Building An Inverted-V(Dipole) - Coax Part 3

[@FranktheDachshund]

What he was trying to say is, Speed of light 186,000 miles per second x 5280 = 984,000,000 feet per second divided by 1,000,000 gives wavelength in MHz -->984 MHz divide by 2 to get half wave = 492 (in a vacuum) speed reduced by .095 in copper --> so 492 x 0.095 = 468

[@kn6jhtradio943]

really appreciate all of Kevin's talks, it's refreshing to hear a charismatic EE . Thank you for filming and posting these

[@kn6jhtradio943]

You know he's a retired Engineer because he approximated 300/80=4 lol. Love it

[@HamRadioTV]

You are very welcome, it's our pleasure! He is a great Ham and person!

[@HamRadioTV]

😀

[@seeharvester]

Shouldn't the half wave multiple for the feed line be included in the coax ugly balun? In other words, the length of the coax used to form the ugly balun be part of the 1/2 wave length?

[@52Warlock]

Just stumbled upon this channel and I gotta say that it is an excellent resource. Great explanations, great content. Keep it up!

[@HamRadioTV]

Thanks! We are glad you're finding it informational.

[@mrmrlee]

@ 16:46 To me this is simply an RF choke, which uses a coil of coax to choke off any stray RF from running back along the shield of the feedline. You can accomplish exactly the same effect by coiling several turns of coax and taping it together to form a circle several inches in diameter. I had RF issues in a mobile HF installation and did this, worked great!

[@HamRadioTV]

Yes, I've herd people doing it that way as well. Thanks for sharing.

[@artgumbus]

Super Super Job.... Kevin on the core windings for the balun what wire gauge is used for 1000W Pep, Teflon or PTFE coating... Great Video... Thanks. W1SWL

[@VoeViking]

This guy is the best.

[@HamRadioTV]

Thank You!! We are lucky to have him and I'm glad people are enjoying the extra work I put into sharing these! ~73 KM6FAK~

[@wallstreettrader1]

The lesson here is...cut an 80m Inverted Vee dipole resonant at 3.575 MHz (allowing for the antenna's location and surrounding influences) and it will also be resonant at 7.15 MHz and 14.3 MHz. Then make sure that the RG-213 coax (velocity factor 0.66) is cut to either 90.83ft., or 181.66ft., ensuring coax impedance unity from feed point to transceiver connection point, keeping in mind the additional line loss with 181.66ft., though I suspect that coax length is not a significant factor. It's worth considering anyway. An Inverted Vee at 37 degrees of slope for each leg, resonant at 3.575 MHz, will be 125.6727ft. in total length (62.8364ft. per leg). If you have a 30ft. mast for the above dipole (higher than 30 would be best), then the preferred distance from the base of the mast to the guy rope stakes should be around 40ft. (for a 37 deg. slope angle) I would still recommend trimming using an MFJ-269 or similar to pinpoint and confirm the resonant frequencies to ensure that they are 3.575, 7.15 and 14.3 MHz, because the great outdoors can bring all kinds of factors into play: Terrain, ground moisture, surrounding tree density, proximity to conductive elements, antenna height, etc. These are, of course, calculations applying only to an antenna standing in free-space. Best example: Even better would be to cut the Inverted Vee dipole to be sloping at about 22 degrees, meaning that the mast to peg length is more like 75ft., but then the dipole will also be slightly longer...128.29ft. / 64.145ft. per leg. Using trigonometrics: Inverse tangent [30ft/75ft]=21.801deg. slope (approximately 22 degrees) 30ft. mast (30×30=900) 75ft. mast to guy peg (75×75=5625) 900+5625=6525, square root=80.777ft. Dipole leg length: 128.29÷2=64.145ft. 80.777-64.145ft.=16.632ft. guy rope per leg. Then trim using an antenna analyzer to suit installation conditions. You won't need to trim much, if anything. Use a taught line hitch knot on the rope at the guy stake. (KZbin videos available to help with that.) This video is an amazing head start, beautifully showing the importance of coax length and velocity factor. Thank you! 73! Erik, NO8Z

[@PatAutrey]

This is a fantastic video thank you so much quick question do you have any suggestions when you're running a doublet antenna in terms of the length of the feed line and the length of the coax after the 4:1 balun thanks in advance

[@trevorhardisty6255]

Hi, very good series of videos, my question is: Would the calculation be different if you used ladder line(window line)? Trevor ZL1UTH

[@HamRadioTV]

The antenna calculations would be the same and you wouldn’t have to mess with a current balun on balanced ladder line. Open ladder line is a great option but can be more difficult to run/install with out couples into things.

[@brentsecord3016]

Thanks for making the video, Here is a question (rookie here), what would you do if you were figuring out coax length for a Alpha Delta DXLBplus Dipole that does 10- 160M? .

[@SMShannon55]

He covers multiple band antennas at 14:30.

[@rickkingsley2407]

A tried and true method to tune a wire antenna, is first to use the 468 constant, as you usually would, but instead of then checking the VSWR at your desired frequency, perform the following steps, which will take into account, the actual surroundings that will effect your antenna's resonance (the 468 value is based on free space, not taking account surrounding buildings...trees...etc.). Do the 468 calculation, raise your antenna into position, and never mind your desired frequency...but instead find the point where VSWR is at its lowest. Remember-the antenna WILL be at resonance SOMEWHERE on that band. When you find that frequency, MULTIPLY that number, by the length of your antenna (the length you calculated at the start, using the 468 constant) and you will discover that the 468 value has changed...THIS IS YOUR NEW CONSTANT which accounts for the actual conditions will effect the way that antenna "behaves" in the only that matters...not in a laboratory "Isotropic" test condition. Now, using your new...call it your "personal" constant, your final length calculation will be different, and after you make that adjustment...viola! Dont take my word for...try it! Bottom line: who likes running out to the back yard a zillion times trimming...and adding....and did I mention in the rain?..73

[@HamRadioTV]

Thank you Rick! This sounds like a great method too.

[@la3dna]

What is this 0.95 value that he multiply with at 3:47 playtime from start? Is it the speed factor in the antenna wire?

[@HamRadioTV]

The short answer is, YES. The long answer is as follows: A half-wavelength is actually 492/f, so where did the number 468 come from? The explanation most often given these days is that a radio wave travels about 5% slower in wire than it does in free space, so the distance that a radio wave would travel in a wire is about 5% less than it would travel in free space. Now, I don’t know about you, but while I’ve used this formula for building dipoles, I’ve never had one tune up perfectly using that number. There are a number of reasons for this, the main one being the height above ground of the dipole. What I’ve found is that the elements of the dipole are usually longer than they need to be. I sometimes joke that whoever came up with that number did so so that hams wouldn’t cut their dipoles too short. After all, it’s much easier to make a length of wire shorter than it is to make it longer.

[@iamhe999]

​@@HamRadioTV.. the velocity factor of the radiators, not the velocity factor of the coax feed line. Yes two different velocity factors... one for the radiators the other for the feed line.

[@bill-2018]

I experimented with coax lengths because CB'ers keep saying about the magical ½λ of coax. In 49 years I've never cut coax to a specific length. I put a 50Ω dummy load on the tx at 21.200 MHz. What better way to check the output than direct? The oscilloscope read 32 Volts. I put it on the end of 20" of coax, then 38" of coax and joined them together to 58", then added 16', a total of 20' 10". Guess what? All read 32 Volts. You're saying I would need a ½λ which is 23 feet mulitiplied by 0.66 Vf for my RG58 = 15.18 feet for 21 MHz. 20" is not 15.18 feet, neither is 38" or 58" or 20' 10". All showed the same result. Where's this magic half wave? I had no problem getting the SWR down on all five 20m to 10m dipoles and my 60m inverted vee all fed with what was the length necessary to bring the coax into my shack, not measured, and they do work. G4GHB·

[@bklynhubby]

Hi Kevin. Great videos. Do these calculations also apply to end fed and off center end fed antennas? Thanks. 🙏

[@norloufowler9074]

Can't make out writing on board

[@stevenlawhon3695]

In this example , what if you need more than 83ft?

[@wallstreettrader1]

166ft. would be the next multiple.

[@thepacketnarc7634]

I'm sorry man, but 3:40 to 4:06 was an out of context bluthering mess of "what was I gonna say here"... Can you please give some context ? Are you converting between Imperial and Metric ? Are you jumping between 40 and 80 meters ? I'm lost

[@richarde735]

Kevin’s 3.9 or 3.9MHz reference is the center for a General class license phone portion of the 80m band (3.5-4.0). the speed of light is 300 000 km/s, or 186 000 mi/s) or 300 in Kevin mathematical calculations. www.electronics-notes dot com/articles/antennas-propagation/dipole-antenna/length-calculations-equation-formula.php

[@Aleziss]

3:43 492X.95 ? Where is that .95 come from ?

[@alanvk5ar436]

.95 is approximately the velocity of the antenna wire

[@Aleziss]

@@alanvk5ar436 thanks Alan !

[@feeatlastfeeatlast5283]

His example is flawed. That is, his matched antenna with ANY length of coax will still be a very low swr and the impedance at the transmitter end will be nearly perfect. IF he was measuring off resonance theory tells us the swr will be the same at any point along the feedline, although the Z at the transmitter will be a function of the length. Remember, off resonance the 1/2 wavelength of coax is no longer a half wavelength either.

[@HamRadioTV]

Hmm, thank you for the feedback. I will speak to Kevin about this. I got my ticket in 2016 and don't have enough standing in this as I spend most of my time with the video end of things and not reading up on theory.

[@amirb715]

exactly. if your antenna impedance is well matched (you are close to center of smith chart) the it absolutely does not matter how long a 50 ohm coax you use for feed line. SWR will remain close to 1 for any length (you remain close to center of smith chart for any length). If this didn't happen, then the antenna was not matched in first place. Still if at the antenna terminal your impedance is close to center of smith chart (low SWR number) as you add ANY length of coax feed, you will not get any further than the center of smith chart. you will only rotate about its center but SWR remains the same

[@mrmrlee]

@Remy Kallen The coax length does not matter, however poor quality coax will be more lossy, as will excessively long coax....Shorter is better.

[@stevenlawhon3695]

300/80 is 3.75 not 4

[@stevenlawhon3695]

4 Mhz is 75m

[@stevenlawhon3695]

BTW what coax do you recommend for runs around 20 ft HF

[@HamRadioTV]

lol, you win the prize! Thanks for the correction. Kevin does those presentations with zero notes, and I think he does great! The newer hams learn the general concepts and the veteran Hams can find little Easter eggs! ~73 KM6FAK~

[@HamRadioTV]

LMR 400 or 213 are my top favorites. But, for only 20 ft I would go with 8X.

[@stevenlawhon3695]

8x is what I have thanks. compromise set up

[@HamRadioDX]

Love this series. Thanks! Keep them coming

[@HamRadioTV]

Thank You! Looking forward to your next videos too!

[@thepacketnarc7634]

So we are not gonna even talk about 8 x 4 = 32 ? HAM Radio test be like "you gotta know and remember all these very specific things"... Also HAM radio be like "ahh, just to the nearest 100s place will work"

[@keanaw2]

300/80 = 3.75 MHz and not 4MHz..use a calculator would help.. 😂