Ferris Wheel Comparison

World’s Fair Ferris Wheel.

Given info: diameter 250ft , height from the ground 14ft, 1 period = 10

minutes.

h - vertical position (height) at time t where t is given in minutes What are the five critical t-axis values ???

250 ft

14 ft

t (min) h (ft)

0 14

2.5 139

5 264

7.5 139

10 14

Characteristics of the World’s Fair Ferris Wheel

Ride procedure: board the Ferris wheel at the bottom, height of 14 feet, go up, reach maximum after 5 minutes, complete one ride (one period) after 10 minutes.

This does not resemble parent graph of sin(x) or cos(x), we must apply a phase shift in the equation

Characteristics of the periodic function: a) Phase shift: - (sine), - (cosine) b) Vertical shift : + 139 c) Amplitude: 125 ft d) Period : 10 minutes 2/10 = 1/5 or 36 e) y-intercept: (0, 14) f) Min. value: 14 ft Max. value: 264 ft g) the height at time = 0 ? h(0) = 14

2

EQUATIONS

Cosine equation for this ride:

Sine equation for this ride:

1395

1cos125

xy 139180cos125 xy

13925

1sin125

xy 13990sin125 xy

or

or

GRAPHS: Cosine Equation

1395

1cos125

xy

Minimum h(0)= 14 h(10)=14

Maximumh(5)=264

a = 125 ft

If the equation is written in radian measure, calculator mode

must be in radian and the window settings must be

appropriate:X Min: 0

X Max: 10Y Min: 0

Y Max: 300

GRAPHS: Sine Equation(s)

Minimum h(0)= 14 h(10)=14

Maximumh(5)=264

a = 125 ft

13925

1sin125

xy 13990sin125 xy

Maximum h(180)=264

a = 125 ft

Calculations for World’s Fair Ferris Wheel

1)What is the circumference of the wheel? C = 250 ft

2.) At what speed is the wheel traveling? Speed = 1.31 ft/sec.

If you begin your ride at the base of the wheel, what is your height after 1 minute? h(1) = 37.873 ft

4 minutes? h(4) = 240.13 ft

At what approximate time(s) will you reach the following heights?a.) 100 ft t(100) 2 min AND t(100) 8 min

b.) 240 ft. t(240) 4 min AND t(240) 6 min

sec/31.1sec)600(

)250(

min)10(

)250(ft

ftft

time

distnacespeed

Navy Pier Ferris Wheel.

Given info:

diameter 140ft , height from the ground 10ft, 1 period = 6 min

h - vertical position (height) at time t where t is given in minutes What are the five critical t-axis values ???

140 ft

10 ft

t (min) h (ft)

0 10

1.5 80

3 150

4.5 80

6 10

Characteristics of the Navy Pier Ferris Wheel

Ride procedure: board the Ferris wheel at the bottom, height of 10 feet, go up, reach maximum after 3 minutes, complete one ride (one period) after 6 minutes.

This does not resemble parent graph of sin(x) or cos(x), we must apply a phase shift in the equation

Characteristics of the periodic function: a) Phase shift: - (sine), - (cosine) b) Vertical shift : +80 c) Amplitude: 70 ft d) Period : 6 minutes 2/6 = 1/3 or 60 e) y-intercept: (0, 10) f) Min. value: 10 ft Max. value: 150 ft g) the height at time = 0 ? h(0) = 10

2

EQUATIONS

Cosine equation for Navy Pier ride:

Sine equation for Navy Pier ride:

or

or

803

1cos70

xy 80180cos70 xy

8023

1sin70

xy 8090sin70 xy

GRAPHS:

803

1cos70

xy 80

23

1sin70

xy

Minimum h(0)= 10 h(10)=10

Maximumh(3)=150

a = 70 ft

Maximum h(3)=150

a = 70 ft

Calculations for Navy Pier Ferris Wheel

1)What is the circumference of the wheel? C = 140 ft

2.) At what speed is the wheel traveling? Speed = 1.2 ft/sec.

If you begin your ride at the base of the wheel, what is your height after 1 minute? h(1) = 45 ft

4 minutes? h(4) = 115 ft

At what approximate time(s) will you reach the following heights?a.) 100 ft t(100) 4.2 min AND t(100)7.8 min

b.) 240 ft. t(240) NEVER

sec/2.1sec)360(

)140(

min)6(

)140(ft

ftft

time

distnacespeed

Ferris Wheel Challenge Imagine the Navy Pierand the Worlds FairFerris Wheel being builtbeside each other. Ifboth wheels beginturning at once, over a20 minute time period, atwhat times are thewheels at the same

height?

World’s Fair Ferris Wheel and Navy Pier Ferris Wheel ride over 20 minutes

h(0.6)=21.6

h(2.5)=141.95

h(7.9)=

108.2

h(11.

1)=40

.95

h(18

.9)=

40.9

5

Ferris Wheel Challenge

What is the length of the arc traveled by the Navy Pier Ferris wheel from the 4 o’clock to the 7 o’clock position?

Circumference = 140 ft From 4 o’clock 7 o’clock = 3 hours = 90 =

/2 ¼ of the circumference Arc length = ( ¼ ) (140) = 109.96 ft