Sidewinder

I. Group Members Kayla Berkebile Steven Warner
 * II. **** The Ride – Sidewinder **



a. **Location**: Hershey Park b. **Type**: Steel-Shuttle c. **Status** : Open d. **Opening Day**: May 11, 1991 e. **Manufacturer**: Vekoma f. **Lift/Launch System**: Chain lift system g. **Height**: 32 meters h. **Length**: 268 meters i. **Max Speed**: 21.8 m/s j. **Number of Cars per train**: 7 k. **Number of loops:** l. **Loops per ride**: 6 (3 back, 3 forward) m. **Ride Duration**: 105 seconds n. **Capacity**: 540 riders per hour o. **Max G**-**force:** 5.2 p. **Mass of Full Car**: 8260 kg q. **Time through loop 1**: 1.36 s r. **Time through Cobra Roll**: 1.47 s s. **Time through loop 2**: 1.12 s t. **Angle of Drop:** 60 degrees (measured from picture)


 * I. **** How It Works **

The train of the Sidewinder initially starts at rest. An external force, from a hook, causes the train to be pulled in a backwards and upwards motion. At this point, the train is at the maximum height of 105 feet. The train is then released from the hook causing the train to fall at an acceleration of 9.8 m/s^2. The train stays in motion due to inertia. While going through the 3 loops, the velocity would be enough to make it over the loop as long as the initial height is higher than the loop. This is because of potential energy. Halfway through the ride, the train approaches its maximum height of 105 ft due to potential energy, The train can never go past the maximum velocity, therefore the train starts to fall back down the slope of the tracks. Once again, it has an acceleration of 9.8 m/s^2. Due to potential energy, the train will go past the loops and would go up to the maximum height, but a mechanical device applies a retarding force to the train therefore causing it to slow while going up the slant. While falling, a second mechanical force stops the ride completely.

**II. Important Sidewinder Calculations**

a. __Velocity after Drop__

1.) Ei = Ef PEi + KEi = PEf + KEf mgh = 1/2mv^2 gh = 1/2v^2 (9.8m/s2)(32 m) = ½(v)^2 313.6 x 2 = v^2
 * v = 25 m/s**

Theoretical velocity: 21.8 m/s Actual velocity: 25 m/s

b. __Velocity Percent Difference__

%diff = [(x1 - x2) / ((x1 + x2)/2)] x 100 %diff = [(21.8 m/s - 25 m/s) / ((21.8 m/s + 25 m/s)/2)] x 100 %diff = [(-3.5 m/s) / (46.8 m/s / 2)] x 100 %diff = 0.15 x 100
 * %diff = 15%**

c. __Force of Hook__

F = mgsin(theta) F = (8260 kg)(9.8 m/s^2)sin(60 deg)
 * F = 70,103.0 N**

d. __Acceleration__

a = gsin(theta) a = 9.8m/s^2(sin(60))
 * a = 8.49 m/s^2**

e. __Angular Velocities__

1.) //Loop 1// w = q /t w = 2 pi / t w = 2 p i / 1.36 s
 * w = 4.62 rad/s**

2.) //Loop 2// w = q /t w = 2 pi / t w = 2 pi / 1.47s
 * w = 4.27 rad/s**

3.) //Loop 3// w = q /t w = 2 pi /t w = 2 p i / 1.12s
 * w = 5.61 rad/s**

f. __Radii__

1.) //Loop 1//

KE = KErot 1/2mv2 = 1/2Iw2 ½(8260 kg)(21.8 m/s)^2 = 1/2(8260 kg)(r^2)(4.62 rad/s)^2 (21.8 m/s)^2 = r^2(4.62 rad/s)^2 21.8 m/s =r(4.62 rad/s)
 * r = 4.72 m**

2.) //Loop 2//

KE = KErot 1/2mv^2 = 1/2Iw^2 ½(8260 kg)(21.8 m/s)^2 = ½(8260 kg)(r^2)(4.27 rad/s)^2 (21.8 m/s)^2 = r^2(4.27 rad/s)^2 21.8 m/s = r(4.27 rad/s)
 * r = 5.11 m**

3.) //Loop 3//

KE = KErot 1/2mv^2 = 1/2Iw^2 ½(8260 kg)(21.8 m/s)^2 = ½(8260 kg)(r^2)(5.61 rad/s)^2 (21.8 m/s)^2 = r^2(5.61 rad/s)^2 21.8 m/s = r(5.61 rad/s)
 * r = 3.89 m**

g. __Total Mechanical Energy__

1.) Theoretical ME = PE + KE ME = mgh + 1/2mv^2 ME = (8260 kg)(9.8 m/s^2)(32 m) + 1/2(8260 kg)(21.8 m/s)^2 ME = 2590336 J + 1962741.2 J
 * ME = 4553077.2 J**

2.) __Actual Mechanical Energy__

ME = PE + KE ME = mgh + 1/2mv^2 ME = (8260 kg)(9.8 m/s^2)(32 m) + 1/2(8260 kg)(25 m/s)^2 ME = 2590336 J + 2581250 J
 * ME = 5171586 J**

3.) __Percent Difference__

%diff = [(x1 - x2)/((x1 + x2)/2)] x 100 %diff = [(4553077.2 J - 5171586 J) / (4553077.2 J + 5171586 / 2)] x 100 %diff = [-618509 J) / 862331.5 J)] x 100
 * %diff = 72%**


 * Large difference mostly due to 14% difference of velocities.**