"Stefan Wolfe" wrote in message
...

"Jerry Martes" wrote in message
news:2Yvyh.37369$5U4.35764@trnddc07...

"Stefan Wolfe" wrote in message
...

"Jerry Martes" wrote in message
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"Stefan Wolfe" wrote in message
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wrote in message
oups.com...
hello
Jeff
i am very thank full to your help, i have got Rx height 101.2

Regards
naqvi

Please show your work.


Hi Stephan

Your posts seem to inply that the receive antenna will "see" the 20
meter high transmitting antenna when the receiver antenna is in the
shadow of the 80 meter hill. It seems that the receiver needs to be
out of the shadow of the hill unless you are able to estimate
refraction from the hill.

But, your aparent confidance in the statement "0 feet" makes me wonder
if I have this problem wrongly analyzed. I have so much confidance in
Richard Fry's data that I had accepted his estimation of 270 meters to
be as close as you can estimate.

Do I misunderstand your post about what minimum height is needed?

Jerry
Hi Jerry,

how many shadows have you seen that are 20Km long?


Hi Stephan

Tell me where I have misunderstood the problem. I assumed the
transmitting antenna was Lower than the top of the hill. But, you seem
to imply that the transmitter can be seen even when the hill is blocking
the "view" to it.

I have actually never measured a shadow longer that a few feet, but I
assumed they continued to exist to infinity when an object blocks them
from view.

Well you see Jerry, the reason you only see it for a few feet is because
the attenuation of the light varies inversely with the distance from the
object that blocks the light. I think you have done a good job in making
my point. Thanks,

Hi Stephan

I now understand now that you do think you can receive 10 GHz signals
while the receiver is in the shadow caused by the mountain between you and
the transmitter. Do I understand you correctly?

Jerry

Richard Clark wrote:

Hi Stefan,

You seem to be simultaneously agreeing and disagreeing. The hill is
either in the way, or it is not. It is in the way. To count on an
intermediary, such as suggested by Jimmie, knife-edge propagation or
bend of the waves, is probably not in our student's syllabus. Besides,
I have seen neither you nor Jimmie offer the attenuation presented by
such refractions (and the attenuation is not marginal). Without
quantifiables, the path budget cannot be calculated.

The problem, as stated, has a clear answer in looking over the hill by
raising one antenna, the question informs us that is the answer and
that is simply resolved with trig (albeit, including the radius of
earth and accounting for its curvature).

73's
Richard Clark, KB7QHC

Might I recommend a program misleadlingly called Radio Mobile. This
piece of software will all let you check many real world situations.
This program is not real easy to use, it's a lot worse than Windows or
Office (ok, not worse than Office), but it is worth learning unlike the
previous 2 mentioned.

The site you need for the software is

http://www.cplus.org/rmw/english1.html

and then you need to download a large amount of terrain data, which is
freely available from NASA. It looks like the way this is handled has
changed since I did it last, so I can't comment on how it is done now.

tom
K0TAR

On Wed, 07 Feb 2007 22:24:26 -0600, Tom Ring
wrote:

The site you need for the software is

http://www.cplus.org/rmw/english1.html

and then you need to download a large amount of terrain data, which is
freely available from NASA. It looks like the way this is handled has
changed since I did it last, so I can't comment on how it is done now.

Hi tom,

I was involved in Beta testing this. The current version allows
dynamic map loading over the Internet, and overlays of Mapquest, Tiger
or many other mapping programs available. I've been using it heavily
for the last few months and its world of variables allows for finely
grained analysis. However, it also allows for massive headaches if
all you are looking for is a simple solution. In short, no
quantifiables are going to follow from your recommendation - unless I
do it. This lil Red Hen isn't interested in cooking that bread.

73's
Richard Clark, KB7QHC

" In truth, i could receive that signal holding a hand held 10GHz receiver
while sitting on the ground. The 80m hill is nothing from an observer 10KM
away...only .006 degree from the top of the transmitter tower. It is part
of the horizon. I love it when you guys talk like you are sol
knowledgeable yet lack the common sense to conceptualize the problem as it
really exists.

So you are saying that you can achieve 30km at 10GHz to a hand held receiver
at ground level with a 300 foot hill in the way!!! I am sure that you could
not do this with any sensible power even at 2m let alone 10GHz.

Perhaps it is your concept of what is going on that is wrong. Have you heard
of Fresnel Zones??
When obstructions come within the first Fresnel zone significant attenuation
occurs.

With the situation that you are describing the path is totally obstructed,
with the path only possible due to diffraction from the hill top. The hill
top impinges to at least the top of the 5th Fresnel Zone, hence the
attenuation is very high.

As the height of the Rx antenna increases the attenuation is still very high
until the hill top only start to intersect with of the first zone (antenna
height~150m). It then drops quite rapidly until there is true line of sight
and bottoms out when the hill top is clear of the second zone.

You make a great deal of the hill only being 0.006 degree at the horizon. If
you plot it accurately and with reference to the Fresnel zones, it does make
a big difference. With the Rx antenna at ground level the top of the Fresnel
zones are never below the horizon, which is completely different to the
situation when the hill is there (-5th Zone obstructed).

Also, without the hill you only have to raise the Rx antenna to about 15m to
achieve line of sight compared to 200m with the hill there!!!!!! Quite a big
difference I think you will agree, and one that your 'conceptualisation
doesn't seem to allow for!!

73
Jeff

"Richard Clark" wrote in message
...
On Wed, 7 Feb 2007 00:08:19 -0500, "Stefan Wolfe"
wrote:


"Richard Clark" wrote in message
. ..
On Tue, 6 Feb 2007 22:35:04 -0500, "Stefan Wolfe"
wrote:

2. The effect of the 80m hill 10Km away is negligible. The arc tan is
only
.008 degrees, thus the transmitter hardly "sees" it.

Actually, Stefan, the transmitter cannot see through it at all.

Exactly . And taking into account the 15 degree bend of the radio horizon
(even at 10 GHz) vs the arctan of .008 degrees for the the hill, reduced
somewhat by the 20 foot transmitter tower to .006 degrees, the hill
itself
is invisible at a 30km far field. The bend of the propagating waves which
extends the radio horizon clearly mitigates any possible effects of the
80m
hill.


Hi Stefan,

You seem to be simultaneously agreeing and disagreeing. The hill is
either in the way, or it is not. It is in the way. To count on an
intermediary, such as suggested by Jimmie, knife-edge propagation or
bend of the waves, is probably not in our student's syllabus. Besides,
I have seen neither you nor Jimmie offer the attenuation presented by
such refractions (and the attenuation is not marginal). Without
quantifiables, the path budget cannot be calculated.

The problem, as stated, has a clear answer in looking over the hill by
raising one antenna, the question informs us that is the answer and
that is simply resolved with trig (albeit, including the radius of
earth and accounting for its curvature).

73's
Richard Clark, KB7QHC

There are lots of programs out here for calculating path loss for LOS
situations but figuring in the path via knife edge defraction is something I
have always tried to avoid.. In real life there are too many variables that
effect this and you could have a signal that would tend to fade. Over the
years I have forgotten or lost interest in figuring the impractical.

Jimmie

On 7 Feb, 15:43, "Richard Fry" wrote:
wrote The total distance between the transmitting and receivingantennaof a
microwave link at 10GHz, is 30 Km. the height of the Txantennais
above ground level is 20 m. the maximum acceptable total path loss is
169 dB. Furthermore there is hill located 10 km away from the
transmitterantennawith a height of 80m.

calculate the height of the receiverantennafor the path loss to be
just equal to the maximum acceptable value?

_____________

The height above mean sea level of the tx and rx sites, and the terrain
profile for the path would be necessary to answer this ~ accurately. But
for a smooth earth model, the graphic at the link below will give some
insight.

It shows that a height of around 270 meters would be needed for the receiveantenna, using a K-factor of 1.33 and 0.6 fresnel clearance for an 80 m hill
10 km downrange. The path loss then would be about 142 dB.

http://i62.photobucket.com/albums/h8.../10GigPath.gif

RF

hi

Richard
i have read ur response very carefully and its perfect 142dB Path
loss but problem is that what's the Rx height
if i suppose Tx and Rx install on the same height
then my Rx height after the earth buldge and knife e calculation
i have got 100m Rx
its totally wrong
because
if i increase that (100m) Rx height its mean get the problem at
fresnel Zone

need help

thanks
regard

naqvi

naqvi wrote
i have read ur response very carefully and its perfect 142dB Path
loss but problem is that what's the Rx height if i suppose Tx and Rx
install on the same height then my Rx height after the earth buldge
and knife e calculation i have got 100m Rx its totally wrong
because if i increase that (100m) Rx height its mean get the problem
at fresnel Zone
_________________

Suggest that you print my graphic, and look for different heights of the
transmit and receive antennas that still clear the 80 meter hill by using a
straight edge that always crosses the location of the hill at the elevations
shown in my plot.

It is best to provide more path clearance at the 80 meter hill than I
showed, because some K-factor variations could steer that 10 GHz beam into
that hill, and cause loss of a usable signal for the receiver.

Also remember that this plot was done over a smooth earth. The true
elevations at the endpoints of the path and at the hill could be
considerably different than I showed.

RF