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Ferris Wheel Comparison

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

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Ferris Wheel Comparison

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  1. Ferris Wheel Comparison

  2. 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

  3. 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

  4. EQUATIONS • Cosine equation for this ride: • Sine equation for this ride: or or

  5. GRAPHS:Cosine Equation 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: 10 Y Min: 0 Y Max: 300 Maximum h(5)=264 a = 125 ft Minimum h(0)= 14 h(10)=14

  6. GRAPHS:Sine Equation(s) Maximum h(5)=264 Maximum h(180)=264 a = 125 ft a = 125 ft Minimum h(0)= 14 h(10)=14

  7. 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

  8. 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

  9. 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

  10. EQUATIONS • Cosine equation for Navy Pier ride: • Sine equation for Navy Pier ride: or or

  11. GRAPHS: Maximum h(3)=150 Maximum h(3)=150 a = 70 ft a = 70 ft Minimum h(0)= 10 h(10)=10

  12. 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

  13. Ferris Wheel Challenge World’s Fair Ferris Wheel and Navy Pier Ferris Wheel ride over 20 minutes Imagine the Navy Pier and the Worlds Fair Ferris Wheel being built beside each other. If both wheels begin turning at once, over a 20 minute time period, at what times are the wheels at the same height? h(2.5)=141.95 h(7.9)=108.2 h(18.9)=40.95 h(11.1)=40.95 h(0.6)=21.6

  14. 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

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