measured at **2.71x** Tangential Speed: v = (2(pi)r)/T ... ... v = m/s Centripetal Acceleration: Ac = v^2/r ... ... Ac = Percent Error of G's:

Calculations/Data:
Velocity (m/s) ||= Centripetal Acceleration (m/s^2) ||
 * =  ||= Time (sec) ||= Period (sec) ||= Tangential
 * Flat Roll ||= n/a ||= 1.65 ||= 8 ||= 30.48 ||
 * First Loop ||= n/a ||= 3.1 ||= n/a ||= n/a ||
 * Zero Gravity Roll ||= n/a ||= 1.8 ||= 7.33 ||= 25.59 ||
 * Point After Helix ||= .99 ||= n/a ||= 12.93 ||= n/a ||
 * Point After 1st Loop ||= .4725 ||= n/a ||= 27.09 ||= n/a ||

Accepted Max Velocity: 58mph; 25.93m/s Calculated Max Velocity: mgy = .5mv^2 mass cancels out gy = .5v^2 (9.8)(37.8) = (.5)(v^2) v = 27.2m/s % Error = ((27.2-25.93)/(25.93))*100 = 4.9%
 * Finding Max Velocity:**

PE' - KE = (9.8)(12336.6)(37.8) - .5(27.2)^2(12336.6) = 6414.928 Joules 6414.93J = Ff µN = 6414.93 12336.6*9.8µ = 6414.93 µ = .053
 * Finding µ:**

Using distance/time: v = (12.8/.99) v = 12.93m/s
 * Velocity after Helix (90 feet; 27.4m):**

Tangential Speed: v = (2(pi)r)/T ... v = (2(pi)2.1)/1.65 ... v = 8.0 m/s Centripetal Acceleration: Ac = (v^2)/r ... Ac = (8.0^2)/2.1 ... Ac = 30.48 m/s^2 ... in G's = 30.48/9.8 = 3.1 G's Percent Error of G's: (3.1 - 2.78)/3.1 = 10% error
 * Velocity and Centripetal Acceleration of Flat Roll:**

Tangential Speed: v = (2(pi)r)/T ... v = (2(pi)2.1)/1.8 ... v = 7.33 m/s Centripetal Acceleration: Ac = (v^2)/r ... Ac = (7.33^2)/2.1 ... Ac = 25.59 m/s^2 ... in G's = 25.59/9.8 = 2.6 G's Percent Error of G's: (2.6 - 2.71)/2.6 = 4% error
 * Velocity and Centripetal Acceleration of Zero Gravity Roll:**

PE' = PE + KE (g)(y) = (g)(y) + .5v^2 (9.8)(37.8) = (9.8)(27.4) + .5(v)^2 v = 14.3 m/s % Error = ((14.3-12.93)/12.93)*100 = 10.6%
 * Using Energy**:

Picture Data [] Data [] Video help [|www.crabbiefan.com/vids.bear.mpg]
 * Works Cited **