Anyway, reading the op carefully, he asked for an exact _explanation_, not
an exact calculation result..
Steve, K;9.D,C'I

"Bill Turner" wrote in message
...
Reg Edwards wrote:

There's no such value as 'exact'.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Nonsense. I have exactly two pencils in my pencil holder. :-)

73, Bill W6WRT

Harry wrote:
I am a newbi in antennas.

Here's my question:

I know that a half-wave dipole in free space has
a feed-point impedance of approximately 73 ohms.

Can anyone tell me **exactly** how this number is calculated.

From "Antenna Theory" by Balanis:

Rr = 2*Prad/|Io^2| = 73 ohms (4-93)

Prad is found by integrating the Poynting Vector over a
certain radius. Io is the current maximum magnitude.

The ASCII limitation prevents much more than this.
--
73, Cecil http://www.qsl.net/w5dxp


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"Cecil Moore" wrote
Prad is found by integrating the Poynting Vector over a
certain radius.

====================================

Cec, what's the Poynting Vector?
---
Reg.

Reg Edwards wrote:

"Cecil Moore" wrote

Prad is found by integrating the Poynting Vector over a
certain radius.

Cec, what's the Poynting Vector?

Same thing as a power flow vector. The dimensions are
energy per second per unit area. If the power flow vector
is integrated over the entire surface of a sphere of radius
r, the result is total radiated power. I thought maybe you
and Mr. Poynting might have worked together at some time in
the 20th century. :-)

Poynting, John Henry (1852-1914)
English physicist, mathematician, and inventor. He devised
an equation by which the rate of flow of electromagnetic energy
(now called the Poynting vector) can be determined.

http://www.cartage.org.lb/en/themes/...oynting/1.html
--
73, Cecil http://www.qsl.net/w5dxp


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Cec, what's the Poynting Vector?

Same thing as a power flow vector. The dimensions are
energy per second per unit area. If the power flow vector
is integrated over the entire surface of a sphere of radius
r, the result is total radiated power. I thought maybe you
and Mr. Poynting might have worked together at some time in
the 20th century. :-)

Poynting, John Henry (1852-1914)
English physicist, mathematician, and inventor. He devised an equation by
which the rate of flow of electromagnetic energy
(now called the Poynting vector) can be determined.

S = E X H

Hi Cecil,

From "Antenna Theory" by Balanis:

Rr = 2*Prad/|Io^2| = 73 ohms (4-93)

I got Balanis' book and found the above formula.
Thanks a lot! (Tom also mentioned the same book.)

4-94 (I have the 1982 edition) is from formula 4-70.
The math is OK to me. I happen to have a PH.D. degree in math.

-- Harry

Harry wrote:
Hi Cecil,

From "Antenna Theory" by Balanis:

Rr = 2*Prad/|Io^2| = 73 ohms (4-93)

I got Balanis' book and found the above formula.
Thanks a lot! (Tom also mentioned the same book.)

4-94 (I have the 1982 edition) is from formula 4-70.
The math is OK to me. I happen to have a PH.D. degree in math.

For your "passing-out" test, calculate the radiation pattern of a
two-radial groundplane antenna directly from Maxwell's equations
(including the two small cross-polarized lobes that nobody else
mentions), using pencil and paper only, in a pub.

I've seen it done by a math professor from W6. (Nah... it must have been
a party trick... he must have rehearsed that a million times... mustn't
he?)

When he'd finished, Charlie sat back, smiled asked for comments. The
rest of us passed out.


--
73 from Ian G/GM3SEK 'In Practice' columnist for RadCom (RSGB)
http://www.ifwtech.co.uk/g3sek

Hi Cecil,

I find another formula in Balnais book (section 4.3, 1982 edition)
for Rr:

Rr = 20 * (pi)^2 * ( l / lamda)^2 (4-37)

If l = lamda/2, then this formula gives Rr .=. 50 Ohms.

(4-93) the one you quoted is

Rr = (120*pi) / (4*pi) * Cin( 2*pi ) = 73 Ohms, (for l = lamda/2)

where 120*pi is the intrinsic impedance for a free-space medium.

Would you please explain the difference between these
two formulas (4-37 and 4-93)?


Thanks!

-- Harry

Harry wrote:

Hi Cecil,

I find another formula in Balnais book (section 4.3, 1982 edition)
for Rr:

Rr = 20 * (pi)^2 * ( l / lamda)^2 (4-37)

If l = lamda/2, then this formula gives Rr .=. 50 Ohms.

(4-93) the one you quoted is

Rr = (120*pi) / (4*pi) * Cin( 2*pi ) = 73 Ohms, (for l = lamda/2)

where 120*pi is the intrinsic impedance for a free-space medium.

Would you please explain the difference between these
two formulas (4-37 and 4-93)?


Thanks!

-- Harry

4-37 should apparently read:

Rr = 30 * (pi)^2 * (1 / lambda)^2.

ac6xg

Harry wrote:
Would you please explain the difference between these
two formulas (4-37 and 4-93)?

Equation 4-37 is from section "4.3 SMALL DIPOLE", i.e. shorter
than 1/2WL. The dipole is so short that its current distribution
is triangular, not sinusoidal. Quoting section 4.3: "The radiation
resistance of the antenna is strongly dependent upon the current
distribution." The "1/2" on the diagram does NOT mean 1/2WL.

Equation 4-93 is from section "4.6 HALF-WAVELENGTH DIPOLE". The
current distribution on the thin-wire 1/2WL dipole is considered
to be sinusoidal.

See Figure 4.8. Note the triangular current distribution for
L = 1/4WL and the sinusoidal current distribution for L = 1/2WL.
--
73, Cecil http://www.qsl.net/w5dxp

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