60m operation is limited to 50w pep relative to a
1/2WL dipole. Since a typical 60m mobile antenna
would be much less than 50% efficient, seems it
would it be OK to run an IC-706 at its normal 100
watt output level. Am I reading the rules right?
--
73, Cecil, http://www.qsl.net/w5dxp

Cecil Moore wrote in news:sFoDh.262$M65.0
@newssvr21.news.prodigy.net:

60m operation is limited to 50w pep relative to a
1/2WL dipole. Since a typical 60m mobile antenna
would be much less than 50% efficient, seems it
would it be OK to run an IC-706 at its normal 100
watt output level. Am I reading the rules right?

What are the actual words used in the rules?

If this is based on ERP, and uses the term Effective Radiated Power to mean
relative to a dipole (rather than say, EIRP), then the ERP would be
calculated from transmitter power output and antenna gain wrt a dipole, or

ERP = TxPower*AntennaDirectivity/DipoleDirectivity*Efficiency.

So the terms you left out were AntennaDirectivity and DipoleDirectivity,
but I should think that most HF whips would have Directivity lower than a
dipole, so you may be ok.

Owen

Cecil Moore wrote:
60m operation is limited to 50w pep relative to a
1/2WL dipole. Since a typical 60m mobile antenna
would be much less than 50% efficient, seems it
would it be OK to run an IC-706 at its normal 100
watt output level. Am I reading the rules right?

I think your interpretation is correct, but the key is:

"A half-wave dipole antenna will be presumed to have a gain of 0 dBd.
Licensees using other antennas must maintain in their station records
either manufacturer data on the antenna gain or calculations of the
antenna gain."

Meaning you probably need to keep an EZNEC printout in your glove
compartment. ;-)

Chuck

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"Cecil Moore" wrote in message
...
60m operation is limited to 50w pep relative to a
1/2WL dipole. Since a typical 60m mobile antenna
would be much less than 50% efficient, seems it
would it be OK to run an IC-706 at its normal 100
watt output level. Am I reading the rules right?
--
73, Cecil, http://www.qsl.net/w5dxp

That is the way I see it. The 1/2 wave dipole would be the 0 db refferance
point. I doubt any practical antenna on a car would be anywhere near 50% of
a dipole.
YOu seem more able than most to calculate the number of db below the dipole
so you can calculate the ammount of power you would have to run to get up to
the same level of the 1/2 wave dipole. I am not sure even the legal limit
of 1500 watts would get you there with some antennas. Maybe the 1500 watt
rule does not apply on this band if you have a very poor antenna.

YES! You are reading the rules correctly.

There are many 60 meter mobiles running 300 to 400 watts into a 10% efficient
antenna relative to a 1/2 wavelength dipole.

Get on 60, It;s is a great band.

Cecil Moore wrote:
60m operation is limited to 50w pep relative to a
1/2WL dipole. Since a typical 60m mobile antenna
would be much less than 50% efficient, seems it
would it be OK to run an IC-706 at its normal 100
watt output level. Am I reading the rules right?

So, hows does an armchair operator (one who doesn't have $10K worth of
test equipment) determine the efficiency of the antenna?

Scott
N0EDV

Dave wrote:

YES! You are reading the rules correctly.

There are many 60 meter mobiles running 300 to 400 watts into a 10%
efficient antenna relative to a 1/2 wavelength dipole.

Get on 60, It;s is a great band.

Cecil Moore wrote:

60m operation is limited to 50w pep relative to a
1/2WL dipole. Since a typical 60m mobile antenna
would be much less than 50% efficient, seems it
would it be OK to run an IC-706 at its normal 100
watt output level. Am I reading the rules right?

Scott wrote:
So, hows does an armchair operator (one who doesn't have $10K worth of
test equipment) determine the efficiency of the antenna?

Radiation resistance/Feedpoint resistance?
--
73, Cecil http://www.w5dxp.com

On Feb 23, 7:06 am, Scott wrote:
So, hows does an armchair operator (one who doesn't have $10K worth of
test equipment) determine the efficiency of the antenna?

Scott
N0EDV



Dave wrote:
YES! You are reading the rules correctly.

There are many 60 meter mobiles running 300 to 400 watts into a 10%
efficient antenna relative to a 1/2 wavelength dipole.

Get on 60, It;s is a great band.

Cecil Moore wrote:

60m operation is limited to 50w pep relative to a
1/2WL dipole. Since a typical 60m mobile antenna
would be much less than 50% efficient, seems it
would it be OK to run an IC-706 at its normal 100
watt output level. Am I reading the rules right?- Hide quoted text -

- Show quoted text -

Hmmm, one could go to Terman, et. al... Lots of heavy math... ughhhh
One could ask the mobile whip manufacturer what the relative gain of
his product is - On second thought that's unlikely to get an answer
other than 42... (See: HitchHikers Guide To the galaxy)

Or one could do the bone simple, farm boy stupid, yet amazingly
effective method of comparing the relative length of the mobile whip
to a quarter wave vertical...
First, the quarter wave vertical is half the length of a dipole so you
immediately have a multiplier of 2...
(Awwww right you nit pickers, DOWN! - Yes I know about vertical-P
loss compared to horizontal-P, but I'm farm boy simple for this one)
Assume the 1/4 Lambda vertical is 45 feet (rough number, I'm a farm
boy, remember) and the whip is 8 feet... Then 45/8 = 5.6 ratio...
So 2 times 5.6 = 11.2 ratio so far...
Therefore 11.2 times 50w = 284 watts...

Now the efficiency of a mobile whip that is just over 20% tall (of a
quarter wave) is roughly 8%-10% (swag) Lets call it 10% for rough
numbers therefore we can expand the 284 watts by dividing 0.1 into
it... or 2840 watts.... Which will make your IC-706 sweat a bit...

OK nitpickers, have fun...

denny / k8do

Scott wrote:

So, hows does an armchair operator (one who doesn't have $10K worth of
test equipment) determine the efficiency of the antenna?

Scott
N0EDV

SNIPPED

There are two easy methods and one rule of thumb method that will get you into
the ballpark.

1) As Cecil replied, divide the radiation resistance by the feedpoint
resistance. EZNEC will give a reasonable value for radiation resistance and an
MFJ 259B [~$250] will give a measure of the feedpoint resistance.

2) A freeware program, mobile antenna, will also calculate the efficiency.

3) [Rule of thumb] The gain is proportional to the effective aperture in square
wavelengths. So, as mentioned in another reply, approximately, the ratio of
length of the mobile antenna [~8 feet] to the length of a 1/2 wavelength antenna
for 60 meters [~93 feet] yields 8.6% efficiency. So, 100 watts from an IC-706
[series] yields ~9 watts effective radiated power. Conclusion, 50 watts ERP on
60 meters with a 9% efficient antenna would require 555 watts into the antenna.

/s/ DD

Dave wrote in
:

....
3) [Rule of thumb] The gain is proportional to the effective aperture
in square wavelengths. So, as mentioned in another reply,
approximately, the ratio of length of the mobile antenna [~8 feet] to
the length of a 1/2 wavelength antenna for 60 meters [~93 feet] yields
8.6% efficiency. So, 100 watts from an IC-706 [series] yields ~9 watts
effective radiated power. Conclusion, 50 watts ERP on 60 meters with a
9% efficient antenna would require 555 watts into the antenna.
....

It certainly is ROT that in the general case, and in this case, that you
can run a tape measure over an antenna to calculate the aperture area and
in turn calculate gain.

Can you explain how your method deals with capturing the effects of a high
loss loading coil vs a low loss loading coil in such an antenna?

Owen