Einthoven's triangle models the heart as a single current dipole at the centre of an…

PHYSICS · Multi-part · Fermi University Physics · Biophysics & Medical Physics · Bioelectricity, the senses and the physics of the body · Bioelectricity, the senses and the physics of the body

Forces, Machines and Signals, Ch Muscle, heart and the fields outside the body

Problem

Einthoven's triangle models the heart as a single current dipole at the centre of an equilateral triangle whose vertices are the right arm, left arm and left leg. The three limb leads are defined as potential differences between pairs of limbs, I=VL−VR,II=VF−VR,III=VF−VL,\mathrm{I} = V_L - V_R,\qquad \mathrm{II} = V_F - V_R,\qquad \mathrm{III} = V_F - V_L, where VRV_R, VLV_L and VFV_F are the potentials of the right arm, left arm and left foot. By the standard convention lead I points along 0∘0^\circ, lead II along +60∘+60^\circ and lead III along +120∘+120^\circ (measured clockwise from the horizontal in the frontal plane, as seen facing the patient, so that positive angles point towards the feet), and each lead reads a signal proportional to the projection of the heart's dipole moment onto its own direction, i.e. leadk∝cos⁡(θ−ϕk)_k \propto \cos(\theta - \phi_k) where θ\theta is the electrical axis and ϕk\phi_k is the direction of lead kk. ((a)) Add the three definitions above to show that II=I+III\mathrm{II} = \mathrm{I} + \mathrm{III} for any dipole, whatever its direction or size (Einthoven's law). ((b)) Using I∝cos⁡θ\mathrm{I}\propto\cos\theta and II∝cos⁡(θ−60∘)\mathrm{II}\propto\cos(\theta-60^\circ), derive tan⁡θ=2 II/I−13\tan\theta = \dfrac{2\,\mathrm{II}/\mathrm{I} - 1}{\sqrt3}. ((c)) An ECG trace reads I=0.60 mV\mathrm{I} = 0.60\,\mathrm{mV} and II=0.90 mV\mathrm{II} = 0.90\,\mathrm{mV} at the R peak. Find the electrical axis θ\theta and the reading of lead III, and check your value of III both from Einthoven's law and from the projection formula directly.

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