Late to the party, but thank you. This is one of the most concise, well delivered antenna vids I have come across as a fairly new ham. I have been doing my research and discovered that half-waves are great, but didn't really grasp why. You have a gift sir, I hope you continue to be a teach.
What a clear and understandable explanation. So glad I came across this video. As a noob to ham I’m on a huge learning curve and this video has probably saved me 1000 hours watching other videos on the topic.
I struggled with math all through school into college until one grad student tutored me. you remind me of him. he was the only person that tapped into the math part of my brain. afterwards I began making A's and B's in math and statistics. the way you write and talk out what you're doing on paper is so helpful. like another guy said, you've got a gift for teaching and explaining things. so many people out there think they can't do math and science it's a shame. if only more teachers had the gift... anyway, glad I found your videos. maybe now I'll find the motivation to study for the next amateur radio license.
Unfortunately there seem to be several errors in Mr. Oliver's presentation. First, the voltage on each side of a dipole will be differential, in other words, when one side (1/4λ element) of the 1/2λ dipole is positive in voltage, the other side will be opposite in phase by 180 degrees, or negative; this will be the case whether or not the 1/2λ dipole is fed by balanced line, or coax with a BalUn. (For those who don't use a BalUn when feeding a dipole with coax, you don't have a dipole, you have a 1/4λ antenna with a counterpoise.) Secondly, the electromagnetic field of an end-fed halfwave element will NOT be canceled out, because both positive and negative voltages are on the wire (at times) at the same time; if that were true, there would be no such thing as an end-fed 1/2λ antenna. Thirdly, you could have a FULL-WAVE dipole, that would garner about a 3dB increased perpendicular radiation pattern; if using 50 Ω coax, you would have to feed the dipole with a 49:1 BalUn since the feed-point impedance of a 1/2λ element is ~2450 Ω. But you'd have to make the 49:1 BalUn yourself since commercially available 49:1 transformers are UnUns, made for end-fed 1/2λ antennas. Also, while there are other ways to do impedance matching, in this case, you'd have to use a BalUn, since you need differential voltages on each side of the dipole. If I'm incorrect, please comment and I'll amend my comment.
This is simplification gone wrong. There is a difference in polarity of each end of a half wave dipole. If there is positive voltage at one end the voltage is negative in the other. And a full wave dipole has radiation in the same direction as a half wave dipole, but with a different BW (47 degrees instead of 68 degrees). There is also one more important difference between half wave and full wave dipoles. The impedance is very different. A half wave dipole is close to 73 ohms in free space (73+j42), but a full wave dipole is several thousands of ohms making it extremely difficult to match.
Some people have perfect pitch. It's a genetic "gift". Those with the gift to sing (or other genetic gift) can develop their gift with the help of professional mentors and teachers. I grew up with a love of music and a love of singing and I was encouraged to sing my heart out by my parents and my church. One day I learned I was tone deaf and did not have the gift. I stopped singing and began a search for my gifts with support from family and friends and teachers. As a licensed amateur radio "ham", I have watched A LOT of videos explaining topics that interest me and came upon this video. Your explanation of the 1/2 wave dipole was an eye opener for me as a direct result of your upbeat, concise, and clear explanation of this topic. You, sir, are one of those amazing folks who have the "gift" to organize and teach. I hope to encounter your videos again and I wrote this note to just give you thanks and to give you encouragement to continue. You have "the gift to teach". Best wishes. N6ATS
It is misleading to describe the voltage distribution as shown in this video and also it is not true that an antenna which is one wavelength long does not radiate. Please revise your understanding.
Carmel Pule' Hi...An explanation from you might have been just a bit kinder & helpful to other readers needing an Elmer, a seasoned operator, no new-comer you, a guy who like you, knows his stuff. You are of course QUITE correct but just a bit vitriolic, not much but a bit! I admire your accuracy but Ham Spirit an' all that! Eh? Big 73 to you.
***** What is "vitriolic" about Carmels post? You should more interested in Political Correctness than accurate information. He used a "soft" word "misleading'. He did not state the TRUTH that the information is inaccurate. Why SHOULD Carmel explain? HE did not post the video.
Standing Bear Hi Bear, I gotta admit that you got me again. Twice now! I need to find to what it was I replied. Would be quite prepared to swallow my words if I goofed but really I DID only say "A BIT" vitriolic and was non aggressive in the rest. However I AM puzzled by your Political Correctness remark. Now..If you and I were ever to have a 'Pistols at Dawn' scrap it could easily be over political correctness. However I don't pick up the point of your remark. Final point..Have I goofed by attributing a video to the wrong person? Help me here please, Bear...It WAS in April! Regards M.O.L PS..Just poured me a scotch..Could fax YOU one..ice?
Standing Bear Ah...Have found it! Carmel Pule did not address his posting to anybody in particular. Bit bewildering. He said that it is not true that a half wave doesn't radiate. He is perfectly correct. It WILL radiate but very badly if at all, when fed in the centre with coax. As far as I know, nobody had suggested that a 1/2 would not radiate, other than myself, when I pointed out that the balanced centre of a full wave would accept NO power from a lo Z, unbalanced feed...coax. I now feel, although not absolutely convinced, that maybe Carmen's remark "Please revise your understanding" may have deserved the lesser word "sarcastic!" rather that "vitriolic" which in the spirit of good will, I beg permission to withdraw! Carmel was not asked to explain the video which was obviously not his but more to explain his remarks. I wish you well..M.O.L.
@@MauriatOttolink Hopefully my comment clears up several errors in Mr. Oliver's presentation. First, unlike what he illustrates, the voltage, and therefore current direction, on each side of a dipole will be differential, in other words, when one side (1/4λ element) of the 1/2λ dipole is positive in voltage, the other side will be opposite in phase by 180 degrees, or negative; this will be the case whether or not the 1/2λ dipole is fed by balanced line, or coax with a BalUn. (For those who don't use a BalUn when feeding a dipole with coax, you don't have a dipole, you have a 1/4λ antenna with a counterpoise.) Secondly, the electromagnetic field of an end-fed halfwave element will NOT be canceled out, because both positive and negative voltages are on the wire (at times) at the same time; if that were true, there would be no such thing as an end-fed 1/2λ antenna. Thirdly, you could have a FULL-WAVE dipole, that would garner about a 3dB increased perpendicular radiation pattern; if using 50 Ω coax, you would have to feed the dipole with a 49:1 BalUn since the feed-point impedance of a 1/2λ element is ~2450 Ω. But you'd have to make the 49:1 BalUn yourself since commercially available 49:1 transformers are UnUns, made for end-fed 1/2λ antennas. Also, while there are other ways to do impedance matching, in this case, you'd have to use a BalUn, since you need differential voltages on each side of the dipole. If I'm incorrect, please comment and I'll amend my comment.
A fullwave antenna does radiate and has more gain than anything smaller. The antenna becomes more directional, in that the nulls on the ends are steeper, resulting in a near perfect figure 8 pattern. The impedance is not 50 ohms at it's feed point however, so those of you trying to match it, it's tough because it's in the range of about 5,000 ohms. Even feed in the middle, this antenna is a very high Z! That's a great bit away from the 50 ohms our transmitters want to see and it's a very long antenna. So, the half wave is the better choice on both space and matching requirements. So, to say a 1 wavelength antenna does not radiate is completely wrong. The RF radiation pattern may not be where you desire it, it may be too directional, it might not fit on your property and you may not want to transform the impedance as needed, bit it will radiate and receive more power than anything shorter if properly matched and oriented.
I know this from trying and sheer experimentation. I took a long strand of insulated copper wire on both sides of a factory made dipole connector, and every inch or so I would snip slowly off of each end equally I would go key the radio and check the SWR meter. It would NEVER approach flat SWRs until I had physically cut it to half the wave length of the radio I was using. Even if you did not know the math and physics , you would find out quickly that it only resonates properly half of the wave length of the radio you are using. I was talking side band on a CB so it was about 18 feet long give or take an inch or two.
Thanx Carl. Nice. But it is incorrect to say that full wavelength antenna doesn't radiate. When the length of the antenna is one wavelength, the polarity of the current in one of the antenna is opposite to that of the other half. Because of these out-of-phase currents, the radiation at right angles from this antenna is zero. The field radiated by one half of the antenna alters the field radiated by the other half. However, a direction of maximum radiation still exists at 54 deg to the antenna.
This video only considers HALF of a radio wave. The wave has an ELECTRIC portion and a MAGNETIC portion. The VOLTAGE is in the electrical portion. The CURRENT is in the magnetic. It requires BOTH for radiation to occur. Remembering the equation that Power = voltage times current we know it takes BOTH in order to actually radiate any power. The suggestion that a full wave results in no radiation is inaccurate at best.
I should start by confessing that the purpose of this video was to explain in simple terms to people new to antenna theory how they radiate and I admit, does not contain the whole story. I will also confess to knowing that there are full wave antennas that radiate very well! There are matching sections or other ways to ensure that the fields add in phase, which complicates things for the newbie. So how about this: "A dipole antenna fed with a balanced 75 ohm line will not radiate efficiently in the direction perpendicular to the antenna." Is that fair to say? Here's what I said about this in a prior comment: Technically, That is true. When the electrons at the top are moving away from the generator, the electrons at the bottom are moving toward the generator. But what makes the fields add in phase is that for the top, away from the generator means UP, while for the bottom, toward the generator mean UP as well. So even though technically they have an opposite charge, the resultant current flow is in the same direction. I "cheated" during my explanation in order to illustrate this point. Others may disagree, but I see it as a question of convention rather than one of accuracy. While I didn't mention magnetic field, i did not means to imply that it did not exist (my apologies if it came across that way). I did mention the motion of the electrons, which indirectly referenced the magnetic field.
Standing Bear: Most textbooks now don't explain EM using simple English like what you did. Thanks. I always suspect that all so-called EM waves, including visible light, is nothing more than the movement of energy (what they call photons for light waves) from a high energy point in space to a low energy point in space (a.k.a. radiation). In other words, energy is relative, not absolute. In other words, a photon is not a particle without mass, but a passage of a fixed amount of energy through space.
Personally, I currently believe that the 'gem' photon is the energy unit of this universe that makes up everything in existence in this universe. Here is my latest TOE idea followed by the 'gravity' test for it: Revised TOE: 3/25/2017a. My Current TOE: THE SETUP: 1. Modern science currently recognizes four forces of nature: The strong nuclear force, the weak nuclear force, gravity, and electromagnetism. 2. In school we are taught that with magnetism, opposite polarities attract and like polarities repel. But inside the arc of a large horseshoe magnet it's the other way around, like polarities attract and opposite polarities repel. (I have proved this to myself with magnets and anybody with a large horseshoe magnet and two smaller bar magnets can easily prove this to yourself too. It occurs at the outer end of the inner arc of the horseshoe magnet.). 3. Charged particles have an associated magnetic field with them. 4. Protons and electrons are charged particles and have their associated magnetic fields with them. 5. Photons also have both an electric and a magnetic component to them. FOUR FORCES OF NATURE DOWN INTO TWO: 6. When an electron is in close proximity to the nucleus, it would basically generate a 360 degree spherical magnetic field. 7. Like charged protons would stick together inside of this magnetic field, while simultaneously repelling opposite charged electrons inside this magnetic field, while simultaneously attracting the opposite charged electrons across the inner portion of the electron's moving magnetic field. 8. There are probably no such thing as "gluons" in actual reality. 9. The strong nuclear force and the weak nuclear force are probably derivatives of the electro-magnetic field interactions between electrons and protons. 10. The nucleus is probably an electro-magnetic field boundary. 11. Quarks also supposedly have a charge to them and then would also most likely have electro-magnetic fields associated with them, possibly a different arrangement for each of the six different type of quarks. 12. The interactions between the quarks EM forces are how and why protons and neutrons formulate as well as how and why protons and neutrons stay inside of the nucleus and do not just pass through as neutrinos do. THE GEM FORCE INTERACTIONS AND QUANTA: 13. Personally, I currently believe that the directional force in photons is "gravity". It's the force that makes the sine wave of EM energy go from a wide (maximum extension) to a point (minimum extension) of a moving photon and acts 90 degrees to the EM forces which act 90 degrees to each other. When the EM gets to maximum extension, "gravity" flips and EM goes to minimum, then "gravity" flips and goes back to maximum, etc, etc. A stationary photon would pulse from it's maximum extension to a point possibly even too small to detect, then back to maximum, etc, etc. 14. I also believe that a pulsating, swirling singularity (which is basically a pulsating, swirling 'gem' photon) is the energy unit in this universe. 15. When these pulsating, swirling energy units interact with other energy units, they tangle together and can interlock at times. Various shapes (strings, spheres, whatever) might be formed, which then create sub-atomic material, atoms, molecules, and everything in existence in this universe. 16. When the energy units unite and interlock together they would tend to stabilize and vibrate. 17. I believe there is probably a Photonic Theory Of The Atomic Structure. 18. Everything is basically "light" (photons) in a universe entirely filled with "light" (photons). THE MAGNETIC FORCE SPECIFICALLY: 19. When the electron with it's associated magnetic field goes around the proton with it's associated magnetic field, internal and external energy oscillations are set up. 20. When more than one atom is involved, and these energy frequencies align, they add together, specifically the magnetic field frequency. 21. I currently believe that this is where a line of flux originates from, aligned magnetic field frequencies. NOTES: 22. The Earth can be looked at as being a massive singular interacting photon with it's magnetic field, electrical surface field, and gravity, all three photonic forces all being 90 degrees from each other. 23. The flat spiral galaxy can be looked at as being a massive singular interacting photon with it's magnetic fields on each side of the plane of matter, the electrical field along the plane of matter, and gravity being directed towards the galactic center's black hole where the gravitational forces would meet, all three photonic forces all being 90 degrees from each other. 24. As below in the singularity, as above in the galaxy and probably universe as well. 25. I believe there are only two forces of nature, Gravity and EM, (GEM). Due to the stability of the GEM with the energy unit, this is also why the forces of nature haven't evolved by now. Of which with the current theory of understanding, how come the forces of nature haven't evolved by now since the original conditions acting upon the singularity aren't acting upon them like they originally were, billions of years have supposedly elapsed, in a universe that continues to expand and cool, with energy that could not be created nor destroyed would be getting less and less dense? My theory would seem to make more sense if in fact it is really true. I really wonder if it is in fact really true. 26. And the universe would be expanding due to these pulsating and interacting energy units and would also allow galaxies to collide, of which, how could galaxies ever collide if they are all speeding away from each other like is currently taught? DISCLAIMER: 27. As I as well as all of humanity truly do not know what we do not know, the above certainly could be wrong. It would have to be proved or disproved to know for more certainty. _______________________________________________________________________________________ Here is the test for the 'gravity' portion of my TOE idea. I do not have the necessary resources to do the test but maybe you or someone else reading this does, will do the test, then tell the world what is found out either way. a. Imagine a 12 hour clock. b. Put a magnetic field across from the 3 to 9 o'clock positions. c. Put an electric field across from the 6 to 12 o'clock positions. (The magnetic field and electric field would be 90 degrees to each other and should be polarized so as to complement each other.) d. Shoot a high powered laser through the center of the clock at 90 degrees to the em fields. e. Do this with the em fields on and off. (The em fields could be varied in size, strength, density and depth. The intent would be to energy frequency match the laser and em fields for optimal results.) f. Look for any gravitational / anti-gravitational effects. (Including the utilization of ferro cells so as to be able to actually see the energy field movements.) (An alternative to the above would be to shoot 3 high powered lasers, or a single high powered laser split into 3 beams, each adjustable to achieve the above set up, all focused upon a single point in space.) 'If' effects are noted, 'then' further research could be done. 'If' effects are not noted, 'then' my latest TOE idea is wrong. But still, we would know what 'gravity' was not, which is still something in the scientific world. Science still wins either way and moves forward.
re: "It requires BOTH for radiation to occur." I do not think so. I have performed experiments with loops and 9A4ZZ "Bipole" antenna. Maybe you do not understand "radiated wave" as Maxwell describes as opposed to near-reactive-field?
Sometimes the matching section on a full-wave dipole can alter the phase such that cancellation of fields doesn't occur; other times they feed full-wave dipoles off-centre so that you get a quarter wave of one side and three quarters on the other.
Thank you for the video, Carl. 🙂 I used to think that the half wavelength is kind of a compromise to keep the overall length shorter, and I was not aware of what would happen when using a full wavelength instead - well, now I know. 👍
Great qualitative analysis of this subject! Was hoping you'd talk a little about standing waves of voltage and current on the antenna in connection with this, but nonetheless, your presentation is something I can remember and reproduce!!!
Thanks for your video! I thought that whole, half or quarter lengths all work because when i blow onto a flute/pipe it resonates good on a given note for a given length (and not in-between values of lengths--im talking about resonance), when it is opened only on one side it sounds the note, when for the same length it is opened on both sides it sounds the same note octave higher (doubles the frequency), if you start to overblow it goes to yet octave higher, then quarter (i think...), etc. When it goes octave higher the wave inside the pipe produces a node in the middle. The whole deal is that the wave inside resonates real strong when it fits into the length of the pipe/antenna--it is selfexplanatory and when seen visualy its very clear. By the same token i can see a full length of wave rod resonating just as good as half or quarter.
Thanks for a great explanation, I am using 2.4ghz and 5.8ghz radio gear for radio controlled quadcopters and video downlink and this really helps understand how it all works. Thanks.
When you draw the full wavelength antenna, you draw currents cancelling each other in each halves of the antenna. With the voltage you correctly indicate, there will be a current going only one way in each part og the dipol, but there will be a current going the other way in the other half part of the dipole. The radiation from those two parts will cancel out each other. There will not be cancellation inside each of the two parts of the dipole.
Good to see that my thinking is congruent to the comments. I'm just learning this stuff and it didn't compute. I like the method of teaching tho. the thing that didnt make sense the most was having both legs of the dipole in sync. The core of the coax cable is the positive and the copper shell is the ground(i think) and the core connects to the one then the copper shell goes to the other so they will never see the same polarity. so the upper and lower would have an opposite +/- sine wave. totaling 1/2 wave.
Someone may have commented on this previously, as I have not read comments. The signal radiated from a half wave dipole is highest at a quarter wave from the end. This is proportional with the current showing on an antenna, and is always highest 90 degrees from the end. Current is lowest on ends where the voltage is highest. Remember an antennas radiation is higher at the center, and the radiated signal past the 60 degree mark (from center) is essentially linear to zero current on ends. A center fed half wave (current) dipole will have strongest signal at the center feed point and minimal radiation from the ends. An end fed (voltage) antenna is the same, except the voltage is highest at feed point, and strongest signal still radiates from the center at the 1/4 wave point from far end. The easy way to calculate the current showing at any point along an antenna is sin(angle)...in excel =sin(radians(angle)) the "angle" is the distance from end in electrical degrees of your desired antenna measurement point. You will get a value between 0 and 1 where 1 is the highest current. I find it easier, when designing a higher gain shortened dipole antenna, to always imagine the current showing on antenna. So if I linear load it, I only fold the low current ends to save space, and never have any folds (opposing fields) within 30 to 60 degrees of antenna center.
You are absolutely right--I should have been more precise. I added a note to the video that will hopefully clear any uncertainties. Thanks for making the video better.
Thank you for taking your time to make this video; however, I'm not seeing any obvious "note" to this video. I also must say, that I'm a little confused, now (originally, I thought your video cleared up why we use 1/2 wave dipoles, and not full wave dipoles) - once I read some of the comments about this video being misleading. Again, thank you for your time - nonetheless.
Smiler Liverpool Unfortunately, it is not true. There is no mention of the Magnetic field which contains the current. BOTH fields are required to for radiation to occur. It also states that full wave antennas will not radiate. You need to inform all the Hams in the world using them that their contacts are imaginary.
All toads are frogs but not all frogs are toads All verticals are dipoles (the ground is the second half of the dipole) but not all dipoles are verticals.
A full wave dipole can and does radiate. If not you couldn't use an 80 meter doublet on 40 meters. Having a tuner that can feed into high voltage is an issue if you're using store bought stuff. It would have to be fed with ladder line and a good balanced tuner. If you used coax there would have to be a matching device at the feedpoint....two half waves in phase.
Good video but I think you should update it and have a BOLD message at beginning saying its been superseded with link to new video. The part about a full wave not radiating is crazy and when watching it and I totally missed the message you added about RAGHU's comments. A full wave (two half waves in phase) radiates quite well and often with gain over a 1/2 wave. Of course gain at the expense of other direction. As I said, good video but is going to mislead many new hams as the message you added is easy to miss. Carl, you do a great job with this videos and I enjoy watching them.
A full wave antenna doesn't cancel radiaton. The impedance is the issue if fed in the center. Just feed it at the end/voltage maximum with a 1:49 or 1:64 transformer. When fed at the end you can use even and odd harmonics without a tuner.
You drew the dipole voltages as if they are the same at both ends,but they're opposite. The current throughout the dipole is going the same direction at any instant in time, you have the top half going one way and the bottom half going the other way.
If we accept that Voltage describes electromotive force, then we can say that the electromotive force (at an instant in time) at both ends of a dipole is in the same direction and therefore the same. As long as the antenna is the correct length, the current flow should be in the same direction; the cancellation of fields occurs when the antenna is not the correct length. The seems paradoxical since even though the current is traveling in the same direction for both poles, it is traveling toward the generator for one of them and away from the generator for the other; by bending the transmission line into a dipole we have, in effect, inverted the phase of one of the conductors with respect to the other.
" If we accept that Voltage describes electromotive force..." We do :-) "...then we can say that the electromotive force (at an instant in time) at both ends of a dipole is in the same direction..." No, direction implies current, if you mean polarity, then no, they are the opposite. When one end is becoming more positive, the other end is becoming more negative. "...and therefore the same." No, the polarities are opposite, The way to draw the voltage max/min curves along the half wave dipole would be to flip either the top or bottom of your drawing to the other side and make it negative. When people are just starting to catch on to these ideas, I think it's important to be simple and accurate because the first thing they learn will stick with them for a long time. I don't mean to give you a hard time, because it was nice of you to post this, but the way you drew this would make it hard to visualize the electric field, which the student needs to do on day 1.
Carl Oliver you speak of current flow---if there is flow, it goes from one side to the other---if both sides are the same voltage---there is no flow===and no current----the way you draw--both input points at the antenna center are the same voltage,, and no flow-----there is only flow if one is positive and the other negative.
sdold Abso bloody lutely! I'm with you Boo Boo. It needed your clarification.. Mine along the same lines was received if not abusively then sceptically!
Thanks for the video. Full wave antennas are not used in hf applications cause there is no affective way to measure wavelength from what i understand & therefore nobody uses them.
oliver good video thanks! chris9364 ac generator connection implies that when top portion of dipole electrons move away from generator; bottom portion dipole electrons move towards generator. --->so geometry of dipole is such that electrons moving down up in top pole and also down-up bottom pole, The radiation is a function i-e- proportional to elecron movement. Hence dipole 1/2 wave lenght max radiation
This answers my confusion over why an 80m dipole can't be stepped down for other higher frequency bands. Cristien Gossi's question is one that I was going to ask also though. I had a 5/8 wave vertical back in the CB days and that seems to contradict the info given in your video although I'm sure there's a valid explanation :-)
Hi . You remember me of a great teacher. Daniel Neugrochel. great explanation. Simple, clear.To be used on ham radio preparation winter course befor exams. 73 de i4oqa.
This is quite interesting ......I have used a full waver dipole on 40 meters (half wave on 80) for 40 years and it been my observation that it easily beats out a half wave diapole in both directions.....it's kind of common knowledge .......that this is the case....
Why is it that youtube can't post audio that is uniform. I can't hear a darn thing and my volume is all the way up on my laptop. I'm sure it would have been a great video if I could have heard it.
In an ocean of You tube crap, your very clear explanation is is like a breath of fresh air and Boy! don't you draw straight lines better I can. I regret, with great respect to you, that I must quarrel with you when you moved to a full wave. It really does seem quite valid when you draw the two half waves and their voltage or current distributions. Were it actually like that, your prediction would come to pass BUT if you consider the feeder, quite accurately depicted by you as a balanced feed and not coaxial, at its application to the centre of the full wave, it's necessary to point out that when one leg goes positive, the other goes negative. This means that whilst your E/I distribution on the lower leg is utterly 100% accurate in magnitude, in reality, it is of reverse polarity down that lower half wave leg. Radiation is additive and not subtractive. Agreed wholeheartedly, radiation wouldn't be very good but NOT by reason of the cancellation of fields, more a problem of poor power transfer. Lambda/2, in your 1st example, would present a low value of impedance..high current, low volts at the centre and would need an appropriately designed low Z feeder. Actual values not needed for this purpose. In your full wave example, the feed point would be of a very high value of Z (voltage feed) and would form an unacceptable reflected voltage standing wave, very lossy on a low Z feeder. Power transfer would be poor, with power travelling backwards, detrimental to the feeder itself and certainly the power generator. In mechanical terms, it would be in the wrong gear! Such a full-wave antenna is known as a full Zepp. antenna and was devised as a safety antenna for Zeppelin Dirigibles during WW1. They wanted the full wave antenna, in order to exploit its increased radiation over a half wave but by manipulating the length of a higher Z feeder (capable of withstanding High VSWR without losses,) to Lambda/4 or odd numbers thereof, they could ensure a low Z, current at the transmitter end of the feeder and keep those naughty, sparky volts well away from the airship whose envelope was full of Hydrogen. This was barely contained in a leaky balloon and would have gratefully accepted any encouragement from sparky voltages, off the TX feeder, to form a dramatic liaison with an amorous molecule of O2 which might be lurking, romantically awaiting its opportunity. The antenna is in use to this day with the gain being achieved by two half waves in phase, re-enforcing each other, not cancelling. When space is lacking, a half wave can be end-fed by the same HI Z type feeder but with one leg of the feeder simply not connected to anything! Funny things antennae, ain't they? I built a 1/2 Zepp for 14 Mhz..yesterday! Thanks for your super description..With your permission, I will plagiarise it when I am Elmering some newcomer. 73 G3NBY
***** The problem is that this video is not as great as you would like it to be. He does not consider the MAGNETIC field which contains the CURRENT. It takes BOTH to radiate.
Standing Bear Hi ...You are perfectly correct ..Both E and Mag fields must be present to create an e.m.field.Can't argue with that.. Problem is that it's ages ago since I posted. Must find what I said. Did I suggest that it would radiate with only one field? It certainly wasn't intended. I'll need to roll back and find it in order to put it right, if I did say that and I will acknowledge your contribution. Good wishes and thank you.
Technically they are opposite charges, yes...but due to the orientation of the conductors in the antenna, the electron flow is in the same direction, resulting in radiation of the same "polarity" of field--in other words the fields from each conductor add constructively.
:20-33 "change in magnetic field creates current" emf = n times (change in magnetic flux). If n (number of loops) is 0, then current is 0. Antennas radiate and receive light.
wavelength is horizontal, both antennas are vertical, the only way this would work is if the vertical amplitude is always equal to wavelength and this has always been my question which no video has answered, because the horizontal plane is time related.
"Batteryless Battery"???? a. Small aluminum cones with an electrical wire running through the center of the cones, cones spaced apart (not touching I'm thinking) but end to end. b. Electromagnetic radiation energy in the atmosphere interacts with the aluminum cones. c. Jostled atoms and molecules in the cone eventually have some electrons try to get away from other electrons of which those electrons gather at the larger end of the cone, of which also creates an area of positive charge at the smaller end of the cone. d. The electron's in the wire are attracted to the positive end of the cone and the positive 'end' in the wire are attracted to the negatively charged end of the cone. e. Basically a 'battery' has been created inside the electrical wire itself, different areas of electrical potential. Basically a 'wire battery' or a 'batteryless battery', however one wanted to call it. f. Numerous cones placed end to end increases the number of 'batteries' in the wire. Then of course, one could take that generated dc electricity to split H2O into H2 and O2 of which could be burned or utilized in a hydrogen fuel cell to get the electricity back with a byproduct of basically pure water. Species need pure water too. With the increased cosmic radiation that is going to be impacting this Earth this century as well as possibly a mini-ice age, might as well put that cosmic radiation to work for us.
Giving the fact that there will be no radiation at full wavelength is completely incorrect. A full wavelength antenna is more directive and has the same radiation pattern. However the radiation resistance is more and the size increases which is not desirable.
Sorry Carl, I don´t like to act as a smart-arse, but this is definitely completely wrong and should be rectified! The main mistake is that what you draw is the voltage but not the current-distribution along the antenna. What makes antenna radiate is current flowing along the conductor and along a dipole with length of lambda there is definitely current flowing! The current-distribution is simple this of two halfwave-dipoles in one line, and additional: currents along the two elements of the lambda-dipole are also still in phase. Therefore this antenna radiates formidable as it proves every day here. I build such a "non-radiating-antenna" and this works superb. If the antenna is elongated over lambda, you find sections along the conductor where the direction of the current has reversed. This results in a diagram that splits into several lobes and has no longer the 8-characteristic of dipoles shorter than lambda. By the way, because of this raghu is also wrong with his statement. The lambda-dipole has still his maximum radiation at 0 degrees at the z-axis (broadside). The gain of the lambda-dipole comes from the "compression" of the vertical diagram (if the antenna is mounted vertical, but thats the common mode) due to the mechanical length and the partial summation of the fields along the antenna. Believe it, im practising this for decades now... ;-)
Howdy. With all due respect. You need to mirror the grapics of one of the dipole legs. When there is plus in one leg there is minus in the other. Also I would argue the radiation pattern of a full wave is like 4 leaf clover whereas it is a 2 leaf clover for a half wave. To match a full wave to low impedance just feed it a quarter wavelength from either end. Regards.
The real reason why we use half wave antennas is because in that length the reactive part almost gets to 0 (42 ohm) and the real part become 70 ohms that make it much easier to match (to a 50 ohm wire), so it will be more efficient, and because the radiation pattern is omnidirectional.
I don't get that. I meant the wavelenght has to do with the lenght until a full cycle is done. And has nothing to do with the amplitute. However, maybe the amplitute has to do with the lenght of a full cycle, assuming we use a nice 'average' clean sinus-wave.
hold it hold it---your input lead ins have two different charges on them--- plus and minus---in which case the charges at the dipole are plus on one end and minus on the other, like a magnet. This can be measured with a voltmeter across the two of them anywhere along the length and at either end..----The way you have drawn this figure, it puts positive voltage on the center conductor of the lead in coax and positive on the shield of the coax also. IN essence, the two are the same. IN which case they are the same all the way up the coax and at the dipole. In which case there is no need for a separation of the two in the middle....esplain please
You're absolutely right: if you take a voltmeter it would confirm that the voltage at either end is opposite to the other. That is true. But what the voltmeter is actually measuring is that at a particular instant in time, electrons in one pole of the dipole are being forced away from the generator while those in the other pole are being forced toward it. Your voltmeter is interpreting this as opposite voltage since the DIRECTION of the charge on each pole is opposite with respect to the generator. But forget about the generator for a minute and look only at the antenna itself. The electrons in each pole of the dipole are being forced IN THE SAME DIRECTION at a particular instant in time, even though for one pole they're leaving the generator and for the other they are moving toward it. Ignoring the generator and focusing only on the antenna and the resultant fields generated, it's my opinion that we can express the voltage of both poles as having the same sign, due to the direction of the force being the same. Unconventional? Yes. But I find that it helps people understand what's actually going on.
Carl, that was great! It might be simple to most but i didn't get it until watching you video. TheCarpentorUnion brings up a good point...what about a 1/8 wave dipole?
Since electrons in each half of the dipole move in the same direction during each wave cycle, wouldn't the voltage distributions be opposite in each half of the dipole transmittor/antenna? ~confused newbie
Thx Carl for your informative video, my question here is what happens if you earthed one of the dipole radiator's element. would it still act like half wavelength dipole ? another question , how do you calculate antennas' separation for same frequency band Ants .
Why can't you make the same argument about the half wave version? Isn't the top accelerating electrons in the opposite direction of the bottom? Why can't you draw a box around those and say there's no net radiation?
Late to the party, but thank you. This is one of the most concise, well delivered antenna vids I have come across as a fairly new ham. I have been doing my research and discovered that half-waves are great, but didn't really grasp why. You have a gift sir, I hope you continue to be a teach.
What a clear and understandable explanation. So glad I came across this video. As a noob to ham I’m on a huge learning curve and this video has probably saved me 1000 hours watching other videos on the topic.
Pro Tip: old youtube videos are better
Most teachers describe. The best teachers also explain. Thanks very much!
I struggled with math all through school into college until one grad student tutored me. you remind me of him. he was the only person that tapped into the math part of my brain. afterwards I began making A's and B's in math and statistics. the way you write and talk out what you're doing on paper is so helpful. like another guy said, you've got a gift for teaching and explaining things. so many people out there think they can't do math and science it's a shame. if only more teachers had the gift... anyway, glad I found your videos. maybe now I'll find the motivation to study for the next amateur radio license.
Unfortunately there seem to be several errors in Mr. Oliver's presentation. First, the voltage on each side of a dipole will be differential, in other words, when one side (1/4λ element) of the 1/2λ dipole is positive in voltage, the other side will be opposite in phase by 180 degrees, or negative; this will be the case whether or not the 1/2λ dipole is fed by balanced line, or coax with a BalUn. (For those who don't use a BalUn when feeding a dipole with coax, you don't have a dipole, you have a 1/4λ antenna with a counterpoise.) Secondly, the electromagnetic field of an end-fed halfwave element will NOT be canceled out, because both positive and negative voltages are on the wire (at times) at the same time; if that were true, there would be no such thing as an end-fed 1/2λ antenna. Thirdly, you could have a FULL-WAVE dipole, that would garner about a 3dB increased perpendicular radiation pattern; if using 50 Ω coax, you would have to feed the dipole with a 49:1 BalUn since the feed-point impedance of a 1/2λ element is ~2450 Ω. But you'd have to make the 49:1 BalUn yourself since commercially available 49:1 transformers are UnUns, made for end-fed 1/2λ antennas. Also, while there are other ways to do impedance matching, in this case, you'd have to use a BalUn, since you need differential voltages on each side of the dipole. If I'm incorrect, please comment and I'll amend my comment.
This is simplification gone wrong. There is a difference in polarity of each end of a half wave dipole. If there is positive voltage at one end the voltage is negative in the other. And a full wave dipole has radiation in the same direction as a half wave dipole, but with a different BW (47 degrees instead of 68 degrees). There is also one more important difference between half wave and full wave dipoles. The impedance is very different. A half wave dipole is close to 73 ohms in free space (73+j42), but a full wave dipole is several thousands of ohms making it extremely difficult to match.
That's what I was thinking
Some people have perfect pitch. It's a genetic "gift". Those with the gift to sing (or other genetic gift) can develop their gift with the help of professional mentors and teachers. I grew up with a love of music and a love of singing and I was encouraged to sing my heart out by my parents and my church. One day I learned I was tone deaf and did not have the gift. I stopped singing and began a search for my gifts with support from family and friends and teachers. As a licensed amateur radio "ham", I have watched A LOT of videos explaining topics that interest me and came upon this video. Your explanation of the 1/2 wave dipole was an eye opener for me as a direct result of your upbeat, concise, and clear explanation of this topic. You, sir,
are one of those amazing folks who have the "gift" to organize and teach. I hope to encounter your videos again and I wrote this note to just give you thanks and to give you encouragement to continue. You have "the gift to teach". Best wishes. N6ATS
That is one of the kindest compliments I have ever been paid. Thank you for taking the time to say that. You made my day!
what the hell r u talking about?
What a sweetheart. This video helped me, as well.
Incredible video. I am new to HAM and this information is beyond anything I have seen before on antennas. Thank you, sir!
That 5.46 minute video was worth every second of my time - thanks !
It is misleading to describe the voltage distribution as shown in this video and also it is not true that an antenna which is one wavelength long does not radiate. Please revise your understanding.
Carmel Pule' Hi...An explanation from you might have been just a bit kinder & helpful to other readers needing an Elmer, a seasoned operator, no new-comer you, a guy who like you, knows his stuff. You are of course QUITE correct but just a bit vitriolic, not much but a bit! I admire your accuracy but Ham Spirit an' all that! Eh? Big 73 to you.
***** What is "vitriolic" about Carmels post? You should more interested in Political Correctness than accurate information. He used a "soft" word "misleading'. He did not state the TRUTH that the information is inaccurate.
Why SHOULD Carmel explain? HE did not post the video.
Standing Bear Hi Bear, I gotta admit that you got me again. Twice now!
I need to find to what it was I replied.
Would be quite prepared to swallow my words if I goofed but really I DID only say "A BIT" vitriolic and was non aggressive in the rest.
However I AM puzzled by your Political Correctness remark. Now..If you and I were ever to have a 'Pistols at Dawn' scrap it could easily be over political correctness.
However I don't pick up the point of your remark.
Final point..Have I goofed by attributing a video to the wrong person?
Help me here please, Bear...It WAS in April!
Regards M.O.L
PS..Just poured me a scotch..Could fax YOU one..ice?
Standing Bear Ah...Have found it!
Carmel Pule did not address his posting to anybody in particular. Bit bewildering.
He said that it is not true that a half wave doesn't radiate.
He is perfectly correct. It WILL radiate but very badly if at all, when fed in the centre with coax.
As far as I know, nobody had suggested that a 1/2 would not radiate, other than myself, when I pointed out that the balanced centre of a full wave would accept NO power from a lo Z, unbalanced feed...coax.
I now feel, although not absolutely convinced, that maybe Carmen's remark "Please revise your understanding" may have deserved the lesser word "sarcastic!" rather that "vitriolic" which in the spirit of good will, I beg permission to withdraw!
Carmel was not asked to explain the video which was obviously not his but more to explain his remarks.
I wish you well..M.O.L.
@@MauriatOttolink Hopefully my comment clears up several errors in Mr. Oliver's presentation. First, unlike what he illustrates, the voltage, and therefore current direction, on each side of a dipole will be differential, in other words, when one side (1/4λ element) of the 1/2λ dipole is positive in voltage, the other side will be opposite in phase by 180 degrees, or negative; this will be the case whether or not the 1/2λ dipole is fed by balanced line, or coax with a BalUn. (For those who don't use a BalUn when feeding a dipole with coax, you don't have a dipole, you have a 1/4λ antenna with a counterpoise.) Secondly, the electromagnetic field of an end-fed halfwave element will NOT be canceled out, because both positive and negative voltages are on the wire (at times) at the same time; if that were true, there would be no such thing as an end-fed 1/2λ antenna. Thirdly, you could have a FULL-WAVE dipole, that would garner about a 3dB increased perpendicular radiation pattern; if using 50 Ω coax, you would have to feed the dipole with a 49:1 BalUn since the feed-point impedance of a 1/2λ element is ~2450 Ω. But you'd have to make the 49:1 BalUn yourself since commercially available 49:1 transformers are UnUns, made for end-fed 1/2λ antennas. Also, while there are other ways to do impedance matching, in this case, you'd have to use a BalUn, since you need differential voltages on each side of the dipole. If I'm incorrect, please comment and I'll amend my comment.
Fantastic explanation of a dipole antenna. Thank you.
Excellent! You are a teacher! A rare gift.
Finally, someone explains it for me. New ham here...and wish I had seen this earlier. Thanks.
A fullwave antenna does radiate and has more gain than anything smaller. The antenna becomes more directional, in that the nulls on the ends are steeper, resulting in a near perfect figure 8 pattern. The impedance is not 50 ohms at it's feed point however, so those of you trying to match it, it's tough because it's in the range of about 5,000 ohms. Even feed in the middle, this antenna is a very high Z! That's a great bit away from the 50 ohms our transmitters want to see and it's a very long antenna. So, the half wave is the better choice on both space and matching requirements.
So, to say a 1 wavelength antenna does not radiate is completely wrong. The RF radiation pattern may not be where you desire it, it may be too directional, it might not fit on your property and you may not want to transform the impedance as needed, bit it will radiate and receive more power than anything shorter if properly matched and oriented.
You have to learn how electrons move and create E and M fields
in a full wave lenght antenna the center voltge keeps at 0 so no radiation and no gain
@@mestertester3150 A 1 or 1.25 wavelength radiates and are used, but with a bit thicker diameter to reduce impedance
I know this from trying and sheer experimentation. I took a long strand of insulated copper wire on both sides of a factory made dipole connector, and every inch or so I would snip slowly off of each end equally I would go key the radio and check the SWR meter. It would NEVER approach flat SWRs until I had physically cut it to half the wave length of the radio I was using. Even if you did not know the math and physics , you would find out quickly that it only resonates properly half of the wave length of the radio you are using. I was talking side band on a CB so it was about 18 feet long give or take an inch or two.
Thanx Carl. Nice. But it is incorrect to say that full wavelength antenna doesn't radiate. When the length of the antenna is one wavelength, the polarity of the current in one of the antenna is opposite to that of the other half. Because of these out-of-phase currents, the radiation at right angles from this antenna is zero. The field radiated by one half of the antenna alters the field radiated by the other half. However, a direction of maximum radiation still exists at 54 deg to the antenna.
This video only considers HALF of a radio wave. The wave has an ELECTRIC portion and a MAGNETIC portion. The VOLTAGE is in the electrical portion. The CURRENT is in the magnetic. It requires BOTH for radiation to occur.
Remembering the equation that Power = voltage times current we know it takes BOTH in order to actually radiate any power.
The suggestion that a full wave results in no radiation is inaccurate at best.
I should start by confessing that the purpose of this video was to explain in simple terms to people new to antenna theory how they radiate and I admit, does not contain the whole story. I will also confess to knowing that there are full wave antennas that radiate very well! There are matching sections or other ways to ensure that the fields add in phase, which complicates things for the newbie. So how about this: "A dipole antenna fed with a balanced 75 ohm line will not radiate efficiently in the direction perpendicular to the antenna." Is that fair to say?
Here's what I said about this in a prior comment:
Technically, That is true. When the electrons at the top are moving away from the generator, the electrons at the bottom are moving toward the generator. But what makes the fields add in phase is that for the top, away from the generator means UP, while for the bottom, toward the generator mean UP as well. So even though technically they have an opposite charge, the resultant current flow is in the same direction. I "cheated" during my explanation in order to illustrate this point. Others may disagree, but I see it as a question of convention rather than one of accuracy.
While I didn't mention magnetic field, i did not means to imply that it did not exist (my apologies if it came across that way). I did mention the motion of the electrons, which indirectly referenced the magnetic field.
+Standing Bear Thank you for preventing me from being mislead .
Standing Bear: Most textbooks now don't explain EM using simple English like what you did. Thanks. I always suspect that all so-called EM waves, including visible light, is nothing more than the movement of energy (what they call photons for light waves) from a high energy point in space to a low energy point in space (a.k.a. radiation). In other words, energy is relative, not absolute. In other words, a photon is not a particle without mass, but a passage of a fixed amount of energy through space.
Personally, I currently believe that the 'gem' photon is the energy unit of this universe that makes up everything in existence in this universe. Here is my latest TOE idea followed by the 'gravity' test for it:
Revised TOE: 3/25/2017a.
My Current TOE:
THE SETUP:
1. Modern science currently recognizes four forces of nature: The strong nuclear force, the weak nuclear force, gravity, and electromagnetism.
2. In school we are taught that with magnetism, opposite polarities attract and like polarities repel. But inside the arc of a large horseshoe magnet it's the other way around, like polarities attract and opposite polarities repel. (I have proved this to myself with magnets and anybody with a large horseshoe magnet and two smaller bar magnets can easily prove this to yourself too. It occurs at the outer end of the inner arc of the horseshoe magnet.).
3. Charged particles have an associated magnetic field with them.
4. Protons and electrons are charged particles and have their associated magnetic fields with them.
5. Photons also have both an electric and a magnetic component to them.
FOUR FORCES OF NATURE DOWN INTO TWO:
6. When an electron is in close proximity to the nucleus, it would basically generate a 360 degree spherical magnetic field.
7. Like charged protons would stick together inside of this magnetic field, while simultaneously repelling opposite charged electrons inside this magnetic field, while simultaneously attracting the opposite charged electrons across the inner portion of the electron's moving magnetic field.
8. There are probably no such thing as "gluons" in actual reality.
9. The strong nuclear force and the weak nuclear force are probably derivatives of the electro-magnetic field interactions between electrons and protons.
10. The nucleus is probably an electro-magnetic field boundary.
11. Quarks also supposedly have a charge to them and then would also most likely have electro-magnetic fields associated with them, possibly a different arrangement for each of the six different type of quarks.
12. The interactions between the quarks EM forces are how and why protons and neutrons formulate as well as how and why protons and neutrons stay inside of the nucleus and do not just pass through as neutrinos do.
THE GEM FORCE INTERACTIONS AND QUANTA:
13. Personally, I currently believe that the directional force in photons is "gravity". It's the force that makes the sine wave of EM energy go from a wide (maximum extension) to a point (minimum extension) of a moving photon and acts 90 degrees to the EM forces which act 90 degrees to each other. When the EM gets to maximum extension, "gravity" flips and EM goes to minimum, then "gravity" flips and goes back to maximum, etc, etc. A stationary photon would pulse from it's maximum extension to a point possibly even too small to detect, then back to maximum, etc, etc.
14. I also believe that a pulsating, swirling singularity (which is basically a pulsating, swirling 'gem' photon) is the energy unit in this universe.
15. When these pulsating, swirling energy units interact with other energy units, they tangle together and can interlock at times. Various shapes (strings, spheres, whatever) might be formed, which then create sub-atomic material, atoms, molecules, and everything in existence in this universe.
16. When the energy units unite and interlock together they would tend to stabilize and vibrate.
17. I believe there is probably a Photonic Theory Of The Atomic Structure.
18. Everything is basically "light" (photons) in a universe entirely filled with "light" (photons).
THE MAGNETIC FORCE SPECIFICALLY:
19. When the electron with it's associated magnetic field goes around the proton with it's associated magnetic field, internal and external energy oscillations are set up.
20. When more than one atom is involved, and these energy frequencies align, they add together, specifically the magnetic field frequency.
21. I currently believe that this is where a line of flux originates from, aligned magnetic field frequencies.
NOTES:
22. The Earth can be looked at as being a massive singular interacting photon with it's magnetic field, electrical surface field, and gravity, all three photonic forces all being 90 degrees from each other.
23. The flat spiral galaxy can be looked at as being a massive singular interacting photon with it's magnetic fields on each side of the plane of matter, the electrical field along the plane of matter, and gravity being directed towards the galactic center's black hole where the gravitational forces would meet, all three photonic forces all being 90 degrees from each other.
24. As below in the singularity, as above in the galaxy and probably universe as well.
25. I believe there are only two forces of nature, Gravity and EM, (GEM). Due to the stability of the GEM with the energy unit, this is also why the forces of nature haven't evolved by now. Of which with the current theory of understanding, how come the forces of nature haven't evolved by now since the original conditions acting upon the singularity aren't acting upon them like they originally were, billions of years have supposedly elapsed, in a universe that continues to expand and cool, with energy that could not be created nor destroyed would be getting less and less dense? My theory would seem to make more sense if in fact it is really true. I really wonder if it is in fact really true.
26. And the universe would be expanding due to these pulsating and interacting energy units and would also allow galaxies to collide, of which, how could galaxies ever collide if they are all speeding away from each other like is currently taught?
DISCLAIMER:
27. As I as well as all of humanity truly do not know what we do not know, the above certainly could be wrong. It would have to be proved or disproved to know for more certainty.
_______________________________________________________________________________________
Here is the test for the 'gravity' portion of my TOE idea. I do not have the necessary resources to do the test but maybe you or someone else reading this does, will do the test, then tell the world what is found out either way.
a. Imagine a 12 hour clock.
b. Put a magnetic field across from the 3 to 9 o'clock positions.
c. Put an electric field across from the 6 to 12 o'clock positions.
(The magnetic field and electric field would be 90 degrees to each other and should be polarized so as to complement each other.)
d. Shoot a high powered laser through the center of the clock at 90 degrees to the em fields.
e. Do this with the em fields on and off.
(The em fields could be varied in size, strength, density and depth. The intent would be to energy frequency match the laser and em fields for optimal results.)
f. Look for any gravitational / anti-gravitational effects.
(Including the utilization of ferro cells so as to be able to actually see the energy field movements.)
(An alternative to the above would be to shoot 3 high powered lasers, or a single high powered laser split into 3 beams, each adjustable to achieve the above set up, all focused upon a single point in space.)
'If' effects are noted, 'then' further research could be done.
'If' effects are not noted, 'then' my latest TOE idea is wrong. But still, we would know what 'gravity' was not, which is still something in the scientific world. Science still wins either way and moves forward.
re: "It requires BOTH for radiation to occur."
I do not think so. I have performed experiments with loops and 9A4ZZ "Bipole" antenna. Maybe you do not understand "radiated wave" as Maxwell describes as opposed to near-reactive-field?
What a great video and a great explanation. You made it so simple. Thank you.
2019 - Just found this and I found it very useful and competently delivered. Much appreciated.
Sometimes the matching section on a full-wave dipole can alter the phase such that cancellation of fields doesn't occur; other times they feed full-wave dipoles off-centre so that you get a quarter wave of one side and three quarters on the other.
Thank you for the video, Carl. 🙂 I used to think that the half wavelength is kind of a compromise to keep the overall length shorter, and I was not aware of what would happen when using a full wavelength instead - well, now I know. 👍
Great qualitative analysis of this subject! Was hoping you'd talk a little about standing waves of voltage and current on the antenna in connection with this, but nonetheless, your presentation is something I can remember and reproduce!!!
Thanks for your video! I thought that whole, half or quarter lengths all work because when i blow onto a flute/pipe it resonates good on a given note for a given length (and not in-between values of lengths--im talking about resonance), when it is opened only on one side it sounds the note, when for the same length it is opened on both sides it sounds the same note octave higher (doubles the frequency), if you start to overblow it goes to yet octave higher, then quarter (i think...), etc. When it goes octave higher the wave inside the pipe produces a node in the middle. The whole deal is that the wave inside resonates real strong when it fits into the length of the pipe/antenna--it is selfexplanatory and when seen visualy its very clear. By the same token i can see a full length of wave rod resonating just as good as half or quarter.
Thanks for a great explanation, I am using 2.4ghz and 5.8ghz radio gear for radio controlled quadcopters and video downlink and this really helps understand how it all works. Thanks.
When you draw the full wavelength antenna, you draw currents cancelling each other in each halves of the antenna. With the voltage you correctly indicate, there will be a current going only one way in each part og the dipol, but there will be a current going the other way in the other half part of the dipole. The radiation from those two parts will cancel out each other. There will not be cancellation inside each of the two parts of the dipole.
Easy to understand. Thank you very much.
Great presentation. Thanks
Good to see that my thinking is congruent to the comments. I'm just learning this stuff and it didn't compute. I like the method of teaching tho. the thing that didnt make sense the most was having both legs of the dipole in sync. The core of the coax cable is the positive and the copper shell is the ground(i think) and the core connects to the one then the copper shell goes to the other so they will never see the same polarity. so the upper and lower would have an opposite +/- sine wave. totaling 1/2 wave.
Excellent presentation! Thank you.
wow ... so simple and so clear!!! great job!!!
Someone may have commented on this previously, as I have not read comments. The signal radiated from a half wave dipole is highest at a quarter wave from the end. This is proportional with the current showing on an antenna, and is always highest 90 degrees from the end. Current is lowest on ends where the voltage is highest. Remember an antennas radiation is higher at the center, and the radiated signal past the 60 degree mark (from center) is essentially linear to zero current on ends. A center fed half wave (current) dipole will have strongest signal at the center feed point and minimal radiation from the ends. An end fed (voltage) antenna is the same, except the voltage is highest at feed point, and strongest signal still radiates from the center at the 1/4 wave point from far end. The easy way to calculate the current showing at any point along an antenna is sin(angle)...in excel =sin(radians(angle)) the "angle" is the distance from end in electrical degrees of your desired antenna measurement point. You will get a value between 0 and 1 where 1 is the highest current. I find it easier, when designing a higher gain shortened dipole antenna, to always imagine the current showing on antenna. So if I linear load it, I only fold the low current ends to save space, and never have any folds (opposing fields) within 30 to 60 degrees of antenna center.
You are absolutely right--I should have been more precise. I added a note to the video that will hopefully clear any uncertainties. Thanks for making the video better.
Thank you for taking your time to make this video; however, I'm not seeing any obvious "note" to this video. I also must say, that I'm a little confused, now (originally, I thought your video cleared up why we use 1/2 wave dipoles, and not full wave dipoles) - once I read some of the comments about this video being misleading. Again, thank you for your time - nonetheless.
Reminds me of the time I built an antenna with longer receivers thinking it would work better. Shock on me, it didn't work. Now I know why.
That's a beautiful explanation , thank you sir ..
Great video :D. Easiest video to understand on antennas that I have found !
Smiler Liverpool Unfortunately, it is not true. There is no mention of the Magnetic field which contains the current. BOTH fields are required to for radiation to occur.
It also states that full wave antennas will not radiate. You need to inform all the Hams in the world using them that their contacts are imaginary.
@@jaytibbles2223 And can you imagine the postage to send all those QSL cards back???
OMG! I'm So Thankful For This Video, I Just Got Mind Blowned !!!
Excellent explanation!! As Kent Smith's comments below. Thanks for the vid 😀
I don't think I've ever seen this explained better. Or, at all. Brilliant video. Cheers Carl. Sub'd.
Excellent explanation, I think that antenna should be look at as circuits and categorized as coupling devices. It will make designing easier
All toads are frogs but not all frogs are toads
All verticals are dipoles (the ground is the second half of the dipole) but not all dipoles are verticals.
Writing and drawing speed is fabulous
A full wave dipole can and does radiate. If not you couldn't use an 80 meter doublet on 40 meters. Having a tuner that can feed into high voltage is an issue if you're using store bought stuff. It would have to be fed with ladder line and a good balanced tuner. If you used coax there would have to be a matching device at the feedpoint....two half waves in phase.
Good video but I think you should update it and have a BOLD message at beginning saying its been superseded with link to new video. The part about a full wave not radiating is crazy and when watching it and I totally missed the message you added about RAGHU's comments. A full wave (two half waves in phase) radiates quite well and often with gain over a 1/2 wave. Of course gain at the expense of other direction. As I said, good video but is going to mislead many new hams as the message you added is easy to miss. Carl, you do a great job with this videos and I enjoy watching them.
Thank you. This was an eye opener.
A full wave antenna doesn't cancel radiaton. The impedance is the issue if fed in the center. Just feed it at the end/voltage maximum with a 1:49 or 1:64 transformer. When fed at the end you can use even and odd harmonics without a tuner.
You drew the dipole voltages as if they are the same at both ends,but they're opposite. The current throughout the dipole is going the same direction at any instant in time, you have the top half going one way and the bottom half going the other way.
If we accept that Voltage describes electromotive force, then we can say that the electromotive force (at an instant in time) at both ends of a dipole is in the same direction and therefore the same. As long as the antenna is the correct length, the current flow should be in the same direction; the cancellation of fields occurs when the antenna is not the correct length. The seems paradoxical since even though the current is traveling in the same direction for both poles, it is traveling toward the generator for one of them and away from the generator for the other; by bending the transmission line into a dipole we have, in effect, inverted the phase of one of the conductors with respect to the other.
" If we accept that Voltage describes electromotive force..."
We do :-)
"...then we can say that the electromotive force (at an instant in time) at both ends of a dipole is in the same direction..."
No, direction implies current, if you mean polarity, then no, they are the opposite. When one end is becoming more positive, the other end is becoming more negative.
"...and therefore the same."
No, the polarities are opposite, The way to draw the voltage max/min curves along the half wave dipole would be to flip either the top or bottom of your drawing to the other side and make it negative. When people are just starting to catch on to these ideas, I think it's important to be simple and accurate because the first thing they learn will stick with them for a long time. I don't mean to give you a hard time, because it was nice of you to post this, but the way you drew this would make it hard to visualize the electric field, which the student needs to do on day 1.
Carl Oliver you speak of current flow---if there is flow, it goes from one side to the other---if both sides are the same voltage---there is no flow===and no current----the way you draw--both input points at the antenna center are the same voltage,, and no flow-----there is only flow if one is positive and the other negative.
sdold Abso bloody lutely! I'm with you Boo Boo.
It needed your clarification..
Mine along the same lines was received if not abusively then sceptically!
Yes, when you drew the full wave example, look at half of that, you got the polarities right there, but the dipole is wrong.
Thanks for the video. Full wave antennas are not used in hf applications cause there is no affective way to measure wavelength from what i understand & therefore nobody uses them.
A full wave loop is the best example of an hf antenna
oliver good video thanks!
chris9364 ac generator connection implies that when top portion of dipole electrons move away from generator; bottom portion dipole electrons move towards generator. --->so geometry of dipole is such that electrons moving down up in top pole and also down-up bottom pole, The radiation is a function i-e- proportional to elecron movement. Hence dipole 1/2 wave lenght max radiation
This answers my confusion over why an 80m dipole can't be stepped down for other higher frequency bands.
Cristien Gossi's question is one that I was going to ask also though. I had a 5/8 wave vertical back in the CB days and that seems to contradict the info given in your video although I'm sure there's a valid explanation :-)
Hi . You remember me of a great teacher. Daniel Neugrochel. great explanation. Simple, clear.To be used on ham radio preparation winter course befor exams. 73 de i4oqa.
*remind me*
thanks. very simple explanation. the best kind
Very nice explanation . I like the video.Thanks for uploading .
Very good Carl. Could you explain us about 5/8 wavelenght antennas? Why a 5/8 antenna has more gain? Thank you
Awesome explanation!!
Very nice video.paper and pencils always work better teach.thanks for sharing.
Excellent video! Thank you!
This is quite interesting ......I have used a full waver dipole on 40 meters (half wave on 80) for 40 years and it been my observation that it easily beats out a half wave diapole in both directions.....it's kind of common knowledge .......that this is the case....
Clearly explained. V helpful. Thank you
Excellent explanation !! Thank you for your time.
Pedro Ortega
Why is it that youtube can't post audio that is uniform. I can't hear a darn thing and my volume is all the way up on my laptop. I'm sure it would have been a great video if I could have heard it.
In an ocean of You tube crap, your very clear explanation is is like a breath of fresh air and Boy! don't you draw straight lines better I can.
I regret, with great respect to you, that I must quarrel with you when you moved to a full wave.
It really does seem quite valid when you draw the two half waves and their voltage or current distributions. Were it actually like that, your prediction would come to pass BUT if you consider the feeder, quite accurately depicted by you as a balanced feed and not coaxial, at its application to the centre of the full wave, it's necessary to point out that when one leg goes positive, the other goes negative.
This means that whilst your E/I distribution on the lower leg is utterly 100% accurate in magnitude, in reality, it is of reverse polarity down that lower half wave leg.
Radiation is additive and not subtractive.
Agreed wholeheartedly, radiation wouldn't be very good but NOT by reason of the cancellation of fields, more a problem of poor power transfer.
Lambda/2, in your 1st example, would present a low value of impedance..high current, low volts at the centre and would need an appropriately designed low Z feeder. Actual values not needed for this purpose.
In your full wave example, the feed point would be of a very high value of Z (voltage feed) and would form an unacceptable reflected voltage standing wave, very lossy on a low Z feeder. Power transfer would be poor, with power travelling backwards, detrimental to the feeder itself and certainly the power generator. In mechanical terms, it would be in the wrong gear!
Such a full-wave antenna is known as a full Zepp. antenna and was devised as a safety antenna for Zeppelin Dirigibles during WW1.
They wanted the full wave antenna, in order to exploit its increased radiation over a half wave but by manipulating the length of a higher Z feeder (capable of withstanding High VSWR without losses,) to Lambda/4 or odd numbers thereof, they could ensure a low Z, current at the transmitter end of the feeder and keep those naughty, sparky volts well away from the airship whose envelope was full of Hydrogen. This was barely contained in a leaky balloon and would have gratefully accepted any encouragement from sparky voltages, off the TX feeder, to form a dramatic liaison with an amorous molecule of O2 which might be lurking, romantically awaiting its opportunity.
The antenna is in use to this day with the gain being achieved by two half waves in phase, re-enforcing each other, not cancelling.
When space is lacking, a half wave can be end-fed by the same HI Z type feeder but with one leg of the feeder simply not connected to anything!
Funny things antennae, ain't they? I built a 1/2 Zepp for 14 Mhz..yesterday!
Thanks for your super description..With your permission, I will plagiarise it when I am Elmering some newcomer.
73 G3NBY
***** The problem is that this video is not as great as you would like it to be. He does not consider the MAGNETIC field which contains the CURRENT. It takes BOTH to radiate.
Standing Bear Hi ...You are perfectly correct ..Both E and Mag fields must be present to create an e.m.field.Can't argue with that.. Problem is that it's ages ago since I posted.
Must find what I said. Did I suggest that it would radiate with only one field?
It certainly wasn't intended.
I'll need to roll back and find it in order to put it right, if I did say that and I will acknowledge your contribution.
Good wishes and thank you.
IMHO this is one of the best videos on TH-cam. Absolutely love it! Thank you, and God Bless you sir.
keøeyh
thank you for making excellent video!
very, very useful video! thank you Man, very much!!
Excellent explained! Finnaly I understand!
Technically they are opposite charges, yes...but due to the orientation of the conductors in the antenna, the electron flow is in the same direction, resulting in radiation of the same "polarity" of field--in other words the fields from each conductor add constructively.
Very well explained!! thank you!
Amazing video sir
:20-33 "change in magnetic field creates current" emf = n times (change in magnetic flux). If n (number of loops) is 0, then current is 0. Antennas radiate and receive light.
Nice video. I think, you should correct the pictures shown at 2:35. Voltage maxima at both ends occur only at multiples of lambda, not lambda/2.
....this fellow needs to do his diploma again....need not be taken seriously....lot of wrong information on internet....this is best example....
wavelength is horizontal, both antennas are vertical, the only way this would work is if the vertical amplitude is always equal to wavelength and this has always been my question which no video has answered, because the horizontal plane is time related.
excellent explanation!
I learned how to draw a happy face!
jokes aside, really good explanation!
Not really. You're not understanding it, otherwise you'd see the flaws. I'm still learning too, but I have enough now to see what's wrong here.
@@ThatEEguy2818 Yes. look at this applet here: they both radiate but the half wave dipole has a bigger one lobe no side lobes? ibb.co/BGPY0yx
"Batteryless Battery"????
a. Small aluminum cones with an electrical wire running through the center of the cones, cones spaced apart (not touching I'm thinking) but end to end.
b. Electromagnetic radiation energy in the atmosphere interacts with the aluminum cones.
c. Jostled atoms and molecules in the cone eventually have some electrons try to get away from other electrons of which those electrons gather at the larger end of the cone, of which also creates an area of positive charge at the smaller end of the cone.
d. The electron's in the wire are attracted to the positive end of the cone and the positive 'end' in the wire are attracted to the negatively charged end of the cone.
e. Basically a 'battery' has been created inside the electrical wire itself, different areas of electrical potential. Basically a 'wire battery' or a 'batteryless battery', however one wanted to call it.
f. Numerous cones placed end to end increases the number of 'batteries' in the wire.
Then of course, one could take that generated dc electricity to split H2O into H2 and O2 of which could be burned or utilized in a hydrogen fuel cell to get the electricity back with a byproduct of basically pure water. Species need pure water too.
With the increased cosmic radiation that is going to be impacting this Earth this century as well as possibly a mini-ice age, might as well put that cosmic radiation to work for us.
Hi, I confused, how could both side will pe positive or negative if considering max Voltage at each corner? Should be + and - at Vmax position.
Giving the fact that there will be no radiation at full wavelength is completely incorrect. A full wavelength antenna is more directive and has the same radiation pattern. However the radiation resistance is more and the size increases which is not desirable.
There is no radiation on full wavelength antennas?
I do have a full wavelength 40 meter ham radio dipole antenna.
Probably gets out like a banshee on steroids, too.
Great video, thank you!
Sorry Carl, I don´t like to act as a smart-arse, but this is definitely completely wrong and should be rectified!
The main mistake is that what you draw is the voltage but not the current-distribution along the antenna. What makes antenna radiate is current flowing along the conductor and along a dipole with length of lambda there is definitely current flowing! The current-distribution is simple this of two halfwave-dipoles in one line, and additional: currents along the two elements of the lambda-dipole are also still in phase. Therefore this antenna radiates formidable as it proves every day here. I build such a "non-radiating-antenna" and this works superb. If the antenna is elongated over lambda, you find sections along the conductor where the direction of the current has reversed. This results in a diagram that splits into several lobes and has no longer the 8-characteristic of dipoles shorter than lambda. By the way, because of this raghu is also wrong with his statement. The lambda-dipole has still his maximum radiation at 0 degrees at the z-axis (broadside). The gain of the lambda-dipole comes from the "compression" of the vertical diagram (if the antenna is mounted vertical, but thats the common mode) due to the mechanical length and the partial summation of the fields along the antenna. Believe it, im practising this for decades now... ;-)
Howdy. With all due respect.
You need to mirror the grapics of one of the dipole legs. When there is plus in one leg there is minus in the other.
Also I would argue the radiation pattern of a full wave is like 4 leaf clover whereas it is a 2 leaf clover for a half wave.
To match a full wave to low impedance just feed it a quarter wavelength from either end.
Regards.
But shouldn't the bottom leg of a half wave dipole have the opposite voltage?
Great tutorial!!
The real reason why we use half wave antennas is because in that length the reactive part almost gets to 0 (42 ohm) and the real part become 70 ohms that make it much easier to match (to a 50 ohm wire), so it will be more efficient, and because the radiation pattern is omnidirectional.
I don't get that. I meant the wavelenght has to do with the lenght until a full cycle is done. And has nothing to do with the amplitute. However, maybe the amplitute has to do with the lenght of a full cycle, assuming we use a nice 'average' clean sinus-wave.
Underrated video
Awesome explanation. ;-)
Good video.
Thank you!!
He is talking about a dipole antenna not a single length type. Big difference. You folks are confusing dipole with a non dipole.
Good video!!!
So set "rabbit ears" just under 34" for channel 7. Too bad my signal area is too weak!
hold it hold it---your input lead ins have two different charges on them--- plus and minus---in which case the charges at the dipole are plus on one end and minus on the other, like a magnet. This can be measured with a voltmeter across the two of them anywhere along the length and at either end..----The way you have drawn this figure, it puts positive voltage on the center conductor of the lead in coax and positive on the shield of the coax also. IN essence, the two are the same. IN which case they are the same all the way up the coax and at the dipole. In which case there is no need for a separation of the two in the middle....esplain please
You're absolutely right: if you take a voltmeter it would confirm that the voltage at either end is opposite to the other. That is true. But what the voltmeter is actually measuring is that at a particular instant in time, electrons in one pole of the dipole are being forced away from the generator while those in the other pole are being forced toward it. Your voltmeter is interpreting this as opposite voltage since the DIRECTION of the charge on each pole is opposite with respect to the generator. But forget about the generator for a minute and look only at the antenna itself. The electrons in each pole of the dipole are being forced IN THE SAME DIRECTION at a particular instant in time, even though for one pole they're leaving the generator and for the other they are moving toward it. Ignoring the generator and focusing only on the antenna and the resultant fields generated, it's my opinion that we can express the voltage of both poles as having the same sign, due to the direction of the force being the same. Unconventional? Yes. But I find that it helps people understand what's actually going on.
Thank you.
Carl, that was great! It might be simple to most but i didn't get it until watching you video. TheCarpentorUnion brings up a good point...what about a 1/8 wave dipole?
Since electrons in each half of the dipole move in the same direction during each wave cycle, wouldn't the voltage distributions be opposite in each half of the dipole transmittor/antenna?
~confused newbie
+chris9364 i think so too!
Would you please do a visual video to explain how electricity and magnetism work together ?? Thanks
What would happen if you have One Complete Cycle on Both Ends? Full Wave for each side.
Thx Carl for your informative video, my question here is what happens if you earthed one of the dipole radiator's element. would it still act like half wavelength dipole ? another question , how do you calculate antennas' separation for same frequency band Ants .
Why can't you make the same argument about the half wave version? Isn't the top accelerating electrons in the opposite direction of the bottom? Why can't you draw a box around those and say there's no net radiation?