homerb.b.

[hotrod]

or anyone else . when building your own antennas theres a math formula to get the lenth .
234didvide by freq. for 1/4 wave
468 divide by freq for half wave
question is what is it for 5/8 wave??
also been said here when building a 5/8 u need a tuning system? would it be nessary if antenna is
correct lenth? and if niot could someone just use a antenna tuner on it?

 

[Captain Kilowatt]

or anyone else . when building your own antennas theres a math formula to get the lenth .
234didvide by freq. for 1/4 wave
468 divide by freq for half wave
question is what is it for 5/8 wave??
also been said here when building a 5/8 u need a tuning system? would it be nessary if antenna is
correct lenth? and if niot could someone just use a antenna tuner on it?

It is a basic mathematical ratio so you can calculate whatever you want. Since 468/F is a 1/2 wave and 234/F is a 1/4 wave you can work the numbers to suit your needs. Basically 936?F is a full wavelength so therefore 936/F multiplied by the desired fractional wavelength is what you would use. Applying that 936?27.2 MHz times 5/8 equals 21.5 feet. You need a tuning and matching system any time an antenna is not 50 ohms and in the case of a 5/8 it is NOT 50 ohms and therefore you DO NEED a method of matching it. Simply thinking that if it is the "correct length" to be a 5/8 then it should be a good 50 ohm match is not correct because if that was the case then any length would be correct and antenna impedance would never change regardless of the length. Using a tuner is about theleast efficient and least desirable method to use unless the tuner itself is located directly at the antenna feedpoint. The SWR is still very high between the antenna and the tuner and the losses in the cable increase dramatically because of this.

 

[Marconi]

Hotrod, that is a good question, and I had to refer to some models to see what the effects were.

Some say a 5/8 wave is not a resonant device to start with, so I suppose it needs a tuner at the antenna.

I don't know for sure in the real world, but my unmatched 5/8 wave Exnec model looks to show pretty stead results with gain, match, angle, and pattern over a good range of frequencies around the CB band,

but the antenna model was set 67.5' feet high.

I lowered the antenna to 18' feet, and expecting about the same trends. The performance results for the models took the expected change (worse performance) one might expect...but here the patterns at the range of frequencies around CB I checked were dramatically different, and of course all the patterns produced higher angles with the antenna installed lower.

Generally speaking I don't consider the mismatch (SWR) to be a good indicator of performance...so whatever is going on here probably will not change much if you add a tuner in line except for the SWR.

So, I would compare the install both low and high to be sure.

Of course your radio won't like such results at any height without the tuner.

Keep us posted if you do the project.

 

[fourstringburn]

or anyone else . when building your own antennas theres a math formula to get the lenth .
234didvide by freq. for 1/4 wave
468 divide by freq for half wave
question is what is it for 5/8 wave??
also been said here when building a 5/8 u need a tuning system? would it be nessary if antenna is
correct lenth? and if niot could someone just use a antenna tuner on it?

Those numbers are a shortcut method that someone came up with which represent nothing in understanding basic antenna theory that's why you are left with questions.

The real method which represents the basics are simple and may give a better understanding for how it works. Then you can figure your needs from there when you get a better understanding. The formula is as simple as this below;

Velocity divided by the frequency equals the wavelength in meters.

Velocity is 300 million meters a second ( same as the speed of light in free space )

Frequency is how many cycles per second ( like waves on an ocean, each wave crest between another is 1 cycle)

Wavelength is how long the wave is between cycles ( like waves on the ocean it's the length between each wave's crest)



27 MHz will use for an example frequency.

Divide 300 ( represents the 300 million but the extra 0's can be omitted) by 27 (represents 27 million times a second the wave cycles for 27 MHz) and you get 11.11 ( this is the wavelength in meters)

Since we Americans still don't use the metric system, convert meters to feet by multiplying meters by 3.28 and it comes out close enough.

11 x 3.28 =36.08 ft.

Now for antenna lengths just divide 36 ft which is a full wavelength for 11 meters by 2 and you get 18 ft. for a 1/2 wave or divide it by 4 for a 1/4 wave at 9 ft. and you can figure out any other fractional wavelength from there if you have basic math skills.

 

[Captain Kilowatt]

Hotrod, that is a good question, and I had to refer to some models to see what the effects were.

Some say a 5/8 wave is not a resonant device to start with, so I suppose it needs a tuner at the antenna.

Resonance has nothing to do with tuning. An antenna is resonant when there is zero reactance present. The impedance can be virtually anything as long as the reactance is zero. A halfwave antenna has an impedance of hundreds if not thousands of ohms yet it is resonant.

I don't know for sure in the real world, but my unmatched 5/8 wave Exnec model looks to show pretty stead results with gain, match, angle, and pattern over a good range of frequencies around the CB band,

but the antenna model was set 67.5' feet high.

As I am sure you have noticed there is often a big difference between models and the real world. Usually the results are pretty good however there are exceptions.

I lowered the antenna to 18' feet, and expecting about the same trends. The performance results for the models took the expected change (worse performance) one might expect...but here the patterns at the range of frequencies around CB I checked were dramatically different, and of course all the patterns produced higher angles with the antenna installed lower.

Pretty much what would be expected.

Generally speaking I don't consider the mismatch (SWR) to be a good indicator of performance...so whatever is going on here probably will not change much if you add a tuner in line except for the SWR.

True, SWR is not a great indicator of overall performance however it will prevent maximum transfer of power which will contribute to a lack of efficiency.

So, I would compare the install both low and high to be sure.

Of course your radio won't like such results at any height without the tuner.

Keep us posted if you do the project.

Why bother trying with different heights? The OP was asking about antenna length not heights. About 99.999% of the time higher is better.

 

[Captain Kilowatt]

Those numbers are a shortcut method that someone came up with which represent nothing in understanding basic antenna theory that's why you are left with questions.

The real method which represents the basics are simple and may give a better understanding for how it works. Then you can figure your needs from there when you get a better understanding. The formula is as simple as this below;

Velocity divided by the frequency equals the wavelength in meters.

Velocity is 300 million meters a second ( same as the speed of light in free space )

Frequency is how many cycles per second ( like waves on an ocean, each wave crest between another is 1 cycle)

Wavelength is how long the wave is between cycles ( like waves on the ocean it's the length between each wave's crest)



27 MHz will use for an example frequency.

Divide 300 ( represents the 300 million but the extra 0's can be omitted) by 27 (represents 27 million times a second the wave cycles for 27 MHz) and you get 11.11 ( this is the wavelength in meters)

Since we Americans still don't use the metric system, convert meters to feet by multiplying meters by 3.28 and it comes out close enough.

11 x 3.28 =36.08 ft.

Now for antenna lengths just divide 36 ft which is a full wavelength for 11 meters by 2 and you get 18 ft. for a 1/2 wave or divide it by 4 for a 1/4 wave at 9 ft. and you can figure out any other fractional wavelength from there if you have basic math skills.

Those numbers in the OP such as 468/F=1/2 wavelength do indeed represent an understanding in antenna theory. They are used to determine feet instead of meters and take the end effect into account. 300/F gives the free space length in meters while 936/F gives a closer approximation to the real world lengths in feet. Either formula will never be 100% precise any way.

 

[fourstringburn]

Those numbers in the OP such as 468/F=1/2 wavelength

So what does "468" mean??? I only see it as a shortcut method because like you said, it keeps the calculations in feet without doing any metric conversions. I realize this method works, but that's not the point I was making.

My post explained the theory and makes perfect sense of all the numbers and I gave relative examples to help explain them so maybe some people may understand it instead of just going thru the motions buy using shortcut methods.

If the Op understood the basics, he wouldn't be asking questions to figure out a 5/8 wavelength. I left it on the table for him and others reading to figure it out by with the theory behind it to help him figure it out on his own.

Either formula will never be 100% precise any way

refer to?

True, mainly because the velocity factor difference, but it's always better to come out a little over then shorter. This sets it up for fine tuning by trimming the antenna.

 

[Captain Kilowatt]

The 468 comes from being half of 936 which is a full wavelength just like 234 is used for a 1/4 wave because it is half of 468. If you mean how were those numbers derived in the first place then that is from the speed of light (and radio waves) traveling at a speed of 984 feet per microsecond. This is equivalent to your 300 meters. These are free space numbers. Since in the real world antennas tend to be shorter due to the end effect, the more precise number is 936 rather than 984 and was determined by applying practice to theory. Note that one meter is 39.37 inches or 3.28 feet. Also note that this yields a ratio of 3.28 :1 for metric conversions. Also note that 984/300 is also a ratio of 3.28:1 Both formulas are the exact same thing except one is in metric and the other not. The final result of 936 takes the end effect into account.

 

[Road Squawker]

It is a basic mathematical ratio so you can calculate whatever you want...

so easy, even a caveman can do it,..................

 

[Heavy Metal]

now what about .64 wave?

 

[vkrules]

Use metric and not that silly antiquated feet and inches. 11m x .64 will get you close but you have to allow for velocity factor of element used.

 

[vkrules]

keep the math simple divide 300 by frequency in mhz (27) =11.1111 x 1/4 wave .25
or .half wave .5 or .64 etc And like said earlier this will be a little long so multiply x velocity factor .98 or .95 if using insulated wire. This is in the ball park but will still vary depending on diameter of element .,if its tapered or not etc.

 

[Captain Kilowatt]

Use metric and not that silly antiquated feet and inches. 11m x .64 will get you close but you have to allow for velocity factor of element used.

I only stated the formula to determine feet because most Americans give blank stares when you talk metric. LOL I grew up going to school when we switched to the metric system here so I am fluent in both systems.

 

[Heavy Metal]

I'm one of those blank stares LOL

 

[vkrules]

I know what it's like CK I was about 9 when Aus changed. Best idea is just use metric and stop trying to convert it all the time, thats when most confuse themselves. 15 cm sounds more imppresive than 6 inches .