You are a great lecturer. I know you discussed the problem with I+II+II =0 somewhere below but many students who watch this video first time may miss it completely. Since lead II is reversed in this Einthoven's format, the correct formula is I+(-II)+III = o or I+III=o. Also what they add is voltage in this case (potential difference), not potentials at each electrode and sometimes you mention electrical information which many find it very confusing.. Also, the zero reference (null) point is in the center of the triangle and your reference to something bisecting I or II or III is confusing; it almost sounds like averaging two points to get the middle point and I don't think that is what is happening with this WCT case.
+Ollie Deans Me too. But I think it was a good thing because this video was the final piece to the puzzle. I saw what it looked like in the end so to speak.
This is the absolute best video about ECG theory currently available on RUclips, thank you so much! Now I fully understand why some deflections are positive and others negative! Greetings from Italy 🇮🇹
you only said that the waves get smaller as the angle between the lead axis and the vector increases. And what about biphasic QRS waves? Isn't it that the waves are biphasic if the vector travels perpendicularly to the lead axis?
Thanks for watching, ksufler! That's a great question. We don't classically refer to a biphasic QRS complex since the expected morphology includes a positive and negative deflection by design. Remember, a vector represents the sum of all activity at a single moment in time (in the context of ECGs). At any moment in time, if the sum of all electrical activity is zero, the resultant tracing is a flat line. A biphasic wave illustrates the vectors of opposing forces occurring at different times and can appear as a P wave with both positive and negative components. In that case, the biphasic shape is due to opposing depolarization angles relative to the lead viewing the activity and plotted against time. To summarize, if a vector is perpendicular to the viewing lead, the resultant tracing is flat. If two distinct depolarization vectors are happening at different times and with opposing force, then a biphasic wave will be recorded.
Excellent observation, Rasdocus and Adriano SC! Check out Kirchhoff's Voltage Law (KVL), though (www.electronics-tutorials.ws/dccircuits/dcp_4.html). KVL states that the algebraic sum of all voltages within the loop must be equal to zero. Another important concept of KVL is that the direction of electrical flow must be the same for all components. Einthoven's triangle creates 3 leads but only leads I and III are "moving" in the same direction; lead II is reversed. Picture a starting point of the right arm, move clockwise to the left arm to create lead I, continue clockwise to create lead III, then, in order to use KVL, you must continue clockwise to create lead II to complete the circuit at the right arm. In doing so, however, the result of lead II is negatively oriented on the tracing so it must be reversed. So it goes, the application of KVL's theory to Einthoven's triangle yields this equation I+II+III=0 - that satisfies the sum of all voltages must equal zero. In order to return a positively oriented lead II in the normal heart, we must "artificially" reverse the difference in voltage potential for lead II (onlinelibrary.wiley.com/doi/10.1002/clc.4960130702/pdf), which is done by software in the ECG machine. Here's what the adjusted equation looks like for practical application of theory: I + III + (-II) = 0, or, I + III = II. This is a rather difficult concept to take on so I decided to simplify and you both astutely discovered that! Thanks for the dialogue.
You are a great lecturer. I know you discussed the problem with I+II+II =0 somewhere below but many students who watch this video first time may miss it completely. Since lead II is reversed in this Einthoven's format, the correct formula is I+(-II)+III = o or I+III=o. Also what they add is voltage in this case (potential difference), not potentials at each electrode and sometimes you mention electrical information which many find it very confusing.. Also, the zero reference (null) point is in the center of the triangle and your reference to something bisecting I or II or III is confusing; it almost sounds like averaging two points to get the middle point and I don't think that is what is happening with this WCT case.
Thanks for something that actually makes sense and covers everything, was getting frustrated trawling through the rest of youtube!
+Ollie Deans Me too. But I think it was a good thing because this video was the final piece to the puzzle. I saw what it looked like in the end so to speak.
Thank you so much! I've had much clarity after watching this. God bless you.
What a great foundation! This opened up so much understanding for me.
This is the absolute best video about ECG theory currently available on RUclips, thank you so much! Now I fully understand why some deflections are positive and others negative! Greetings from Italy 🇮🇹
Thanks for making all my concepts clear ...🙂🙂
Totally clarified everything for me, and put it into layman's terms. Thank you so much!
judyshin1220 Glad you found it helpful. Thanks for your kind words.
Brilliant! Just what I needed, Thank you for sharing!
Excellent talk , thank you
Very detailed! Exactly what I am looking for!
Excellent video.
This one is really conceptual. Thanks a lot.
thank you so much , exactly what i was searching for. completely understood.
This is fabulous teaching !!
Thanku so much 👏
Earned a subscriber
Thank you so much for this vedio it was really help me to understand the basic of EKG
Thank you! This is very informative!
Waaah I LOVE ALL YOUR VIDEOS 💚 Thank you muaaah!
great explanation... thank you!
you only said that the waves get smaller as the angle between the lead axis and the vector increases. And what about biphasic QRS waves? Isn't it that the waves are biphasic if the vector travels perpendicularly to the lead axis?
Thanks for watching, ksufler! That's a great question. We don't classically refer to a biphasic QRS complex since the expected morphology includes a positive and negative deflection by design. Remember, a vector represents the sum of all activity at a single moment in time (in the context of ECGs). At any moment in time, if the sum of all electrical activity is zero, the resultant tracing is a flat line. A biphasic wave illustrates the vectors of opposing forces occurring at different times and can appear as a P wave with both positive and negative components. In that case, the biphasic shape is due to opposing depolarization angles relative to the lead viewing the activity and plotted against time. To summarize, if a vector is perpendicular to the viewing lead, the resultant tracing is flat. If two distinct depolarization vectors are happening at different times and with opposing force, then a biphasic wave will be recorded.
thanks very nicely explained
Great video!
The vectors you are referring to at 16 min, are they extracellular dipole?
I think lead I + lead III = lead II and not -lead II as we can see from vector diagram
Rasdocus D Yes. That's a vector summing.
You're correct man!
Excellent observation, Rasdocus and Adriano SC!
Check out Kirchhoff's Voltage Law (KVL), though (www.electronics-tutorials.ws/dccircuits/dcp_4.html). KVL states that the algebraic sum of all voltages within the loop must be equal to zero. Another important concept of KVL is that the direction of electrical flow must be the same for all components. Einthoven's triangle creates 3 leads but only leads I and III are "moving" in the same direction; lead II is reversed. Picture a starting point of the right arm, move clockwise to the left arm to create lead I, continue clockwise to create lead III, then, in order to use KVL, you must continue clockwise to create lead II to complete the circuit at the right arm. In doing so, however, the result of lead II is negatively oriented on the tracing so it must be reversed. So it goes, the application of KVL's theory to Einthoven's triangle yields this equation I+II+III=0 - that satisfies the sum of all voltages must equal zero. In order to return a positively oriented lead II in the normal heart, we must "artificially" reverse the difference in voltage potential for lead II (onlinelibrary.wiley.com/doi/10.1002/clc.4960130702/pdf), which is done by software in the ECG machine. Here's what the adjusted equation looks like for practical application of theory: I + III + (-II) = 0, or, I + III = II.
This is a rather difficult concept to take on so I decided to simplify and you both astutely discovered that! Thanks for the dialogue.
Good explain
Can u be answering something is
Confusing me ??
Thank you sir
in kershof law lead1 - lead2 +lead3 =0
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