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32.4 Balancing of Rotating Machinery

Back

Many practical devices that move contain rotating components. Examples are wheels of vehicles, shafts,

gear transmissions of machinery, belt drives, motors, turbines, compressors, fans, and rollers. An

unbalance (imbalance) is created in a rotating part when its center of mass does not coincide with the axis

0.0 10.0

Nondimensional Frequency r

Transmissibility

Magnitude

Ratio

Tinertial

Tflexible

FIGURE 32.9 The effect of system flexibility on the transmissibility magnitude in the undamped case (mass

ratio ¼ 1.0; natural frequency ratio ¼ 10.0).

Vibration Design and Control 32-15

of rotation. The reasons for this eccentricity include the following:

1. Inaccurate production procedures (machining, casting, forging, assembly, etc.)

2. Wear and tear

3. Loading conditions (mechanical)

4. Environmental conditions (thermal loads and deformation)

5. Use of inhomogeneous and anisotropic material (which does not have a uniform density

distribution)

6. Component failure

7. Addition of new components to a rotating device

For a component of mass m; eccentricity e; and rotating at angular speed v; the centrifugal force that is

generated is mev2: Note the quadratic variation with v: This rotating force may be resolved into two

orthogonal components, which will be sinusoidal with frequency v: It follows that harmonic forcing

excitations are generated due to the unbalance, which can generate undesirable vibrations and associated

problems.

Problems caused by unbalance include wear and tear, malfunction and failure of components, poor

quality of products, and undesirable noise. The problem becomes increasingly important given the

present trend of developing high-speed machinery. It is estimated that the speed of operation of

machinery has doubled during the past 50 years. This means that the level of unbalance forces may have

quadrupled during the same period, causing more serious vibration problems.

An unbalanced rotating component may be balanced by adding or removing material to or from the

component. We need to know both the magnitude and location of the balancing masses to be added to,

or removed. The present section will address the problem of component balancing for vibration

suppression.

Note that the goal to remove the source of vibration (namely, the mass eccentricity) typically by adding

one or more balancing mass elements. Two methods are available:

1. Static (single-plane) balancing

2. Dynamic (two-plane) balancing

The first method concerns balancing of planar objects (e.g., pancake motors, disks) whose longitudinal

dimension about the axis of rotation is not significant. The second method concerns balancing of objects

that have a significant longitudinal dimension. We will discuss both methods.

32.4.1 Static Balancing

Consider a disk rotating at angular velocity v about a fixed axis. Suppose that the mass center of the disk

has an eccentricity e from the axis of rotation, as shown in Figure 32.10(a). Place a fixed coordinate frame

Center (Axis)

of Rotation

Center

of Mass

x = real axis

y = imaginary axis

fo = mw2e

= centrifugal forcing

(a) (b) amplitude

mw2e

w fo cos wt

fo sin wt

Eccentricity = e

(vector e

or complex

number in

the (x-y plane) e

q = wt

FIGURE 32.10 (a) Unbalance in a rotating disk due to mass eccentricity; (b) rotating vector (phasor) of centrifugal

force due to unbalance.

32-16 Vibration and Shock Handbook

x – y at the center of rotation. The position ke of the mass center in this coordinate frame may be

represented as:

1. A position vector rotating at angular speed v;.

2. A complex number, with x-coordinate denoting the real part and y-coordinate denoting the

imaginary part.

The centrifugal force due to the mass eccentricity is also a vector in the direction of ke; but with a

magnitude fo ¼ mv2e; as shown in Figure 32.10(b). It is seen that harmonic excitations result in both x

and y directions, given by fo cos vt and fo sin vt; respectively, where u ¼ vt ¼ orientation of the rotating

vector with respect to the x-axis. To balance the disk, we should add a mass m at 2~e: However, we do not

know the value of m and the location of ~e:

32.4.1.1 Balancing Approach

1. Measure the amplitude Vu and the phase angle f1 (e.g., by the signal from an accelerometer

mounted on the bearing of the disk) of the unbalance centrifugal force with respect to some

reference.

2. Mount a known mass (trial mass) Mt at a known location on the disk. Suppose that its own

centrifugal force is given by the rotating vector Vkw; and the resultant centrifugal force due to both

the original unbalance and the final mass is Vkr:

3. Measure the amplitude Vr and the phase angle f2 of the resultant centrifugal force as in step 1,

with respect to the same phase reference.

A vector diagram showing the centrifugal forces Vku and Vkw due to the original unbalance and the

trial-mass unbalance, respectively, is shown in Figure 32.11. The resultant unbalance is Vkr ¼ Vku þ Vkw:

Note that 2Vku represents the centrifugal force due to the balancing mass. Therefore, if we

determine the angle fb in Figure 32.11, it will give the orientation of the balancing mass. Suppose

also that the balancing mass is Mb and it is mounted at an eccentricity equal to that of the trial

mass Mt: Then,

Mt ¼

(Balancing Load)

Vu + Vw = Vr

(Unbalance Load)

(Resultant)

(Load Due to

Trial Mass Only)

−Vu

FIGURE 32.11 A vector diagram of the single-plane (static) balancing problem.

Vibration Design and Control 32-17

We need to determine the ratio Vu=Vw and the angle fb: These values can be derived as follows:

f ¼ f2 2 f1 ð32:26Þ

The cosine rule gives

V 2

w ¼ V 2

u þ V 2

r 2 2VuVr cos f ð32:27Þ

This will provide Vw since Vu; Vr; and f are known. Apply the cosine rule again:

V 2

r ¼ V 2

u þ V 2

w 2 2VuVw cos fb

Hence,

fb ¼ cos21 V 2

u þ V 2

w 2 V 2

2VuVw

" #

ð32:28Þ

Note: One may think that since we measure f1; we know exactly where Vku is. This is not the case

because we do not know the reference line from which f1 is measured. We only know that this reference is

kept fixed (through strobe synchronization of the body rotation) during measurements. Hence, we need

to know fb; which gives the location of 2Vku with respect to the known location of Vkw on the disk.

32.4.2 Complex Number/Vector Approach

Again, suppose that the imbalance is equivalent to

a mass of Mb that is located at the same

eccentricity (radius) r as the trial mass Mt: Define

complex numbers (mass location vectors in a body

frame)

Mk b ¼ Mb/ub ð32:29Þ

Mk t ¼ Mt/ut ð32:30Þ

as shown in Figure 32.12.

Associated force vectors are

Vku ¼ v2re jvt Mb/ub ð32:31Þ

Vkw ¼ v2re jvt Mt/ut ð32:32Þ

Vku ¼ AkMk b ð32:33Þ

Vkw ¼ AkMk t ð32:34Þ

where Ak ¼ v2re jvt is the conversion factor (complex) from the mass to the resulting dynamic force

(rotating). This factor is the same for both cases since r is the same. We need to determine Mk b:

From Equation 32.33

Mk b ¼

Vku

Ak ð32:35Þ

Substitute Equation 32.34:

Mk b ¼

Vku

Vkw

·Mk t ð32:36Þ

However, since

Vkr ¼ Vku þ Vkw ð32:37Þ

Body Coordinate

Reference

(Attached to body)

World Coordinate

Reference

Mt Mb

Unbalance

Mass

Trial

Mass

FIGURE 32.12 Rotating vectors of mass location.

32-18 Vibration and Shock Handbook

we have

Mk b ¼

Vku

ðVkr 2 VkuÞ

·Mk t ð32:38Þ

Since we know Mk t and we measure Vku and Vkr to the same scaling factor, we can compute Mk b: Locate the

balancing mass at 2Mk b (with respect to the body frame).

Example 32.3

Consider the following experimental steps:

Measured: Accelerometer amplitude (oscilloscope reading) of 6.0 with a phase lead (with respect to

strobe signal reference, which is synchronized with the rotating body frame) of 508.

Added: Trial mass Mt ¼ 20 g at angle 1808 with respect to a body reference radius.

Measured: Accelerometer amplitude of 8.0 with a phase lead of 608 (with respect synchronized strobe

signal).

Determine the magnitude and location of the balancing mass.

Solution

Method 1:

We have the data

f ¼ 60 2 508 ¼ 108

Vu ¼ 6:0; Vr ¼ 8:0

Hence, from Equation 32.27:

Vw ¼

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

62 þ 82 2 2 £ 6 £ 8 cos 108

¼ 2:37

Balancing mass:

Mb ¼

6:0

2:37 £ 20 ¼ 50:63 g

Equation 32.28 gives

fb ¼ cos21 62 þ 2:372 2 82

2 £ 6 £ 2:37

" #

¼ cos21ð20:787Þ ¼ 1428 or 2188

Pick the result 08 # fb # 1808; as clear from the vector diagram shown in Figure 32.11.

Hence,

fb ¼ 1428

However,

Mk t ¼ 20/1808 g

It follows that

2Mk b ¼ 50:63/ð1808 þ 1428Þ g ¼ 50:63/3228 g

Method 2:

We have

Mk t ¼ 20/1808 g

Vku ¼ 6:0/508

Vkr ¼ 8:0/608

Vibration Design and Control 32-19

Then, from Equation 32.38 we obtain

Mk b ¼

6:0/508

ð8:0/608 2 6:0/508Þ

20/1808 g

First, we compute

8:0/608 2 6:0/508 ¼ ð8:0 cos 608 þ j8:0 sin 6:08Þ 2 ð6:0 cos 508 þ j6:0 sin 508Þ

¼ ð8:0 cos 608 2 6:0 cos 508Þ þ jð8:0 sin 608 2 6:0 sin 508Þ ¼ 0:1433 þ j2:332

¼ 2:336/86:488

Hence,

Mk b ¼

6:0/508

2:336/86:488

20/1808 ¼

6:0 £ 20

2:336

/ð508 þ 1808 2 86:488Þ ¼ 51/143:58 g

The balancing mass should be located at

2Mk b ¼ 51/323:58 g

Note: This angle is measured from the same body reference as for the trial mass.

32.4.3 Dynamic (Two-Plane) Balancing

Instead of an unbalanced disk, consider an elongated rotating object supported at two bearings, as shown

in Figure 32.13. In this case, in general, there may not be an equivalent single unbalance force at a single

plane normal to the shaft axis. To show this, recall that a system of forces may be represented by a

single force at a specified location and a couple (two parallel forces that are equal and opposite).

If this single force (resultant force) is zero, we are left with only a couple. The couple cannot be balanced

by a single force.

All the unbalance forces at all the planes along the shaft axis can be represented by an equivalent single

unbalance force at a specified plane and a couple. If this equivalent force is zero, then to balance the

couple we will need two equal and opposite forces at two different planes.

Bearing

Accelerometer

Fu1

Bearing

Accelerometer

F Fu2 u = Fu1 + Fu2

Unbalance

Plane

l1 l2

l = l1 + l2

Resultant

Unbalance Force

Unbalance

Couple

= l2Fu2 − l1Fu1

FIGURE 32.13 A dynamic (two-plane) balancing problem.

32-20 Vibration and Shock Handbook

On the other hand, if the couple is zero, then a single force in the opposite direction at the same plane

of the resultant unbalance force will result in complete balancing. However, this unbalance plane may not

be reachable, even if it is known, for the purpose of adding the balancing mass.

In the present (two-plane) balancing problem, the balancing masses are added at the two bearing

planes so that both the resultant unbalance force and couple are balanced, in general. It is clear from

Figure 32.13 that even a sole unbalance mass Mk b at a single unbalance plane may be represented by two

unbalance masses Mk b1 and Mk b2 at the bearing planes 1 and 2. Likewise, in the presence of an unbalance

couple, we can simply add two equal and opposite forces at the planes 1 and 2 so that its couple is equal

to the unbalance couple. Hence, a general unbalance can be represented by the two unbalancemassesMk b1

and Mk b2 at planes 1 and 2, as shown in Figure 32.13. As for the single-plane balancing

problem, the resultant unbalance forces at the two bearings (which would be measured by the

accelerometers at 1 and 2) are

Vku1 ¼ Ak11 Mk b1 þ Ak12 Mk b2 ð32:39Þ

Vku2 ¼ Ak21 Mk b1 þ Ak22 Mk b2 ð32:40Þ

Suppose that a trial mass of Mk t1 (at a known location with respect to the body reference line) was added

at plane 1. The resulting unbalance forces at the two bearings are

Vkr11 ¼ Ak11ð Mk b1 þ Mk t1Þ þ Ak12 Mk b2 ð32:41Þ

Vkr21 ¼ Ak21ð Mk b1 þ Mk t1Þ þ Ak22 Mk b2 ð32:42Þ

Next, suppose that a trial mass of Mk t2 (at a known location with respect to the body

reference line) was added at plane 2, after removing Mk t1: The resulting unbalance forces at the two

bearings are

Vkr12 ¼ Ak11 Mk b1 þ Ak12ð Mk b2 þ Mk t2Þ ð32:43Þ

Vkr22 ¼ Ak21 Mk b1 þ Ak22ð Mk b2 þ Mk t2Þ ð32:44Þ

The following subtractions of equations are now made.

Equation 32.41 minus Equation 32.39:

Vkr11 2 Vku1 ¼ Ak11 Mk t1 or Ak11 ¼

Vkr11 2 Vku1

Mk t1 ð32:45Þ

Equation 32.42 minus Equation 32.40:

Vkr21 2 Vku2 ¼ Ak21 Mk t1 or Ak21 ¼

Vkr21 2 Vku2

Mk t1 ð32:46Þ

Equation 32.43 minus Equation 32.39:

Vkr12 2 Vku1 ¼ Ak12 Mk t2 or Ak12 ¼

Vkr12 2 Vku1

Mk t2 ð32:47Þ

Equation 32.44 minus Equation 32.40:

Vkr22 2 Vku2 ¼ Ak22 Mk t2 or Ak22 ¼

Vkr22 2 Vku2

Mk t2 ð32:48Þ

Hence, generally

Akij ¼

Vkrij 2 Vkui

Mk tj ð32:49Þ

These parameters Aij are called influence coefficients.

Vibration Design and Control 32-21

Next, in Equation 32.39 and Equation 32.40 eliminate Mk b2 and Mk b1 separately to determine the other.

Thus,

Ak22 Vku1 2 Ak12 Vku2 ¼ ðAk22

Ak11 2 Ak12

Ak21Þ Mk b1

Ak21 Vku1 2 Ak11 Vku2 ¼ ðAk21

Ak12 2 Ak11

Ak22Þ Mk b2

Mk b1 ¼

Ak22 Vku1 2 Ak12 Vku2

ðAk22 Ak11 2 Ak12 Ak21Þ ð32:50Þ

Mk b2 ¼

Ak21 Vku1 2 Ak11 Vku2

ðAk21 Ak12 2 Ak11 Ak22Þ ð32:51Þ

Substitute Equation 32.45 to Equation 32.48 into Equation 32.50 and Equation 32.51 to determine Mk b1

and Mk b2: Balancing masses that should be added are 2Mk b1 and 2Mk b2 in planes 1 and 2, respectively.

The single-plane and two-plane balancing approaches are summarized in Box 32.3.

Example 32.4

Suppose that the following measurements are obtained.

Without trial mass:

Accelerometer at 1: amplitude ¼ 10.0; phase lead ¼ 558.

Accelerometer at 2: amplitude ¼ 7.0; phase lead ¼ 1208

With trial mass 20 g at location 2708 of plane 1:

Accelerometer at 1: amplitude ¼ 7.0; phase lead ¼ 1208

Accelerometer at 2: amplitude ¼ 5.0; phase lead ¼ 2258

With trial mass 25 g at location 1808 of plane 2:

Accelerometer at 1: amplitude ¼ 6.0; phase lead ¼ 1208

Accelerometer at 2: amplitude ¼ 12.0; phase lead ¼ 1708

Determine the magnitude and orientation of the necessary balancing masses in planes 1 and 2 in order

to completely balance (dynamic) the system.

Solution

In the phasor notation, we can represent the given data as follows:

Vku1 ¼ 10:0/558; Vku2 ¼ 7:0/1208

Vkr11 ¼ 7:0/1208; Vkr21 ¼ 5:0/2258

Vkr12 ¼ 6:0/1208; Vkr22 ¼ 12:0/1708

Mk t1 ¼ 20/2708 g; Mk t2 ¼ 25/1808 g

From Equation 32.45 to Equation 32.48, we have

Ak11 ¼

7:0/1208 2 10:0/558

20/2708

; Ak21 ¼

5:0/2258 2 7:0/1208

20/2708

Ak12 ¼

6:0/1208 2 10:0/558

25/1808

; Ak22 ¼

12:0/1708 2 7:0/1208

25/1808

These phasors are computed as below:

Ak11 ¼ ð7:0 cos 1208 2 10 cos 558Þ þ jð7 sin 1208 2 10 sin 558Þ

20/2708 ¼

29:235 2 j2:129

20/2708 ¼

9:477/1938

20/2708

¼ 0:474/2778

32-22 Vibration and Shock Handbook

Ak21 ¼ ð5 cos 2258 2 7 cos 1208Þ þ jð5 sin 2258 2 7 sin 1208Þ

20/2708 ¼

27:036 2 j9:6

20/2708 ¼

11:9/2348

20/2708

¼ 0:595/2368

Ak12 ¼ ð6 cos 1208 2 10 cos 558Þ þ jð6 sin 1208 2 10 sin 558Þ

25/1808 ¼

28:736 2 j3:0

25/1808 ¼

9:237/1998

25/1808

¼ 0:369/198

Box 32.3

BALANCING OF ROTATING COMPONENTS

Static or single-plane balancing (balances a single equivalent dynamic force)

Experimental approach:

1. With respect to a body reference line of accelerometer signal at bearing, measure magnitude

(V) and phase (f ):

(a) Without trial mass: ðVu;f1Þ or Vku ¼ Vu/f1

(b) With trial mass Mt: ðVr;f2Þ or Vkr ¼ Vr/f2.

2. Compute balancing mass Mb and its location with respect to Mt.

3. Remove Mt and add Mb at determined location.

Computation approach 1:

Vw ¼ ½V 2

u þ V 2

r 2 2VuVw cosðf2 2 f1Þ􀀉1=2

fb ¼ cos21 V 2

u þ V 2

w 2 V 2

2VuVw

" #

and Mb ¼

Locate Mb at fb from Mt:

Computation approach 2:

Unbalance mass phasor

Mk b ¼

Vku

ð Vkr 2 VkuÞ

Mk t

where Mk t ¼ Mt/ut (trial mass phasor).

Locate balancing mass at 2Mk b.

Dynamic or two-plane balancing (balances an equivalent dynamic force and a couple)

Experimental approach:

1. Measure Vkui at bearings i ¼ 1; 2, with a trial mass.

2. Measure Vkrij at bearings i ¼ 1; 2, with only one trial mass Mk tj at j ¼ 1; 2.

3. Compute unbalance mass phasor Mk bi in planes i ¼ 1; 2.

4. Remove trial mass and place balancing masses 2Mk bi in planes i ¼ 1; 2.

Computations:

Influence coefficients: Akij ¼ ð Vkrij 2 VkuiÞ=Mk tj

Unbalance mass phasors:

Mk b1 ¼

Ak22 Vku1 2 Ak12 Vku2

ðAk22

Ak11 2 Ak12

Ak21Þ

and Mk b2 ¼

Ak21 Vku1 2 Ak11 Vku2

ðAk21

Ak12 2 Ak11

Ak22Þ

Vibration Design and Control 32-23

Ak22 ¼ ð12 cos 1708 2 7 cos 1208Þ þ jð12 sin 1708 2 7 sin 1208Þ

25/1808 ¼

28:318 2 j4:0

25/1808 ¼

9:23/205:78

25/1808

¼ 0:369/25:78

Next, the denominators of the balancing mass phasors (in Equation 32.50 and Equation 32.51) are

computed as

Ak22

Ak11 2 Ak12

Ak21 ¼ ð0:369/25:78 £ 0:474/2778Þ 2 ð0:369/198 £ 0:595/2368Þ

¼ 0:1749/251:38 2 0:2196/2178

¼ ð0:1749 cos 51:38 2 0:2196 cos 178Þ 2 jð0:1749 sin 51:38 2 0:2196 sin 178Þ

¼ 20:1 2 j0:0723 ¼ 0:1234/2168

and, hence

2ðAk22 Ak11 2 Ak12 Ak21Þ ¼ 0:1234/368

Finally, the balancing mass phasors are computed using Equation 32.50 and Equation 32.51 as

Mk b1 ¼

0:369/25:78 £ 10/558 2 0:369/198 £ 7:0/1208

0:1234/2168 ¼

3:69/80:78 2 2:583/1398

0:1234/2168

¼ ð3:69 cos 80:78 2 2:583 cos 1398Þ þ jð3:69 sin 80:78 2 2:5838 sin 1398Þ

0:1234/2168 ¼

2:546 þ j1:947

0:1234/2168

3:205/37:48

0:1234/2168 ¼ 26/2178:68

Mk b2 ¼

0:595/2368 £ 10/558 2 0:474/2778 £ 7:0/1208

0:1234/368 ¼

5:95/198 2 3:318/438

0:1234/368

¼ ð5:95 cos 198 2 3:318 cos 438Þ 2 jð5:95 sin 198 2 3:318 sin 438Þ

0:1234/368 ¼

3:2 þ j0:326

0:1234/368

1:043/5:88

0:1234/368 ¼ 8:45/230:08

Finally, we have

2Mk b1 ¼ 26/1:48 g; 2Mk b2 ¼ 8:45/1508 g

32.4.4 Experimental Procedure of Balancing

The experimental procedure for determining the balancing masses and locations for a rotating system

should be clear from the analytical developments and examples given above. The basic steps are:

1. Determine the magnitude and the phase angle of accelerometer signals at the bearings with and

without trial masses at the bearing planes.

2. Using this data, compute the necessary balancing masses (magnitude and location) at the bearing

planes.

3. Place the balancing masses.

4. Check whether the system is balanced. If not, repeat the balancing cycle.

A laboratory experimental setup for two-plane balancing is shown schematically in Figure 32.14. A

view of the system is shown in Figure 32.15. The two disks rigidly mounted on the shaft are driven by a

DC motor. The drive speed of the motor is adjusted by the manual speed controller. The (two) shaft

32-24 Vibration and Shock Handbook

bearings are located very close to the disks, as shown in Figure 32.14. Two accelerometers are mounted on

the top of the bearing housing so that the resulting vertical accelerations can be measured. The

accelerometer signals are conditioned using the two-channel charge amplifier and read and displayed

through two channels of the digital oscilloscope. The output of the stroboscope (tachometer) is used as

the reference signal with respect to the phase angles of the accelerometer signals that are measured.

In Figure 32.15, the items of equipment are seen from left to right. The first item is the two-channel

digital oscilloscope. The manual speed controller with control knob for the DC motor follows. Next is the

pair of charge amplifiers for the accelerometers. The strobe-light unit (strobe-tacho) is placed on top of

the common housing of the charge amplifier pair. The two-disk rotor system with the drive motor is

shown as the last item to the right. Also, note the two accelerometers (seen as small vertical projections)

mounted on the bearing frame of the shaft directly above the two bearings.

Because this reference always has to be fixed prior to reading the oscilloscope data, the strobe-tacho is

synchronized with the disk rotation. This is achieved as follows (note that all the readings are taken with

Charge Amplifier

Pair

Digital

Oscilloscope

Disk 1 Disk 2

Accelerometer 1 Accelerometer 2

ch 1 ch 2

Support

Frame

DC Motor

Speed Control Unit

for DC Motor Stroboscope

Strobe Signal

FIGURE 32.14 Schematic arrangement of a rotor balancing experiment.

FIGURE 32.15 A view of the experimental setup for two-plane balancing at the University of British Columbia.

Vibration Design and Control 32-25

the same rotating speed, which is adjusted by the manual speed controller): First, make a physical mark

(e.g., a black spot in a white background) on one of the disks. Aim the strobe flash at this disk. As the

motor speed is adjusted to the required fixed value, the strobe flash is synchronized such that the mark on

the disk “appears” stationary at the same location (e.g., at the uppermost location of the circle of

rotation). This ensures not only that the strobe frequency is equal to the rotating speed of the disk, but

also that the same phase angle reference is used for all readings of accelerometer signals.

The two disks have slots at locations whose radius is known, and whose angular positions in relation to

a body reference line (a radius representing the 08 reference line) are clearly marked. Known masses

(typically, bolts and nuts of known mass) can be securely mounted in these slots. Readings obtained

through the oscilloscope are:

1. Amplitude of each accelerometer signal

2. Phase lead of the accelerometer signal with respect to the synchronized and reference-fixed strobe

signal (note: a phase lag should be represented by a negative sign in the data)

The measurements taken and the computations made in the experimental procedure should be clear

from Example 32.4.

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