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

I'll rely on the calculation from the freeware program "Mobile Antenna". That
program allows variation in coil designs to be specifically input as a variable.

If you have a better answer or method then please provide it.

The original post is basically asking if 100 watts can be run from a 706 in a 60
meter mobile. The answer is YES. Why? Because an eight feet long antenna is
substantially less than the reference dipole. [Typically it is -10dBd].

/s/ DD, W1MCE

Owen Duffy wrote:
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

OK, I take that answer...now, why did the FCC go with an ERP as that
would require calculation rather than a measurement. Hard to measure
ERP as far as I know...

Oh well, go figure!

Scott
N0EDV

Dave wrote:

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

On Feb 23, 7:57 pm, Scott wrote:
OK, I take that answer...now, why did the FCC go with an ERP as that
would require calculation rather than a measurement. Hard to measure
ERP as far as I know...


Because ERP (or EIRP, depending on the application) is how most
licensing works, because that's what's important for spectrum
management in a shared band. Ham radio is somewhat unique in that the
rules are applied on power somewhere in the middle of the system (i.e.
in the feedline from Tx to antenna), with no limits (except for RF
safety) on the radiated field strength. If you can build a 60dBi
antenna and radiate 1.5 GW EIRP, you can do it. (part of that
'encouraging experimentation and advances in the radio art').

yes, commercial broadcast is regulated in terms of power output, but,
since you have to do a "proof of performance" and demonstrate that you
have a certain field strength at a distance, it's really regulating
ERP. Change your antenna gain, and they can conceivably make you
change your Tx power.

On Sat, 24 Feb 2007 03:57:24 +0000, Scott
wrote:

OK, I take that answer...now, why did the FCC go with an ERP as that
would require calculation rather than a measurement. Hard to measure
ERP as far as I know...

Hi Scott,

This never intimidated the FCC when they handed me tests for
RadioTelephone Second and First Class. In this group, we can account
for ERP by using modeling tools such as EZNEC. This and other
modeling tools generally agree with measurements found in the field.

73's
Richard Clark, KB7QHC