Dave, measured standing-wave RF voltage is highest at a dipole's ends and RF current is lowest (zero). Your hand-drawing shows both RF voltage and current as lowest (zero) at the ends. To show a dipole's standing-wave RF voltage distribution without polarity, a U shaped voltage curve would be appropriate. Recall that a half-wave dipole is essentially an unterminated quarter-wave parallel transmission line spread apart 180 degrees -- including its standing-wave voltages and currents. Ken, WA8FCI
Also, apart from RF power absorption due to an antenna's conductor loss, a dipole antenna does NOT absorb RF power -- it radiates applied RF power as an EM wave. In other words, a dipole antenna is not an RF power dissipative load -- although, it's mathematically viewed as such. Rather, it's a combination of an impedance matching transformer (matching transmission line impedance to free-space impedance) and a transducer that converts applied RF power into an EM wave. Ken WA8FCI
I think you mix up phase angle with respect to time and phase angle over distance at a given time. Obviously the voltage will make a jump at the feedpoint.
Thank you, Dave. You gave me something to ponder and cogitate about. Some may consider this video as throwing a satchel charge into a powder magazine, but it gets us thinking about what is really happening. Thanks again.
Instead of a ground you need a counterpoise. In many installations, the outside of the coax shield serves as the counterpoise. Otherwise you can add a counterpoise, like ten or 20 feet of wire to the counterpoise connection and leave that dangling.
Hello Dave. Thanks for the video! The waveform graph which shows the voltage and current out of phase is utilized in several ARRL publications. Since you work with the ARRL it would be helpful if you would follow up on this video's topic in an effort to bring clarity to the theory.
With respect.... a radiating dipole is not a closed circuit thus one cannot apply circuit theory. From Physics we know the current at the ends of the dipole must be zero. There is a radiation resistance where the power is dissipated.
Think of an antenna from a Physics point of view ... it is not a circuit..... it is radiating energy into space . This is what makes Ham Radio so interesting..... the physics is radiating energy into space... cool stuff Maxwell's Equations... and information is being sent through space.
Well, I am back for a review. The older electronics technicians I worked with did not like working with RF- they regarded it as strange, weird, magical, etc. Can this "strangeness" be a consequence of the Faraday-Maxwell-Heaviside electromagnetism equations? N0QFT
Given that your presentation on the voltage and current distribution vs location on the dipole is contrary to numerous, should I say just about all, sources that explain the subject as "Current is maximum at feedpoint, while voltage is maximum (and opposite sign) at ends", could you please provide the source for your presentation of voltage and current vs. location on the dipole?
Great video, good math too! This is the first time I grasped this, which means that you can teach ANYBODY!!! Thanks for turning on the light bulb for me!
I thought he was about to draw a rainbow there for a bit. Look, this is about as wrong as pride month. The diagram of the voltage and current distribution on a resonant, centre fed 'half' wave dipole is correct. What's going on?!
Thank you. I have learned a lot in the last couple minutes. I will watch this again tomorrow to replace what I learned over night. I eat up your videos like candy. Thanks again.
Dave, measured standing-wave RF voltage is highest at a dipole's ends and RF current is lowest (zero). Your hand-drawing shows both RF voltage and current as lowest (zero) at the ends. To show a dipole's standing-wave RF voltage distribution without polarity, a U shaped voltage curve would be appropriate. Recall that a half-wave dipole is essentially an unterminated quarter-wave parallel transmission line spread apart 180 degrees -- including its standing-wave voltages and currents.
Ken, WA8FCI
Also, apart from RF power absorption due to an antenna's conductor loss, a dipole antenna does NOT absorb RF power -- it radiates applied RF power as an EM wave. In other words, a dipole antenna is not an RF power dissipative load -- although, it's mathematically viewed as such. Rather, it's a combination of an impedance matching transformer (matching transmission line impedance to free-space impedance) and a transducer that converts applied RF power into an EM wave.
Ken WA8FCI
I think you mix up phase angle with respect to time and phase angle over distance at a given time. Obviously the voltage will make a jump at the feedpoint.
The voltage at the feedpoint is lowest. It is highest on each end. The current is the opposite, highest at the center
Thank you, Dave.
You gave me something to ponder and cogitate about.
Some may consider this video as throwing a satchel charge into a powder magazine, but it gets us thinking about what is really happening.
Thanks again.
So I have 2 ham stick dipoles for 11 meters can I run 2 of them in parallel together?
How is an end fed dipole connected. Looks like you have only one wire at the end. What do you do with the ground?
Instead of a ground you need a counterpoise. In many installations, the outside of the coax shield serves as the counterpoise. Otherwise you can add a counterpoise, like ten or 20 feet of wire to the counterpoise connection and leave that dangling.
Hello Dave. Thanks for the video! The waveform graph which shows the voltage and current out of phase is utilized in several ARRL publications. Since you work with the ARRL it would be helpful if you would follow up on this video's topic in an effort to bring clarity to the theory.
With respect.... a radiating dipole is not a closed circuit thus one cannot apply circuit theory. From Physics we know the current at the ends of the dipole must be zero. There is a radiation resistance where the power is dissipated.
Think of an antenna from a Physics point of view ... it is not a circuit..... it is radiating energy into space . This is what makes Ham Radio so interesting..... the physics is radiating energy into space... cool stuff Maxwell's Equations... and information is being sent through space.
Well, I am back for a review.
The older electronics technicians I worked with did not like working with RF- they regarded it as strange, weird, magical, etc.
Can this "strangeness" be a consequence of the Faraday-Maxwell-Heaviside electromagnetism equations? N0QFT
Given that your presentation on the voltage and current distribution vs location on the dipole is contrary to numerous, should I say just about all, sources that explain the subject as "Current is maximum at feedpoint, while voltage is maximum (and opposite sign) at ends", could you please provide the source for your presentation of voltage and current vs. location on the dipole?
Great video, good math too! This is the first time I grasped this, which means that you can teach ANYBODY!!! Thanks for turning on the light bulb for me!
I thought he was about to draw a rainbow there for a bit.
Look, this is about as wrong as pride month. The diagram of the voltage and current distribution on a resonant, centre fed 'half' wave dipole is correct. What's going on?!
Thank you. I have learned a lot in the last couple minutes. I will watch this again tomorrow to replace what I learned over night. I eat up your videos like candy. Thanks again.
Now lets see if you can use the same presentation to explain to folks why full wave dipoles dont work! Youre half wave there!
Wrong explanation.
This graph has nothing to phase and waves. It's voltage and current amplitudes.
So the graph is correct.
Mind blown... Yeah, over my head :) I'm going to have to watch this a few times.