Antenna Height Calculations
Matthew Moran
Line-of-Site transmissions ?
Rick Matthews
>Does anyone know of a tool to calculate the height to two Antennas for
>Line-of-Site transmissions ?
>
Ignoring man-made and natural obstacles, the answer would be the
"curvature-of-the-earth-thing", wouldn't it?
I'd like to know how to figure that out, too.
Rick
K5WLS
Richard Harrison
"Does anyone know of a tool to calculate the height of two antennas for
line-of-sight transmissions?
Easy to remember. The distance to the horizon in miles is equat to the
square root of twice the height of the antenna in feet.
So, the distance two antennas can span is equal to the sum of their
distances to a common point on the shared horizon between them.
Example: an antenna at 200 feet can see a horizon the square root of 400
, or 20 miles away. Two of these could span 40 miles. Path reliability
would not be very good as there is no Fresnel zone clearance. There is
no allowance for periods of atmospheric refractions when rays are
actually bent away from the earth. To allow for bad air the earth is
often assumed only to be 2/3 its actual size and the antenna heights
adjusted accordingly. 0.6 Fresnel zone clearance is usually added at the
path`s earth grazing point to avoid attenuation from destructive
interference between the direct ray and the ray reflected from the
earth`s surface. The graging point and the specifics of the Fresnel zone
are functions of the relative antenna heights and path length,etc. For
commercial microwave paths the design gets detailled due to the cost of
producing the paths and the reliability required. For amateur purposes,
D=sq. rt. of 2H is probably all that needs be remembered.
Best regards, Richard Harrison, KB5WZI
Richard Harrison
"Does anyone know of a tool to calculate the height of two antennas for
line-of-sight transmissions?"
The distance (D) to the horizon in miles from an antenna (H) feet high
is:
D= sq. rt.(2H)
Between two antennas you get the sum of the distances each antenna can
reach.
Cortland Richmond
database was used to calculate the necessary point-to-point commo paths. But
when they were tried, they didn't work!
The map was old, and someone had forgotten to tell the computer that trees
_grow_ in 20 years!
Cortland
(ka...@saber.net)
Richard Harrison wrote:
> Matthew Moran wrote:
> "Does anyone know of a tool to calculate the height of two antennas for
> line-of-sight transmissions?
>
> Easy to remember. The distance to the horizon in miles is equat to the
> square root of twice the height of the antenna in feet.
>
> So, the distance two antennas can span is equal to the sum of their
> distances to a common point on the shared horizon between them.
>
> Example: an antenna at 200 feet can see a horizon the square root of 400
> , or 20 miles away. Two of these could span 40 miles. Path reliability
> would not be very good as there is no Fresnel zone clearance. There is
> no allowance for periods of atmospheric refractions when rays are
> actually bent away from the earth. To allow for bad air the earth is
> often assumed only to be 2/3 its actual size and the antenna heights
> adjusted accordingly. 0.6 Fresnel zone clearance is usually added at the
> path`s earth grazing point to avoid attenuation from destructive
> interference between the direct ray and the ray reflected from the
> earth`s surface. The graging point and the specifics of the Fresnel zone
> are functions of the relative antenna heights and path length,etc. For
> commercial microwave paths the design gets detailled due to the cost of
> producing the paths and the reliability required. For amateur purposes,
> D=sq. rt. of 2H is probably all that needs be remembered.
>
Matthew Moan
distance I want to go and so I wouldn't normally start with the height of
the antenna but rather the required distance and then figure antenna
height. Since I am a math idiot how do you reverse the formula ?
Richard Harrison wrote:
> Matthew Moran wrote:
> "Does anyone know of a tool to calculate the height of two antennas for
> line-of-sight transmissions?"
>
> The distance (D) to the horizon in miles from an antenna (H) feet high
> is:
> D= sq. rt.(2H)
>
> Between two antennas you get the sum of the distances each antenna can
> reach.
>
Richard Harrison
"The map was old, and soneone had forgotten to tell the computer that
trees_grow_in 20 years.
An up-to-date path profile is a big help. It also helps to predict as
well as you can what might develop on each path in 20 years. Progress
shouldn`t get in your way. I`ve had to work around new sky scrapers in
my paths at times.
Trees don`t grow to the sky, so it is possible to put those repeaters
closer together and add to those towers to beat the trees, but
biological warfare may be cheaper if you don`t get caught.
The original questioner said to ignore possible obstructions and just
give a tool for plainslands, or something of the sort.
Really, for commercial paths you gather site and path data from a
physical survey, complete with altitude checks of any spots that might
be close to grazing. Then a detailled path profile is made using
whatever "K" factor shrinks the earth enough to provide the reliability
you think you will need. Fade margin in longer systems exceeds that
needed for reliability because strong signals are needed to nearly
eliminate any noise on an individual hop. The cumulative effect of so
many individual contributions can be surprising at times.
Path surveying has to be easier now that GPS is available. The maps are
good but in some cases as Cortland noted the trees have grown and it may
be hard to exactly position yourself by recognizing map features.
Best regards, Rchard Harrison, KB5WZI
Andrew Koch
<mmo...@cisco.com> wrote:
> Thank you very much and I am starting to get it but I usually know the
> distance I want to go and so I wouldn't normally start with the height of
> the antenna but rather the required distance and then figure antenna
> height. Since I am a math idiot how do you reverse the formula ?
if d is greater than 0 (Duh!) than:
h= 0.5 d squared
(h=0.5d^2)
Andrew
Richard Harrison
"----how do you reverse the formula?"
D=sq.rt.(2H)
To solve for H:
Square both sides
Dsquared=2H
Then: H=Dsquared/2
Example: You need to reach 20 miles.
20 squared is 400.
400/2=200
So, you need to hang your antenna at 200 feet to reach 20 miles.
Cortland Richmond
commercial microwave PP links need to plan on moving every day.
Cortland
MC 10kW Jesus
Yes - my brain. Provide for me the following information and I'll show you
how to do it.
Transmit Tower
- antenna gain
- collective losses of transmission line & connectors
- if you don't know, tell me what you plan on using and I will
estimate
- distance in miles to receiving site
- transmitter output power
- transmitter frequency
Receive Tower
- antenna gain
- collective losses of transmission line & connectors
- minimum receiver signal strength necessary to maintain communication
- ie: 0.25uV or -119dBm for 12dB SINAD on an FM signal, much higher on
high-speed data
General
- type of terrain you will be traversing
- is the path clear?
From this data I will be able to give you some useful information - and
I'll show you how I derive it.
I'll give you a hint. Convert Tx power and Rx sensitivity into dBm. Anyone
who still uses uV and W has their feet firmly planted in the past.
MC