"In regards to the electrocardiogram, there is often a Statement made that,
'The a' in aVF means 'augmented' by a factor of 1.5 for sensitivity adjustment.
If the above statement is true, following equation should be true.
Here, $VF$ represents the original voltage in the VF direction, VI represents lead I.
If the above Statement is correct, the argument of Heart vector, $\theta$ should be
$\theta = \arctan \left(\frac{VF}{V_I}\right)
= arctan\left(\frac{2 VF}{ 3 V_I}\right)$
However, the correct formula by Novosel is as follows, as described below in the British Physiological Society's resource intended to "Beginner" level Medical School students . https://www.physoc.org/magazine-articles/trigonometry-of-the-ecg/
$\theta = arctan \left(\frac{2 VF}{\sqrt{3} VI}\right)$
and my own calculations also lead to the same conclusion.
Therefore, in order to align with this formula:
$\frac{2}{\sqrt3}\frac{aVF}{VI} = \frac{VF}{VI}$
It follows that:
$aVF = \frac{aVF}{2 VF} \approx 0.85 VF$
Hence, aVF seems to be smaller than VF.
My question;
So the Statement that aVF is 1.5 times the true voltage of VF seems to contradict Novosel's formula. What do you think?
After several hours of study after post this question, I actually understood the answer. As a result, I realized that this question can be chunked down to the following question, which can probably be answered by anyone who understands the beginner's course in EKG.
- What is aVF and where is the potential difference?
- What is aVF 1.5x?
- How does the Vectorcardiogram relate to aVF?
References;
- https://www.physoc.org/magazine-articles/trigonometry-of-the-ecg/
- https://www.bem.fi/book/15/15.htm Written in Japanese
- http://www.hoku-iryo-u.ac.jp/~kurahasi/kurahasi/103.3.pdf
- https://detail.chiebukuro.yahoo.co.jp/qa/question_detail/q11285430786
- https://detail.chiebukuro.yahoo.co.jp/qa/question_detail/q11285430786