Notes from the Therapist

Spinal Motion and Facet Joints

2011-07-19T02:02:00.009+08:00

To determine the exact location of a facet injury, it is very important to understand the underlying principles of the spinal movement and how the facets are oriented in different regions. Thus, this blog entry only focuses on Fryette's laws, facets' anatomical orientation, and how the trunk movement influences the opening and closing of facet joints.Facet joint syndrome will be discussed in the next blog entry.

Note that Fryette only presented three laws. I don't know where Sir Gerard Martin, PTRP, RPT

(lecturer) got the fourth law

I. Fryette’s Laws of Physiologic Spinal Motion

“Coupled motions in the spine originate with concepts brought forth by an osteopathic physician and are named after him. Although they are not actually “laws” of motion as much as they are observations and ideas, they are called Fryette’s Laws of Spinal Physiologic Motion,” or more simply, “Fryette’s Laws of Motion.” Fryette presented his ideas of coupled motions between facets in both a neutral position and out of a neutral position. He presented three laws that determine coupling motions of the spine.”(Houglum, 563).

Gerard Martin speaks of these laws as that which “explain the movement of the spine.”

Law 1: If the upper segments of vertebrae are moving, the lower segments will have an opposite direction of movement provided that the spine is neutral.

“Fryette’s First Law states that when the lumbar or thoracic spine is neutral, side-bending occurs to the opposite side of the vertebral rotation. For example, in neutral position, if the spine is laterally flexed to the right, it will also rotate to the left. This law does not address the cervical spine since he defined the neutral position as when the facet joint surfaces are not in contact with each other, but the adjacent cervical facet are always in contact with each other” (Therapeutic Exercise for Musculoskeletal Injuries by Peggy A. Houglum, 563).

Law 2: If the trunk is in flexion or hyperextention, the movements of all the segments are the same.

“Fryette’s Second Law deals with pathological positions and coupled motions. Fryette indicated that when the spinal alignment is either in flexion or extension, the side-bending and rotation of the vertebrae will be toward the same direction. For example, if the lumbar spine is placed into a lordotic position, and side-bending to the left is performed, rotation to the left will also occur at the vertebral level.” (Houglum, 563).

Law 3: If only a segment of the vertebra is moving the movements of the other segments are reduced.

“Fryette’s Third Law states that if motion in one place occurs in the spine, motion in the other direction is diminished. For an example of this, stand erect, rotate the spine to the left, and note the amount of rotation that occurs. Next, forward flex at the trunk and then in that position, repeat the left rotation motion. You will find that the amount of rotation you are able to perform is less than in erect standing.” (Houglum, 563).

Law 4: If a segment of the vertebra becomes dysfunctional, other segments tend to compensate.

Example: When you have stiff neck, compensatory movement is more manifest in the thoracic area.

II. Zygopophyseal joints (Facet joints): responsible for directing the movement of the spine.

Anatomical orientation

Facets joints have different orientations depending on the area.

Fig. 1The cervical, thoracic, and lumbar vertebrae differ from each other. From the cervical to the lumbar region the bodies of the vertebrae become larger, and the transverse processes, spinous processes, and apophyseal joints all changetheir orientation. (Biomechanical Basis of Human Movement by Joseph Hamill; Kathleen M. Knutzen).

Cervical Region = 45 degrees; frontal plane; all movements are possible such as flexion, extension, lateral flexion, and rotation.

The articulating facets in the cervical vertebrae face 45 degrees to the transverse plane and
lie parallel to the frontal plane (82), with the superior articulating process facing posterior and up and
the inferior articulating processes facing anterior and down. In contrast to other regions of the
vertebral column, the intervertebral discs are smaller laterally than bodies of the vertebrae. The
cervical discs are thicker ventrally than dorsally, producing a wedge shape and contributing to the
lordotic curvature in the cervical region.

Because of the short spinous processes, the shape of the discs, and the backward and
downward orientation of the articulating facets, movement in the cervical region is greater than in
any other region of the vertebral column. The cervical vertebrae can rotate through approximately
90degress, flex 20 to 45 degrees to each side, flex through 80 to 90°, and extend through 70 degrees
(87). Maximum rotation in the cervical vertebrae occurs at C1-C2, maximum lateral flexion at C2
C4, and maximum flexion and extension at C1- C3 and C7-T1. Also, all cervical vertebrae move
simultaneously in flexion. (Hamill & Knutzen, 268.)

Thoracic Region = 60 degrees; frontal plane; lateral flexion and rotation; no flexion/extension

The apophyseal joints between adjacent thoracic vertebrae are angled at 60° to the transverse plane and 20° to the frontal plane, with the superior facets facing posterior and a little up and
laterally and the inferior facets facing anteriorly, down, and medially (Fig. 1 above). Compare with the cervical vertebrae, the thoracic intervertebral joints are oriented more in the vertical plane.

The movements in the thoracic region are limited primarily by the connection with the ribs, the orientation of the facets, and the long spinous processes that overlap in the back. Range of motion in the thoracic region for flexion and extension combined is 3° to 12°, with very limited motion in the upper thoracic (2° to 4°) that increases in the lower thoracic to 20° at the thoracolumbar junction (10,97).

Lateral flexion is also limited in the thoracic vertebrae, ranging from 2° to 9° and again increasing as one progresses down through the thoracic vertebrae. Whereas in the upper thoracic vertebrae, lateral flexion is limited to 2° to 4°, in the lower thoracic vertebrae, it may be as high as 9° (10,97).

Rotation in the thoracic vertebrae ranges from 2° to 9°. Rotation range of motion is opposite to that of flexion and lateral flexion because it is maximum at the upper levels (9°) and is reduced at the lower levels (2°) (10,97) (Hamill & Knutzen, 268)

Lumbar Region = 90 degrees; sagittal plane; only flexion and extension.

The apophyseal joints in the lumbar region lie in the sagittal plane; the articulating facets are at right angles to the transverse plane and 45° to the frontal plane (97). The superior facets face medially, and the inferior facets face laterally. This changes at the lumbosacral junction, where the apophyseal joint moves into the frontal plane and the inferior facet on L5 faces front. This change in orientation keeps the vertebral column from sliding forward on the sacrum. (Hamill & Knutzen, 269)

The range of motion in the lumbar region is large in flexion and extension, ranging from 8° to 20° at the various levels of the vertebrae (10,97). Lateral flexion at the various levels of the lumbar vertebrae is limited, ranging from 3° to 6°, and there is also very little rotation (1° to 2°) at each levels of the lumbar vertebrae (10,97). However, the collective range of motion in the lumbar region ranges from 52° to 59° for flexion, 15° to 37° for extension, 14° to 26° for lateral flexion and 9° to 18° of rotation (93).

The lumbosacral joint is the most mobile of the lumbar joints, accounting for a large proportion of the flexion and extension in the region. Of the flexion and extension in the lumbar vertebrae, 75% may occur at this joint, with 20% of the remaining flexion at L4- L5 and 5% at the other lumbar levels (77).

B. Different orientations of facets as spine moves

- Facets are parts of the posterior pillar

- When spine moves, facets will have different orientation.

Trunk motion:

Flexion: facets open

Extension: facets closed

Left rotation: left facets open; right facets closed

Right rotation: right facets open; left facets closed

Right lateral flexion: right facets closed; left facets elongate

Left lateral flexion: left facets closed; right facets elongate

Examples:

Anterior ankle rock (tip toe): facets closed

Heel-walk: facets open

Anterior perturbation (pushed anteriorly/pushed from back to front): facets closed

Posterior perturbation: facets open

Posterior ankle rock: facets open

Standing toe-reach: facets open

Backward lurching: facets closed

Arms move upward from a horizontal forward reach: facets closed

Horizontal forward reach with a forward lean (erector spinae muscles): facets closed

Keep horizontal forward reach with backward movement (abdominals contracting): facets open

Arms reach side-ward into a horizontal side-ward reach to the right: right facets elongated and left facets closed.

Holding a wand, then rotate to the right: right facets open; left facets closed.

References

Hamill, Joseph, and Kathleen M. Knutzen. Biomechanical Basis of Human Movement, 3^rd Ed., USA: Lippincott Williams & Wilkins, 2009.

Houglum, Peggy A. Therapeutic Exercise for Musculoskeletal Injuries, 3^rd Ed., USA: Human Kinetics, 2010.

Martin, Gerard L. “Spinal Motion and Facet Joints.” Class lecture, SLRC, Sampaloc, Manila, July 12, 2011.

Martin, Gerard L. “Spinal Motion and Facet Joints.” Class lecture, SLRC, Sampaloc, Manila, July 13, 2011.

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CARDIAC PHYSIOLOGY AND ECG INTERPRETATION (Last of Two Parts)

2011-07-13T21:42:00.003+08:00

II. The Electrocardiogram

“When the cardiac impulse passes through the heart, electrical current also spreads from the heart into the adjacent tissues surrounding the heart. A small portion of the current spreads all the way to the surface of the body. If electrodes are placed on the skin on opposite sides of the heart, electrical potentials generated by the current can be recorded; the recording is known as an electrocardiogram” (Guyton and Hall, 123).
Defines the graphic representation of the electrical activity of the heart.
The printed record of the electrical activity of the heart is called a rhythm strip or an ECG strip.
Normal electrocardiogram (Figures 1 and 2). Please take note that Guyton and Hall (Fig 1) give a specific P-R interval which is 0.16 sec.
In other sources, the P-R interval is between 0.12 and 0.20 sec. (Fig 2)

Fig. 1. Normal ECG. (From Guyton and Hall, Textbook of Medical Physiology 11^th Ed.,, 124)

Fig. 2

Fig. 3.A block (0.20 sec) and a small square/box (0.04 sec)

X and Y axes (coordinate system)

Any wave that goes up is a positive deflection or up sloping (+X vertical axis).

Any wave that goes down is a negative deflection or down sloping (-X vertical axis)

III. How are the (normal) waves/intervals/segment of the electrocardiogram produced? (Breakdown of ECG)

A. Normal waves

P = (Pacemaker) SA node fires, sends the electrical impulse outward to stimulate both atria and manifests as a P wave (Fig.4). “The P wave is caused by electrical potentials generated when the atria depolarize before atrial contraction begins”(Guyton and Hall, 123). P-Wave is the first positive deflection or the first up sloping which is responsible for atrial depolarization (Martin).

“The P wave represents the spread of the electrical impulse through both atria.The electrical impulse begins in the SA node and depolarizes the right atrium and then the left atrium. Thus, the first part of the P wave reflects right atrial activity, and the late portion of the P wave represents electrical potential generated by the left atrium” (Kahn, 81).

Fig. 4. P Wave

Ø Wave duration: approximately 0.10 second in length. (But Gerard Martin said its wave duration is less than 0.10 sec, which is approximately 2 small squares, that is, 0.08 sec)

Ø Wave amplitude: 1-3mm. So, it is only 0.1 to 0.3mV.

Q = Q wave follows the P wave (atrial depolarization or atrial contraction ). It is a negative deflection or the first down sloping after the P wave.
R = R wave follows the Q wave. This is the 2^nd positive deflection.
S =  S wave follows the Q wave. This is the 2^nd negative deflection.        Three waves combine into one: The QRS Complex(Fig. 5)
Ø Ventricular depolarization (contraction)
“The QRS complex is caused by potentials generated when the ventricles depolarize before contraction, that is, as the depolarization wave spreads through the ventricles” (Guyton and Hall, 123).
Ø Impulse from the Bundle of HIS throughout the ventricular muscles
“The QRS complex represents the spread of electrical activation through the ventricular myocardium; the resultant electrical forces generated from ventricular depolarization is recorded on the ECG as spiky deflection. The sharp, pointed deflections are labeled QRS regardless of whether they are positive (upward) or negative (downward)” (Kahn, 9).
Ø Wave duration of ventricular depolarization: 0.06 to 0.10 sec
·         Small squares: 11/2 to 21/2 or 1.5 to 2.5
·         (This is a review. Remember that a small square has a wave duration of 0.04)
Ø Wave amplitude: 25- 30mm
·         2.5 – 3mV.
·         5-6 big blocks or 25 to 30 small squares
·         (Always remember that a block has wave amplitude of 0.5mV and its vertical length is 5mm.)
Fig. 5. QRS Complex
T

Fig. 6. T Wave

T wave follows the QRS complex
It is the 3^rd positive deflection or the 3^rd up sloping.
The T wave represents repolarization, the recovery period of the ventricles (Kahn, 193).
No associated activity of the ventricular muscle; the resting phase of the cardiac cycle.
“The T wave is caused by potentials generated as the ventricles recover from the state of depolarization. This process normally occurs in ventricular muscle 0.25 to 0.35 second after depolarization, and the T wave is known as a repolarization wave” (Guyton and Hall, 123).

Its wave duration is 0.16 sec ( 4 small squares)
Its wave amplitude is 5-10mm ( 0.5 to 1 mV; 1-2 big blocks or 5-10 small squares)

Is there such a thing as atrial repolarization? Yes. But it is being overshadowed by tall QRS.
Absence of Q on ECG is still normal. Q appears on an ECG to gain momentum (like a pre-lengthening stage) as the electrical current continues to spread. But if the ventricles are healthy, they do not need to gain momentum. And so, on an ECG only P, R, S, T appears. In others words, since healthy ventricles are strong, they can easily contract without going through a pre-lengthening stage, so to speak. Hence, Q is absent.

B. Normal intervals
PR Interval (PRI)
- The P is the atrial depolarization and the R is part of the ventricular depolarization.
-  P to R is the conduction of the heart.
-  Time which impulse travels from the SA node to the atria and downward to the ventricles.
  “The PR interval involves the time required for the electrical impulse to advance from the atria through the AV node, bundle of His, bundle branches, and Purkinje fibers until the ventricular muscle begins to depolarize”(Khan, 9)

Fig. 7. PR interval (also see figures 1 and 2)

-Normal PR interval is 0.12 to 0.20 sec. comprises of 3 to 5 small squares.(see fig. 2). If there is a conduction interference, the patient will have heart block or AV block.

- a wave duration of 0.38 sec. is a prolonged PR interval, which leads to first-degree heart block; only light exercises are allowed.

QT Interval

Q is the ventricular depolarization

- T is also ventricular depolarization

- So, Q-T is ventricular contraction.

“Contraction of the ventricle lasts almost from the beginning of the Q wave (or R wave, if the Q wave is absent) to the end of the T wave. This interval is called the Q-T interval and ordinarily is about 0.35 second” (Guyton and Hall, 125-126).

- Normal QT interval is 0.32 to 0.40 sec (Martin).

C. Normal segment

- ST segment (see figures 1 and 2)

J point (Junctional point) interconnects S and T. This point is significant. If this point goes up, the ST segment elevates. And if it goes down, ST depression occurs. ST segment wave duration is 0.12 sec. (3 small squares). Its wave amplitude is – .1/2 to +1mm; -0,05 to +0.1mV.

IV. Determining Heart Rate on an ECG

A. Heart rate computation: 6-Second Method: Have a six second strip, count the QRS complexes and multipled by 10.

Formula: H.R. = Number of QRS complexes in 6 second strip x 10.

6 second strip = 30 big blocks

= 30 x 10 = 300 big blocks (beats per minute)

Example 1:

- Calculate the heart rate using 6 second rule (method)

1. 6 second strip is selected between two arrows

2. Number of QRS complexes counted in 6 seconds is 13.

3. Heart rate = 13 x 10 = 130 beats per minute

Example 2: (This is the example Sir Gerard used in his lecture. Please check your notebook. I presume you wrote down the examples and ECG strips he gave during the lecture.)

Given:

§ There are 3 QRS complexes in 3 second strip

§ Big blocks: 15

§ 3 x 20 = 60 beats per minute

or: 3 x 2 x 10 = 60 beats per minute

3 (QRS complexes) is multiplied by 2 to make it a 6 second strip. It is presumed that the rhythm in the second example is regular.

References

Gabriel Khan, M. Rapid ECG Interpretation, 3^rd Ed. Totowa, New Jersey: Human Press, 2008.

Guyton, Arthur C., and John E. Hall. Text Book of Medical Physiology, 11^th Ed. Philadelphia, Pennsylvania: Elsevier Inc., 2006.

Martin, Gerard L. “Cardiac Physiology and ECG Interpretation.” Class lecture, SLRC, Sampaloc, Manila, May 6, 2011.

Martin, Gerard L. “Cardiac Physiology and ECG Interpretation.” Class lecture, SLRC, Sampaloc, Manila, May 9, 2011.

“Understanding ECG’s.” Flight Medic Course (ACLS). http://usasam.amedd.army.mil/_fm_course/Study/UnderstandingECG.pdf (accessed July10, 2011)

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CARDIAC PHYSIOLOGY AND ECG INTERPRETATION (First of Two Parts)

2011-07-12T00:15:00.008+08:00

I. Basic Concept

A. Electrical Activity of the Heart

- Each contraction of the heart is preceded by excitation waves ofelectrical activity that originate in the sinoatrial (SA) node.

ü SA node

· is a small, flattened, ellipsoid strip of specialized cardiac muscle about 3 millimeters wide, 15 millimeters long, and 1 millimeter thick.

· located near the junction of the superior vena cava and RA.

· located in the superior posterolateral wall of the right

atrium immediately below and slightly lateral to the opening of the superior vena cava.

· The fibers of this node have almost no contractile muscle filaments

and are each only 3 to 5 micrometers in diameter, in contrast to a diameter of 10 to 15 micrometers for the surrounding atrial muscle fibers.

· However, the sinus nodal fibers connect directly with the atrial muscle fibers, so that any action potential that begins in the sinus node spreads immediately into the atrial muscle wall.

· Consist of 2 cell types

1. pacemaker of P cells – ovoid in shaped thought to perform the actual pacemaker function.

2. Transitional or T cells – elongated lie between P cells and typical atrial muscle cells.

· has an intrinsic firing rate of 90-120 beats/min being higher in young patients.

- The waves of electrical activity spread through the atria and reach the atrioventricular (AV) node.

ü AV node

· located in the posterior wall of the right atrium immediately behind the tricuspid valve.

· located beneath the endocardium

· right side of the interatrial septum

· only pathway for excitation to proceed from atria to ventricles.

Note that the SA node tracing shows no steady resting potential, as does the ventricular muscle tracing.

The SA node’s spontaneous depolarization and repolarization provides a unique and miraculous automatic pacemaker stimulus that activates the atria and the AV node, which conducts the activation current down the bundle branches to activate the ventricular muscle mass.

ü Bundle of His (Fig. 1)
·   divides into R and L bundle branches
·   lie beneath the endocardium on the two respective sides of the ventricular septum.
·   Each branch spreads downward toward the apex of the ventricle, progressively dividing into smaller branches. These branches in turn course sidewise around each ventricular chamber and back toward the base of the heart.
·   The ends of the Purkinje fibers penetrate about one third the way into the muscle mass and finally become continuous with the cardiac muscle fibers.

Fig. 1. Sinus node, and the Purkinje system of the heart, showing also the AV node,

atrial internodal pathways, and ventrical bundle branches.

  Cardiac cells outside the SA node normally do not exhibit spontaneous depolarization; thus they must be activated.

Here then is the conduction pathways of the heart:
SA node-->AV node --> Right and Left Bundle Branches --> Purkinje Fibers --> Ventricle

       Physiologic delay at AV node
·     To allow for ventricular filling. Without this normal delay at AV node and if there is an immediate ejection, air embolus may happen (Martin).
·      The AV node provides a necessary physiologic delay of the electrical currents, which allows the atria to fill the ventricles with blood before ventricular systole.
·     (Mechanism) The reason why there is a physiologic delay at AV node is because the fibers of AV node are extremely small. Thus, conduction velocity is slow. Also, the Purkinje fibers conduct very slowly (Martin)
·     The transient halt and slowing of conduction through the specialized AV
node fibers play an important protective role in patients with atrial flutterand atrial fi brillation. In these common conditions, a rapid atrial rate of approximately 300 to 600 beats/min reaches the AV node; this AV “tollgate” reduces the electrical traffi c that reaches the superhighway that traverses the ventricles to approximately 120 to 180 beats/min, and serious life-threatening events are prevented.

       Heart block (AV block): Cardiac dromotropic incompetence; conduction interference
  ·   There is discrepancy in P-R interval; prolonged P-R interval.
  ·   Normal P-R interval is 0.12 to 0.20 sec.
a.   First degree heart block
·     involves Purkinje fibers and ventricles
·     can still perform exercise (light aerobic exercise)

b. Second degree heart block (Mobitz type 1)
·   involves AV Bundle of His, right and left bundle branches, Purkinje Fibers, and Ventricles.
· P on P; absence of QRS complex:
Ø Wenckebach phenomenon (named after the Dutch Internist Karl Wenckebach) describes a disturbance in conduction where the atrioventricular node conducts each successive impulse earlier and earlier. Eventually, an impulse arrives when the node is not able to conduct it. The following impulse will then be conducted normally, causing the cycle to begin again.
Ø The QRS complex and the P wave are normal, but not all P waves are followed by a QRS complex. The PR interval gets longer until an impulse is not conducted.
Ø The ECG trace below shows sinus rhythm at a rate of 67 bpm (arrowed), with a lengthening PR interval (shaded) preceding a nonconducted P wave.

(Fig.2) Electrocardiogram: Second Degree Block — Mobitz Type I (Wenckebach

· exercise is contraindicated

c. Third degree heart block (Mobitz type II)

· complete heartblock; from SA node to ventricles are affected

· More on P on P phenomenon than Mobitz type I

· artificial cardiac pacemaker needed

Pharmacology: Effects of meds on heart block

· Heart block may lead to cardiac arrest. Both HR and BP decrease.

· Meds should be administered to increase both the HR and BP.

· Pharmacologic effect: Abrupt;

§ Cardiac terminology:

a) Abrupt: rapidly increasing

b) Shunted-Blunted: gradually decreasing

c) Plateau: steady state

§ Abrupt increase in HR and BP

ü Anticholinergic drug: sympathomimetic and parasympatholytic

Ø Atropine SO4: antidote to cholinergic crisis; it blocks the acethylcholine release in the parasympathetic.

Artificial cardiac pacemaker (for third degree heart block)

· Poorly insulated pace maker (in the past): very sensitive; when a pt is grounded, it reverts to default settings. Ultrasound should be applied 6 inches away from the pacemaker. Use of electrical modalities is relatively contraindicated.

· Well-insulated pacemaker (2011): nothing to worry about. ES, US, and other electrical modalities, except MRI, are not contraindicated. These electrical modalities can be used as long as they are more than 2 inches from the apparatus.

Cardiac Action Potential

- Our heart is excitable tissue.

- Normal resting membrane potential of the heart (RMP): -60 to -80 mV

- resting membrane potential is in the polarized state.

- In a resting cardiac muscle cell, molecules dissociate into positively charged ions on the outer surface and negatively charged ions on the inner surface of the cell membrane; the cell is in an electrically balanced or polarized resting state.

- Cardiac action potential has 5 phases.

- Phase 4 is an rmp phase which is brought about by sodium-potassium pump and by the presence of the negatively charged protein which is imperpeable. Thus, rmp phase is in the negative state.

- Phase 0 is a rapid depolarization. At this point, the membrane is less negative brought about by the sodium (Na+) influx. Sodium then gets inside the membrane making it less negative until it reaches +30mV, a peak potential, which then undergoes an overshoot.

- This peak potential immediately goes down losing its positivity which is now the initial repolarization phase (phase 1). Thus, phase 1 is the initial repolarization.

- And since the membrane loses positivity, it then loses positive ion because potassium (K+) goes outside (potassium efflux). (Sodium displaces potassium, hence potassium efflux occurs.)

- Once potassium is outside the membrane, calcium then goes inside since both potassium and calcium are positively charged ions (cat ion) They cannot stay together. There is repulsion.

- The influx of Calcium then creates a steady state which is the plateau phase. (Nerve action potential doesn’t have a plateau phase.)

- This plateau phase (phase 2) is only momentary because Calcium has only a scanty amount of concentration, and therefore it is easily used up while potassium efflux continues

- As the potassium continues to go outside the membrane, repolarization (phase 3, the real repolarization) occurs

- Next depolarization does not happen if sodium (and calcium) remain inside while potassium stays outside the membrane. They have to be brought back to where they came from. Thus, sodium-potassium happens again. It pumps 3 Na+ from the intracellular to the extracellular fluid compartment. It also pumps 2 K+ from extracellular to the intracellular fluid compartment. So the net diffusion is 1, that is, 1 Na+. If you have five cycles of pump, then you have 15 Na+ pumped to the outside and 10 K+ to the inside. The net diffusion is 5 Na+ outside, thus the outer surface of the cell membrane is positively charged. The inner surface is negative. Hence, the rmp is always negative.

fig. 3. Action potential

Fig. 4. Sodium infl ux, potassium eff ux, the action potential, and the electrocardiogram.(From Khan, M. Gabriel: On Call Cardiology, 3rd ed., Philadelphia, 2006, WB Saunders, Elsevier

Science.

- When the cardiac impulse passes through the heart, electrical current also spreads from the heart into the adjacent tissues surrounding the heart. A small portion of the current spreads all the way to the surface of the body. If electrodes are placed on the skin on opposite sides of the heart, electrical potentials generated by the current can be recorded; the recording is known as an electrocardiogram.

Fig. 5(From Guyto and Hall). A normal electrocardiogram for two beats of the heart.

12 Leads to record ECG
a. Limb leads
1) Unipolar = aVleads: aVR, aVL, aVF

2) Bipolar = I(right/left UE), II(right UE/left LE), III(left UE/right LE)\

b. Chest Leads:
V1 = red (4^th interscostal space right from the sternum)
V2 = yellow (4^th intercostals space left from the sternum)
V3 = grey (between V2 and V4)
V4 = brown (apex of the heart; 5^th intercostals space left mid-clavicular line)
V5 = black (5^th intercostals space left anterior axillary line)
V6 = purple (5^th intercostals space left mid axillary line) – hidden when shoulder is adducted.

Leads represent different parts of the heart
  aVR = superior wall of the heart
I, aVL, V5, V6 = lateral wall of the heart
II, III, aVF= inferior wall of the heart
V1 & V2 = septal regions of the heart; also send reflection to the posterior wall of the heart.
V1, V2, V3, V4 = anterior wall of the heart

Leads changes may indicate wall infarct
WALL INFARCTS LEADS AFFECTED
Lateral   I, aVL, V5, V6
Inferior   II, III, aVF
Posterior   V1, V2
Anterior   V1, V2, V3, V4

       Posterior wall infarct: a tall T wave, a tall R wave, and a persistent ST depression.
       Components of an ECG waveform
·   ECG strip is like a graphing paper
·   traces of evoked potential could be seen.
·    There are two axes:
  §     X axis = horizontal axis
  §      Y axis = vertical axis

Each of the strip (graphing-like paper) has a small square

· The area of this small square is 1mm x 1mm. or one square mm.

· The horizontal line of this square is the wave duration which is 0.04 sec.

· The vertical line of this square is the wave amplitude which is 0.1 mV.

· So, each small square has 0.1 mV and 0.04 sec.

· A small is too small to read on an ECG strip. A block makes reading an ECG easier.

· A block is composed of small squares.

The dimension of a block:

· Its area is 25 square mm; thus, it is a 5x5 mm.

· It comprises of one isoelectric line.

· It is easy to recognize the block because it is highlighted.

· The wave duration of one isoelectric line (horizontal) is 0.20 sec.

· It is 0.20 sec because the wave duration in each one small square is 0.04 sec. And since the block has a measurement of 5mm horizontally, 0.04 sec then is multiplied by 5. The product is 0.20 sec.

· The wave amplitude of one isoelectric line is 0.5mV.

· It is 0.5 mV because the vertical length of a small square is 0.1 mV. And since a block has a vertical length of 5mm, then 0.1 mV is multiplied by 5 which is equal to 0.5 mV.

· Therefore, one isoelectric line is 0.5 mV @ 0.20 sec

Examples:

v If you are reading a 1 sec strip, how many small squares are there? Answer: 25 small small squares. Or 5 (big) blocks because five 0.20 sec is equal to 1 sec.

It looks like this (elementary math J):

0.20 (1 block)

0.20 (1block)

0.20 (1 block)

----------------------

1.00 sec

Thus, there are 5 blocks in 1 sec. strip.

v How many big blocks if you are reading a 5 sec strip?

Answer: 25 big blocks or 125 small squares

v How many big blocks if you are reading a 1 min. strip?

Answer: 300 big blocks

1 min. is equal to 60 sec. And when you multiply 60 sec by 5, the product is 300.

v How many big blocks in a 45 sec strip?

Answer: 225 big blocks

45 sec multiplied by 5 is equal to 225.

Always remember that 1 sec = 5 block.

v How many big blocks in a 30 sec strip.

Answer: 150 big blocks

v How many blocks in a 15 sec strip?

Answer: 75 blocks.

v This then is the sequence:

0 à 150 à 225 à 300

(15s) (30s) (1min)

v How many small squares in 3 mV?

Answer: 30 small squares or 6 big blocks

Each block has a wave amplitude of 0.5 mV. 1 mV then is equal to 2 blocks. So, if you have 3 mV, you have to multiply 2 blocks by 3 mV which is equal to 6 blocks.

To get the number of small squares, multiply 6 blocks by 5. The answer is 30 small squares.

v If you have 13 small squares, how many millivolts?

Answer: 1.3 mV

There is 1 mV in 2 big blocks or 10 small squares. And there are 3 remaining squares since what is given in the problem above is 13 small squares. So, 3 small squares are equal to 0.3 mV, since each small square has a wave amplitude of 0.1 mV.

Therefore, the answer is 1.3 mV. Simply put, 1 mV plus 0.3 mV is equal to 1.3 mV.

v How many millivolts in 15 small squares?

Answer: 1.5 mV

15 small squares or 3 big blocks.

3 big blocks are multiplied by 0.5 mV. The product is 1.5 mV.

v How many big blocks in 2.3 mV?

Answer: 4 blocks and 3 small squares

Again as explained above, each 1mV is equal to 2 blocks. So, if you have 2 mV, then you have 4 big blocks.

The remaining 0.3 mV corresponds to 3 small squares, since each small square is 0.1 mV.

v How many big blocks in wave amplitude of 0.9 mV?

Answer: 1 big block and 4 small squares

0.5 mV = 1 block

0.4 mV = 4 small squares

v If you have 0.65 mV, how many big blocks?

Answer: 1 block and 1 ½ small squares

A wave amplitude of 0.50 mV is confined to 1 block. The remaining 0.15 mV includes 1 small square and a half. Hence, 1 block and 11/2 small squares.

0.15 mV contains 1 small square and a half, since each small square has a wave amplitude of 0.10 mV and the remaining 0.05 is half of 0.10 mV. So the answer is 1 small square and a half.

v Given: 0.20 mV @ 0.16 sec.

How many big blocks and small squares?

Answer: 0.20 = 2 small squares; 0.16 sec = 4 small squares

There are two small squares in 0.20 mV, because each small square has 0.1 mV (or 0.10 mV). Since the give wave amplitude above is 0.20 mV, the answer is 2 small squares.

There are 4 small squares in 0.16 sec, because each small square has a wave duration of 0.04 sec.

0.16 sec divided by 0.04 sec is equal to 4 small squares. Or you may use multiplication to make it easier for a manual computation. So, 4 x 0.04 sec is equal to 0.16 sec.

v Given:

a) Wave amplitude: 6mm

b) Wave duration: 31/2 mm

Answer:

a) 0.6 mV

Remember, the vertical length of each block is 5mm which is equal to 0.5 mV. And each small square has a vertical length of 0.1 mV. So, add 0.5 mV to 0.1 mV, then you have a wave amplitude of 0.6 mV.

b) 0.14 sec.

Again, the horizontal length of each small square is 1 mm with a wave duration of 0.04 sec. So, if you have horizontal length of 3mm, then you have to multiply 0.04 sec by 3mm. The product is 0.12 sec. The remaining ½ mm is 0.02, which is half of 0.04 sec.

It looks like this:

0.12 sec (3mm)

+ 0.02 sec (1/2mm)

------------------------

0.14 sec

v Given:

a) 12 small squares going up (vertical direction)

b) 2 big blocks and 21/2 small squares going horizontally

Answer:

a) 1.2 mV

b) 0.50 sec

It is 1.2 mV that is confined to 12 small squares going up, because each block has wave amplitude of 0.5 mV and each small square has wave amplitude of 0.1 mV. So, if you have 12 small squares, then you have two big blocks and two small squares. There is 1 mV in two big blocks and 0.2 mV in two small squares. Therefore, the answer is 1.2 mV.

The wave duration is 0.50sec which corresponds to two big blocks and 21/2 small squares in the horizontal line, because each big block corresponds to 0.20 sec. And since, there are two big blocks, the wave duration then is 0.40 sec, that is, 0.20 plus .20.

The 21/2 remaining small squares correspond to 0.10 sec. It is so because each small square has wave duration of 0.04 sec. So, two small squares have a wave duration of 0.08 sec. And half of the small square is 0.02. Therefore, the answer is 0.50 sec.

Simply put:

0.40 sec (two big blocks) + 0.08 (two small squares) + 0.02 (half square) = 0.50 sec

References

Gabriel Khan, M. Rapid ECG Interpretation, 3^rd Ed. Totowa, New Jersey: Human Press, 2008.

Guyton, Arthur C., and John E. Hall. Text Book of Medical Physiology, 11^th Ed. Philadelphia, Pennsylvania: Elsevier Inc., 2006.

Martin, Gerard L. “Cardiac Physiology and ECG Interpretation.” Class lecture, SLRC, Sampaloc, Manila, May 6, 2011.

Pablo-Santos, Ramona Luisa, PT-OT Reviewer, Manila, Philippines: Ramona Luisa P. Santos, M.D., 1995.

Paz, Jaime C. and Michelle P. West. Acute Care Handbook for Physical Therapist, 3^rd Ed. Boston: Butterworth-Heinemann, 2009.

_________. “Rhythms and Arrythmias.” Cardionetics. http://www.cardionetics.com/cardiology/second-degree-block.php (accessed June 30, 2011).

R.O.L.