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Файл:Ординатура / Хирургия / Библиотека им академика М.И. Перельмана / Книга_2818_Библиотеки_им_академика_М_И_Перельмана
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66
Lead V6
4
5
3
2
1
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4 ECG Intervals andSegment
R-R
1
P-P
Lead II
Fig. 4.19 Answer of Case Study Question 2 (Fig.4.11)
The fth R-R interval is 22×0.04s=0.88s.
Hence, it can be seen that there is varying R-R intervals (as well as P-P
intervals). This is seen in sinus arrhythmia, where the heart rate varies with
the various phases of respiration. The most important point to be noted is that
one must not stop after calculating only the rst P-P and R-R interval. One
has to calculate all the intervals. You will read about sinus arrhythmias in
Chapter 18.
3. The QRS duration is 4×0.04s=0.16s. This is a wide QRS interval. It is seen in
bundle branch block. Also note the notch in the descending limb of R wave
(Fig.4.20). The P-R interval is 4×0.04s=0.16s. The J point is marked with arrow.
Fig. 4.20 Answer of Case
Study Question 3
(Fig.4.12)
2
3
4
5
J Point
P-R
QRS
segment
4.9 Ventricular Activation Time
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4. The QRS duration is 5×0.04s=0.20s (Fig.4.21). This is another example of
wide QRS interval, which is seen in bundle branch block. Actually this is a RSR’
pattern, which is typically seen in bundle branch block.
5. The Q-T interval is 13×0.04s=0.52s (Fig.4.22). This is prolonged Q-T interval. At a glance one can detect, that, Q-T interval is prolonged, because the Q-T
interval is more than half of R-R interval (T wave ends closer to next P wave).
This condition is typically seen in hypocalcaemia.
Fig. 4.21 Answer of Case
Study Question 4
(Fig.4.13)
R’
67
R
P
S
S-T
T
QRS
Lead II
Fig. 4.22 Answer of Case Study Question 5 (Fig.4.14)
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4 ECG Intervals andSegment
6. The Q-T interval is 7×0.04s=0.28s (Fig.4.23). This is shortened Q-T interval.
At a glance one can detect, that, Q-T interval is shortened, because the Q-T interval is much less than half of R-R interval (T wave ends closer to QRS complex).
This condition is typically seen in hypercalcaemia.
7. a. The P-R interval is 3.5×0.04s=0.14s.
b. The QRS duration is 2.5×0.04s=0.10s.
c. P wave: Smooth, round, upright and every P wave is followed by QRS complex. Hence, sinus rhythm is present.
d. Q-T interval: 9×0.04s=0.36s.
8. a. The P-R interval is 4.5×0.04s=0.18s.
b. The QRS duration is 2×0.04s=0.08s.
c. Every QRS complex is preceded by P wave. The P wave is smooth, round and
upright. Hence, normal sinus rhythm is present.
d. The S-T segment is elevated with concavity upwards.
e. Q-T interval: 9×0.04s=0.36s.
Lead II
Fig. 4.23 Answer of Case Study Question 6 (Fig.4.15)
Chapter 5
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Calculation ofHeart Rate
Learning Objectives
After studying this chapter, the reader will learn about:
• Card method of checking cardiac rhythm
• Caliper method of checking cardiac rhythm
• Six second method of calculation of heart rate
Calculation of heart rate in ECG means calculation of ventricular rate. This is done
by calculating the number of QRS complexes per minute. Ventricular rate can be
calculated from R-R interval and atrial rate can be calculated from P-P interval. In
sinus rhythm both are same, but, during arrhythmia they have to be calculated separately. Before proceeding to the actual calculation of heart rate, one must check
whether the rhythm is regular or irregular. To check rhythm, two methods are used:
card method and caliper method.
5.1 Card Method (Paper andPencil Method)
In card method, the ECG strip is placed on a at surface. Next, the straight edge of
a card is placed along the baseline of the ECG strip. Gradually, the card is moved up
near the peak of the R wave and peak of three consecutive R waves is marked on the
card. Next, the card is moved and placed over the next three R waves and so on. If
the rhythm is regular, the marks on the card will coincide with the peak of the R
waves. If the rhythm is irregular, the marks on the card will not coincide with the
peak of the R waves.
Ltd. 2024
T. K. Koley, Rapid Review of ECG,
https://doi.org/10.1007/978-981-99-9116-7_5
69© The Author(s), under exclusive license to Springer Nature Singapore Pte
70
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5 Calculation ofHeart Rate
5.2 Caliper Method
In this method, the two points of ECG caliper are placed on the peak of two consecutive R waves (Fig.5.1). This is the R-R interval. Next, by pivoting the rst point
of the caliper towards the third R wave, it is checked if it falls on the peak of that
wave. By proceeding from left to right, the succeeding R-R intervals are checked. If
they are all the same, the ventricular rhythm is regular, otherwise irregular.
5.2.1 Regular Rhythm
Heart rate is calculated by calculating the R-R interval in presence of regular rhythm.
Heart rate=1500/R–R interval. This is true when the paper speed is at 25mm/s. For
example, if the R-R interval is 10 (10 smallest squares), then heart rate is
1500/10=150bpm (beats per minute).
There are 5 large squares per second and 300 per minute. Hence, simply count
the number of larger squares (5mm squares) in the R-R interval and divide 300 by
the number of bigger squares if the R-R interval is such that the peak of the R wave
corresponds with the dark lines on the ECG paper. This is true when the rhythm is
regular and paper speed is 25mm/s (Figs.5.2 and 5.3). For example, if the number
of bigger squares in the R-R interval is 5, then heart rate is 300/5=60bpm.
Fig. 5.1 ECG caliper is used to see regularity of R waves. This helps in checking the rhythm of
the ECG waves
5.2 Caliper Method
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Fig. 5.2 Calculation of heart rate. In this ECG, the number of small squares between the two successive R waves is 20. Hence, the heart rate (ventricular rate) is 1500÷20=75bpm. In the other
method, the calculation will be 300÷4=75bpm, because there are four big squares between the
two ‘R’ waves
Fig. 5.3 Calculation of
heart rate by counting the
number of large squares
between two successive R
waves
Two large squares (boxes) in R-R interval.
Heart rate is 300/2 = 150 beats/minute.
Three large squares (boxes) in R-R interval.
Heart rate is 300/3 = 100 beats/minute.
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5.2.2 Irregular Rhythm
During irregular rhythm, rapid rate calculation is done by 6-second method. While
using this method, the number of QRS complexes in a 6 second strip (thirty 5mm
squares) is counted, and, then it is multiplied by 10. This will give the heart rate per
minute. It is easy to calculate the 6s period, as majority of the ECG papers are
scored with a vertical mark every 3s (Fig.5.4). The 6second strip is equal to 15cm
(at 25mm/s paper speed, 1s=2.5cm, thus, 6s=6×2.5cm=15cm). For example,
if the number of QRS complexes is 15 in a 6 second strip, heart rate is
15×10=150bpm. See Fig.5.5 also.
Four large squares (boxes) in R-R interval.
Heart rate is 300/4 = 75 beats/minute.
Five large squares (boxes) in R-R interval.
Heart rate is 300/5 = 60 beats/minute.
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Hash marks
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3 second 3 second
Fig. 5.4 ECG paper with hash mark at the top of the ECG paper. These hash marks help to mark
time. The distance from one hash mark to the next represents 3s; the distance from the rst to the
third hash mark represents 6s. 15 big boxes take up the space between two hash marks
5 Calculation ofHeart Rate
Fig. 5.5 6-second method of calculation of heart rate. In this strip, the number of QRS complexes
in 6s (thirty 5mm squares) is 10. Thus, heart rate is 10×10=100bpm
Tips and Tricks
• First check the R-R interval to determine whether the rhythm is regular or
irregular.
• If regular, count the number of small boxes between two R waves and divide
1500 by this number to get the heart rate.
• If irregular, use the 6-second method.
Self-Assessment Questions
1. The heart rate can be calculated by counting the number of QRS complexes in a
6-second strip of ECG and multiplying it by 10. True or false?
2. The heart rate calculation is affected by irregular heart rhythms such as atrial
brillation. True or false?
3. The heart rate can also be calculated by dividing 1500 by the number of small
squares between two R waves. True or false?
4. If the R-R interval in an ECG is 0.8s, the heart rate is 75bpm. True or false?
5. To calculate the heart rate from an ECG, one must count the number of P waves
on an ECG strip over a 10-s interval and multiply by 6. True or false?
5.2 Caliper Method
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6. While calculating heart rate, 300 is divided by:
a. The number of large boxes between two R waves, when the R waves fall on
the dark vertical lines.
b. The number of small boxes between two R waves, when the R waves fall on
the dark vertical lines.
c. The number of small boxes between beginning of P wave and end of T wave.
d. The number of large boxes between beginning of P wave and end of T wave.
7. While calculating heart rate, by 6-second method, which of the following is
multiplied by 10?
a. Total number of QRS complexes in 6s.
b. Total number of P waves in 6s.
c. Total number of P waves and QRS complexes in 6s.
d. Total number of T waves in 6s.
Case Studies
1. Calculate the heart rate in the ECG of Fig.5.6. The paper speed is 25mm/s and
10mm=1mV.
2. Calculate the heart rate in the ECG of Fig.5.7. The paper speed is 25mm/s and
10mm=1mV.
3. Calculate the heart rate in the ECG of Fig.5.8. The paper speed is 25mm/s and
10mm=1mV.
4. Calculate the heart rate in the ECG of Fig.5.9. The paper speed is 25mm/s and
10mm=1mV.
5. Calculate the heart rate in the ECG of Fig.5.10. The paper speed is 25mm/s and
10mm=1mV.
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Lead II
Fig. 5.6 Calculate the heart rate
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Lead II
Fig. 5.7 Calculate the heart rate
Lead II
Fig. 5.8 Calculate the heart rate
5 Calculation ofHeart Rate
Lead V1
Fig. 5.9 Calculate the heart rate
Lead II
Fig. 5.10 Calculate the heart rate
5.2 Caliper Method
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75
Answers
1. True 2. True 3. True 4. True 5. False 6. a. 7. a
Case Studies
1. Figure 5.6 is a rhythm strip. To calculate the heart rate, rst check whether the
rhythm is regular or irregular. In this strip, the rhythm is regular (R-R intervals
are equal).
Next, calculate the number of small squares between two successive R waves.
Here, the number of small squares is 20.
Next, divide 1500 by 20 to get the heart rate per minute, i.e. 1500÷20=75bpm.
This is normal heart rate.
2. Figure 5.7 is another rhythm strip. First check the regularity of the rhythm. In
this strip, the rhythm is regular and the number of small squares is 11.5 between
two successive R waves. Next, divide 1500 by 11.5 to get the heart rate per minute, i.e. 1500÷11.5=130.43bpm. This may be considered as heart rate 130bpm.
3. Figure 5.8 is also a rhythm strip. To calculate the heart rate, the rst step is to
check the regularity of the rhythm. In this strip, the rhythm is regular (equal R-R
intervals) and the number of small squares is 29 between two successive R
waves. Next, divide 1500 by 29 to get the heart rate per minute, i.e.
1500÷29=51.72bpm. This may be considered as heart rate 52bpm.
4. Figure 5.9 is a rhythm strip as well. Here also at the rst step check the regularity
of the complexes. In this strip, the rhythm is irregular (unequal R-R intervals).
Hence, you have to use the 6-second method. Look at the hash marks present at
the top of the ECG paper at intervals of 15 big squares. This is a 3s interval. So
to get a 6s interval, you have to consider three hash marks, which in turn mean
30 big squares. Next, calculate the number of QRS complexes in these 30 big
squares. In this strip, there are 8 QRS complexes inside the 30 big squares.
Carefully note that the ninth QRS complex is beyond the third hash mark. The
next step is to multiply the number of QRS complexes by 10 to get the heart rate
per minute. Hence, the heart rate is 8×10=80bpm.
5. To calculate the heart rate in the rhythm strip given in Fig.5.10, rst check the
regularity of the QRS complexes. Here, the rhythm is irregular (unequal R-R
interval). Hence you have to use the 6-second method. The number of QRS complexes in the 6s interval is 16. Hence, the heart rate is 16×10=160bpm.
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