Lets assume a single hop 40m signal from 400 miles away. What
elevation angle does it arrive at? Both stations are using inverted V's
at nominal height. There are no large bodies of water in between.

Daytime and/or nighttime.

--
73's de Ken KG0WX - Kadiddlehopper #11808,
Flying Pigs #-1055, Grid EM17io, TS-850SAT,
Elecraft XG2, 4SQRP Tenna Dipper, Heath GD-1B

"Ken Bessler" wrote in message
news:VEmag.22577$4H.10017@dukeread03...
Lets assume a single hop 40m signal from 400 miles away. What
elevation angle does it arrive at? Both stations are using inverted V's
at nominal height. There are no large bodies of water in between.

Daytime and/or nighttime.

--
73's de Ken KG0WX -

depends on the height of the reflecting/refracting layer, time of day,
sunspot number and associated geometry. Intensity of signals will depend on
how the radiation angle - pattern of the antennas fits the propagation
angles.
There are some propagation programs that will provide good answers based on
above data.

Yuri, K3BU

"Ken Bessler" wrote in message
news:VEmag.22577$4H.10017@dukeread03...
Lets assume a single hop 40m signal from 400 miles away. What
elevation angle does it arrive at? Both stations are using inverted
V's
at nominal height. There are no large bodies of water in between.

Daytime and/or nighttime.

--
73's de Ken KG0WX
===================================

The type of antenna or its radiation pattern has nothing whatever to
do with the path taken by the radio wave through the ionosphere. The
take-off angle and its name, generated by EZNEC, can be very
misleading.

The radio path is simply a matter of trigonometry involving only the
groundpath distance between transmitter and receiver and the height of
the reflecting layer.

The height of the reflecting layer changes between day and night. And
there may be more than one layer present in daylight. The layer
actually used depends on frequency.

If the Tx and Rx stations are far apart, the trigonometry becomes a
little bit complicated because of the curvature of the Earth's
surface. But for groundpath distances up to 500 miles a flat earth can
be assumed. Get a sheet of paper and a pencil and sketch the triangle
to be solved. The average height of the F-Layer in darkness is about
200 miles. In daylight it is about 300 miles.

To do the actual calculations download program SKYTRIG from website
below in a few seconds and run immediately. SKYTRIG is near the bottom
of the list on the "Download Progs From Here" page. Just left-click
on it.
----
.................................................. ..........
Regards from Reg, G4FGQ
For Free Radio Design Software go to
http://www.btinternet.com/~g4fgq.regp
.................................................. ..........

Ken Bessler wrote:
Lets assume a single hop 40m signal from 400 miles away. What
elevation angle does it arrive at? Both stations are using inverted V's
at nominal height. There are no large bodies of water in between.

Daytime and/or nighttime.

That chart is in the ARRL Antenna Book. Assuming F2 layer
reflection, the arriving angle is reported to be in excess
of 50 degrees.
--
73, Cecil http://www.qsl.net/w5dxp

Cec, there you go again, quoting the bibles!

If the F2 is present then it must be daylight.

And if its daylight then the E-Layer must also be present.

For a groundpath distance of 400 miles the most likely reflector is
the E-Layer. In daylight the E-Layer is an excellent, stable
reflector. Furthermore, the radio path distance is considerably
shorter than via the higher F1 or F2-Layers.

The height of the E-Layer is about 70 miles.

Assuming a flat Earth, the elevation angle is -

ArcTan(70/200) = 19 degrees.

It depends to some extent on the E-Layer critical frequency and the
MUF. If the MUF is low enough and the transmit frequency is high
enough, e.g., 14 or 21 MHz, then the wave may pass right through the
E-Layer and be reflected most likely by the F1-Layer at a height of
roughly 300 miles.

Signals received via the F-Layers, if received at all, will be weaker
than via the E-Layer, if only because the path length is greater.

The elevation angle via the F1-Layer will be about -

ArcTan(300/200) = 56 degrees.

At night the E and F2 layers disappear. There remains only the
nighttime F at a height of about 200 miles to give an elevation angle
of 45 degrees.

That is, of course, if the nighttime-F critical frequency and MUF
allows propagation. There are such things as skip distances.

( Roy, no, I havn't made arrangements with high power broadcasters at
known distances to make tests to prove the foregoing predictions. And
yes, I know it's refraction and not reflection.)
----
Reg, G4FGQ.
========================================

"Cecil Moore" wrote in message
. net...
Ken Bessler wrote:
Lets assume a single hop 40m signal from 400 miles away. What
elevation angle does it arrive at? Both stations are using
inverted V's
at nominal height. There are no large bodies of water in between.

Daytime and/or nighttime.

That chart is in the ARRL Antenna Book. Assuming F2 layer
reflection, the arriving angle is reported to be in excess
of 50 degrees.
--
73, Cecil http://www.qsl.net/w5dxp

That is, of course, if the nighttime-F critical frequency and MUF
allows propagation. There are such things as skip distances.

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

Simplified Critical Frequencies, MUF and Skip Distances.

At a vertical elevation angle of 90 degrees, the Critical Frequency is
that frequency above which the radio wave passes right through the
layer and is not reflected.

It may be reflected back to Earth by a higher layer if there is one.
If the critical frequency of the higher layer is not high enough then
the wave may pass through that layer too and be lost forever.

Critical frequencies are generally at the lower HF frequencies and
depend on geographical latitude, the angle of the sun, time of day,
winter or summer, and on the state of the sun. That's why Near
Vertical Incidence transmissions are at low frequencies in the 80m and
sometimes in the 40m bands and are uncertain.

As the transmit elevation angle changes from vertical, the angle of
incidence of the wave with the ionospheric layer becomes less than 90
degrees and frequencies greater than critical begin to be reflected.

The Maximum Usable Frequency (MUF), that is the highest frequency
which is reflected, is aways higher than the critical frequency. It
is given by MUF = Fcrit/Sin(Phi) where Phi is the angle of incidence
of the wave with the layer. The MUF can be several times critical
frequency - conditions which occur at very low transmit elevation
angles.

This explains how best DX is obtained on the 15m and 10m amateur bands
at the height of the sun-spot cycle when critical frequencies are at
their maximum but still relatively low. Nobody points their 10m beams
up into the sky to work DX. Very low angle radiation is called for.
Yagi beam booms are horizontal, pointed at the horizon.

The lower the elevation angle of the radio path the higher is the MUF.
There is a skip distance. At distances less than the skip distance
nothing can be heard. This is because the operating frequency is too
high. It is greater than the MUF and the wave passes through the layer
without reflection. The lower the operating frequency the shorter is
the skip distance. Eventually, at MF, there is only Near Vertical
Incidence radiation and groundwave.

I trust the foregoing makes sense.

There is a short table of typical critical frequencies in the notes
attached to program SKYTRIG. They have been collected over the years
from various sources. MUF = Fcrit/Sin(Phi). To find Phi use the
program. It's only trigonometry.
----
.................................................. ..........
Regards from Reg, G4FGQ
For Free Radio Design Software go to
http://www.btinternet.com/~g4fgq.regp
.................................................. ..........

Reg Edwards wrote:
"Ken Bessler" wrote in message
news:VEmag.22577$4H.10017@dukeread03...

Lets assume a single hop 40m signal from 400 miles away. What
elevation angle does it arrive at? Both stations are using inverted

V's

at nominal height. There are no large bodies of water in between.

Daytime and/or nighttime.

--
73's de Ken KG0WX

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

The type of antenna or its radiation pattern has nothing whatever to
do with the path taken by the radio wave through the ionosphere. The
take-off angle and its name, generated by EZNEC, can be very
misleading.

It is hard to look at a radiation pattern, conclude that the take-off
angle is the only angle of radiation, and then blame it on EZNEC! Most
of the antennas that I have modeled seem to have radiation in lots of
directions. 8^) Otherwise you are correct.



The radio path is simply a matter of trigonometry involving only the
groundpath distance between transmitter and receiver and the height of
the reflecting layer.

The height of the reflecting layer changes between day and night. And
there may be more than one layer present in daylight. The layer
actually used depends on frequency.

If the Tx and Rx stations are far apart, the trigonometry becomes a
little bit complicated because of the curvature of the Earth's
surface. But for groundpath distances up to 500 miles a flat earth can
be assumed. Get a sheet of paper and a pencil and sketch the triangle
to be solved. The average height of the F-Layer in darkness is about
200 miles. In daylight it is about 300 miles.

To do the actual calculations download program SKYTRIG from website
below in a few seconds and run immediately. SKYTRIG is near the bottom
of the list on the "Download Progs From Here" page. Just left-click
on it.


- 73 de Mike KB3EIA -

Mike Coslo wrote:
It is hard to look at a radiation pattern, conclude that the
take-off angle is the only angle of radiation, and then blame it on
EZNEC!

I knew a ham in Chandler, AZ who would adjust not only
the direction but the height of his beam for maximum signal.
I assume by adjusting the height, he was changing his TOA.
--
73, Cecil http://www.qsl.net/w5dxp

Reg wrote:
The type of antenna or its radiation pattern has nothing whatever to
do with the path taken by the radio wave through the ionosphere. The
take-off angle and its name, generated by EZNEC, can be very
misleading.


It has to do. It allows us to direct the RF under desired angle to hit the
layer or region that supports the propagation to the chosen target.
I have seen situations when signals to Europe were coming under low angle
and in the same direction, signals to deep Asia were coming under higher
angle at the same time.

There are different propagation modes and paths and be able to control
radiation pattern of antenna is important (for serious hams, like
contesters). I am not talking about skewed path, long path and other modes
of propagation when horizontal and vertical control of the antenna radiation
pattern is of huge benefit.

So making blank statements like above is not proper.

73 Yuri K3BU

"Yuri Blanarovich" wrote in message
...
Reg wrote:
The type of antenna or its radiation pattern has nothing whatever
to
do with the path taken by the radio wave through the ionosphere.
The
take-off angle and its name, generated by EZNEC, can be very
misleading.


It has to do. It allows us to direct the RF under desired angle to
hit the
layer or region that supports the propagation to the chosen target.
=======================================
Yuri,

But you can't "direct" it.

You have to do your best with whatever elevation angle Eznec dictates.
----
Reg.