rollercoaster story board

8/13/2019 Rollercoaster Story Board

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The Grid

As done by:Layton Funk

Kristen ReichertFariha Hamid


2/18

Element I consists ofthe lift hill, whose

down hill leads intoElement II.

Element II is anuphill segment thatleads into Element

III.

Element III is made up of aseries of angled banked hills

that appear to follow along aninclined half-pipe shape. The

Buzzball will exit in adirection opposite to which it

entered the element.

Element IV will havethe Buzzball enter a

loop-de-loop, ofwhich the enteringand exit height are

the same.

Element V is asecond uphill

segment which willlead the Buzzballinto Element VI.

Element VI is a turn that isslightly inclined down so that

the Buzzball will movesmoothly into Element VII.

From the turn, the Buzzballslides off the track into a funnel

at an angle in order tominimize the gs felt and then

swirls about the funnel(Element VII).

Element VIII is the finale ofthe ride. The Buzzball

splashes down into the Seaof Simulation (water). There

is a small current running,which will bring the Buzzballto the exit dock, where the it

will then be sent up by waysof a pulley/conveyor system.


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1 cm = 3 meters

r = 7.5 m

H=30m

Element I (Side-view)

= 53

Beginning of Element II

h1 = 2.98 m

h2 = 23.44 m

= 6.94 m

l = 16.65 m

r = 9m

= 53

h3 = 3.58 m

= 53

Total distance : 52.94 m

X North

O SouthEastWest


4/18

Element I (Front, Top, andcalculations)

Front

X West

O EastNorthSouth

Top

North

South

EastWest

H = 30 m

x = 38.66 m

hlift=30 mm=200 kgE0=mghlift=58800 JE(l)=E0-E0[1-(2/100)(l/10)]

mgh-(E(52.94))=1/2*(1+keff)mv

2

venter=0 m/svleave=13.592 m/sFN=mgcos(53)=1179.56Nng=FN/mg=.602 gs on slopesFN=mac=m*vtrough

2/rcircleFN=200*14.737

2/9=4826.2Nng=FN/mg=2.462 gs intrough


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Element II (Side-view)

Beginning of Element III

X North

O SouthEastWest

r = 6.75 m

H = 23.42 m

h = 5.38m l = 12.49 m

= 53

1 cm = 3 meters Total distance : 42.466 m


6/18

Element II (Front, Top, andcalculations)

Front

X West

O EastNorthSouth

Top

North

South

EastWest

H = 23.42 m

l = 24.04 m

hhill2=27mm=200 kgE0=mghlift=58800 JE(l)=E0-E0[1-(2/100)(l/10)]

200*9.8*3-(E(95.406))=1/2*(1+1.466)*200*v2

venter=13.592 m/svleave=4.653 m/sFN=mgcos(53)=1179.56N

ng=FN/mg=.602 gs on slopesmgcos(53)-FN=m*v

2/rcircle=377.756Nng=FN/mg=.193 gs on curvestartF

N=200*5.2022/6.75=801.802N

ng=.409 gs on hill


7/18

Element III (Side-view)

1 cm = 3 meters Total distance : 152 m

Beginning of Element IV

X North

O SouthEastWest

(A change in direction. Rollercoaster entered Element III

going South, now exits leadingNorth.)


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Element III (Front, Top)

Front

X West

O EastNorthSouth

Top

North

Southt

EastWest

The energy lost, E(l)=E0-E0[1-(2/100)(l/10)], during this partof the ride and long turns andlow angle of the wavy-banked-

hills prevents a dangerousamount of gs. (The math wasREAL ugly with this, I gave up,but I did 5 other elementsinstead, so its all good in thehood)


9/18

Element IV (Side-view)


X West

O East

NorthSouth

r = 10 m

H = 20 m

x = 10 m x = 10 m

Beginning of Element V


10/18

Element IV (Front, Top, andcalculations)

Front

X South

O NorthWestEast

Top

North

South

EastWest

H = 30 m L = 30 m

hloop=20mm=200 kgE0=mghlift=58800 JE(l)=E0-E0[1-(2/100)(l/10)]

200*9.8*30-(E(330.236))=1/2*(1+1.466)*200*v2

venter=10.975 m/svleave=8.998 m/svinversion=7.632 m/s

FN=mac=m*vinversion2/rinversionFN=200*7.632

2/10=634.388Nng=FN/mg=.324 gs on inversion


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Element V (Side-view)


X West

O East

NorthSouth

r1=7.5

m

r2 = 4 m

(A change indirection.Element V

begins in theNorth

direction andthen endsgoing West

into ElementVI.)

Beginning of Element VI

= 53= 53

l = 8.33 m

l = 4.63 m

H=8m

h=7.5m


12/18

Element V (Front, Top, andcalculations)

Front

X South

O NorthWestEast

Top

North

South

EastWest

H = 10 m

L = 11 m

hhill3=8mm=200 kgE0=mghlift=58800 JE(l)=E0-E0[1-(2/100)(l/10)]

200*22*-(E(352.586))=1/2*(1+1.466)*200*v2

venter=8.998m/svleave=2.591 m/sFN=mgcos(53)=1179.56N

ng=FN/mg=.602 gs on slopesmgcos(53)-FN=m*v

2/rcircle=534.793Nng=FN/mg=.0965 gs on curvestartn

g

=.205 gs on hill


13/18


14/18

Element VI (Front, Top, andcalculations)

Front

X East

O WestSouthNorth

Top

North

South

EastWest

h=8mm=200 kgE0=mghlift=58800 JE(l)=E0-E0[1-(2/100)(l/10)]

200*22*-(E(359.216))=1/2*(1+1.466)*200*v2

venter=2.591 m/svleave=5.601 m/sFN=mgcos(53)=1179.56N

ng=FN/mg=.602 gs on slopesng=.205 gs on hillng-horizontal=4.2^2/4 /9.8=..45 gson turn


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Element VII (Side-view)

Beginning of Element VIIIH = 4.5 m

1 cm = 3 meters Total distance : N/A


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Element VII (Front, Top)

Front

X West

O EastNorthSouth

Top

North

South

EastWest

Smooth transition into thefunnel instead of a simple dropinto water will prevent too manygs from happening.


17/18

Element VIII (Side-view)

1 cm = 3 meters Total distance : N/A m


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Element VIII (Front, Top)

Front

X West

O EastNorthSouth

Top

North

South

EastWest

Splash.

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