COMMUNICATIONS RANGE vs. PAYLOAD ALTITUDE and
DISTANCE
Ralph Wallio, WØRPK WØRPK at
netINS.net
As a
high altitude balloon/payload gradually ascends to higher and higher altitudes
its horizon is gradually extended further and further away in all directions.
This increasing distance to the horizon defines the payload's line-of-sight
VHF/UHF radio communication footprint. A ground station within this footprint
will perceive the payload to be above its horizon and radio communication is
possible.
A first approximation of the relationship between payload altitude and maximum
straight line-of-sight distance can be calculated with an easy formula that is
a reduction of required trigonometry:
Distance [miles] = 1.23 * SQRT(Altitude [feet]) |
(See below for the trigonometric heritage of this formula.)
This formula can be applied to selected altitudes to compile a table of maximum
distances:
ALTITUDE [feet] |
DISTANCE [miles] [km] |
Distance
values from this table can be transferred to a map by using a service available
from the FCC at http://www.fcc.gov/mb/audio/bickel/circleplot.html . Enter a set of payload coordinates such as the
launch site (NWS/Valley in this example, 41d19m11sN 96d21m58sW). Use DISTANCE
[km] for the six circle radii. Turn off all Map Features Options. Submit
Initial Data and then Go Get The Map. These inputs should result in this map. Resulting
distance circles on this map predict the maximum line-of-sight communications coverage
area when the payload is at 20,000, 40,000, 60,000, 80,000, 90,000 and
100,000ft (red).
(See High Altitude Balloon RADIO COMMUNICATIONS COVERAGE -
Second Approximation based on Topographic Terrain and Radio Mobile Deluxe, for a more arduous prediction
method.)
This process has sufficient accuracy for the vast majority of uses. However, it
needs to be understood that the payload will drift with winds aloft and the center
of the coverage circles needs to move accordingly. During the summer this drift
is minimal, on the order of 20 miles or less, but during the winter drift can
be 100 miles and more. One strategy for understanding the impact of drift is to
make a map centered on launch coordinates, another map centered on predicted
burst coordinates, a third map centered on predicted touchdown coordinates and
then interpolate (gauge by eye) as necessary.
The formula constant 1.23 is for straight line-of-sight distances. However, VHF
and UHF radio signals will bend slightly toward the earth's surface. This
bending effectively increases the radio horizon and therefore slightly
increases the formula constant. The ARRL Antenna Book (page 23-5) gives a
constant of 1.415 for weak signals during normal tropo conditions. (This
constant can increase significantly during rare VHF/UHF tropo and other openings.)
I have carefully observed signal strengths during distant missions that were
transmitting on an otherwise quiet frequency. These observations were matched
with payload coordinates from which distances were calculated. These distances
were matched with payload altitudes so that actual formula constants could be
calculated:
Very
weak signal via product detector |
1.38 |
Packet
reception never reaches perfect reliability (see Packet Radio Bit Error Rate) but a
formula constant of 1.25 represents roughly 24dB of quieting (Eb/No) required
for almost 100% packet reception. Therefore, it is only slightly conservative
for normal tropo conditions to use a value of 1.23 as in the table above.
Ground station look angle (elevation above the
horizon) also changes with payload distance and altitude. Line-of-sight VHF and
UHF paths yield very strong signals so antenna gain and resulting beamwidth are
usually not issues. Beamwidth of low gain antennas is wide enough that most
stations do not need to worry about using antenna elevation mounts. However,
stations very close to the payload, e.g., chase vehicles, may have look angles
approaching 90 degrees (straight up).
A Ground Station Look Angle table can be
calculated with increasing distance and altitudes:
A look angle of 0.0 degrees (the station's horizon) corresponds with a formula constant of 1.23, e.g., 123 miles at 10,000ft.
Courtney Duncan, N5BF,
offers the following trigonometric solution to derive this simple formula:
Range (miles) = 1.23 * sqrt (height in feet).
The whole problem is over a region of the earth and heights above it where the
approximation
re >> ht and the approximation that the earth is spherical are
reasonable. (re is the radius of the earth, ht is height above the
surface.)
For a spherical earth and no non-geometric propagation effects, the correct
way to calculate geometric range is
range = re * acos (re / (re + ht))
where 'acos' is "arc cosine" or "inverse cosine" in radians
and re and ht are in the same units of length.
This represents a right triangle with the right angle at the maximum reception
range (target at altitude "ht" is on the horizon), the hypotenuse
connects the center of the earth with the target, the "opposite" side
is the slant range from observer to target, and the "adjacent" side
connects the observer with the center of the earth. The angle at the
center of the earth (in radians) times earth radius is the desired great-circle
range. In this problem, the angle is rather small, on the order of a
tenth of a radian (5-6 degrees).
For small x, (i.e. x << 1) we can approximate
cos x ~ 1 - 1 / x^2
meaning
x ~ acos ( 1 - 1 / x^2)
substituting
u = 1 - 1 / x^2
and thus
x = sqrt(2) * sqrt(1 - u)
we have
acos u ~ sqrt(2) * sqrt(1 - u)
So now we have a good approximation for range:
range ~ re * sqrt(2) * sqrt( 1 - (re / (re + ht)))
Let's work on that last radical for a minute:
1 - (re / (re + ht)) = (re + ht - re) / (re + ht) = ht / (re + ht)
and further, since re >> ht
~ ht / re
That's better. So now we have
range ~ re * sqrt(2) * sqrt( ht / re ) = sqrt ( 2 * re) * sqrt( ht )
Taking re as the WGS-84 earth equatorial radius, 6378.137 km, for all values in
km, this gives us:
range (km) ~ 112.94 * sqrt( ht ) (ht in km)
Now we just need to convert units. You want this in statute miles with ht
in feet.
The conversion from km to sm is sm = km * 0.62137.
The conversion from ft to km is km = ft / 3280.83.
So we have
range (sm) ~ 112.94 * 0.62137 * sqrt( ht (ft) / 3280.83)
= 70.178 * sqrt( 1 / 3280.83 ) * sqrt( ht ) (ht in feet)
= 1.225 * sqrt( ht ).
Pretty close to 1.23.