(C) 2021 Vernon R Stanley, MD, PhD, Author | Courtney Stanley, PA-C Co-editor | ECGcourse.com, LLC | All rights reserved.

Question: True or False? The computer interpretation of the 12-lead ECG is very reliable and should be used regularly in clinical practice. (See bottom of post for answer.)

The electrical axis refers to the frontal plane.  It is a vector (an arrow with direction and magnitude) pointing in the direction of the resultant depolarization wavefront (net summation) of the ventricles.  Since the Left Ventricle supplies the dominate wavefront voltage and we know that this wavefront is usually in a leftward-downward direction, we would expect the electrical axis to be in this same direction. This is indeed the case as is demonstrated in Diagram #52 below.  It is certainly true that the axis can be found by many different approaches; but the most straightforward approach arises from the field of mathematics, and it is the approach that I recommend.

This approach relies on the following law of mathematics: If two vectors are perpendicular to each other, the resultant vector is found by drawing the diagonal of the rectangle formed by these two vectors (demonstrated in Diagram #52 below).

Diagram #52: Normal Axis

The salient points to be gleaned from this diagram are as follows:

•  A NORMAL axis will have a dominantly positive Lead I and Lead avF

and the converse…

•  If Lead I and Lead avF are dominantly positive…..the axis is NORMAL

If available to the student, please see pages 68-69 of the 12-lead ECG Textbook for the other three quadrants of axes and for further discussion of axis. Diagram #53: Demonstration of Normal, Left, Right and Severe Right Axis

For those inclined to review the details, here is an example of axis calculation:

Net Area Lead I = 1/2 b (2 * 3 – 2 * 1) = 1/2 * 4 = 2

Net Area Lead avF = 1/2 b = 1/2 (-2 * 1 + 2 * 8) = 1/2 (14) = 7

The vector diagram then becomes:

B = tan -1 3.5 = 74 degrees = Normal Axis

Note: For other Example Axis Calculations see the 12-lead ECG Workbook page 28 and Text Appendix A-2 in back of the 12-lead ECG Textbook.

Net Area Lead I = 1/2 b (2 * 3 – 2 * 1) = 1/2 * 4 = 2 Net Area Lead avF = 1/2 b = 1/2 (-2 * 1 + 2 * 8) = 1/2 (14) = 7

## CASE STUDY

You are a member of a jeopardy team from your alma mater competing with a long-time rival university.  The question (final jeopardy) is: “What is the approximate QRS axis of the tracing below?”

Approximate axis = – 30 degrees

TOPIC OF DISCUSSION

This lesson is specifically directed to the topic of the electrical axis, more particularly the QRS axis of the frontal plane.  This axis is determined by the limb leads I, II, III, avR, avL, avF.  For emphasis I have sampled and magnified each of these six leads and will refer to select ones in my following discussion.

GENERAL CONCEPTS

The electrical axis is best calculated by borrowing from the mathematics concepts of vector analysis.  The electrical axis is described as the vector direction of the resultant depolarization wave of the free  right, left ventricle and the septum.  This vector can be found by the resultant of two vectors directed at 90 degrees to each other.

We will agree to use the following two vectors:

Lead avF (pointing straight downward)

and

Lead I (pointing horizontal and to the patient’s left)

The resultant of these two vectors is found as the diagonal of the rectangle formed by the two vectors:

Equation 1:  Axis Vector = avF Vector + Lead I Vector
Equation 1………………….Axis Vector = avF Vector + Lead I Vector

The magnitude of the axis vector is found using the Pythagorean Theorem as follows:

Equation 2………………….. | Axis Vector | squared =………| avF Vector | squared + |  I Vector | squared

Taking the square root of each side, Equation 2 becomes:

Equation 3……… | Axis Vector | = Square Root of [   (avF Vector) squared      +         (I Vector) squared    ]

The angle direction of the electrical axis is calculated using the principles of trigonometry as follows:

Equation 4………. axis angle   =       arc tangent    ( | avF vector |  /   |  I  vector |  )

where…… tan -1   =     arc tangent.

EXAMPLE AXIS CALCULATION

For the sake of example let us consider the special case of

| avF  vector |   =   1

|  I  vector  |     =   square root of 3

Equation 4 becomes:

Angle of Axis      =   arc tangent     ( 1   /  square root of 3  )       =   arc tangent     (  square root of 3 / 3  )

Angle of axis  = 30 degrees

It can be shown that, if the electrical axis is – 30 degrees, Lead II will be equiphasic, i.e. the positive area of the QRS complex will equal the negative area.

SPECIAL CASE OF AXIS  =  – 30 degrees

The key to the calculation of the axis lies in Lead I and Lead avF.  It can be shown that if the following triad holds true:

1. Lead I  = dominantly positive
2. Lead avF = dominantly negative
3. Lead II = Equiphasic (positive area = negative area)

We can conclude

Equation 4   axis  =  – 30 degrees

SPECIAL CASE OF AXIS =  – 45 degrees

It can be shown that if the following triad holds true:

1. Lead I  =  dominantly positive
2. Lead avF  =  dominantly negative
3. | Net area Lead I | = |  Net area Lead avF |

Equation 5      axis   =  – 45 degrees.

The Left Anterior Fascicular Block (LAFB) is characterized by the following triad:

1.  Axis  more negative than -45 degrees.

2.  Small R-wave in Leads II, III, avF.

3.  Small Q-wave in Leads I, avL.

From Equation 5 above we can clearly see that the Triad of the LAFB is equivalently stated as follows:

1.  Lead I = positive
Lead avF = negative
| Lead I | = | Lead avF |

2.  Small R-wave in Leads II, III, avF.

3.  Small Q-wave in Leads I, avL.

See Dr. Stanley’s H*E*A*R*T Rule for clinical tool reference.

PEARL AND CONCLUSION

Normal Axis (between 0 degrees and 90 degrees)

Lead I = dominantly positive

Lead avF = dominantly positive

____

Axis – 30 degrees

Lead I = dominantly positive.

Lead avF = dominantly negative.

Lead II = equiphasic (positive = negative).

____

Axis – 45 degrees

Lead I = dominantly positive.

Lead avF = dominantly negative.

| Lead I  | = | Lead avF |.

_____
LAFB

1.  R-wave Lead II, III, avF.

2.  Q-wave Lead I, avL.

3.  Axis < or = – 45 degrees.

From this Case Study tracing, please note that I have magnified all six limb leads and will ask you to particularly notice the characteristics of Lead I, Lead avF and Lead II:

Lead I is dominantly positive.

Lead avF is dominantly negative.

Lead II is characterized by:  Positive area = negative area (equiphasic)

We therefore conclude from Equation 4 that the axis  =  -30 degrees.

COROLLARY:   If Lead II is predominantly negative, you can conclude that the axis is more negative than – 30 degrees.