Ron wrote:
I figured another Hustler sure beats running radials out in a portable
situation.

Maybe so, and I figured a slightly different approach would
triple the radiated power. Anyone who uses a mobile antenna
for portable operation on 75m is simply wasting power since
portable antennas invariably can be longer/higher.

If someone was to ask you how to make their Chevy run better would you
tell him to get a Ford? ;-)

No, but if someone asks me how to make their Ford run better,
I would tell him to get a Chevy. :-)
--
73, Cecil http://www.qsl.net/w5dxp



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Taking Cecil's example a step further. The radiation resistance [Rr] of
an 8 foot antenna at 3.9 MHz is in the range of 1 to 2 ohms depending
upon coil location. The antenna losses, I^2*R, can be in the range of 10
to 20 ohms [Ra], and the ground losses can be in the range of 10 to 20
ohms {Rg]. So, your nominal '50 ohm load' is comprised of Rr + Ra + Rg = RL.

Efficiency is (Rr/RL)*100%. So, 2/50 = 4% or 1/50 = 2%. Take your pick!

The energy is NOT reflected back. It is wasted as heat in the antenna
and ground losses!

Deacon Dave, W1MCE
+ + +

Ron wrote:

On Sat, 08 May 2004 09:12:50 -0500, Cecil Moore
wrote:


Ron wrote:

I have a Hustler amateur mobile antenna model RM-80 for 80 Meter 25-30 KHz.
Although it was designed as a mobile I want to use it as a portable, can
anybody tell me the length of the counterpoise needed for this model? or can
I just use an earth wire

Use another RM-80 as the counterpoise.

Hmmmmm, a 2% efficient antenna with a 2% efficient counterpoise. :-)



Hmmmm, I wonder what 2% efficient means. 2% radiated power with 98%
reflected back into the radio? Ability to contact only 2% of the
stations that you can receive? What?

Ron, W1WBV

On Sun, 09 May 2004 11:35:42 -0500, Cecil Moore
wrote:

Ron wrote:
I figured another Hustler sure beats running radials out in a portable
situation.

Maybe so, and I figured a slightly different approach would
triple the radiated power. Anyone who uses a mobile antenna
for portable operation on 75m is simply wasting power since
portable antennas invariably can be longer/higher.

If someone was to ask you how to make their Chevy run better would you
tell him to get a Ford? ;-)

No, but if someone asks me how to make their Ford run better,
I would tell him to get a Chevy. :-)

Well, now, that's more like it :-)

Ron

On Sun, 09 May 2004 17:26:05 GMT, Dave Shrader
wrote:

Taking Cecil's example a step further. The radiation resistance [Rr] of
an 8 foot antenna at 3.9 MHz is in the range of 1 to 2 ohms depending
upon coil location. The antenna losses, I^2*R, can be in the range of 10
to 20 ohms [Ra], and the ground losses can be in the range of 10 to 20
ohms {Rg]. So, your nominal '50 ohm load' is comprised of Rr + Ra + Rg = RL.

So you can use the formula I^2*R (which is for finding P) without
knowing the variable "I" to find the resistance (Ra)?

I don't think so. Unless you have a magic wand, Harry Potter :-)

Ron


Efficiency is (Rr/RL)*100%. So, 2/50 = 4% or 1/50 = 2%. Take your pick!

The energy is NOT reflected back. It is wasted as heat in the antenna
and ground losses!

Deacon Dave, W1MCE
+ + +

Ron wrote:

On Sun, 09 May 2004 17:26:05 GMT, Dave Shrader
wrote:

SNIP

So you can use the formula I^2*R (which is for finding P) without
knowing the variable "I" to find the resistance (Ra)?

I don't think so. Unless you have a magic wand, Harry Potter :-)

Ron

You certainly can!!

Total power to the antenna SYSTEM = I^2*(Rr + Ra + Rg).

Total power radiated = I^2*(Rr).

When you divide the two terms to get efficiency the I^2 term cancels.
This leaves Rr/(Rr + Ra + Rg). The current terms drops right out!!

DD

Dave,

I didn't come here to fence with you guys. I know nothing about the
relative efficiency of antenna systems.

You could be right and the effiency of a tuned dipole "may" be 2%,
which in my mind means if it were replaced with a regular dipole, the
signal strength at a remote location would increase 50 times, but that
certainly cannot be determined that with that brand of math.

Sorry,

Ron

On Mon, 10 May 2004 00:20:55 GMT, Dave Shrader
wrote:

Ron wrote:

On Sun, 09 May 2004 17:26:05 GMT, Dave Shrader
wrote:

SNIP

So you can use the formula I^2*R (which is for finding P) without
knowing the variable "I" to find the resistance (Ra)?

I don't think so. Unless you have a magic wand, Harry Potter :-)

Ron

You certainly can!!

Total power to the antenna SYSTEM = I^2*(Rr + Ra + Rg).

Total power radiated = I^2*(Rr).

When you divide the two terms to get efficiency the I^2 term cancels.
This leaves Rr/(Rr + Ra + Rg). The current terms drops right out!!

DD

Yes it can! Gain improvement = 10*Log{50) = +16 dB or about 3 S-units.

Ron wrote:

Dave,

I didn't come here to fence with you guys. I know nothing about the
relative efficiency of antenna systems.

You could be right and the effiency of a tuned dipole "may" be 2%,
which in my mind means if it were replaced with a regular dipole, the
signal strength at a remote location would increase 50 times, but that
certainly cannot be determined that with that brand of math.

Sorry,

Ron

On Mon, 10 May 2004 00:20:55 GMT, Dave Shrader
wrote:


Ron wrote:


On Sun, 09 May 2004 17:26:05 GMT, Dave Shrader
wrote:


SNIP

So you can use the formula I^2*R (which is for finding P) without
knowing the variable "I" to find the resistance (Ra)?

I don't think so. Unless you have a magic wand, Harry Potter :-)

Ron

You certainly can!!

Total power to the antenna SYSTEM = I^2*(Rr + Ra + Rg).

Total power radiated = I^2*(Rr).

When you divide the two terms to get efficiency the I^2 term cancels.
This leaves Rr/(Rr + Ra + Rg). The current terms drops right out!!

DD

On Mon, 10 May 2004 11:41:10 GMT, Dave Shrader
wrote:

Yes it can! Gain improvement = 10*Log{50) = +16 dB or about 3 S-units.

I already gave up. You know more about the subject than me.

Just a question, is W1MCE your original call?

Ron W1WBV (1952)