I had to ask rather than assume. My inclination where it came from
which you didn't say was that since the field produced by an actual
antenna is twice as great as the field produced by the isotropic
antenna the gain RATIO is two and the power gain is 2 squared which is
4. this means that to produce the same field strength at the same
distance, four times as much power would have to be supplied to an
isotropic radiator as to the actual antenna under consideration. But as
I stated many times RATIO as you are using it has not interest to me as
it is not relevant.
What you are doing is based on a RATIO at a given plane and that RATIO
changes with the plane examined. That is why the yagi is termed a
planar array In other words a reflector is used to affect a single
plane of radiation it is not all encompassing of the total rear
radiation. On top of all that the plane chosen is along the plane of
the main lobe only and does not in anyway include the ratio of the
second lobe to the rear or any nulls that are made. The rear radiation
fields is no way a mirror image of the forward radiated field. You are
supplying a conventional answer to a convential question which revolves
around a single plane where I am speaking of the total radiation field.
You can't keep trotting out the conventional answer to the question
that you want to be posed. I am sure glad I didn't guess where you were
getting the figure 4 from otherwise the thread would have been 200
posts long plus a lot of accusations as to who said what.Get back to
basics and stop trying to section the field of a dipole to make it
easier to simplify for newcomers, it does not represent factually
everything.
it is just a means to an end. without involvement in the toital "wave
and fields" subject
As I have oft times stated I am looking at the whole pattern in three
dimensional form and you keep trotting out answers based on a two
dimensional format
Art


,
Roy Lewallen wrote:
art wrote:
Roy Lewallen wrote:

Since there's no "typical" Yagi, I presented one which most people would
consider to be worse than average -- one having only a 6 dB front/back
ratio. I also assumed for a starting point that the shape of the rear
lobe (that is, beamwidth and height) is the same as the front lobe.

See, you are building a house on the basis of an assumption such that
instead of a rock
that you guessed was there it really was just sand.


The
first calculation is to determine just what the ratio is of the powers
in the front and rear lobe. The answer is 4:1. That is, the front lobe
contains four times the power of the rear lobe.

I am not interested in front to back for what I am looking for but this
4:1 has my interest
What does it represent

I can't think how I can state it any more clearly than I did in the last
sentence of the text just above which you quoted.

and how did you get it?

When dealing with a power ratio, dB = 10 * log(ratio). Solving for ratio:

ratio = 10^(dB/10)

Here's where you'll probably need to get out that pocket calculator. dB
is 6 (see above text), so ratio = 10^(0.6) ~ 4.

Conversion between ratios and dB is a skill that anyone interested in
antennas should develop. I had assumed that it was part of the knowledge
required to pass a general class amateur exam, but apparently I was
mistaken. If the calculator operations are too complex for you, get a
chart of conversion factors which have already been calculated.

The rear usually has
more than one lobe

True. See the remainder of my previous posting for a discussion of this.

and the reflector ndestructs or deflects the energy to 90 degrees of
impact.

That's more nonsense. You'd develop a much better understanding of
antennas (or any physical system) by developing and learning to apply
some basic math skills than by dreaming up alternate explanations for
well-known physical phenomena.

I don't believe I can help you any more -- if indeed I've helped you at
all --, and think (or at least hope) that most other readers have
understood what I'm saying. So I'll bow out here.

Roy Lewallen, W7EL