End effect, velocity propagation question
 

I know free space half wave dipole length in feet is 492/f mhz.
I also know 468/f mhz is for "antenna cross section...stray
capacitance...called end effect because the ends of the antenna are
made farther apart electrically than they are physically..." as laid
out in the "Practical Antenna Handbook 3rd edition p.138".

This is suspiciously close to the 95% velocity propagation of parallel
line.
My question is, for a given coax with 66% velocity propagation, is the
formula 66% of 492 (325/f) or
66% of 468 (309/f)?

Thanks in advance

Gee what are these "feet" things? grin (I use 300/f for free space -
for metres)

Velocity factor in coax is vs free-space. If you are using these numbers
for calculating tuned lengths keep in mind that the manufacturing
tolerance for velocity factor is not real accurate and you should
cut/tune with a GDO/TDR etc.

From my reading I have seen dipole end effect vary between 90 and 98%.
Almost always a function of length/diameter ratio.

Cheers Bob VK2YQA

Tac wrote:
I know free space half wave dipole length in feet is 492/f mhz.
I also know 468/f mhz is for "antenna cross section...stray
capacitance...called end effect because the ends of the antenna are
made farther apart electrically than they are physically..." as laid
out in the "Practical Antenna Handbook 3rd edition p.138".

This is suspiciously close to the 95% velocity propagation of parallel
line.
My question is, for a given coax with 66% velocity propagation, is the
formula 66% of 492 (325/f) or
66% of 468 (309/f)?

Thanks in advance

Tac wrote:
"My question is, for a given coax with 66% velocity propagation, is the
formula 66% of 492 (325/f) or 66% of 468 (309/f)?"

VF, velocity factor, is % of the free-space velocity of a radio wave
which is identical to that of light.

Velocity of light is about 300,000,000 m/sec.

Wavelength in free-space is:

WL= 300,000,000 / f meters

There are about 3.28 feet in a meter.

Polyethylene insulated flexible coax cables usually have a velocity of
propagation of about 66% that of free-space. Velocity depends on cable
dimensions as well as its insulating material and distribution.

In my ARRL Antenna Book, the length in feet for a 1/2-wave dipole is
given as 468/f (MHz).
As 1/2-wave= 150/f (MHz) x 3.28 ft/mtr=492/f (MHz), it is clear the
handbook formula has been discounted by 5% for "end effect", which is
actually a function of how fat the wire used is, and relates to wave
velocity reduction.

Best regards, Richard Harrison, KB5WZI

A half wavelength in a transmission line with velocity factor of 66% is
approximately 0.66 * 492/f(MHz).

Here's why. Wavelength = v / f where v is the velocity of
electromagnetic waves and f is the frequency in Hz. v = c * vf where c
is the velocity of light in free space and vf is the velocity factor.
The velocity of light in free space (c) is approximately 984,000,000
feet/second. So one wavelength, in feet, in any medium is approximately
984,000,000 * vf / f(Hz), or 984 * vf / f(MHz). A half wavelength is
then half of this, or about 492 * vf / f(MHz) feet.

A resonant "half wave" antenna is always shorter than a free-space half
wavelength for a variety of reasons. Its actual length depends on wire
diameter, end insulators, the effect of ground, and so forth. 468/f is a
useful approximation that's usually in the ballpark but seldom exactly
correct.

Roy Lewallen, W7EL

Tac wrote:
I know free space half wave dipole length in feet is 492/f mhz.
I also know 468/f mhz is for "antenna cross section...stray
capacitance...called end effect because the ends of the antenna are
made farther apart electrically than they are physically..." as laid
out in the "Practical Antenna Handbook 3rd edition p.138".

This is suspiciously close to the 95% velocity propagation of parallel
line.
My question is, for a given coax with 66% velocity propagation, is the
formula 66% of 492 (325/f) or
66% of 468 (309/f)?

Thanks in advance