Precalculus

Height on a turning wheel

A rider boards an observation wheel at the bottom and the wheel turns at a steady rate. What formula gives the rider’s height at every moment of the ride, and when is the rider above a given height?

You need

Angular and linear speedPrecalculus

Where you are 3 in Modeling things that repeat

Trigonometry7

Trigonometric functions and their graphs8

Modeling things that repeat3

Height on a turning wheel

Simple harmonic motion Tides, daylight and sound

When you reach for it

Reach for this model whenever a point rides a wheel that turns at a steady rate and you need its height as a function of time: a rider on an observation wheel, a paddle on a millwheel, a bucket on a water lift. The height rises and falls on a repeating cycle, so it is a sinusoid, and every number in its formula is a measurement of the wheel:

the amplitude is the radius of the wheel,
the midline is the height of the hub,
the period is the time for one turn,
the phase, the choice of sine or cosine and any sideways shift, comes from where the point is when the clock starts.

Given a time, the model gives a height; given a height, inverse trigonometry gives the times.

The procedure

Height from the angle turned

Put the hub at height k above the ground and let the wheel have radius r. Measure the angle a spoke makes the way the unit circle does: from the horizontal ray pointing right from the hub, counterclockwise as you watch. A point on the rim at angle \theta sits r\sin\theta above the hub (below it when \sin\theta is negative), because on a circle of radius r the vertical coordinate of the point at angle \theta is r times the one on the unit circle. So its height above the ground is

h = k + r\sin\theta

The wheel turns at a steady Angular and linear speed \omega radians per unit of time, so after time t a point that started at angle \theta_0 has angle \theta = \theta_0 + \omega t (when the wheel turns counterclockwise as you watch it). Substituting,

h(t) = k + r\sin(\omega t + \theta_0)

One full turn is 2\pi radians, so a wheel that takes time T per turn has \displaystyle \omega = \frac{2\pi}{T}. That is exactly the rule of Finding the period from the equation read backward: the coefficient of t is \displaystyle b = \frac{2\pi}{T}.

Boarding at the bottom gives a negative cosine

A rider who boards at the bottom starts at \displaystyle \theta_0 = -\frac{\pi}{2}, straight down from the hub. Then

\sin\left(\omega t - \frac{\pi}{2}\right) = -\cos(\omega t)

because sliding the sine graph \displaystyle \frac{\pi}{2} to the right and turning it upside down gives the cosine graph. So the rider’s height is

h(t) = k - r\cos(bt), \qquad b = \frac{2\pi}{T}

Check it at the start: h(0) = k - r, the lowest point of the wheel. Half a turn later, bt = \pi and h = k + r, the top. The figure shows the same thing: after turning through \theta past the bottom, the rider is r\cos\theta below the hub.

Figure 1The rider has turned through \theta past the bottom. The rider sits 32\cos\theta below the hub, so the height is h = 34 - 32\cos\theta.

Turning clockwise instead changes the angle to \displaystyle -\frac{\pi}{2} - \omega t, and since cosine is even the height is still k - r\cos(\omega t).

The four quarter starts

The same substitution with the other three quarter points of the rim gives a short table. Each line is k + r\sin(bt + \theta_0) simplified, for a wheel turning counterclockwise as you watch.

The clock starts with the point	\theta_0	Height
at the bottom	\displaystyle -\frac{\pi}{2}	k - r\cos(bt)
level with the hub, rising	0	k + r\sin(bt)
at the top	\displaystyle \frac{\pi}{2}	k + r\cos(bt)
level with the hub, falling	\pi	k - r\sin(bt)

Any other start is a phase shift. If the rider is at the bottom at time c rather than at time 0, the whole graph slides c to the right and h(t) = k - r\cos(b(t - c)); the shift is read from the factored form, as in Phase shift. The amplitude r is the subject of Amplitude and reflection, and the midline y = k, halfway between the lowest height k - r and the highest k + r, is the subject of Midline and vertical shift.

From a height back to the times

To find when the rider boarding at the bottom is at height H, solve k - r\cos(bt) = H for the cosine:

\cos(bt) = \frac{k - H}{r}

This has a solution only when H lies between the lowest height k - r and the highest k + r. Write u = bt for the angle turned. The inverse cosine returns the one angle \displaystyle u_1 = \cos^{-1}\frac{k - H}{r} in [0, \pi], the slice of the cosine that was chosen to be inverted. In the first turn, 0 \le u < 2\pi, there is a second angle with the same cosine, u_2 = 2\pi - u_1, because \cos(2\pi - u) = \cos u. During the first half turn the rider climbs, so the rider passes height H going up at u_1 and coming down at u_2, and is above H for every angle between them. Divide each angle by b to turn it into a time. Later turns repeat these times plus T, 2T, and so on. Solving basic trig equations treats the same two-solution pattern for any trigonometric equation.

Figure 2Kerubo’s height over two turns. The wave is above the dashed line h = 50 between each pair of dots, for \displaystyle \frac{20}{3} minutes of every 20.

Worked examples

Example 1

Kerubo’s ride

Kerubo boards an observation wheel of diameter 64 meters at the bottom. The hub is 34 meters above the ground and the wheel makes one turn every 20 minutes. Write her height h in meters as a function of the time t in minutes since she boarded, and find her height after 3 minutes.

Solution

The radius is half the diameter, r = 32, so the amplitude is 32. The hub is at k = 34, so the midline is h = 34. One turn takes T = 20 minutes, so \displaystyle b = \frac{2\pi}{20} = \frac{\pi}{10} radians per minute. She boards at the bottom, which gives a negative cosine:

h(t) = 34 - 32\cos\left(\frac{\pi}{10}t\right)

Check the extremes: h(0) = 34 - 32 = 2, the lowest car, 2 meters up, and h(10) = 34 - 32\cos\pi = 66, the top, half a turn later. After 3 minutes,

\begin{aligned} h(3) &= 34 - 32\cos\frac{3\pi}{10} \\ &\approx 34 - 32(0.5878) \approx 15.19 \end{aligned}

so she is about 15.19 meters up. The calculator must be in radian mode, since b is in radians per minute.

Example 2

When Kerubo is above 50 meters

On the same ride, when does Kerubo first reach 50 meters, and for how many minutes of each turn is she above 50 meters?

Solution

Set the height equal to 50 and solve for the cosine:

\begin{aligned} 34 - 32\cos\left(\frac{\pi}{10}t\right) &= 50 \\ \cos\left(\frac{\pi}{10}t\right) &= -\frac{1}{2} \end{aligned}

With \displaystyle u = \frac{\pi}{10}t, the inverse cosine gives \displaystyle u_1 = \cos^{-1}\left(-\frac{1}{2}\right) = \frac{2\pi}{3}. The other angle in the first turn with the same cosine is \displaystyle u_2 = 2\pi - \frac{2\pi}{3} = \frac{4\pi}{3}. Since \displaystyle t = \frac{10}{\pi}u,

\begin{aligned} t_1 &= \frac{10}{\pi} \cdot \frac{2\pi}{3} = \frac{20}{3} \\ t_2 &= \frac{10}{\pi} \cdot \frac{4\pi}{3} = \frac{40}{3} \end{aligned}

She first reaches 50 meters after \displaystyle \frac{20}{3} \approx 6.67 minutes, 6 minutes 40 seconds, on the way up, and comes back down through 50 meters at \displaystyle \frac{40}{3} minutes. She is above 50 meters for \displaystyle \frac{40}{3} - \frac{20}{3} = \frac{20}{3} minutes of every turn, one third of the ride. The graph above shows these crossings and the same pair one period later, at \displaystyle \frac{80}{3} and \displaystyle \frac{100}{3}.

Example 3

Starting the clock partway round

Wambui is already riding the same wheel. When she starts her stopwatch, her car is level with the hub and rising. Write her height as a sine, and as a negative cosine with a phase shift.

Solution

Level with the hub and rising is \theta_0 = 0, so

h(t) = 34 + 32\sin\left(\frac{\pi}{10}t\right)

For the cosine form, ask when she is at the bottom. She is a quarter turn past the bottom already, so she reaches the bottom after the remaining three quarters of a turn, at \displaystyle t = \frac{3}{4} \cdot 20 = 15. The bottom-start model slid 15 minutes to the right is

h(t) = 34 - 32\cos\left(\frac{\pi}{10}(t - 15)\right)

The two agree: \displaystyle \frac{\pi}{10}(t - 15) = \frac{\pi}{10}t - \frac{3\pi}{2}, and \displaystyle -\cos\left(x - \frac{3\pi}{2}\right) = \sin x for every x. At t = 5 both give 66, the top.

Example 4

A view that needs height

From Kerubo’s wheel the skyline shows only above 40 meters. For how many minutes of each turn can she see it, to the nearest hundredth of a minute?

Solution

Solve \displaystyle 34 - 32\cos\left(\frac{\pi}{10}t\right) = 40:

\cos\left(\frac{\pi}{10}t\right) = \frac{34 - 40}{32} = -0.1875

This is not a value from the unit circle, so the calculator supplies the angle: u_1 = \cos^{-1}(-0.1875) \approx 1.7594 radians, which lies in [0, \pi] as every inverse cosine does. Then \displaystyle t_1 = \frac{10}{\pi}u_1 \approx 5.6004 minutes. The second time in the turn is t_2 = 20 - t_1 \approx 14.3996, because u_2 = 2\pi - u_1 and \displaystyle \frac{10}{\pi} \cdot 2\pi = 20. She sees the skyline from about 5.60 to 14.40 minutes into each turn, for about 14.3996 - 5.6004 \approx 8.80 minutes.

Example 5

Nyokabi’s millwheel

Nyokabi watches a millwheel of radius 2.5 meters whose hub is 1.5 meters above the water. Seen from her bank it turns counterclockwise at 5 revolutions per minute. A painted paddle tip is level with the hub on the right when she starts timing. Write the tip’s height above the water in meters after t seconds, and find when, during the first turn, the tip is under water.

Solution

Five revolutions per minute is 5 \cdot 2\pi = 10\pi radians per minute, which is \displaystyle \frac{10\pi}{60} = \frac{\pi}{6} radians per second, so one turn takes 12 seconds. The tip starts level with the hub and rising, \theta_0 = 0:

h(t) = 1.5 + 2.5\sin\left(\frac{\pi}{6}t\right)

The lowest height is 1.5 - 2.5 = -1, so the tip dips 1 meter below the surface. It is under water when h(t) < 0, that is, when \displaystyle \sin\left(\frac{\pi}{6}t\right) < -0.6. The inverse sine gives \sin^{-1}(0.6) \approx 0.6435. Sine equals -0.6 at two angles of the first turn, both in the lower half of the circle: \pi + 0.6435 \approx 3.7851 and 2\pi - 0.6435 \approx 5.6397. Sine is below -0.6 between them. Multiplying by \displaystyle \frac{6}{\pi}, the tip enters the water at about 7.23 seconds and leaves it at about 10.77 seconds, under water for about 3.54 seconds of each 12.

Example 6

A start that is not a quarter point

On the same millwheel, a second paddle tip starts at angle \displaystyle \frac{\pi}{6}, above the horizontal on the right, and rises. Write its height and find when it first enters the water.

Solution

With \displaystyle \theta_0 = \frac{\pi}{6},

\begin{aligned} h(t) &= 1.5 + 2.5\sin\left(\frac{\pi}{6}t + \frac{\pi}{6}\right) \\ &= 1.5 + 2.5\sin\left(\frac{\pi}{6}(t + 1)\right) \end{aligned}

Factoring shows a phase shift of -1: this tip does everything the first tip does 1 second sooner, because it starts where the first tip will be after 1 second. It starts at \displaystyle 1.5 + 2.5\sin\frac{\pi}{6} = 2.75 meters. The first tip entered the water at about 7.23 seconds, so this one enters at about 7.23 - 1 = 6.23 seconds. Solving directly gives the same: \displaystyle \frac{\pi}{6}(t + 1) \approx 3.7851, so t + 1 \approx 7.23 and t \approx 6.23.

Boundaries

Steady turning only. The model needs a constant angular speed. A wheel that pauses to load riders, or speeds up from rest, does not have \theta = \theta_0 + \omega t, and its height is not a sinusoid of time.

A point fixed on the wheel. The formula follows a point at distance r from the hub. A car hanging a fixed distance below the rim lowers every height by that distance: the midline moves, the amplitude and period do not.

Heights outside the range. h(t) = H has no solution when H is below k - r or above k + r, because no cosine is outside [-1, 1]. The lowest and highest heights are each reached once per turn; every height strictly between them is reached twice per turn, once rising and once falling.

Direction matters off the vertical. Starting at the top or the bottom, the direction of turning does not change the formula. Starting level with the hub, it does: counterclockwise from the right side rises, k + r\sin(bt), and clockwise from the same place falls, k - r\sin(bt).

Where this goes next

The wheel is the first of the periodic models. The same four readings, amplitude, midline, period and phase, describe the tide at a harbor and a weight bouncing on a spring, where no wheel is visible but the motion repeats in the same way. When the data come as a graph rather than as a wheel’s measurements, Writing an equation from a wave reads the same four numbers off the picture. The step from a height back to the times grows into solving trigonometric equations.

Exercises

Building the model

An observation wheel of diameter 80 meters has its hub 43 meters above the ground and turns once every 24 minutes. A rider boards at the bottom. Write the rider’s height h in meters after t minutes.
For the rider in Exercise 1, find the height after 4, 12 and 18 minutes.
A wheel of radius 20 meters has its hub 23 meters up and turns once every 15 minutes. Write the height of a car that is at the top when the clock starts.
A millwheel of radius 1.8 meters has its hub 1.2 meters above the water and turns counterclockwise, as seen, at 3 revolutions per minute. A paddle tip starts level with the hub on the right. Write its height above the water after t seconds.
A rider’s height in meters after t minutes is \displaystyle h(t) = 18 - 16\cos\left(\frac{\pi}{6}t\right). Give the radius of the wheel, the height of the hub, the lowest and highest heights, the time per turn, and where the rider is at t = 0.
The wheel of Exercise 3 turns clockwise as you watch it. Write the height of a car that is level with the hub on the right when the clock starts.
A wheel’s highest point is 70 meters up and its lowest point is 4 meters up. It turns once every 18 minutes. Write the height of a rider who boards at the bottom.

Phase shifts

Write the height of a car on the wheel of Exercise 1 that is at the top at t = 0, both as a cosine with no shift and as a negative cosine with a phase shift.
On the wheel of Exercise 1, a car’s height is \displaystyle h(t) = 43 - 40\cos\left(\frac{\pi}{12}(t - 6)\right). Where is the car at t = 0, and is it rising or falling?
A car on the wheel of Exercise 3 reaches the bottom at t = 4 minutes. Write its height.
On the millwheel of Exercise 4, a second paddle tip is at the top at t = 0. Write its height as a cosine, and show it equals a sine with a phase shift of -5 seconds.

When the rider is above a height

When does the rider of Exercise 1 first reach a height of \displaystyle 43 + 20\sqrt{2} meters?
For how many minutes of each turn is the rider of Exercise 1 above 63 meters?
When is the car of Exercise 3 first below 10 meters, and for how many minutes of each turn is it below 10 meters? Give both to the nearest hundredth.
For how many seconds of each turn is the paddle tip of Exercise 4 under water, to the nearest hundredth?
Find every time in the first 24 minutes at which the rider of Exercise 5 is 26 meters up.
From the wheel of Exercise 1 a lake is visible above 30 meters. For how many minutes of each turn is it visible, to the nearest hundredth?

From a speed

The rim of an observation wheel of radius 30 meters moves at 0.4 meters per second, and its hub is 32 meters up. Write the height of a rider who boards at the bottom, with t in seconds, and find the time per turn to the nearest hundredth of a minute.

Reasoning

Explain, from the angle a spoke makes, why a rider who boards at the bottom has a height of the form k - r\cos(bt).
Kerubo’s height rises 64 meters from the bottom of her ride to the top. What does the amplitude of her model measure, and how is it related to that 64?
Explain why a rider who boards at the bottom of any steadily turning wheel spends exactly half of each turn above the height of the hub, and less than half above any greater height.
Explain why the direction of turning does not affect the model for a rider who boards at the bottom, but does affect it for a point that starts level with the hub.

Answers

\displaystyle h(t) = 43 - 40\cos\left(\frac{\pi}{12}t\right). The radius is 40, the hub is at 43, and \displaystyle b = \frac{2\pi}{24} = \frac{\pi}{12}; boarding at the bottom gives the negative cosine.
23, 83 and 43 meters. \displaystyle h(4) = 43 - 40\cos\frac{\pi}{3} = 43 - 20; h(12) = 43 - 40\cos\pi = 83; \displaystyle h(18) = 43 - 40\cos\frac{3\pi}{2} = 43.
\displaystyle h(t) = 23 + 20\cos\left(\frac{2\pi}{15}t\right). Starting at the top gives a positive cosine, and \displaystyle b = \frac{2\pi}{15}.
\displaystyle h(t) = 1.2 + 1.8\sin\left(\frac{\pi}{10}t\right). Three revolutions per minute is 6\pi radians per minute, \displaystyle \frac{6\pi}{60} = \frac{\pi}{10} radians per second; level and rising gives sine.
Radius 16 meters, hub 18 meters up, lowest 2 and highest 34 meters, 12 minutes per turn (\displaystyle \frac{2\pi}{\pi/6} = 12), and the rider starts at the bottom, since h(0) = 2.
\displaystyle h(t) = 23 - 20\sin\left(\frac{2\pi}{15}t\right). Clockwise the angle is \displaystyle -\frac{2\pi}{15}t, and \sin(-x) = -\sin x: the car starts down.
\displaystyle h(t) = 37 - 33\cos\left(\frac{\pi}{9}t\right). The hub is at \displaystyle \frac{70 + 4}{2} = 37, the radius is \displaystyle \frac{70 - 4}{2} = 33, and \displaystyle b = \frac{2\pi}{18} = \frac{\pi}{9}.
\displaystyle h(t) = 43 + 40\cos\left(\frac{\pi}{12}t\right), which equals \displaystyle 43 - 40\cos\left(\frac{\pi}{12}(t - 12)\right). The car reaches the bottom half a turn later, at t = 12, and -\cos(x - \pi) = \cos x.
Level with the hub, 43 meters up, and falling. At t = 0 the cosine is \displaystyle \cos\left(-\frac{\pi}{2}\right) = 0; the car reaches the bottom at t = 6, a quarter turn later, so it is on its way down.
\displaystyle h(t) = 23 - 20\cos\left(\frac{2\pi}{15}(t - 4)\right), the bottom-start model slid 4 minutes right.
\displaystyle h(t) = 1.2 + 1.8\cos\left(\frac{\pi}{10}t\right), which equals \displaystyle 1.2 + 1.8\sin\left(\frac{\pi}{10}(t + 5)\right), since \displaystyle \frac{\pi}{10}(t + 5) = \frac{\pi}{10}t + \frac{\pi}{2} and \displaystyle \sin\left(x + \frac{\pi}{2}\right) = \cos x. The second tip is 5 seconds, a quarter turn, ahead of the first.
After 9 minutes. \displaystyle \cos\left(\frac{\pi}{12}t\right) = -\frac{20\sqrt{2}}{40} = -\frac{\sqrt{2}}{2}, so \displaystyle \frac{\pi}{12}t = \frac{3\pi}{4} and t = 9.
8 minutes. \displaystyle \cos\left(\frac{\pi}{12}t\right) = -\frac{1}{2} at \displaystyle \frac{\pi}{12}t = \frac{2\pi}{3} and \displaystyle \frac{4\pi}{3}, so t = 8 and t = 16.
First below 10 meters at about 5.44 minutes, for about 4.12 minutes of each turn. \displaystyle \cos\left(\frac{2\pi}{15}t\right) = -0.65 gives \displaystyle \frac{2\pi}{15}t \approx 2.2784, t \approx 5.4392; the car rises back through 10 meters at 15 - 5.4392 = 9.5608, and 9.5608 - 5.4392 \approx 4.12.
About 5.35 seconds. h < 0 when \displaystyle \sin\left(\frac{\pi}{10}t\right) < -\frac{2}{3}; with \displaystyle \sin^{-1}\frac{2}{3} \approx 0.7297 the angles are \pi + 0.7297 and 2\pi - 0.7297, so t \approx 12.32 to 17.68 seconds.
t = 4, 8, 16 and 20. \displaystyle \cos\left(\frac{\pi}{6}t\right) = -\frac{1}{2} at \displaystyle \frac{\pi}{6}t = \frac{2\pi}{3}, \displaystyle \frac{4\pi}{3}, and those plus 2\pi; the period is 12, so add 12 to 4 and 8.
About 14.53 minutes. \displaystyle \cos\left(\frac{\pi}{12}t\right) = \frac{43 - 30}{40} = 0.325 gives \displaystyle \frac{\pi}{12}t \approx 1.2398, t \approx 4.7356; it is visible until 24 - 4.7356 \approx 19.2644.
\displaystyle h(t) = 32 - 30\cos\left(\frac{t}{75}\right), and about 7.85 minutes per turn. From Angular and linear speed, \displaystyle \omega = \frac{v}{r} = \frac{0.4}{30} = \frac{1}{75} radians per second, so T = 2\pi \cdot 75 = 150\pi \approx 471.24 seconds.
A rim point at angle \theta from the rightward horizontal is r\sin\theta above the hub. Boarding at the bottom, the angle is \displaystyle -\frac{\pi}{2} + bt after time t, and \displaystyle \sin\left(bt - \frac{\pi}{2}\right) = -\cos(bt), so h = k - r\cos(bt).
The amplitude is how far the height moves above and below the midline, the hub’s height: one radius, 32 meters. The 64 meters is the whole swing, twice the amplitude.
h > k exactly when \cos(bt) < 0, which is the middle half of each turn, \displaystyle \frac{\pi}{2} < bt < \frac{3\pi}{2}. For H > k, \displaystyle \cos(bt) < \frac{k - H}{r} < 0 holds only on a smaller arc around bt = \pi, so less than half.
From the bottom, turning either way through the same angle gives mirror-image positions across the vertical through the hub, at the same height (cosine is even). From the level position one direction goes up and the other down, so the sine changes sign.